Optimal. Leaf size=94 \[ \frac{4 a (d x)^{m+3} \text{Hypergeometric2F1}\left (1,\frac{m+3}{2},\frac{m+5}{2},a x^2\right )}{d^3 (m+1)^2 (m+3)}+\frac{(d x)^{m+1} \text{PolyLog}\left (2,a x^2\right )}{d (m+1)}+\frac{2 \log \left (1-a x^2\right ) (d x)^{m+1}}{d (m+1)^2} \]
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Rubi [A] time = 0.0549094, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {6591, 2455, 16, 364} \[ \frac{(d x)^{m+1} \text{PolyLog}\left (2,a x^2\right )}{d (m+1)}+\frac{4 a (d x)^{m+3} \, _2F_1\left (1,\frac{m+3}{2};\frac{m+5}{2};a x^2\right )}{d^3 (m+1)^2 (m+3)}+\frac{2 \log \left (1-a x^2\right ) (d x)^{m+1}}{d (m+1)^2} \]
Antiderivative was successfully verified.
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Rule 6591
Rule 2455
Rule 16
Rule 364
Rubi steps
\begin{align*} \int (d x)^m \text{Li}_2\left (a x^2\right ) \, dx &=\frac{(d x)^{1+m} \text{Li}_2\left (a x^2\right )}{d (1+m)}+\frac{2 \int (d x)^m \log \left (1-a x^2\right ) \, dx}{1+m}\\ &=\frac{2 (d x)^{1+m} \log \left (1-a x^2\right )}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_2\left (a x^2\right )}{d (1+m)}+\frac{(4 a) \int \frac{x (d x)^{1+m}}{1-a x^2} \, dx}{d (1+m)^2}\\ &=\frac{2 (d x)^{1+m} \log \left (1-a x^2\right )}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_2\left (a x^2\right )}{d (1+m)}+\frac{(4 a) \int \frac{(d x)^{2+m}}{1-a x^2} \, dx}{d^2 (1+m)^2}\\ &=\frac{4 a (d x)^{3+m} \, _2F_1\left (1,\frac{3+m}{2};\frac{5+m}{2};a x^2\right )}{d^3 (1+m)^2 (3+m)}+\frac{2 (d x)^{1+m} \log \left (1-a x^2\right )}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_2\left (a x^2\right )}{d (1+m)}\\ \end{align*}
Mathematica [A] time = 0.0454423, size = 72, normalized size = 0.77 \[ \frac{x (d x)^m \left (4 a x^2 \text{Hypergeometric2F1}\left (1,\frac{m+3}{2},\frac{m+5}{2},a x^2\right )+(m+3) \left ((m+1) \text{PolyLog}\left (2,a x^2\right )+2 \log \left (1-a x^2\right )\right )\right )}{(m+1)^2 (m+3)} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.236, size = 177, normalized size = 1.9 \begin{align*} -{\frac{ \left ( dx \right ) ^{m}{x}^{-m}}{2} \left ( -a \right ) ^{-{\frac{1}{2}}-{\frac{m}{2}}} \left ( 2\,{\frac{{x}^{1+m} \left ( -a \right ) ^{3/2+m/2} \left ( -12-4\,m \right ) }{ \left ( 1+m \right ) ^{3} \left ( 3+m \right ) a}}-2\,{\frac{{x}^{1+m} \left ( -a \right ) ^{3/2+m/2} \left ( -6-2\,m \right ) \ln \left ( -a{x}^{2}+1 \right ) }{ \left ( 1+m \right ) ^{2} \left ( 3+m \right ) a}}+2\,{\frac{{x}^{1+m} \left ( -a \right ) ^{3/2+m/2}{\it polylog} \left ( 2,a{x}^{2} \right ) }{ \left ( 1+m \right ) a}}+2\,{\frac{{x}^{1+m} \left ( -a \right ) ^{3/2+m/2} \left ( 6+2\,m \right ){\it LerchPhi} \left ( a{x}^{2},1,1/2+m/2 \right ) }{ \left ( 1+m \right ) ^{2} \left ( 3+m \right ) a}} \right ) } \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*} -4 \, a d^{m} \int \frac{x^{2} x^{m}}{{\left (a m^{2} + 2 \, a m + a\right )} x^{2} - m^{2} - 2 \, m - 1}\,{d x} + \frac{{\left (d^{m} m + d^{m}\right )} x x^{m}{\rm Li}_2\left (a x^{2}\right ) + 2 \, d^{m} x x^{m} \log \left (-a x^{2} + 1\right )}{m^{2} + 2 \, m + 1} \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 (d x\right )^{m}{\rm Li}_2\left (a x^{2}\right ), 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 (d x\right )^{m}{\rm Li}_2\left (a x^{2}\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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