Optimal. Leaf size=243 \[ -\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a^3}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a^3}-\frac {\tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{8 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^2}-\frac {1}{6} a^2 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {7}{24} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{72 a^3}+\frac {\sqrt {1-a^2 x^2}}{16 a^3} \]
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Rubi [A] time = 0.57, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6014, 6010, 6016, 261, 5950, 266, 43} \[ -\frac {i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a^3}+\frac {i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a^3}-\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{72 a^3}+\frac {\sqrt {1-a^2 x^2}}{16 a^3}-\frac {1}{6} a^2 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {7}{24} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^2}-\frac {\tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{8 a^3} \]
Antiderivative was successfully verified.
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Rule 43
Rule 261
Rule 266
Rule 5950
Rule 6010
Rule 6014
Rule 6016
Rubi steps
\begin {align*} \int x^2 \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x) \, dx &=-\left (a^2 \int x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x) \, dx\right )+\int x^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x) \, dx\\ &=\frac {1}{4} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{6} a^2 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {1}{4} \int \frac {x^2 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx-\frac {1}{4} a \int \frac {x^3}{\sqrt {1-a^2 x^2}} \, dx-\frac {1}{6} a^2 \int \frac {x^4 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx+\frac {1}{6} a^3 \int \frac {x^5}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^2}+\frac {7}{24} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{6} a^2 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{8} \int \frac {x^2 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx+\frac {\int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{8 a^2}+\frac {\int \frac {x}{\sqrt {1-a^2 x^2}} \, dx}{8 a}-\frac {1}{24} a \int \frac {x^3}{\sqrt {1-a^2 x^2}} \, dx-\frac {1}{8} a \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )+\frac {1}{12} a^3 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {1-a^2 x^2}}{8 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^2}+\frac {7}{24} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{6} a^2 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {\tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{4 a^3}-\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a^3}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a^3}-\frac {\int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{16 a^2}-\frac {\int \frac {x}{\sqrt {1-a^2 x^2}} \, dx}{16 a}-\frac {1}{48} a \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )-\frac {1}{8} a \operatorname {Subst}\left (\int \left (\frac {1}{a^2 \sqrt {1-a^2 x}}-\frac {\sqrt {1-a^2 x}}{a^2}\right ) \, dx,x,x^2\right )+\frac {1}{12} a^3 \operatorname {Subst}\left (\int \left (\frac {1}{a^4 \sqrt {1-a^2 x}}-\frac {2 \sqrt {1-a^2 x}}{a^4}+\frac {\left (1-a^2 x\right )^{3/2}}{a^4}\right ) \, dx,x,x^2\right )\\ &=\frac {\sqrt {1-a^2 x^2}}{48 a^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{36 a^3}-\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^2}+\frac {7}{24} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{6} a^2 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {\tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{8 a^3}-\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a^3}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a^3}-\frac {1}{48} a \operatorname {Subst}\left (\int \left (\frac {1}{a^2 \sqrt {1-a^2 x}}-\frac {\sqrt {1-a^2 x}}{a^2}\right ) \, dx,x,x^2\right )\\ &=\frac {\sqrt {1-a^2 x^2}}{16 a^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{72 a^3}-\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a^3}-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^2}+\frac {7}{24} x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{6} a^2 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {\tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{8 a^3}-\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a^3}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a^3}\\ \end {align*}
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Mathematica [A] time = 0.97, size = 224, normalized size = 0.92 \[ \frac {38 a^2 x^2 \sqrt {1-a^2 x^2}+31 \sqrt {1-a^2 x^2}-45 a x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-120 a^5 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-24 a^4 x^4 \sqrt {1-a^2 x^2}+210 a^3 x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-45 i \text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )+45 i \text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )-45 i \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )+45 i \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )}{720 a^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a^{2} x^{4} - x^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right ), x\right ) \]
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 \[ \int {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2} \operatorname {artanh}\left (a x\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.36, size = 195, normalized size = 0.80 \[ -\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (120 \arctanh \left (a x \right ) x^{5} a^{5}+24 x^{4} a^{4}-210 a^{3} x^{3} \arctanh \left (a x \right )-38 a^{2} x^{2}+45 a x \arctanh \left (a x \right )-31\right )}{720 a^{3}}-\frac {i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{16 a^{3}}+\frac {i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{16 a^{3}}-\frac {i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a^{3}}+\frac {i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a^{3}} \]
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 \[ \int {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2} \operatorname {artanh}\left (a x\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,\mathrm {atanh}\left (a\,x\right )\,{\left (1-a^2\,x^2\right )}^{3/2} \,d x \]
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 \[ \int x^{2} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \operatorname {atanh}{\left (a x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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