Optimal. Leaf size=101 \[ \frac {8 a d q^2 x^{2+q} \sqrt {d x} \, _2F_1\left (1,\frac {\frac {5}{2}+q}{q};\frac {1}{2} \left (4+\frac {5}{q}\right );a x^q\right )}{25 (5+2 q)}+\frac {4 q (d x)^{5/2} \log \left (1-a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \text {PolyLog}\left (2,a x^q\right )}{5 d} \]
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Rubi [A]
time = 0.04, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6726, 2505, 20,
371} \begin {gather*} \frac {8 a d q^2 \sqrt {d x} x^{q+2} \, _2F_1\left (1,\frac {q+\frac {5}{2}}{q};\frac {1}{2} \left (4+\frac {5}{q}\right );a x^q\right )}{25 (2 q+5)}+\frac {2 (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{5 d}+\frac {4 q (d x)^{5/2} \log \left (1-a x^q\right )}{25 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 20
Rule 371
Rule 2505
Rule 6726
Rubi steps
\begin {align*} \int (d x)^{3/2} \text {Li}_2\left (a x^q\right ) \, dx &=\frac {2 (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{5 d}+\frac {1}{5} (2 q) \int (d x)^{3/2} \log \left (1-a x^q\right ) \, dx\\ &=\frac {4 q (d x)^{5/2} \log \left (1-a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{5 d}+\frac {\left (4 a q^2\right ) \int \frac {x^{-1+q} (d x)^{5/2}}{1-a x^q} \, dx}{25 d}\\ &=\frac {4 q (d x)^{5/2} \log \left (1-a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{5 d}+\frac {\left (4 a d q^2 \sqrt {d x}\right ) \int \frac {x^{\frac {3}{2}+q}}{1-a x^q} \, dx}{25 \sqrt {x}}\\ &=\frac {8 a d q^2 x^{2+q} \sqrt {d x} \, _2F_1\left (1,\frac {\frac {5}{2}+q}{q};\frac {1}{2} \left (4+\frac {5}{q}\right );a x^q\right )}{25 (5+2 q)}+\frac {4 q (d x)^{5/2} \log \left (1-a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{5 d}\\ \end {align*}
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Mathematica [A]
time = 0.09, size = 82, normalized size = 0.81 \begin {gather*} \frac {2 x (d x)^{3/2} \left (4 a q^2 x^q \, _2F_1\left (1,\frac {\frac {5}{2}+q}{q};2+\frac {5}{2 q};a x^q\right )+(5+2 q) \left (2 q \log \left (1-a x^q\right )+5 \text {PolyLog}\left (2,a x^q\right )\right )\right )}{25 (5+2 q)} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
5.
time = 0.49, size = 121, normalized size = 1.20
method | result | size |
meijerg | \(-\frac {\left (d x \right )^{\frac {3}{2}} \left (-a \right )^{-\frac {5}{2 q}} \left (-\frac {4 q^{2} x^{\frac {5}{2}} \left (-a \right )^{\frac {5}{2 q}} \ln \left (1-a \,x^{q}\right )}{25}-\frac {2 q \,x^{\frac {5}{2}} \left (-a \right )^{\frac {5}{2 q}} \left (1+\frac {2 q}{5}\right ) \polylog \left (2, a \,x^{q}\right )}{5+2 q}-\frac {4 q^{2} x^{\frac {5}{2}+q} a \left (-a \right )^{\frac {5}{2 q}} \Phi \left (a \,x^{q}, 1, \frac {5+2 q}{2 q}\right )}{25}\right )}{x^{\frac {3}{2}} q}\) | \(121\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int {\left (d\,x\right )}^{3/2}\,\mathrm {polylog}\left (2,a\,x^q\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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