### 3.499 $$\int \frac{(d+e x)^3}{a+c x^2} \, dx$$

Optimal. Leaf size=90 $\frac{e \left (3 c d^2-a e^2\right ) \log \left (a+c x^2\right )}{2 c^2}+\frac{d \left (c d^2-3 a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a} c^{3/2}}+\frac{3 d e^2 x}{c}+\frac{e^3 x^2}{2 c}$

[Out]

(3*d*e^2*x)/c + (e^3*x^2)/(2*c) + (d*(c*d^2 - 3*a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(3/2)) + (e*(3*
c*d^2 - a*e^2)*Log[a + c*x^2])/(2*c^2)

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Rubi [A]  time = 0.0745295, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.235, Rules used = {702, 635, 205, 260} $\frac{e \left (3 c d^2-a e^2\right ) \log \left (a+c x^2\right )}{2 c^2}+\frac{d \left (c d^2-3 a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a} c^{3/2}}+\frac{3 d e^2 x}{c}+\frac{e^3 x^2}{2 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^3/(a + c*x^2),x]

[Out]

(3*d*e^2*x)/c + (e^3*x^2)/(2*c) + (d*(c*d^2 - 3*a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(3/2)) + (e*(3*
c*d^2 - a*e^2)*Log[a + c*x^2])/(2*c^2)

Rule 702

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{(d+e x)^3}{a+c x^2} \, dx &=\int \left (\frac{3 d e^2}{c}+\frac{e^3 x}{c}+\frac{c d^3-3 a d e^2+e \left (3 c d^2-a e^2\right ) x}{c \left (a+c x^2\right )}\right ) \, dx\\ &=\frac{3 d e^2 x}{c}+\frac{e^3 x^2}{2 c}+\frac{\int \frac{c d^3-3 a d e^2+e \left (3 c d^2-a e^2\right ) x}{a+c x^2} \, dx}{c}\\ &=\frac{3 d e^2 x}{c}+\frac{e^3 x^2}{2 c}+\frac{\left (d \left (c d^2-3 a e^2\right )\right ) \int \frac{1}{a+c x^2} \, dx}{c}+\frac{\left (e \left (3 c d^2-a e^2\right )\right ) \int \frac{x}{a+c x^2} \, dx}{c}\\ &=\frac{3 d e^2 x}{c}+\frac{e^3 x^2}{2 c}+\frac{d \left (c d^2-3 a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a} c^{3/2}}+\frac{e \left (3 c d^2-a e^2\right ) \log \left (a+c x^2\right )}{2 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0565024, size = 80, normalized size = 0.89 $\frac{e \left (\left (3 c d^2-a e^2\right ) \log \left (a+c x^2\right )+c e x (6 d+e x)\right )}{2 c^2}+\frac{d \left (c d^2-3 a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a} c^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^3/(a + c*x^2),x]

[Out]

(d*(c*d^2 - 3*a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(3/2)) + (e*(c*e*x*(6*d + e*x) + (3*c*d^2 - a*e^2
)*Log[a + c*x^2]))/(2*c^2)

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Maple [A]  time = 0.045, size = 99, normalized size = 1.1 \begin{align*}{\frac{{e}^{3}{x}^{2}}{2\,c}}+3\,{\frac{d{e}^{2}x}{c}}-{\frac{\ln \left ( c{x}^{2}+a \right ) a{e}^{3}}{2\,{c}^{2}}}+{\frac{3\,\ln \left ( c{x}^{2}+a \right ){d}^{2}e}{2\,c}}-3\,{\frac{ad{e}^{2}}{c\sqrt{ac}}\arctan \left ({\frac{cx}{\sqrt{ac}}} \right ) }+{{d}^{3}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(c*x^2+a),x)

[Out]

1/2*e^3*x^2/c+3*d*e^2*x/c-1/2/c^2*ln(c*x^2+a)*a*e^3+3/2/c*ln(c*x^2+a)*d^2*e-3/c/(a*c)^(1/2)*arctan(x*c/(a*c)^(
1/2))*a*d*e^2+1/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*d^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.77174, size = 414, normalized size = 4.6 \begin{align*} \left [\frac{a c e^{3} x^{2} + 6 \, a c d e^{2} x +{\left (c d^{3} - 3 \, a d e^{2}\right )} \sqrt{-a c} \log \left (\frac{c x^{2} + 2 \, \sqrt{-a c} x - a}{c x^{2} + a}\right ) +{\left (3 \, a c d^{2} e - a^{2} e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, a c^{2}}, \frac{a c e^{3} x^{2} + 6 \, a c d e^{2} x + 2 \,{\left (c d^{3} - 3 \, a d e^{2}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} x}{a}\right ) +{\left (3 \, a c d^{2} e - a^{2} e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, a c^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+a),x, algorithm="fricas")

[Out]

[1/2*(a*c*e^3*x^2 + 6*a*c*d*e^2*x + (c*d^3 - 3*a*d*e^2)*sqrt(-a*c)*log((c*x^2 + 2*sqrt(-a*c)*x - a)/(c*x^2 + a
)) + (3*a*c*d^2*e - a^2*e^3)*log(c*x^2 + a))/(a*c^2), 1/2*(a*c*e^3*x^2 + 6*a*c*d*e^2*x + 2*(c*d^3 - 3*a*d*e^2)
*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + (3*a*c*d^2*e - a^2*e^3)*log(c*x^2 + a))/(a*c^2)]

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Sympy [B]  time = 0.969667, size = 308, normalized size = 3.42 \begin{align*} \left (- \frac{e \left (a e^{2} - 3 c d^{2}\right )}{2 c^{2}} - \frac{d \sqrt{- a c^{5}} \left (3 a e^{2} - c d^{2}\right )}{2 a c^{4}}\right ) \log{\left (x + \frac{- a^{2} e^{3} - 2 a c^{2} \left (- \frac{e \left (a e^{2} - 3 c d^{2}\right )}{2 c^{2}} - \frac{d \sqrt{- a c^{5}} \left (3 a e^{2} - c d^{2}\right )}{2 a c^{4}}\right ) + 3 a c d^{2} e}{3 a c d e^{2} - c^{2} d^{3}} \right )} + \left (- \frac{e \left (a e^{2} - 3 c d^{2}\right )}{2 c^{2}} + \frac{d \sqrt{- a c^{5}} \left (3 a e^{2} - c d^{2}\right )}{2 a c^{4}}\right ) \log{\left (x + \frac{- a^{2} e^{3} - 2 a c^{2} \left (- \frac{e \left (a e^{2} - 3 c d^{2}\right )}{2 c^{2}} + \frac{d \sqrt{- a c^{5}} \left (3 a e^{2} - c d^{2}\right )}{2 a c^{4}}\right ) + 3 a c d^{2} e}{3 a c d e^{2} - c^{2} d^{3}} \right )} + \frac{3 d e^{2} x}{c} + \frac{e^{3} x^{2}}{2 c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(c*x**2+a),x)

[Out]

(-e*(a*e**2 - 3*c*d**2)/(2*c**2) - d*sqrt(-a*c**5)*(3*a*e**2 - c*d**2)/(2*a*c**4))*log(x + (-a**2*e**3 - 2*a*c
**2*(-e*(a*e**2 - 3*c*d**2)/(2*c**2) - d*sqrt(-a*c**5)*(3*a*e**2 - c*d**2)/(2*a*c**4)) + 3*a*c*d**2*e)/(3*a*c*
d*e**2 - c**2*d**3)) + (-e*(a*e**2 - 3*c*d**2)/(2*c**2) + d*sqrt(-a*c**5)*(3*a*e**2 - c*d**2)/(2*a*c**4))*log(
x + (-a**2*e**3 - 2*a*c**2*(-e*(a*e**2 - 3*c*d**2)/(2*c**2) + d*sqrt(-a*c**5)*(3*a*e**2 - c*d**2)/(2*a*c**4))
+ 3*a*c*d**2*e)/(3*a*c*d*e**2 - c**2*d**3)) + 3*d*e**2*x/c + e**3*x**2/(2*c)

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Giac [A]  time = 1.34387, size = 105, normalized size = 1.17 \begin{align*} \frac{{\left (c d^{3} - 3 \, a d e^{2}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{\sqrt{a c} c} + \frac{{\left (3 \, c d^{2} e - a e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac{c x^{2} e^{3} + 6 \, c d x e^{2}}{2 \, c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+a),x, algorithm="giac")

[Out]

(c*d^3 - 3*a*d*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*c) + 1/2*(3*c*d^2*e - a*e^3)*log(c*x^2 + a)/c^2 + 1/2*(c*
x^2*e^3 + 6*c*d*x*e^2)/c^2
The text passage contains a mathematical question in the form of an integral and provides a detailed solution, including the antiderivative and the steps used to derive it.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
### 3.499 $$\int \frac{(d+e x)^3}{a+c x^2} \, dx$$

Optimal. Leaf size=90 $\frac{e \left (3 c d^2-a e^2\right ) \log \left (a+c x^2\right )}{2 c^2}+\frac{d \left (c d^2-3 a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a} c^{3/2}}+\frac{3 d e^2 x}{c}+\frac{e^3 x^2}{2 c}$

[Out]

(3*d*e^2*x)/c + (e^3*x^2)/(2*c) + (d*(c*d^2 - 3*a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(3/2)) + (e*(3*
c*d^2 - a*e^2)*Log[a + c*x^2])/(2*c^2)

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Rubi [A]  time = 0.0745295, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.235, Rules used = {702, 635, 205, 260} $\frac{e \left (3 c d^2-a e^2\right ) \log \left (a+c x^2\right )}{2 c^2}+\frac{d \left (c d^2-3 a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a} c^{3/2}}+\frac{3 d e^2 x}{c}+\frac{e^3 x^2}{2 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^3/(a + c*x^2),x]

[Out]

(3*d*e^2*x)/c + (e^3*x^2)/(2*c) + (d*(c*d^2 - 3*a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(3/2)) + (e*(3*
c*d^2 - a*e^2)*Log[a + c*x^2])/(2*c^2)

Rule 702

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{(d+e x)^3}{a+c x^2} \, dx &=\int \left (\frac{3 d e^2}{c}+\frac{e^3 x}{c}+\frac{c d^3-3 a d e^2+e \left (3 c d^2-a e^2\right ) x}{c \left (a+c x^2\right )}\right ) \, dx\\ &=\frac{3 d e^2 x}{c}+\frac{e^3 x^2}{2 c}+\frac{\int \frac{c d^3-3 a d e^2+e \left (3 c d^2-a e^2\right ) x}{a+c x^2} \, dx}{c}\\ &=\frac{3 d e^2 x}{c}+\frac{e^3 x^2}{2 c}+\frac{\left (d \left (c d^2-3 a e^2\right )\right ) \int \frac{1}{a+c x^2} \, dx}{c}+\frac{\left (e \left (3 c d^2-a e^2\right )\right ) \int \frac{x}{a+c x^2} \, dx}{c}\\ &=\frac{3 d e^2 x}{c}+\frac{e^3 x^2}{2 c}+\frac{d \left (c d^2-3 a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a} c^{3/2}}+\frac{e \left (3 c d^2-a e^2\right ) \log \left (a+c x^2\right )}{2 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0565024, size = 80, normalized size = 0.89 $\frac{e \left (\left (3 c d^2-a e^2\right ) \log \left (a+c x^2\right )+c e x (6 d+e x)\right )}{2 c^2}+\frac{d \left (c d^2-3 a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a} c^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^3/(a + c*x^2),x]

[Out]

(d*(c*d^2 - 3*a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(3/2)) + (e*(c*e*x*(6*d + e*x) + (3*c*d^2 - a*e^2
)*Log[a + c*x^2]))/(2*c^2)

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Maple [A]  time = 0.045, size = 99, normalized size = 1.1 \begin{align*}{\frac{{e}^{3}{x}^{2}}{2\,c}}+3\,{\frac{d{e}^{2}x}{c}}-{\frac{\ln \left ( c{x}^{2}+a \right ) a{e}^{3}}{2\,{c}^{2}}}+{\frac{3\,\ln \left ( c{x}^{2}+a \right ){d}^{2}e}{2\,c}}-3\,{\frac{ad{e}^{2}}{c\sqrt{ac}}\arctan \left ({\frac{cx}{\sqrt{ac}}} \right ) }+{{d}^{3}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(c*x^2+a),x)

[Out]

1/2*e^3*x^2/c+3*d*e^2*x/c-1/2/c^2*ln(c*x^2+a)*a*e^3+3/2/c*ln(c*x^2+a)*d^2*e-3/c/(a*c)^(1/2)*arctan(x*c/(a*c)^(
1/2))*a*d*e^2+1/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*d^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.77174, size = 414, normalized size = 4.6 \begin{align*} \left [\frac{a c e^{3} x^{2} + 6 \, a c d e^{2} x +{\left (c d^{3} - 3 \, a d e^{2}\right )} \sqrt{-a c} \log \left (\frac{c x^{2} + 2 \, \sqrt{-a c} x - a}{c x^{2} + a}\right ) +{\left (3 \, a c d^{2} e - a^{2} e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, a c^{2}}, \frac{a c e^{3} x^{2} + 6 \, a c d e^{2} x + 2 \,{\left (c d^{3} - 3 \, a d e^{2}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} x}{a}\right ) +{\left (3 \, a c d^{2} e - a^{2} e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, a c^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+a),x, algorithm="fricas")

[Out]

[1/2*(a*c*e^3*x^2 + 6*a*c*d*e^2*x + (c*d^3 - 3*a*d*e^2)*sqrt(-a*c)*log((c*x^2 + 2*sqrt(-a*c)*x - a)/(c*x^2 + a
)) + (3*a*c*d^2*e - a^2*e^3)*log(c*x^2 + a))/(a*c^2), 1/2*(a*c*e^3*x^2 + 6*a*c*d*e^2*x + 2*(c*d^3 - 3*a*d*e^2)
*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + (3*a*c*d^2*e - a^2*e^3)*log(c*x^2 + a))/(a*c^2)]

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Sympy [B]  time = 0.969667, size = 308, normalized size = 3.42 \begin{align*} \left (- \frac{e \left (a e^{2} - 3 c d^{2}\right )}{2 c^{2}} - \frac{d \sqrt{- a c^{5}} \left (3 a e^{2} - c d^{2}\right )}{2 a c^{4}}\right ) \log{\left (x + \frac{- a^{2} e^{3} - 2 a c^{2} \left (- \frac{e \left (a e^{2} - 3 c d^{2}\right )}{2 c^{2}} - \frac{d \sqrt{- a c^{5}} \left (3 a e^{2} - c d^{2}\right )}{2 a c^{4}}\right ) + 3 a c d^{2} e}{3 a c d e^{2} - c^{2} d^{3}} \right )} + \left (- \frac{e \left (a e^{2} - 3 c d^{2}\right )}{2 c^{2}} + \frac{d \sqrt{- a c^{5}} \left (3 a e^{2} - c d^{2}\right )}{2 a c^{4}}\right ) \log{\left (x + \frac{- a^{2} e^{3} - 2 a c^{2} \left (- \frac{e \left (a e^{2} - 3 c d^{2}\right )}{2 c^{2}} + \frac{d \sqrt{- a c^{5}} \left (3 a e^{2} - c d^{2}\right )}{2 a c^{4}}\right ) + 3 a c d^{2} e}{3 a c d e^{2} - c^{2} d^{3}} \right )} + \frac{3 d e^{2} x}{c} + \frac{e^{3} x^{2}}{2 c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(c*x**2+a),x)

[Out]

(-e*(a*e**2 - 3*c*d**2)/(2*c**2) - d*sqrt(-a*c**5)*(3*a*e**2 - c*d**2)/(2*a*c**4))*log(x + (-a**2*e**3 - 2*a*c
**2*(-e*(a*e**2 - 3*c*d**2)/(2*c**2) - d*sqrt(-a*c**5)*(3*a*e**2 - c*d**2)/(2*a*c**4)) + 3*a*c*d**2*e)/(3*a*c*
d*e**2 - c**2*d**3)) + (-e*(a*e**2 - 3*c*d**2)/(2*c**2) + d*sqrt(-a*c**5)*(3*a*e**2 - c*d**2)/(2*a*c**4))*log(
x + (-a**2*e**3 - 2*a*c**2*(-e*(a*e**2 - 3*c*d**2)/(2*c**2) + d*sqrt(-a*c**5)*(3*a*e**2 - c*d**2)/(2*a*c**4))
+ 3*a*c*d**2*e)/(3*a*c*d*e**2 - c**2*d**3)) + 3*d*e**2*x/c + e**3*x**2/(2*c)

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Giac [A]  time = 1.34387, size = 105, normalized size = 1.17 \begin{align*} \frac{{\left (c d^{3} - 3 \, a d e^{2}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{\sqrt{a c} c} + \frac{{\left (3 \, c d^{2} e - a e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac{c x^{2} e^{3} + 6 \, c d x e^{2}}{2 \, c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+a),x, algorithm="giac")

[Out]

(c*d^3 - 3*a*d*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*c) + 1/2*(3*c*d^2*e - a*e^3)*log(c*x^2 + a)/c^2 + 1/2*(c*
x^2*e^3 + 6*c*d*x*e^2)/c^2
user
The text passage contains a mathematical question in the form of an integral and provides a detailed solution, including the antiderivative and the steps used to derive it.

TYPE: SINGLE_QUESTION_WITH_ANSWER