Optimal. Leaf size=392 \[ -\frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a^2 e^4-2 a b d^2 e^2 (3 p+4)-2 b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{b e^4 (2 p+3) \left (a e^2+b d^2\right )}-\frac{2 d^2 x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (2 a e^2+b d^2 (p+2)\right ) F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{e^4 \left (a e^2+b d^2\right )}+\frac{d^3 \left (a+b x^2\right )^{p+1} \left (2 a e^2+b d^2 (p+2)\right ) \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{e^3 (p+1) \left (a e^2+b d^2\right )^2}-\frac{d^4 \left (a+b x^2\right )^{p+1}}{e^3 (d+e x) \left (a e^2+b d^2\right )}-\frac{d (3 p+4) \left (a+b x^2\right )^{p+1}}{b e^3 (p+1) (2 p+3)}+\frac{(d+e x) \left (a+b x^2\right )^{p+1}}{b e^3 (2 p+3)} \]
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Rubi [A] time = 0.888099, antiderivative size = 392, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {1651, 1654, 844, 246, 245, 757, 430, 429, 444, 68} \[ -\frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a^2 e^4-2 a b d^2 e^2 (3 p+4)-2 b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^2}{a}\right )}{b e^4 (2 p+3) \left (a e^2+b d^2\right )}-\frac{2 d^2 x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (2 a e^2+b d^2 (p+2)\right ) F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{e^4 \left (a e^2+b d^2\right )}+\frac{d^3 \left (a+b x^2\right )^{p+1} \left (2 a e^2+b d^2 (p+2)\right ) \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{e^3 (p+1) \left (a e^2+b d^2\right )^2}-\frac{d^4 \left (a+b x^2\right )^{p+1}}{e^3 (d+e x) \left (a e^2+b d^2\right )}-\frac{d (3 p+4) \left (a+b x^2\right )^{p+1}}{b e^3 (p+1) (2 p+3)}+\frac{(d+e x) \left (a+b x^2\right )^{p+1}}{b e^3 (2 p+3)} \]
Antiderivative was successfully verified.
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Rule 1651
Rule 1654
Rule 844
Rule 246
Rule 245
Rule 757
Rule 430
Rule 429
Rule 444
Rule 68
Rubi steps
\begin{align*} \int \frac{x^4 \left (a+b x^2\right )^p}{(d+e x)^2} \, dx &=-\frac{d^4 \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right ) (d+e x)}-\frac{\int \frac{\left (a+b x^2\right )^p \left (\frac{a d^3}{e^2}-\frac{d^2 \left (a e^2+2 b d^2 (1+p)\right ) x}{e^3}+d \left (a+\frac{b d^2}{e^2}\right ) x^2-\frac{\left (b d^2+a e^2\right ) x^3}{e}\right )}{d+e x} \, dx}{b d^2+a e^2}\\ &=-\frac{d^4 \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right ) (d+e x)}+\frac{(d+e x) \left (a+b x^2\right )^{1+p}}{b e^3 (3+2 p)}-\frac{\int \frac{\left (a+b x^2\right )^p \left (a d e \left (a e^2+2 b d^2 (2+p)\right )+\left (a^2 e^4-4 b^2 d^4 (1+p)^2\right ) x+2 b d e \left (b d^2+a e^2\right ) (4+3 p) x^2\right )}{d+e x} \, dx}{b e^3 \left (b d^2+a e^2\right ) (3+2 p)}\\ &=-\frac{d (4+3 p) \left (a+b x^2\right )^{1+p}}{b e^3 (1+p) (3+2 p)}-\frac{d^4 \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right ) (d+e x)}+\frac{(d+e x) \left (a+b x^2\right )^{1+p}}{b e^3 (3+2 p)}-\frac{\int \frac{\left (2 a b d e^3 (1+p) \left (a e^2+2 b d^2 (2+p)\right )+2 b e^2 (1+p) \left (a^2 e^4-2 a b d^2 e^2 (4+3 p)-2 b^2 d^4 \left (6+7 p+2 p^2\right )\right ) x\right ) \left (a+b x^2\right )^p}{d+e x} \, dx}{2 b^2 e^5 \left (b d^2+a e^2\right ) (1+p) (3+2 p)}\\ &=-\frac{d (4+3 p) \left (a+b x^2\right )^{1+p}}{b e^3 (1+p) (3+2 p)}-\frac{d^4 \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right ) (d+e x)}+\frac{(d+e x) \left (a+b x^2\right )^{1+p}}{b e^3 (3+2 p)}-\frac{\left (2 d^3 \left (2 a e^2+b d^2 (2+p)\right )\right ) \int \frac{\left (a+b x^2\right )^p}{d+e x} \, dx}{e^4 \left (b d^2+a e^2\right )}-\frac{\left (a^2 e^4-2 a b d^2 e^2 (4+3 p)-2 b^2 d^4 \left (6+7 p+2 p^2\right )\right ) \int \left (a+b x^2\right )^p \, dx}{b e^4 \left (b d^2+a e^2\right ) (3+2 p)}\\ &=-\frac{d (4+3 p) \left (a+b x^2\right )^{1+p}}{b e^3 (1+p) (3+2 p)}-\frac{d^4 \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right ) (d+e x)}+\frac{(d+e x) \left (a+b x^2\right )^{1+p}}{b e^3 (3+2 p)}-\frac{\left (2 d^3 \left (2 a e^2+b d^2 (2+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^4 \left (b d^2+a e^2\right )}-\frac{\left (\left (a^2 e^4-2 a b d^2 e^2 (4+3 p)-2 b^2 d^4 \left (6+7 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}{b e^4 \left (b d^2+a e^2\right ) (3+2 p)}\\ &=-\frac{d (4+3 p) \left (a+b x^2\right )^{1+p}}{b e^3 (1+p) (3+2 p)}-\frac{d^4 \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right ) (d+e x)}+\frac{(d+e x) \left (a+b x^2\right )^{1+p}}{b e^3 (3+2 p)}-\frac{\left (a^2 e^4-2 a b d^2 e^2 (4+3 p)-2 b^2 d^4 \left (6+7 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 )}{b e^4 \left (b d^2+a e^2\right ) (3+2 p)}-\frac{\left (2 d^4 \left (2 a e^2+b d^2 (2+p)\right )\right ) \int \frac{\left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx}{e^4 \left (b d^2+a e^2\right )}-\frac{\left (2 d^3 \left (2 a e^2+b d^2 (2+p)\right )\right ) \int \frac{x \left (a+b x^2\right )^p}{-d^2+e^2 x^2} \, dx}{e^3 \left (b d^2+a e^2\right )}\\ &=-\frac{d (4+3 p) \left (a+b x^2\right )^{1+p}}{b e^3 (1+p) (3+2 p)}-\frac{d^4 \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right ) (d+e x)}+\frac{(d+e x) \left (a+b x^2\right )^{1+p}}{b e^3 (3+2 p)}-\frac{\left (a^2 e^4-2 a b d^2 e^2 (4+3 p)-2 b^2 d^4 \left (6+7 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 )}{b e^4 \left (b d^2+a e^2\right ) (3+2 p)}-\frac{\left (d^3 \left (2 a e^2+b d^2 (2+p)\right )\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^p}{-d^2+e^2 x} \, dx,x,x^2\right )}{e^3 \left (b d^2+a e^2\right )}-\frac{\left (2 d^4 \left (2 a e^2+b d^2 (2+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^4 \left (b d^2+a e^2\right )}\\ &=-\frac{d (4+3 p) \left (a+b x^2\right )^{1+p}}{b e^3 (1+p) (3+2 p)}-\frac{d^4 \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right ) (d+e x)}+\frac{(d+e x) \left (a+b x^2\right )^{1+p}}{b e^3 (3+2 p)}-\frac{2 d^2 \left (2 a e^2+b d^2 (2+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^4 \left (b d^2+a e^2\right )}-\frac{\left (a^2 e^4-2 a b d^2 e^2 (4+3 p)-2 b^2 d^4 \left (6+7 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 )}{b e^4 \left (b d^2+a e^2\right ) (3+2 p)}+\frac{d^3 \left (2 a e^2+b d^2 (2+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^3 \left (b d^2+a e^2\right )^2 (1+p)}\\ \end{align*}
Mathematica [F] time = 0.726362, size = 0, normalized size = 0. \[ \int \frac{x^4 \left (a+b x^2\right )^p}{(d+e x)^2} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.664, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{4} \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^{4}}{{\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^{4}}{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^{4}}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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