Optimal. Leaf size=118 \[ -\frac{d^2 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a (p+1)}+2 d e x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )+\frac{e^2 \left (a+b x^2\right )^{p+1}}{2 b (p+1)} \]
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Rubi [A] time = 0.093041, antiderivative size = 118, normalized size of antiderivative = 1., 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, 12, 246, 245} \[ -\frac{d^2 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{2 a (p+1)}+2 d e x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )+\frac{e^2 \left (a+b x^2\right )^{p+1}}{2 b (p+1)} \]
Antiderivative was successfully verified.
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Rule 1652
Rule 446
Rule 80
Rule 65
Rule 12
Rule 246
Rule 245
Rubi steps
\begin{align*} \int \frac{(d+e x)^2 \left (a+b x^2\right )^p}{x} \, dx &=\int 2 d e \left (a+b x^2\right )^p \, dx+\int \frac{\left (a+b x^2\right )^p \left (d^2+e^2 x^2\right )}{x} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^p \left (d^2+e^2 x\right )}{x} \, dx,x,x^2\right )+(2 d e) \int \left (a+b x^2\right )^p \, dx\\ &=\frac{e^2 \left (a+b x^2\right )^{1+p}}{2 b (1+p)}+\frac{1}{2} d^2 \operatorname{Subst}\left (\int \frac{(a+b x)^p}{x} \, dx,x,x^2\right )+\left (2 d e \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{e^2 \left (a+b x^2\right )^{1+p}}{2 b (1+p)}+2 d e 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^2 \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 = 0.101324, size = 101, normalized size = 0.86 \[ \frac{1}{2} \left (a+b x^2\right )^p \left (\frac{\left (a+b x^2\right ) \left (a e^2-b d^2 \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )\right )}{a b (p+1)}+4 d e x \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )\right ) \]
Antiderivative was successfully verified.
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Maple [F] time = 0.492, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{2} \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 )}^{2}{\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^{2} x^{2} + 2 \, d e x + d^{2}\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 = 19.2026, size = 109, normalized size = 0.92 \begin{align*} 2 a^{p} d 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{b^{p} d^{2} 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 )} + 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 )}^{2}{\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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