Optimal. Leaf size=99 \[ -\frac {\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2}{2 c}-\frac {\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2}{2 x^2}+\frac {b \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right ) \log \left (\frac {2}{1-\frac {c}{x^2}}\right )}{c}+\frac {b^2 \text {PolyLog}\left (2,1-\frac {2}{1-\frac {c}{x^2}}\right )}{2 c} \]
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Rubi [A]
time = 0.10, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6039, 6021,
6131, 6055, 2449, 2352} \begin {gather*} -\frac {\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2}{2 c}-\frac {\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2}{2 x^2}+\frac {b \log \left (\frac {2}{1-\frac {c}{x^2}}\right ) \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )}{c}+\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-\frac {c}{x^2}}\right )}{2 c} \end {gather*}
Antiderivative was successfully verified.
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Rule 2352
Rule 2449
Rule 6021
Rule 6039
Rule 6055
Rule 6131
Rubi steps
\begin {align*} \int \frac {\left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )^2}{x^3} \, dx &=\int \left (\frac {\left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2}{4 x^3}-\frac {b \left (-2 a+b \log \left (1-\frac {c}{x^2}\right )\right ) \log \left (1+\frac {c}{x^2}\right )}{2 x^3}+\frac {b^2 \log ^2\left (1+\frac {c}{x^2}\right )}{4 x^3}\right ) \, dx\\ &=\frac {1}{4} \int \frac {\left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2}{x^3} \, dx-\frac {1}{2} b \int \frac {\left (-2 a+b \log \left (1-\frac {c}{x^2}\right )\right ) \log \left (1+\frac {c}{x^2}\right )}{x^3} \, dx+\frac {1}{4} b^2 \int \frac {\log ^2\left (1+\frac {c}{x^2}\right )}{x^3} \, dx\\ &=-\left (\frac {1}{8} \text {Subst}\left (\int (2 a-b \log (1-c x))^2 \, dx,x,\frac {1}{x^2}\right )\right )+\frac {1}{4} b \text {Subst}\left (\int (-2 a+b \log (1-c x)) \log (1+c x) \, dx,x,\frac {1}{x^2}\right )-\frac {1}{8} b^2 \text {Subst}\left (\int \log ^2(1+c x) \, dx,x,\frac {1}{x^2}\right )\\ &=-\frac {b \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) \log \left (1+\frac {c}{x^2}\right )}{4 x^2}+\frac {\text {Subst}\left (\int (2 a-b \log (x))^2 \, dx,x,1-\frac {c}{x^2}\right )}{8 c}-\frac {b^2 \text {Subst}\left (\int \log ^2(x) \, dx,x,1+\frac {c}{x^2}\right )}{8 c}-\frac {1}{4} (b c) \text {Subst}\left (\int \frac {x (-2 a+b \log (1-c x))}{1+c x} \, dx,x,\frac {1}{x^2}\right )+\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {x \log (1+c x)}{1-c x} \, dx,x,\frac {1}{x^2}\right )\\ &=\frac {\left (1-\frac {c}{x^2}\right ) \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2}{8 c}-\frac {b \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) \log \left (1+\frac {c}{x^2}\right )}{4 x^2}-\frac {b^2 \left (1+\frac {c}{x^2}\right ) \log ^2\left (1+\frac {c}{x^2}\right )}{8 c}+\frac {b \text {Subst}\left (\int (2 a-b \log (x)) \, dx,x,1-\frac {c}{x^2}\right )}{4 c}+\frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1+\frac {c}{x^2}\right )}{4 c}-\frac {1}{4} (b c) \text {Subst}\left (\int \left (\frac {-2 a+b \log (1-c x)}{c}-\frac {-2 a+b \log (1-c x)}{c (1+c x)}\right ) \, dx,x,\frac {1}{x^2}\right )+\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \left (-\frac {\log (1+c x)}{c}-\frac {\log (1+c x)}{c (-1+c x)}\right ) \, dx,x,\frac {1}{x^2}\right )\\ &=-\frac {a b}{2 x^2}-\frac {b^2}{4 x^2}+\frac {\left (1-\frac {c}{x^2}\right ) \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2}{8 c}+\frac {b^2 \left (1+\frac {c}{x^2}\right ) \log \left (1+\frac {c}{x^2}\right )}{4 c}-\frac {b \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) \log \left (1+\frac {c}{x^2}\right )}{4 x^2}-\frac {b^2 \left (1+\frac {c}{x^2}\right ) \log ^2\left (1+\frac {c}{x^2}\right )}{8 c}-\frac {1}{4} b \text {Subst}\left (\int (-2 a+b \log (1-c x)) \, dx,x,\frac {1}{x^2}\right )+\frac {1}{4} b \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{1+c x} \, dx,x,\frac {1}{x^2}\right )-\frac {1}{4} b^2 \text {Subst}\left (\int \log (1+c x) \, dx,x,\frac {1}{x^2}\right )-\frac {1}{4} b^2 \text {Subst}\left (\int \frac {\log (1+c x)}{-1+c x} \, dx,x,\frac {1}{x^2}\right )-\frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1-\frac {c}{x^2}\right )}{4 c}\\ &=-\frac {b^2}{2 x^2}-\frac {b^2 \left (1-\frac {c}{x^2}\right ) \log \left (1-\frac {c}{x^2}\right )}{4 c}+\frac {\left (1-\frac {c}{x^2}\right ) \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2}{8 c}-\frac {b \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) \log \left (\frac {1}{2} \left (1+\frac {c}{x^2}\right )\right )}{4 c}+\frac {b^2 \left (1+\frac {c}{x^2}\right ) \log \left (1+\frac {c}{x^2}\right )}{4 c}-\frac {b^2 \log \left (\frac {1}{2} \left (1-\frac {c}{x^2}\right )\right ) \log \left (1+\frac {c}{x^2}\right )}{4 c}-\frac {b \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) \log \left (1+\frac {c}{x^2}\right )}{4 x^2}-\frac {b^2 \left (1+\frac {c}{x^2}\right ) \log ^2\left (1+\frac {c}{x^2}\right )}{8 c}+\frac {1}{4} b^2 \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1-c x)\right )}{1+c x} \, dx,x,\frac {1}{x^2}\right )-\frac {1}{4} b^2 \text {Subst}\left (\int \log (1-c x) \, dx,x,\frac {1}{x^2}\right )+\frac {1}{4} b^2 \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1+c x)\right )}{1-c x} \, dx,x,\frac {1}{x^2}\right )-\frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1+\frac {c}{x^2}\right )}{4 c}\\ &=-\frac {b^2}{4 x^2}-\frac {b^2 \left (1-\frac {c}{x^2}\right ) \log \left (1-\frac {c}{x^2}\right )}{4 c}+\frac {\left (1-\frac {c}{x^2}\right ) \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2}{8 c}-\frac {b \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) \log \left (\frac {1}{2} \left (1+\frac {c}{x^2}\right )\right )}{4 c}-\frac {b^2 \log \left (\frac {1}{2} \left (1-\frac {c}{x^2}\right )\right ) \log \left (1+\frac {c}{x^2}\right )}{4 c}-\frac {b \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) \log \left (1+\frac {c}{x^2}\right )}{4 x^2}-\frac {b^2 \left (1+\frac {c}{x^2}\right ) \log ^2\left (1+\frac {c}{x^2}\right )}{8 c}-\frac {b^2 \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1-\frac {c}{x^2}\right )}{4 c}+\frac {b^2 \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1+\frac {c}{x^2}\right )}{4 c}+\frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1-\frac {c}{x^2}\right )}{4 c}\\ &=\frac {\left (1-\frac {c}{x^2}\right ) \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right )^2}{8 c}-\frac {b \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) \log \left (\frac {1}{2} \left (1+\frac {c}{x^2}\right )\right )}{4 c}-\frac {b^2 \log \left (\frac {1}{2} \left (1-\frac {c}{x^2}\right )\right ) \log \left (1+\frac {c}{x^2}\right )}{4 c}-\frac {b \left (2 a-b \log \left (1-\frac {c}{x^2}\right )\right ) \log \left (1+\frac {c}{x^2}\right )}{4 x^2}-\frac {b^2 \left (1+\frac {c}{x^2}\right ) \log ^2\left (1+\frac {c}{x^2}\right )}{8 c}+\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1-\frac {c}{x^2}\right )\right )}{4 c}-\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1+\frac {c}{x^2}\right )\right )}{4 c}\\ \end {align*}
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Mathematica [A]
time = 0.12, size = 114, normalized size = 1.15 \begin {gather*} -\frac {a^2}{2 x^2}-\frac {a b \left (\frac {c \tanh ^{-1}\left (\frac {c}{x^2}\right )}{x^2}-\log \left (\frac {1}{\sqrt {1-\frac {c^2}{x^4}}}\right )\right )}{c}-\frac {b^2 \left (\tanh ^{-1}\left (\frac {c}{x^2}\right ) \left (-\tanh ^{-1}\left (\frac {c}{x^2}\right )+\frac {c \tanh ^{-1}\left (\frac {c}{x^2}\right )}{x^2}-2 \log \left (1+e^{-2 \tanh ^{-1}\left (\frac {c}{x^2}\right )}\right )\right )+\text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}\left (\frac {c}{x^2}\right )}\right )\right )}{2 c} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.64, size = 137, normalized size = 1.38
method | result | size |
derivativedivides | \(-\frac {\frac {c \,a^{2}}{x^{2}}+\frac {\arctanh \left (\frac {c}{x^{2}}\right )^{2} b^{2} c}{x^{2}}+b^{2} \arctanh \left (\frac {c}{x^{2}}\right )^{2}-2 \arctanh \left (\frac {c}{x^{2}}\right ) \ln \left (1+\frac {\left (1+\frac {c}{x^{2}}\right )^{2}}{1-\frac {c^{2}}{x^{4}}}\right ) b^{2}-\polylog \left (2, -\frac {\left (1+\frac {c}{x^{2}}\right )^{2}}{1-\frac {c^{2}}{x^{4}}}\right ) b^{2}+\frac {2 a b c \arctanh \left (\frac {c}{x^{2}}\right )}{x^{2}}+a b \ln \left (1-\frac {c^{2}}{x^{4}}\right )}{2 c}\) | \(137\) |
default | \(-\frac {\frac {c \,a^{2}}{x^{2}}+\frac {\arctanh \left (\frac {c}{x^{2}}\right )^{2} b^{2} c}{x^{2}}+b^{2} \arctanh \left (\frac {c}{x^{2}}\right )^{2}-2 \arctanh \left (\frac {c}{x^{2}}\right ) \ln \left (1+\frac {\left (1+\frac {c}{x^{2}}\right )^{2}}{1-\frac {c^{2}}{x^{4}}}\right ) b^{2}-\polylog \left (2, -\frac {\left (1+\frac {c}{x^{2}}\right )^{2}}{1-\frac {c^{2}}{x^{4}}}\right ) b^{2}+\frac {2 a b c \arctanh \left (\frac {c}{x^{2}}\right )}{x^{2}}+a b \ln \left (1-\frac {c^{2}}{x^{4}}\right )}{2 c}\) | \(137\) |
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 (\frac {c}{x^{2}} \right )}\right )^{2}}{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 (\frac {c}{x^2}\right )\right )}^2}{x^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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