Optimal. Leaf size=276 \[ -\frac{e (g x)^{m+2} \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \left (a e^2 (m+2)-3 c d^2 (m+2 p+4)\right ) \, _2F_1\left (\frac{m+2}{2},-p;\frac{m+4}{2};-\frac{c x^2}{a}\right )}{c g^2 (m+2) (m+2 p+4)}-\frac{d (g x)^{m+1} \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \left (3 a e^2 (m+1)-c d^2 (m+2 p+3)\right ) \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};-\frac{c x^2}{a}\right )}{c g (m+1) (m+2 p+3)}+\frac{3 d e^2 (g x)^{m+1} \left (a+c x^2\right )^{p+1}}{c g (m+2 p+3)}+\frac{e^3 (g x)^{m+2} \left (a+c x^2\right )^{p+1}}{c g^2 (m+2 p+4)} \]
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Rubi [A] time = 0.468204, antiderivative size = 254, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {1809, 808, 365, 364} \[ \frac{e (g x)^{m+2} \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \left (\frac{3 d^2}{m+2}-\frac{a e^2}{c (m+2 p+4)}\right ) \, _2F_1\left (\frac{m+2}{2},-p;\frac{m+4}{2};-\frac{c x^2}{a}\right )}{g^2}+\frac{d (g x)^{m+1} \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \left (\frac{d^2}{m+1}-\frac{3 a e^2}{c (m+2 p+3)}\right ) \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};-\frac{c x^2}{a}\right )}{g}+\frac{3 d e^2 (g x)^{m+1} \left (a+c x^2\right )^{p+1}}{c g (m+2 p+3)}+\frac{e^3 (g x)^{m+2} \left (a+c x^2\right )^{p+1}}{c g^2 (m+2 p+4)} \]
Antiderivative was successfully verified.
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Rule 1809
Rule 808
Rule 365
Rule 364
Rubi steps
\begin{align*} \int (g x)^m (d+e x)^3 \left (a+c x^2\right )^p \, dx &=\frac{e^3 (g x)^{2+m} \left (a+c x^2\right )^{1+p}}{c g^2 (4+m+2 p)}+\frac{\int (g x)^m \left (a+c x^2\right )^p \left (c d^3 (4+m+2 p)-e \left (a e^2 (2+m)-3 c d^2 (4+m+2 p)\right ) x+3 c d e^2 (4+m+2 p) x^2\right ) \, dx}{c (4+m+2 p)}\\ &=\frac{3 d e^2 (g x)^{1+m} \left (a+c x^2\right )^{1+p}}{c g (3+m+2 p)}+\frac{e^3 (g x)^{2+m} \left (a+c x^2\right )^{1+p}}{c g^2 (4+m+2 p)}+\frac{\int (g x)^m \left (-c d (4+m+2 p) \left (3 a e^2 (1+m)-c d^2 (3+m+2 p)\right )-c e (3+m+2 p) \left (a e^2 (2+m)-3 c d^2 (4+m+2 p)\right ) x\right ) \left (a+c x^2\right )^p \, dx}{c^2 (3+m+2 p) (4+m+2 p)}\\ &=\frac{3 d e^2 (g x)^{1+m} \left (a+c x^2\right )^{1+p}}{c g (3+m+2 p)}+\frac{e^3 (g x)^{2+m} \left (a+c x^2\right )^{1+p}}{c g^2 (4+m+2 p)}+\left (d \left (d^2-\frac{3 a e^2 (1+m)}{c (3+m+2 p)}\right )\right ) \int (g x)^m \left (a+c x^2\right )^p \, dx+\frac{\left (e \left (3 d^2-\frac{a e^2 (2+m)}{c (4+m+2 p)}\right )\right ) \int (g x)^{1+m} \left (a+c x^2\right )^p \, dx}{g}\\ &=\frac{3 d e^2 (g x)^{1+m} \left (a+c x^2\right )^{1+p}}{c g (3+m+2 p)}+\frac{e^3 (g x)^{2+m} \left (a+c x^2\right )^{1+p}}{c g^2 (4+m+2 p)}+\left (d \left (d^2-\frac{3 a e^2 (1+m)}{c (3+m+2 p)}\right ) \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p}\right ) \int (g x)^m \left (1+\frac{c x^2}{a}\right )^p \, dx+\frac{\left (e \left (3 d^2-\frac{a e^2 (2+m)}{c (4+m+2 p)}\right ) \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p}\right ) \int (g x)^{1+m} \left (1+\frac{c x^2}{a}\right )^p \, dx}{g}\\ &=\frac{3 d e^2 (g x)^{1+m} \left (a+c x^2\right )^{1+p}}{c g (3+m+2 p)}+\frac{e^3 (g x)^{2+m} \left (a+c x^2\right )^{1+p}}{c g^2 (4+m+2 p)}+\frac{d \left (d^2-\frac{3 a e^2 (1+m)}{c (3+m+2 p)}\right ) (g x)^{1+m} \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} \, _2F_1\left (\frac{1+m}{2},-p;\frac{3+m}{2};-\frac{c x^2}{a}\right )}{g (1+m)}+\frac{e \left (3 d^2-\frac{a e^2 (2+m)}{c (4+m+2 p)}\right ) (g x)^{2+m} \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} \, _2F_1\left (\frac{2+m}{2},-p;\frac{4+m}{2};-\frac{c x^2}{a}\right )}{g^2 (2+m)}\\ \end{align*}
Mathematica [A] time = 0.203106, size = 182, normalized size = 0.66 \[ x (g x)^m \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} \left (e x \left (\frac{3 d^2 \, _2F_1\left (\frac{m+2}{2},-p;\frac{m+4}{2};-\frac{c x^2}{a}\right )}{m+2}+e x \left (\frac{3 d \, _2F_1\left (\frac{m+3}{2},-p;\frac{m+5}{2};-\frac{c x^2}{a}\right )}{m+3}+\frac{e x \, _2F_1\left (\frac{m+4}{2},-p;\frac{m+6}{2};-\frac{c x^2}{a}\right )}{m+4}\right )\right )+\frac{d^3 \, _2F_1\left (\frac{m+1}{2},-p;\frac{m+3}{2};-\frac{c x^2}{a}\right )}{m+1}\right ) \]
Antiderivative was successfully verified.
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Maple [F] time = 0.557, size = 0, normalized size = 0. \begin{align*} \int \left ( gx \right ) ^{m} \left ( ex+d \right ) ^{3} \left ( c{x}^{2}+a \right ) ^{p}\, 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{\left (e x + d\right )}^{3}{\left (c x^{2} + a\right )}^{p} \left (g x\right )^{m}\,{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 ({\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )}{\left (c x^{2} + a\right )}^{p} \left (g x\right )^{m}, 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{\left (e x + d\right )}^{3}{\left (c x^{2} + a\right )}^{p} \left (g x\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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