Optimal. Leaf size=125 \[ \frac {1}{2} c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{2 x^2}+\frac {3}{2} b c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \log \left (2-\frac {2}{1+c x^2}\right )-\frac {3}{2} b^2 c \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {PolyLog}\left (2,-1+\frac {2}{1+c x^2}\right )-\frac {3}{4} b^3 c \text {PolyLog}\left (3,-1+\frac {2}{1+c x^2}\right ) \]
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Rubi [A]
time = 0.23, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6039, 6037,
6135, 6079, 6095, 6203, 6745} \begin {gather*} -\frac {3}{2} b^2 c \text {Li}_2\left (\frac {2}{c x^2+1}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac {1}{2} c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{2 x^2}+\frac {3}{2} b c \log \left (2-\frac {2}{c x^2+1}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-\frac {3}{4} b^3 c \text {Li}_3\left (\frac {2}{c x^2+1}-1\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 6037
Rule 6039
Rule 6079
Rule 6095
Rule 6135
Rule 6203
Rule 6745
Rubi steps
\begin {align*} \int \frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{x^3} \, dx &=\int \left (\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{8 x^3}+\frac {3 b \left (-2 a+b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{8 x^3}-\frac {3 b^2 \left (-2 a+b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{8 x^3}+\frac {b^3 \log ^3\left (1+c x^2\right )}{8 x^3}\right ) \, dx\\ &=\frac {1}{8} \int \frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{x^3} \, dx+\frac {1}{8} (3 b) \int \frac {\left (-2 a+b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{x^3} \, dx-\frac {1}{8} \left (3 b^2\right ) \int \frac {\left (-2 a+b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{x^3} \, dx+\frac {1}{8} b^3 \int \frac {\log ^3\left (1+c x^2\right )}{x^3} \, dx\\ &=\frac {1}{16} \text {Subst}\left (\int \frac {(2 a-b \log (1-c x))^3}{x^2} \, dx,x,x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )+\frac {1}{16} b^3 \text {Subst}\left (\int \frac {\log ^3(1+c x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )+\frac {1}{16} (3 b c) \text {Subst}\left (\int \frac {(2 a-b \log (1-c x))^2}{x} \, dx,x,x^2\right )+\frac {1}{16} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\log ^2(1+c x)}{x} \, dx,x,x^2\right )\\ &=\frac {3}{16} b c \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}+\frac {3}{16} b^3 c \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^2 c^2\right ) \text {Subst}\left (\int \frac {\log (c x) (2 a-b \log (1-c x))}{1-c x} \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^3 c^2\right ) \text {Subst}\left (\int \frac {\log (-c x) \log (1+c x)}{1+c x} \, dx,x,x^2\right )\\ &=\frac {3}{16} b c \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}+\frac {3}{16} b^3 c \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )+\frac {1}{8} \left (3 b^2 c\right ) \text {Subst}\left (\int \frac {(2 a-b \log (x)) \log \left (c \left (\frac {1}{c}-\frac {x}{c}\right )\right )}{x} \, dx,x,1-c x^2\right )-\frac {1}{8} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\log (x) \log \left (-c \left (-\frac {1}{c}+\frac {x}{c}\right )\right )}{x} \, dx,x,1+c x^2\right )\\ &=\frac {3}{16} b c \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}+\frac {3}{16} b^3 c \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}-\frac {3}{8} b^2 c \left (2 a-b \log \left (1-c x^2\right )\right ) \text {Li}_2\left (1-c x^2\right )+\frac {3}{8} b^3 c \log \left (1+c x^2\right ) \text {Li}_2\left (1+c x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x^2\right )-\frac {1}{8} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1+c x^2\right )\\ &=\frac {3}{16} b c \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}+\frac {3}{16} b^3 c \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}-\frac {3}{8} b^2 c \left (2 a-b \log \left (1-c x^2\right )\right ) \text {Li}_2\left (1-c x^2\right )+\frac {3}{8} b^3 c \log \left (1+c x^2\right ) \text {Li}_2\left (1+c x^2\right )-\frac {3}{8} b^3 c \text {Li}_3\left (1-c x^2\right )-\frac {3}{8} b^3 c \text {Li}_3\left (1+c x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 0.29, size = 222, normalized size = 1.78 \begin {gather*} \frac {1}{4} \left (-\frac {2 a^3}{x^2}-\frac {6 a^2 b \tanh ^{-1}\left (c x^2\right )}{x^2}+12 a^2 b c \log (x)-3 a^2 b c \log \left (1-c^2 x^4\right )+6 a b^2 c \left (\tanh ^{-1}\left (c x^2\right ) \left (\left (1-\frac {1}{c x^2}\right ) \tanh ^{-1}\left (c x^2\right )+2 \log \left (1-e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right )-\text {PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right )+2 b^3 c \left (\frac {i \pi ^3}{8}-\tanh ^{-1}\left (c x^2\right )^3-\frac {\tanh ^{-1}\left (c x^2\right )^3}{c x^2}+3 \tanh ^{-1}\left (c x^2\right )^2 \log \left (1-e^{2 \tanh ^{-1}\left (c x^2\right )}\right )+3 \tanh ^{-1}\left (c x^2\right ) \text {PolyLog}\left (2,e^{2 \tanh ^{-1}\left (c x^2\right )}\right )-\frac {3}{2} \text {PolyLog}\left (3,e^{2 \tanh ^{-1}\left (c x^2\right )}\right )\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arctanh \left (c \,x^{2}\right )\right )^{3}}{x^{3}}\, dx\]
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 \frac {\left (a + b \operatorname {atanh}{\left (c x^{2} \right )}\right )^{3}}{x^{3}}\, dx \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 \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^2\right )\right )}^3}{x^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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