Optimal. Leaf size=233 \[ \frac {5 \sqrt {1-a^2 x^2}}{16 a}+\frac {5 \left (1-a^2 x^2\right )^{3/2}}{72 a}+\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a}+\frac {5}{16} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {5}{24} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)-\frac {5 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{8 a}-\frac {5 i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a}+\frac {5 i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a} \]
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Rubi [A]
time = 0.09, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {6089, 6097}
\begin {gather*} \frac {\left (1-a^2 x^2\right )^{5/2}}{30 a}+\frac {5 \left (1-a^2 x^2\right )^{3/2}}{72 a}+\frac {5 \sqrt {1-a^2 x^2}}{16 a}+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)+\frac {5}{24} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac {5}{16} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {5 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{8 a}-\frac {5 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a}+\frac {5 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{16 a} \end {gather*}
Antiderivative was successfully verified.
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Rule 6089
Rule 6097
Rubi steps
\begin {align*} \int \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x) \, dx &=\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a}+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)+\frac {5}{6} \int \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x) \, dx\\ &=\frac {5 \left (1-a^2 x^2\right )^{3/2}}{72 a}+\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a}+\frac {5}{24} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)+\frac {5}{8} \int \sqrt {1-a^2 x^2} \tanh ^{-1}(a x) \, dx\\ &=\frac {5 \sqrt {1-a^2 x^2}}{16 a}+\frac {5 \left (1-a^2 x^2\right )^{3/2}}{72 a}+\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a}+\frac {5}{16} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {5}{24} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)+\frac {5}{16} \int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {5 \sqrt {1-a^2 x^2}}{16 a}+\frac {5 \left (1-a^2 x^2\right )^{3/2}}{72 a}+\frac {\left (1-a^2 x^2\right )^{5/2}}{30 a}+\frac {5}{16} x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {5}{24} x \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)+\frac {1}{6} x \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)-\frac {5 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{8 a}-\frac {5 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a}+\frac {5 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a}\\ \end {align*}
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Mathematica [A]
time = 0.84, size = 224, normalized size = 0.96 \begin {gather*} \frac {299 \sqrt {1-a^2 x^2}-98 a^2 x^2 \sqrt {1-a^2 x^2}+24 a^4 x^4 \sqrt {1-a^2 x^2}+495 a x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-390 a^3 x^3 \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)-225 i \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )+225 i \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-225 i \text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )+225 i \text {PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )}{720 a} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 5.23, size = 193, normalized size = 0.83
method | result | size |
default | \(\frac {\left (120 \arctanh \left (a x \right ) a^{5} x^{5}+24 a^{4} x^{4}-390 a^{3} x^{3} \arctanh \left (a x \right )-98 a^{2} x^{2}+495 a x \arctanh \left (a x \right )+299\right ) \sqrt {-a^{2} x^{2}+1}}{720 a}-\frac {5 i \arctanh \left (a x \right ) \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a}+\frac {5 i \arctanh \left (a x \right ) \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a}-\frac {5 i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a}+\frac {5 i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a}\) | \(193\) |
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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}} \operatorname {atanh}{\left (a x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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 = -0.00 \begin {gather*} \int \mathrm {atanh}\left (a\,x\right )\,{\left (1-a^2\,x^2\right )}^{5/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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