Optimal. Leaf size=129 \[ -\frac {a x}{16 \left (1-a^2 x^2\right )^2}-\frac {11 a x}{32 \left (1-a^2 x^2\right )}-\frac {11}{32} \tanh ^{-1}(a x)+\frac {\tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac {\tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+\frac {1}{2} \tanh ^{-1}(a x)^2+\tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )-\frac {1}{2} \text {PolyLog}\left (2,-1+\frac {2}{1+a x}\right ) \]
[Out]
________________________________________________________________________________________
Rubi [A]
time = 0.19, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps
used = 12, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6177, 6135,
6079, 2497, 6141, 205, 212} \begin {gather*} -\frac {11 a x}{32 \left (1-a^2 x^2\right )}-\frac {a x}{16 \left (1-a^2 x^2\right )^2}+\frac {\tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}-\frac {1}{2} \text {Li}_2\left (\frac {2}{a x+1}-1\right )+\frac {1}{2} \tanh ^{-1}(a x)^2-\frac {11}{32} \tanh ^{-1}(a x)+\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 205
Rule 212
Rule 2497
Rule 6079
Rule 6135
Rule 6141
Rule 6177
Rubi steps
\begin {align*} \int \frac {\tanh ^{-1}(a x)}{x \left (1-a^2 x^2\right )^3} \, dx &=a^2 \int \frac {x \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^3} \, dx+\int \frac {\tanh ^{-1}(a x)}{x \left (1-a^2 x^2\right )^2} \, dx\\ &=\frac {\tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}-\frac {1}{4} a \int \frac {1}{\left (1-a^2 x^2\right )^3} \, dx+a^2 \int \frac {x \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx+\int \frac {\tanh ^{-1}(a x)}{x \left (1-a^2 x^2\right )} \, dx\\ &=-\frac {a x}{16 \left (1-a^2 x^2\right )^2}+\frac {\tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac {\tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+\frac {1}{2} \tanh ^{-1}(a x)^2-\frac {1}{16} (3 a) \int \frac {1}{\left (1-a^2 x^2\right )^2} \, dx-\frac {1}{2} a \int \frac {1}{\left (1-a^2 x^2\right )^2} \, dx+\int \frac {\tanh ^{-1}(a x)}{x (1+a x)} \, dx\\ &=-\frac {a x}{16 \left (1-a^2 x^2\right )^2}-\frac {11 a x}{32 \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac {\tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+\frac {1}{2} \tanh ^{-1}(a x)^2+\tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )-\frac {1}{32} (3 a) \int \frac {1}{1-a^2 x^2} \, dx-\frac {1}{4} a \int \frac {1}{1-a^2 x^2} \, dx-a \int \frac {\log \left (2-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac {a x}{16 \left (1-a^2 x^2\right )^2}-\frac {11 a x}{32 \left (1-a^2 x^2\right )}-\frac {11}{32} \tanh ^{-1}(a x)+\frac {\tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}+\frac {\tanh ^{-1}(a x)}{2 \left (1-a^2 x^2\right )}+\frac {1}{2} \tanh ^{-1}(a x)^2+\tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )-\frac {1}{2} \text {Li}_2\left (-1+\frac {2}{1+a x}\right )\\ \end {align*}
________________________________________________________________________________________
Mathematica [A]
time = 0.12, size = 81, normalized size = 0.63 \begin {gather*} \frac {1}{128} \left (64 \tanh ^{-1}(a x)^2+4 \tanh ^{-1}(a x) \left (12 \cosh \left (2 \tanh ^{-1}(a x)\right )+\cosh \left (4 \tanh ^{-1}(a x)\right )+32 \log \left (1-e^{-2 \tanh ^{-1}(a x)}\right )\right )-64 \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(a x)}\right )-24 \sinh \left (2 \tanh ^{-1}(a x)\right )-\sinh \left (4 \tanh ^{-1}(a x)\right )\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(233\) vs.
\(2(115)=230\).
time = 0.92, size = 234, normalized size = 1.81
method | result | size |
derivativedivides | \(\frac {\arctanh \left (a x \right )}{16 \left (a x -1\right )^{2}}-\frac {5 \arctanh \left (a x \right )}{16 \left (a x -1\right )}-\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{2}+\arctanh \left (a x \right ) \ln \left (a x \right )+\frac {\arctanh \left (a x \right )}{16 \left (a x +1\right )^{2}}+\frac {5 \arctanh \left (a x \right )}{16 \left (a x +1\right )}-\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{2}-\frac {\dilog \left (a x \right )}{2}-\frac {\dilog \left (a x +1\right )}{2}-\frac {\ln \left (a x \right ) \ln \left (a x +1\right )}{2}+\frac {\dilog \left (\frac {a x}{2}+\frac {1}{2}\right )}{2}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{4}-\frac {\ln \left (a x -1\right )^{2}}{8}+\frac {\ln \left (a x +1\right )^{2}}{8}-\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{4}-\frac {1}{64 \left (a x -1\right )^{2}}+\frac {11}{64 \left (a x -1\right )}+\frac {11 \ln \left (a x -1\right )}{64}+\frac {1}{64 \left (a x +1\right )^{2}}+\frac {11}{64 \left (a x +1\right )}-\frac {11 \ln \left (a x +1\right )}{64}\) | \(234\) |
default | \(\frac {\arctanh \left (a x \right )}{16 \left (a x -1\right )^{2}}-\frac {5 \arctanh \left (a x \right )}{16 \left (a x -1\right )}-\frac {\arctanh \left (a x \right ) \ln \left (a x -1\right )}{2}+\arctanh \left (a x \right ) \ln \left (a x \right )+\frac {\arctanh \left (a x \right )}{16 \left (a x +1\right )^{2}}+\frac {5 \arctanh \left (a x \right )}{16 \left (a x +1\right )}-\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{2}-\frac {\dilog \left (a x \right )}{2}-\frac {\dilog \left (a x +1\right )}{2}-\frac {\ln \left (a x \right ) \ln \left (a x +1\right )}{2}+\frac {\dilog \left (\frac {a x}{2}+\frac {1}{2}\right )}{2}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{4}-\frac {\ln \left (a x -1\right )^{2}}{8}+\frac {\ln \left (a x +1\right )^{2}}{8}-\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{4}-\frac {1}{64 \left (a x -1\right )^{2}}+\frac {11}{64 \left (a x -1\right )}+\frac {11 \ln \left (a x -1\right )}{64}+\frac {1}{64 \left (a x +1\right )^{2}}+\frac {11}{64 \left (a x +1\right )}-\frac {11 \ln \left (a x +1\right )}{64}\) | \(234\) |
risch | \(\frac {11 \ln \left (a x -1\right )}{128}-\frac {5 \ln \left (a x +1\right ) \left (a x +1\right )}{64 \left (a x -1\right )}+\frac {1}{64 a x -64}-\frac {\ln \left (a x +1\right ) \left (a x +1\right ) \left (a x -3\right )}{128 \left (a x -1\right )^{2}}-\frac {\ln \left (a x +1\right )^{2}}{8}+\frac {\ln \left (a x +1\right )}{32 \left (a x +1\right )^{2}}+\frac {1}{64 \left (a x +1\right )^{2}}-\frac {\dilog \left (a x +1\right )}{2}+\frac {5 \ln \left (a x +1\right )}{32 \left (a x +1\right )}+\frac {5}{32 \left (a x +1\right )}-\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{4}+\frac {\dilog \left (\frac {a x}{2}+\frac {1}{2}\right )}{4}-\frac {11 \ln \left (-a x -1\right )}{128}+\frac {5 \ln \left (-a x +1\right ) \left (-a x +1\right )}{64 \left (-a x -1\right )}-\frac {1}{64 \left (-a x -1\right )}+\frac {\ln \left (-a x +1\right ) \left (-a x +1\right ) \left (-a x -3\right )}{128 \left (-a x -1\right )^{2}}+\frac {\ln \left (-a x +1\right )^{2}}{8}-\frac {\ln \left (-a x +1\right )}{32 \left (-a x +1\right )^{2}}-\frac {1}{64 \left (-a x +1\right )^{2}}+\frac {\dilog \left (-a x +1\right )}{2}-\frac {5 \ln \left (-a x +1\right )}{32 \left (-a x +1\right )}-\frac {5}{32 \left (-a x +1\right )}+\frac {\left (\ln \left (-a x +1\right )-\ln \left (-\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{4}-\frac {\dilog \left (-\frac {a x}{2}+\frac {1}{2}\right )}{4}\) | \(344\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 268 vs.
\(2 (110) = 220\).
time = 0.27, size = 268, normalized size = 2.08 \begin {gather*} \frac {1}{64} \, a {\left (\frac {2 \, {\left (11 \, a^{3} x^{3} + 4 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} - 8 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 4 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} - 13 \, a x\right )}}{a^{5} x^{4} - 2 \, a^{3} x^{2} + a} + \frac {32 \, {\left (\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a} - \frac {32 \, {\left (\log \left (a x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-a x\right )\right )}}{a} + \frac {32 \, {\left (\log \left (-a x + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (a x\right )\right )}}{a} - \frac {11 \, \log \left (a x + 1\right )}{a} + \frac {11 \, \log \left (a x - 1\right )}{a}\right )} - \frac {1}{4} \, {\left (\frac {2 \, a^{2} x^{2} - 3}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1} + 2 \, \log \left (a^{2} x^{2} - 1\right ) - 2 \, \log \left (x^{2}\right )\right )} \operatorname {artanh}\left (a x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\operatorname {atanh}{\left (a x \right )}}{a^{6} x^{7} - 3 a^{4} x^{5} + 3 a^{2} x^{3} - x}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {\mathrm {atanh}\left (a\,x\right )}{x\,{\left (a^2\,x^2-1\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________