Optimal. Leaf size=281 \[ -\frac{2 x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a e^2+b d^2 (p+1)\right ) F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{e^2 \left (a e^2+b d^2\right )}+\frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a e^2+2 b d^2 (p+1)\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{e^2 \left (a e^2+b d^2\right )}+\frac{d \left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 (p+1)\right ) \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{e (p+1) \left (a e^2+b d^2\right )^2}-\frac{d^2 \left (a+b x^2\right )^{p+1}}{e (d+e x) \left (a e^2+b d^2\right )} \]
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Rubi [A] time = 0.331684, antiderivative size = 277, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {1651, 844, 246, 245, 757, 430, 429, 444, 68} \[ -\frac{2 x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a e^2+b d^2 (p+1)\right ) F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{e^2 \left (a e^2+b d^2\right )}+\frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a+\frac{2 b d^2 (p+1)}{e^2}\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{a e^2+b d^2}+\frac{d \left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 (p+1)\right ) \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{e (p+1) \left (a e^2+b d^2\right )^2}-\frac{d^2 \left (a+b x^2\right )^{p+1}}{e (d+e x) \left (a e^2+b d^2\right )} \]
Antiderivative was successfully verified.
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Rule 1651
Rule 844
Rule 246
Rule 245
Rule 757
Rule 430
Rule 429
Rule 444
Rule 68
Rubi steps
\begin{align*} \int \frac{x^2 \left (a+b x^2\right )^p}{(d+e x)^2} \, dx &=-\frac{d^2 \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right ) (d+e x)}-\frac{\int \frac{\left (a d-\frac{\left (a e^2+2 b d^2 (1+p)\right ) x}{e}\right ) \left (a+b x^2\right )^p}{d+e x} \, dx}{b d^2+a e^2}\\ &=-\frac{d^2 \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right ) (d+e x)}-\frac{\left (2 d \left (a e^2+b d^2 (1+p)\right )\right ) \int \frac{\left (a+b x^2\right )^p}{d+e x} \, dx}{e^2 \left (b d^2+a e^2\right )}+\frac{\left (a+\frac{2 b d^2 (1+p)}{e^2}\right ) \int \left (a+b x^2\right )^p \, dx}{b d^2+a e^2}\\ &=-\frac{d^2 \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right ) (d+e x)}-\frac{\left (2 d \left (a e^2+b d^2 (1+p)\right )\right ) \int \left (\frac{d \left (a+b x^2\right )^p}{d^2-e^2 x^2}+\frac{e x \left (a+b x^2\right )^p}{-d^2+e^2 x^2}\right ) \, dx}{e^2 \left (b d^2+a e^2\right )}+\frac{\left (\left (a+\frac{2 b d^2 (1+p)}{e^2}\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}{b d^2+a e^2}\\ &=-\frac{d^2 \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right ) (d+e x)}+\frac{\left (a+\frac{2 b d^2 (1+p)}{e^2}\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 )}{b d^2+a e^2}-\frac{\left (2 d^2 \left (a e^2+b d^2 (1+p)\right )\right ) \int \frac{\left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx}{e^2 \left (b d^2+a e^2\right )}-\frac{\left (2 d \left (a e^2+b d^2 (1+p)\right )\right ) \int \frac{x \left (a+b x^2\right )^p}{-d^2+e^2 x^2} \, dx}{e \left (b d^2+a e^2\right )}\\ &=-\frac{d^2 \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right ) (d+e x)}+\frac{\left (a+\frac{2 b d^2 (1+p)}{e^2}\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 )}{b d^2+a e^2}-\frac{\left (d \left (a e^2+b d^2 (1+p)\right )\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^p}{-d^2+e^2 x} \, dx,x,x^2\right )}{e \left (b d^2+a e^2\right )}-\frac{\left (2 d^2 \left (a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p}\right ) \int \frac{\left (1+\frac{b x^2}{a}\right )^p}{d^2-e^2 x^2} \, dx}{e^2 \left (b d^2+a e^2\right )}\\ &=-\frac{d^2 \left (a+b x^2\right )^{1+p}}{e \left (b d^2+a e^2\right ) (d+e x)}-\frac{2 \left (a e^2+b d^2 (1+p)\right ) x \left (a+b x^2\right )^p \left (1+\frac{b x^2}{a}\right )^{-p} F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{e^2 \left (b d^2+a e^2\right )}+\frac{\left (a+\frac{2 b d^2 (1+p)}{e^2}\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 )}{b d^2+a e^2}+\frac{d \left (a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;\frac{e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{e \left (b d^2+a e^2\right )^2 (1+p)}\\ \end{align*}
Mathematica [A] time = 0.33191, size = 300, normalized size = 1.07 \[ \frac{\left (a+b x^2\right )^p \left (\frac{d^2 \left (\frac{e \left (x-\sqrt{-\frac{a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac{e \left (\sqrt{-\frac{a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (1-2 p;-p,-p;2-2 p;\frac{d-\sqrt{-\frac{a}{b}} e}{d+e x},\frac{d+\sqrt{-\frac{a}{b}} e}{d+e x}\right )}{(2 p-1) (d+e x)}-\frac{d \left (\frac{e \left (x-\sqrt{-\frac{a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac{e \left (\sqrt{-\frac{a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (-2 p;-p,-p;1-2 p;\frac{d-\sqrt{-\frac{a}{b}} e}{d+e x},\frac{d+\sqrt{-\frac{a}{b}} e}{d+e x}\right )}{p}+e x \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )\right )}{e^3} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.648, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2} \left ( b{x}^{2}+a \right ) ^{p}}{ \left ( ex+d \right ) ^{2}}}\, 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 (b x^{2} + a\right )}^{p} x^{2}}{{\left (e x + d\right )}^{2}}\,{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 (b x^{2} + a\right )}^{p} x^{2}}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 (b x^{2} + a\right )}^{p} x^{2}}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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