Optimal. Leaf size=142 \[ \frac{16 a (d x)^{m+3} \text{Hypergeometric2F1}\left (1,\frac{m+3}{2},\frac{m+5}{2},a x^2\right )}{d^3 (m+1)^4 (m+3)}+\frac{4 (d x)^{m+1} \text{PolyLog}\left (2,a x^2\right )}{d (m+1)^3}-\frac{2 (d x)^{m+1} \text{PolyLog}\left (3,a x^2\right )}{d (m+1)^2}+\frac{(d x)^{m+1} \text{PolyLog}\left (4,a x^2\right )}{d (m+1)}+\frac{8 \log \left (1-a x^2\right ) (d x)^{m+1}}{d (m+1)^4} \]
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Rubi [A] time = 0.0934271, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {6591, 2455, 16, 364} \[ \frac{4 (d x)^{m+1} \text{PolyLog}\left (2,a x^2\right )}{d (m+1)^3}-\frac{2 (d x)^{m+1} \text{PolyLog}\left (3,a x^2\right )}{d (m+1)^2}+\frac{(d x)^{m+1} \text{PolyLog}\left (4,a x^2\right )}{d (m+1)}+\frac{16 a (d x)^{m+3} \, _2F_1\left (1,\frac{m+3}{2};\frac{m+5}{2};a x^2\right )}{d^3 (m+1)^4 (m+3)}+\frac{8 \log \left (1-a x^2\right ) (d x)^{m+1}}{d (m+1)^4} \]
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}_4\left (a x^2\right ) \, dx &=\frac{(d x)^{1+m} \text{Li}_4\left (a x^2\right )}{d (1+m)}-\frac{2 \int (d x)^m \text{Li}_3\left (a x^2\right ) \, dx}{1+m}\\ &=-\frac{2 (d x)^{1+m} \text{Li}_3\left (a x^2\right )}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_4\left (a x^2\right )}{d (1+m)}+\frac{4 \int (d x)^m \text{Li}_2\left (a x^2\right ) \, dx}{(1+m)^2}\\ &=\frac{4 (d x)^{1+m} \text{Li}_2\left (a x^2\right )}{d (1+m)^3}-\frac{2 (d x)^{1+m} \text{Li}_3\left (a x^2\right )}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_4\left (a x^2\right )}{d (1+m)}+\frac{8 \int (d x)^m \log \left (1-a x^2\right ) \, dx}{(1+m)^3}\\ &=\frac{8 (d x)^{1+m} \log \left (1-a x^2\right )}{d (1+m)^4}+\frac{4 (d x)^{1+m} \text{Li}_2\left (a x^2\right )}{d (1+m)^3}-\frac{2 (d x)^{1+m} \text{Li}_3\left (a x^2\right )}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_4\left (a x^2\right )}{d (1+m)}+\frac{(16 a) \int \frac{x (d x)^{1+m}}{1-a x^2} \, dx}{d (1+m)^4}\\ &=\frac{8 (d x)^{1+m} \log \left (1-a x^2\right )}{d (1+m)^4}+\frac{4 (d x)^{1+m} \text{Li}_2\left (a x^2\right )}{d (1+m)^3}-\frac{2 (d x)^{1+m} \text{Li}_3\left (a x^2\right )}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_4\left (a x^2\right )}{d (1+m)}+\frac{(16 a) \int \frac{(d x)^{2+m}}{1-a x^2} \, dx}{d^2 (1+m)^4}\\ &=\frac{16 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)^4 (3+m)}+\frac{8 (d x)^{1+m} \log \left (1-a x^2\right )}{d (1+m)^4}+\frac{4 (d x)^{1+m} \text{Li}_2\left (a x^2\right )}{d (1+m)^3}-\frac{2 (d x)^{1+m} \text{Li}_3\left (a x^2\right )}{d (1+m)^2}+\frac{(d x)^{1+m} \text{Li}_4\left (a x^2\right )}{d (1+m)}\\ \end{align*}
Mathematica [C] time = 0.0995597, size = 166, normalized size = 1.17 \[ \frac{2 x \text{Gamma}\left (\frac{m+3}{2}\right ) (d x)^m \left (4 a (m+1) x^2 \text{Gamma}\left (\frac{m+1}{2}\right ) \, _2\tilde{F}_1\left (1,\frac{m+3}{2};\frac{m+5}{2};a x^2\right )+m^3 \text{PolyLog}\left (4,a x^2\right )-2 m^2 \text{PolyLog}\left (3,a x^2\right )+3 m^2 \text{PolyLog}\left (4,a x^2\right )-4 m \text{PolyLog}\left (3,a x^2\right )+3 m \text{PolyLog}\left (4,a x^2\right )+4 (m+1) \text{PolyLog}\left (2,a x^2\right )-2 \text{PolyLog}\left (3,a x^2\right )+\text{PolyLog}\left (4,a x^2\right )+8 \log \left (1-a x^2\right )\right )}{(m+1)^5 \text{Gamma}\left (\frac{m+1}{2}\right )} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.792, size = 259, normalized size = 1.8 \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 ( -48-16\,m \right ) }{ \left ( 1+m \right ) ^{5} \left ( 3+m \right ) a}}-2\,{\frac{{x}^{1+m} \left ( -a \right ) ^{3/2+m/2} \left ( -24-8\,m \right ) \ln \left ( -a{x}^{2}+1 \right ) }{ \left ( 1+m \right ) ^{4} \left ( 3+m \right ) a}}+2\,{\frac{{x}^{1+m} \left ( -a \right ) ^{3/2+m/2} \left ( 12+4\,m \right ){\it polylog} \left ( 2,a{x}^{2} \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 ){\it polylog} \left ( 3,a{x}^{2} \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 ( 4,a{x}^{2} \right ) }{ \left ( 1+m \right ) a}}+2\,{\frac{{x}^{1+m} \left ( -a \right ) ^{3/2+m/2} \left ( 24+8\,m \right ){\it LerchPhi} \left ( a{x}^{2},1,1/2+m/2 \right ) }{ \left ( 1+m \right ) ^{4} \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*} -16 \, a d^{m} \int -\frac{x^{2} x^{m}}{m^{4} + 4 \, m^{3} -{\left (a m^{4} + 4 \, a m^{3} + 6 \, a m^{2} + 4 \, a m + a\right )} x^{2} + 6 \, m^{2} + 4 \, m + 1}\,{d x} + \frac{4 \,{\left (d^{m} m + d^{m}\right )} x x^{m}{\rm Li}_2\left (a x^{2}\right ) + 8 \, d^{m} x x^{m} \log \left (-a x^{2} + 1\right ) +{\left (d^{m} m^{3} + 3 \, d^{m} m^{2} + 3 \, d^{m} m + d^{m}\right )} x x^{m}{\rm Li}_{4}(a x^{2}) - 2 \,{\left (d^{m} m^{2} + 2 \, d^{m} m + d^{m}\right )} x x^{m}{\rm Li}_{3}(a x^{2})}{m^{4} + 4 \, m^{3} + 6 \, m^{2} + 4 \, 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 polylog}\left (4, a x^{2}\right ), x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{m} \operatorname{Li}_{4}\left (a x^{2}\right )\, dx \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}_{4}(a x^{2})\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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