Optimal. Leaf size=322 \[ \frac{x \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^3}+\frac{e^2 x^3 \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{3}{2};-p,3;\frac{5}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}-\frac{3 c^2 d^2 e \left (a+c x^2\right )^{p+1} \, _2F_1\left (3,p+1;p+2;\frac{e^2 \left (c x^2+a\right )}{c d^2+a e^2}\right )}{2 (p+1) \left (a e^2+c d^2\right )^3}+\frac{c e \left (a+c x^2\right )^{p+1} \left (2 a e^2+c d^2 (p+1)\right ) \, _2F_1\left (2,p+1;p+2;\frac{e^2 \left (c x^2+a\right )}{c d^2+a e^2}\right )}{4 (p+1) \left (a e^2+c d^2\right )^3}-\frac{d^2 e \left (a+c x^2\right )^{p+1}}{4 \left (d^2-e^2 x^2\right )^2 \left (a e^2+c d^2\right )} \]
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Rubi [A] time = 0.315816, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.529, Rules used = {757, 430, 429, 444, 68, 511, 510, 446, 78} \[ \frac{x \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^3}+\frac{e^2 x^3 \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{3}{2};-p,3;\frac{5}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}-\frac{3 c^2 d^2 e \left (a+c x^2\right )^{p+1} \, _2F_1\left (3,p+1;p+2;\frac{e^2 \left (c x^2+a\right )}{c d^2+a e^2}\right )}{2 (p+1) \left (a e^2+c d^2\right )^3}+\frac{c e \left (a+c x^2\right )^{p+1} \left (2 a e^2+c d^2 (p+1)\right ) \, _2F_1\left (2,p+1;p+2;\frac{e^2 \left (c x^2+a\right )}{c d^2+a e^2}\right )}{4 (p+1) \left (a e^2+c d^2\right )^3}-\frac{d^2 e \left (a+c x^2\right )^{p+1}}{4 \left (d^2-e^2 x^2\right )^2 \left (a e^2+c d^2\right )} \]
Antiderivative was successfully verified.
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Rule 757
Rule 430
Rule 429
Rule 444
Rule 68
Rule 511
Rule 510
Rule 446
Rule 78
Rubi steps
\begin{align*} \int \frac{\left (a+c x^2\right )^p}{(d+e x)^3} \, dx &=\int \left (\frac{d^3 \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}-\frac{3 d^2 e x \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}+\frac{3 d e^2 x^2 \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}+\frac{e^3 x^3 \left (a+c x^2\right )^p}{\left (-d^2+e^2 x^2\right )^3}\right ) \, dx\\ &=d^3 \int \frac{\left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx-\left (3 d^2 e\right ) \int \frac{x \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx+\left (3 d e^2\right ) \int \frac{x^2 \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx+e^3 \int \frac{x^3 \left (a+c x^2\right )^p}{\left (-d^2+e^2 x^2\right )^3} \, dx\\ &=-\left (\frac{1}{2} \left (3 d^2 e\right ) \operatorname{Subst}\left (\int \frac{(a+c x)^p}{\left (d^2-e^2 x\right )^3} \, dx,x,x^2\right )\right )+\frac{1}{2} e^3 \operatorname{Subst}\left (\int \frac{x (a+c x)^p}{\left (-d^2+e^2 x\right )^3} \, dx,x,x^2\right )+\left (d^3 \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p}\right ) \int \frac{\left (1+\frac{c x^2}{a}\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx+\left (3 d e^2 \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p}\right ) \int \frac{x^2 \left (1+\frac{c x^2}{a}\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx\\ &=-\frac{d^2 e \left (a+c x^2\right )^{1+p}}{4 \left (c d^2+a e^2\right ) \left (d^2-e^2 x^2\right )^2}+\frac{x \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^3}+\frac{e^2 x^3 \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} F_1\left (\frac{3}{2};-p,3;\frac{5}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}-\frac{3 c^2 d^2 e \left (a+c x^2\right )^{1+p} \, _2F_1\left (3,1+p;2+p;\frac{e^2 \left (a+c x^2\right )}{c d^2+a e^2}\right )}{2 \left (c d^2+a e^2\right )^3 (1+p)}+\frac{\left (e \left (2 a e^2+c d^2 (1+p)\right )\right ) \operatorname{Subst}\left (\int \frac{(a+c x)^p}{\left (-d^2+e^2 x\right )^2} \, dx,x,x^2\right )}{4 \left (c d^2+a e^2\right )}\\ &=-\frac{d^2 e \left (a+c x^2\right )^{1+p}}{4 \left (c d^2+a e^2\right ) \left (d^2-e^2 x^2\right )^2}+\frac{x \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^3}+\frac{e^2 x^3 \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} F_1\left (\frac{3}{2};-p,3;\frac{5}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}+\frac{c e \left (2 a e^2+c d^2 (1+p)\right ) \left (a+c x^2\right )^{1+p} \, _2F_1\left (2,1+p;2+p;\frac{e^2 \left (a+c x^2\right )}{c d^2+a e^2}\right )}{4 \left (c d^2+a e^2\right )^3 (1+p)}-\frac{3 c^2 d^2 e \left (a+c x^2\right )^{1+p} \, _2F_1\left (3,1+p;2+p;\frac{e^2 \left (a+c x^2\right )}{c d^2+a e^2}\right )}{2 \left (c d^2+a e^2\right )^3 (1+p)}\\ \end{align*}
Mathematica [A] time = 0.121307, size = 142, normalized size = 0.44 \[ \frac{\left (a+c x^2\right )^p \left (\frac{e \left (x-\sqrt{-\frac{a}{c}}\right )}{d+e x}\right )^{-p} \left (\frac{e \left (\sqrt{-\frac{a}{c}}+x\right )}{d+e x}\right )^{-p} F_1\left (2-2 p;-p,-p;3-2 p;\frac{d-\sqrt{-\frac{a}{c}} e}{d+e x},\frac{d+\sqrt{-\frac{a}{c}} e}{d+e x}\right )}{2 e (p-1) (d+e x)^2} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.58, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+a \right ) ^{p}}{ \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 (c x^{2} + a\right )}^{p}}{{\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 (c x^{2} + a\right )}^{p}}{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 (c x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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