Optimal. Leaf size=171 \[ -\frac{e x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a e^2-3 b d^2 (2 p+3)\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{b (2 p+3)}-\frac{d^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a (p+1)}+\frac{3 d e^2 \left (a+b x^2\right )^{p+1}}{2 b (p+1)}+\frac{e^3 x \left (a+b x^2\right )^{p+1}}{b (2 p+3)} \]
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Rubi [A] time = 0.132944, antiderivative size = 165, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1652, 446, 80, 65, 388, 246, 245} \[ e x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 d^2-\frac{a e^2}{2 b p+3 b}\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )-\frac{d^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a (p+1)}+\frac{3 d e^2 \left (a+b x^2\right )^{p+1}}{2 b (p+1)}+\frac{e^3 x \left (a+b x^2\right )^{p+1}}{b (2 p+3)} \]
Antiderivative was successfully verified.
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Rule 1652
Rule 446
Rule 80
Rule 65
Rule 388
Rule 246
Rule 245
Rubi steps
\begin{align*} \int \frac{(d+e x)^3 \left (a+b x^2\right )^p}{x} \, dx &=\int \frac{\left (a+b x^2\right )^p \left (d^3+3 d e^2 x^2\right )}{x} \, dx+\int \left (a+b x^2\right )^p \left (3 d^2 e+e^3 x^2\right ) \, dx\\ &=\frac{e^3 x \left (a+b x^2\right )^{1+p}}{b (3+2 p)}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^p \left (d^3+3 d e^2 x\right )}{x} \, dx,x,x^2\right )+\left (e \left (3 d^2-\frac{a e^2}{3 b+2 b p}\right )\right ) \int \left (a+b x^2\right )^p \, dx\\ &=\frac{3 d e^2 \left (a+b x^2\right )^{1+p}}{2 b (1+p)}+\frac{e^3 x \left (a+b x^2\right )^{1+p}}{b (3+2 p)}+\frac{1}{2} d^3 \operatorname{Subst}\left (\int \frac{(a+b x)^p}{x} \, dx,x,x^2\right )+\left (e \left (3 d^2-\frac{a e^2}{3 b+2 b p}\right ) \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p}\right ) \int \left (1+\frac{b x^2}{a}\right )^p \, dx\\ &=\frac{3 d e^2 \left (a+b x^2\right )^{1+p}}{2 b (1+p)}+\frac{e^3 x \left (a+b x^2\right )^{1+p}}{b (3+2 p)}+e \left (3 d^2-\frac{a e^2}{3 b+2 b p}\right ) x \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )-\frac{d^3 \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;1+\frac{b x^2}{a}\right )}{2 a (1+p)}\\ \end{align*}
Mathematica [A] time = 0.13129, size = 170, normalized size = 0.99 \[ \frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (18 a b d^2 e (p+1) x \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )-3 b d^3 \left (a+b x^2\right ) \left (\frac{b x^2}{a}+1\right )^p \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )+a e^2 \left (9 d \left (a+b x^2\right ) \left (\frac{b x^2}{a}+1\right )^p+2 b e (p+1) x^3 \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )\right )\right )}{6 a b (p+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.513, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{3} \left ( b{x}^{2}+a \right ) ^{p}}{x}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{3}{\left (b x^{2} + a\right )}^{p}}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )}{\left (b x^{2} + a\right )}^{p}}{x}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 28.2809, size = 144, normalized size = 0.84 \begin{align*} 3 a^{p} d^{2} e x{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )} + \frac{a^{p} e^{3} x^{3}{{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, - p \\ \frac{5}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{3} - \frac{b^{p} d^{3} x^{2 p} \Gamma \left (- p\right ){{}_{2}F_{1}\left (\begin{matrix} - p, - p \\ 1 - p \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (1 - p\right )} + 3 d e^{2} \left (\begin{cases} \frac{a^{p} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\begin{cases} \frac{\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (a + b x^{2} \right )} & \text{otherwise} \end{cases}}{2 b} & \text{otherwise} \end{cases}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{3}{\left (b x^{2} + a\right )}^{p}}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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