Optimal. Leaf size=161 \[ \frac {64 d^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{343 a^{7/4}}-\frac {64 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{343 a^{7/4}}-\frac {8 (d x)^{7/2} \text {Li}_2\left (a x^2\right )}{49 d}+\frac {2 (d x)^{7/2} \text {Li}_3\left (a x^2\right )}{7 d}-\frac {32 (d x)^{7/2} \log \left (1-a x^2\right )}{343 d}+\frac {128 d (d x)^{3/2}}{1029 a}+\frac {128 (d x)^{7/2}}{2401 d} \]
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Rubi [A] time = 0.13, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {6591, 2455, 16, 321, 329, 298, 205, 208} \[ -\frac {8 (d x)^{7/2} \text {PolyLog}\left (2,a x^2\right )}{49 d}+\frac {2 (d x)^{7/2} \text {PolyLog}\left (3,a x^2\right )}{7 d}+\frac {64 d^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{343 a^{7/4}}-\frac {64 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{343 a^{7/4}}-\frac {32 (d x)^{7/2} \log \left (1-a x^2\right )}{343 d}+\frac {128 d (d x)^{3/2}}{1029 a}+\frac {128 (d x)^{7/2}}{2401 d} \]
Antiderivative was successfully verified.
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Rule 16
Rule 205
Rule 208
Rule 298
Rule 321
Rule 329
Rule 2455
Rule 6591
Rubi steps
\begin {align*} \int (d x)^{5/2} \text {Li}_3\left (a x^2\right ) \, dx &=\frac {2 (d x)^{7/2} \text {Li}_3\left (a x^2\right )}{7 d}-\frac {4}{7} \int (d x)^{5/2} \text {Li}_2\left (a x^2\right ) \, dx\\ &=-\frac {8 (d x)^{7/2} \text {Li}_2\left (a x^2\right )}{49 d}+\frac {2 (d x)^{7/2} \text {Li}_3\left (a x^2\right )}{7 d}-\frac {16}{49} \int (d x)^{5/2} \log \left (1-a x^2\right ) \, dx\\ &=-\frac {32 (d x)^{7/2} \log \left (1-a x^2\right )}{343 d}-\frac {8 (d x)^{7/2} \text {Li}_2\left (a x^2\right )}{49 d}+\frac {2 (d x)^{7/2} \text {Li}_3\left (a x^2\right )}{7 d}-\frac {(64 a) \int \frac {x (d x)^{7/2}}{1-a x^2} \, dx}{343 d}\\ &=-\frac {32 (d x)^{7/2} \log \left (1-a x^2\right )}{343 d}-\frac {8 (d x)^{7/2} \text {Li}_2\left (a x^2\right )}{49 d}+\frac {2 (d x)^{7/2} \text {Li}_3\left (a x^2\right )}{7 d}-\frac {(64 a) \int \frac {(d x)^{9/2}}{1-a x^2} \, dx}{343 d^2}\\ &=\frac {128 (d x)^{7/2}}{2401 d}-\frac {32 (d x)^{7/2} \log \left (1-a x^2\right )}{343 d}-\frac {8 (d x)^{7/2} \text {Li}_2\left (a x^2\right )}{49 d}+\frac {2 (d x)^{7/2} \text {Li}_3\left (a x^2\right )}{7 d}-\frac {64}{343} \int \frac {(d x)^{5/2}}{1-a x^2} \, dx\\ &=\frac {128 d (d x)^{3/2}}{1029 a}+\frac {128 (d x)^{7/2}}{2401 d}-\frac {32 (d x)^{7/2} \log \left (1-a x^2\right )}{343 d}-\frac {8 (d x)^{7/2} \text {Li}_2\left (a x^2\right )}{49 d}+\frac {2 (d x)^{7/2} \text {Li}_3\left (a x^2\right )}{7 d}-\frac {\left (64 d^2\right ) \int \frac {\sqrt {d x}}{1-a x^2} \, dx}{343 a}\\ &=\frac {128 d (d x)^{3/2}}{1029 a}+\frac {128 (d x)^{7/2}}{2401 d}-\frac {32 (d x)^{7/2} \log \left (1-a x^2\right )}{343 d}-\frac {8 (d x)^{7/2} \text {Li}_2\left (a x^2\right )}{49 d}+\frac {2 (d x)^{7/2} \text {Li}_3\left (a x^2\right )}{7 d}-\frac {(128 d) \operatorname {Subst}\left (\int \frac {x^2}{1-\frac {a x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{343 a}\\ &=\frac {128 d (d x)^{3/2}}{1029 a}+\frac {128 (d x)^{7/2}}{2401 d}-\frac {32 (d x)^{7/2} \log \left (1-a x^2\right )}{343 d}-\frac {8 (d x)^{7/2} \text {Li}_2\left (a x^2\right )}{49 d}+\frac {2 (d x)^{7/2} \text {Li}_3\left (a x^2\right )}{7 d}-\frac {\left (64 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{d-\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{343 a^{3/2}}+\frac {\left (64 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{d+\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{343 a^{3/2}}\\ &=\frac {128 d (d x)^{3/2}}{1029 a}+\frac {128 (d x)^{7/2}}{2401 d}+\frac {64 d^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{343 a^{7/4}}-\frac {64 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{343 a^{7/4}}-\frac {32 (d x)^{7/2} \log \left (1-a x^2\right )}{343 d}-\frac {8 (d x)^{7/2} \text {Li}_2\left (a x^2\right )}{49 d}+\frac {2 (d x)^{7/2} \text {Li}_3\left (a x^2\right )}{7 d}\\ \end {align*}
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Mathematica [C] time = 0.10, size = 89, normalized size = 0.55 \[ -\frac {11 d \Gamma \left (\frac {11}{4}\right ) (d x)^{3/2} \left (448 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};a x^2\right )+588 a x^2 \text {Li}_2\left (a x^2\right )-1029 a x^2 \text {Li}_3\left (a x^2\right )-192 a x^2+336 a x^2 \log \left (1-a x^2\right )-448\right )}{14406 a \Gamma \left (\frac {15}{4}\right )} \]
Warning: Unable to verify antiderivative.
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fricas [C] time = 0.62, size = 237, normalized size = 1.47 \[ \frac {2 \, {\left (1029 \, \sqrt {d x} a d^{2} x^{3} {\rm polylog}\left (3, a x^{2}\right ) - 1344 \, \left (\frac {d^{10}}{a^{7}}\right )^{\frac {1}{4}} a \arctan \left (-\frac {\left (\frac {d^{10}}{a^{7}}\right )^{\frac {1}{4}} \sqrt {d x} a^{2} d^{7} - \sqrt {d^{15} x + \sqrt {\frac {d^{10}}{a^{7}}} a^{3} d^{10}} \left (\frac {d^{10}}{a^{7}}\right )^{\frac {1}{4}} a^{2}}{d^{10}}\right ) - 336 \, \left (\frac {d^{10}}{a^{7}}\right )^{\frac {1}{4}} a \log \left (32768 \, \sqrt {d x} d^{7} + 32768 \, \left (\frac {d^{10}}{a^{7}}\right )^{\frac {3}{4}} a^{5}\right ) + 336 \, \left (\frac {d^{10}}{a^{7}}\right )^{\frac {1}{4}} a \log \left (32768 \, \sqrt {d x} d^{7} - 32768 \, \left (\frac {d^{10}}{a^{7}}\right )^{\frac {3}{4}} a^{5}\right ) - 4 \, {\left (147 \, a d^{2} x^{3} {\rm Li}_2\left (a x^{2}\right ) + 84 \, a d^{2} x^{3} \log \left (-a x^{2} + 1\right ) - 48 \, a d^{2} x^{3} - 112 \, d^{2} x\right )} \sqrt {d x}\right )}}{7203 \, a} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d x\right )^{\frac {5}{2}} {\rm Li}_{3}(a x^{2})\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 155, normalized size = 0.96 \[ -\frac {\left (d x \right )^{\frac {5}{2}} \left (\frac {4 x^{\frac {3}{2}} \left (-a \right )^{\frac {11}{4}} \left (2112 a \,x^{2}+4928\right )}{79233 a^{2}}+\frac {64 x^{\frac {3}{2}} \left (-a \right )^{\frac {11}{4}} \left (\ln \left (1-\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (a \,x^{2}\right )^{\frac {1}{4}}\right )+2 \arctan \left (\left (a \,x^{2}\right )^{\frac {1}{4}}\right )\right )}{343 a^{2} \left (a \,x^{2}\right )^{\frac {3}{4}}}-\frac {64 x^{\frac {7}{2}} \left (-a \right )^{\frac {11}{4}} \ln \left (-a \,x^{2}+1\right )}{343 a}-\frac {16 x^{\frac {7}{2}} \left (-a \right )^{\frac {11}{4}} \polylog \left (2, a \,x^{2}\right )}{49 a}+\frac {4 x^{\frac {7}{2}} \left (-a \right )^{\frac {11}{4}} \polylog \left (3, a \,x^{2}\right )}{7 a}\right )}{2 x^{\frac {5}{2}} \left (-a \right )^{\frac {7}{4}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 178, normalized size = 1.11 \[ \frac {2 \, {\left (\frac {336 \, d^{4} {\left (\frac {2 \, \arctan \left (\frac {\sqrt {d x} \sqrt {a}}{\sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d} \sqrt {a}} + \frac {\log \left (\frac {\sqrt {d x} \sqrt {a} - \sqrt {\sqrt {a} d}}{\sqrt {d x} \sqrt {a} + \sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d} \sqrt {a}}\right )}}{a} - \frac {588 \, \left (d x\right )^{\frac {7}{2}} a {\rm Li}_2\left (a x^{2}\right ) + 336 \, \left (d x\right )^{\frac {7}{2}} a \log \left (-a d^{2} x^{2} + d^{2}\right ) - 1029 \, \left (d x\right )^{\frac {7}{2}} a {\rm Li}_{3}(a x^{2}) - 96 \, \left (d x\right )^{\frac {7}{2}} {\left (7 \, a \log \relax (d) + 2 \, a\right )} - 448 \, \left (d x\right )^{\frac {3}{2}} d^{2}}{a}\right )}}{7203 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {polylog}\left (3,a\,x^2\right )\,{\left (d\,x\right )}^{5/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d x\right )^{\frac {5}{2}} \operatorname {Li}_{3}\left (a x^{2}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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