Optimal. Leaf size=416 \[ -\frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 a^2 e^4+a b d^2 e^2 (7 p+6)+b^2 d^4 \left (2 p^2+5 p+3\right )\right ) F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{e^3 \left (a e^2+b d^2\right )^2}+\frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a^2 e^4+a b d^2 e^2 (6 p+5)+b^2 d^4 \left (2 p^2+5 p+3\right )\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{e^3 \left (a e^2+b d^2\right )^2}+\frac{d \left (a+b x^2\right )^{p+1} \left (3 a^2 e^4+a b d^2 e^2 (7 p+6)+b^2 d^4 \left (2 p^2+5 p+3\right )\right ) \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 e^2 (p+1) \left (a e^2+b d^2\right )^3}-\frac{d^2 \left (a+b x^2\right )^{p+1} \left (3 a e^2+b d^2 (p+2)\right )}{e^2 (d+e x) \left (a e^2+b d^2\right )^2}+\frac{d^3 \left (a+b x^2\right )^{p+1}}{2 e^2 (d+e x)^2 \left (a e^2+b d^2\right )} \]
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Rubi [A] time = 0.591918, antiderivative size = 416, normalized size of antiderivative = 1., number of steps used = 11, 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} \[ -\frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 a^2 e^4+a b d^2 e^2 (7 p+6)+b^2 d^4 \left (2 p^2+5 p+3\right )\right ) F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{e^3 \left (a e^2+b d^2\right )^2}+\frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a^2 e^4+a b d^2 e^2 (6 p+5)+b^2 d^4 \left (2 p^2+5 p+3\right )\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{e^3 \left (a e^2+b d^2\right )^2}+\frac{d \left (a+b x^2\right )^{p+1} \left (3 a^2 e^4+a b d^2 e^2 (7 p+6)+b^2 d^4 \left (2 p^2+5 p+3\right )\right ) \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 e^2 (p+1) \left (a e^2+b d^2\right )^3}-\frac{d^2 \left (a+b x^2\right )^{p+1} \left (3 a e^2+b d^2 (p+2)\right )}{e^2 (d+e x) \left (a e^2+b d^2\right )^2}+\frac{d^3 \left (a+b x^2\right )^{p+1}}{2 e^2 (d+e x)^2 \left (a e^2+b d^2\right )} \]
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^3 \left (a+b x^2\right )^p}{(d+e x)^3} \, dx &=\frac{d^3 \left (a+b x^2\right )^{1+p}}{2 e^2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac{\int \frac{\left (a+b x^2\right )^p \left (-\frac{2 a d^2}{e}+2 d \left (a+\frac{b d^2 (1+p)}{e^2}\right ) x-2 \left (\frac{b d^2}{e}+a e\right ) x^2\right )}{(d+e x)^2} \, dx}{2 \left (b d^2+a e^2\right )}\\ &=\frac{d^3 \left (a+b x^2\right )^{1+p}}{2 e^2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac{d^2 \left (3 a e^2+b d^2 (2+p)\right ) \left (a+b x^2\right )^{1+p}}{e^2 \left (b d^2+a e^2\right )^2 (d+e x)}+\frac{\int \frac{\left (-\frac{2 a d \left (2 a e^2+b d^2 (1+p)\right )}{e}+\frac{2 \left (a^2 e^4+a b d^2 e^2 (5+6 p)+b^2 d^4 \left (3+5 p+2 p^2\right )\right ) x}{e^2}\right ) \left (a+b x^2\right )^p}{d+e x} \, dx}{2 \left (b d^2+a e^2\right )^2}\\ &=\frac{d^3 \left (a+b x^2\right )^{1+p}}{2 e^2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac{d^2 \left (3 a e^2+b d^2 (2+p)\right ) \left (a+b x^2\right )^{1+p}}{e^2 \left (b d^2+a e^2\right )^2 (d+e x)}+\frac{\left (a^2 e^4+a b d^2 e^2 (5+6 p)+b^2 d^4 \left (3+5 p+2 p^2\right )\right ) \int \left (a+b x^2\right )^p \, dx}{e^3 \left (b d^2+a e^2\right )^2}-\frac{\left (d \left (3 a^2 e^4+a b d^2 e^2 (6+7 p)+b^2 d^4 \left (3+5 p+2 p^2\right )\right )\right ) \int \frac{\left (a+b x^2\right )^p}{d+e x} \, dx}{e^3 \left (b d^2+a e^2\right )^2}\\ &=\frac{d^3 \left (a+b x^2\right )^{1+p}}{2 e^2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac{d^2 \left (3 a e^2+b d^2 (2+p)\right ) \left (a+b x^2\right )^{1+p}}{e^2 \left (b d^2+a e^2\right )^2 (d+e x)}-\frac{\left (d \left (3 a^2 e^4+a b d^2 e^2 (6+7 p)+b^2 d^4 \left (3+5 p+2 p^2\right )\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^3 \left (b d^2+a e^2\right )^2}+\frac{\left (\left (a^2 e^4+a b d^2 e^2 (5+6 p)+b^2 d^4 \left (3+5 p+2 p^2\right )\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}{e^3 \left (b d^2+a e^2\right )^2}\\ &=\frac{d^3 \left (a+b x^2\right )^{1+p}}{2 e^2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac{d^2 \left (3 a e^2+b d^2 (2+p)\right ) \left (a+b x^2\right )^{1+p}}{e^2 \left (b d^2+a e^2\right )^2 (d+e x)}+\frac{\left (a^2 e^4+a b d^2 e^2 (5+6 p)+b^2 d^4 \left (3+5 p+2 p^2\right )\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 )}{e^3 \left (b d^2+a e^2\right )^2}-\frac{\left (d^2 \left (3 a^2 e^4+a b d^2 e^2 (6+7 p)+b^2 d^4 \left (3+5 p+2 p^2\right )\right )\right ) \int \frac{\left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx}{e^3 \left (b d^2+a e^2\right )^2}-\frac{\left (d \left (3 a^2 e^4+a b d^2 e^2 (6+7 p)+b^2 d^4 \left (3+5 p+2 p^2\right )\right )\right ) \int \frac{x \left (a+b x^2\right )^p}{-d^2+e^2 x^2} \, dx}{e^2 \left (b d^2+a e^2\right )^2}\\ &=\frac{d^3 \left (a+b x^2\right )^{1+p}}{2 e^2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac{d^2 \left (3 a e^2+b d^2 (2+p)\right ) \left (a+b x^2\right )^{1+p}}{e^2 \left (b d^2+a e^2\right )^2 (d+e x)}+\frac{\left (a^2 e^4+a b d^2 e^2 (5+6 p)+b^2 d^4 \left (3+5 p+2 p^2\right )\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 )}{e^3 \left (b d^2+a e^2\right )^2}-\frac{\left (d \left (3 a^2 e^4+a b d^2 e^2 (6+7 p)+b^2 d^4 \left (3+5 p+2 p^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^p}{-d^2+e^2 x} \, dx,x,x^2\right )}{2 e^2 \left (b d^2+a e^2\right )^2}-\frac{\left (d^2 \left (3 a^2 e^4+a b d^2 e^2 (6+7 p)+b^2 d^4 \left (3+5 p+2 p^2\right )\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^3 \left (b d^2+a e^2\right )^2}\\ &=\frac{d^3 \left (a+b x^2\right )^{1+p}}{2 e^2 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac{d^2 \left (3 a e^2+b d^2 (2+p)\right ) \left (a+b x^2\right )^{1+p}}{e^2 \left (b d^2+a e^2\right )^2 (d+e x)}-\frac{\left (3 a^2 e^4+a b d^2 e^2 (6+7 p)+b^2 d^4 \left (3+5 p+2 p^2\right )\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^3 \left (b d^2+a e^2\right )^2}+\frac{\left (a^2 e^4+a b d^2 e^2 (5+6 p)+b^2 d^4 \left (3+5 p+2 p^2\right )\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 )}{e^3 \left (b d^2+a e^2\right )^2}+\frac{d \left (3 a^2 e^4+a b d^2 e^2 (6+7 p)+b^2 d^4 \left (3+5 p+2 p^2\right )\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 )}{2 e^2 \left (b d^2+a e^2\right )^3 (1+p)}\\ \end{align*}
Mathematica [F] time = 0.633978, size = 0, normalized size = 0. \[ \int \frac{x^3 \left (a+b x^2\right )^p}{(d+e x)^3} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.695, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( b{x}^{2}+a \right ) ^{p}{x}^{3}}{ \left ( ex+d \right ) ^{3}}}\, 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^{3}}{{\left (e x + d\right )}^{3}}\,{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^{3}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, 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^{3}}{{\left (e x + d\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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