Optimal. Leaf size=114 \[ \frac {3}{2} b^2 \text {Li}_3\left (-e^{2 \text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )-\frac {3}{2} b \text {Li}_2\left (-e^{2 \text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )^2+\frac {\left (a+b \text {sech}^{-1}(c x)\right )^4}{4 b}-\log \left (e^{2 \text {sech}^{-1}(c x)}+1\right ) \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {3}{4} b^3 \text {Li}_4\left (-e^{2 \text {sech}^{-1}(c x)}\right ) \]
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Rubi [A] time = 0.14, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6285, 3718, 2190, 2531, 6609, 2282, 6589} \[ \frac {3}{2} b^2 \text {PolyLog}\left (3,-e^{2 \text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )-\frac {3}{2} b \text {PolyLog}\left (2,-e^{2 \text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )^2-\frac {3}{4} b^3 \text {PolyLog}\left (4,-e^{2 \text {sech}^{-1}(c x)}\right )+\frac {\left (a+b \text {sech}^{-1}(c x)\right )^4}{4 b}-\log \left (e^{2 \text {sech}^{-1}(c x)}+1\right ) \left (a+b \text {sech}^{-1}(c x)\right )^3 \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3718
Rule 6285
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int \frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{x} \, dx &=-\operatorname {Subst}\left (\int (a+b x)^3 \tanh (x) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=\frac {\left (a+b \text {sech}^{-1}(c x)\right )^4}{4 b}-2 \operatorname {Subst}\left (\int \frac {e^{2 x} (a+b x)^3}{1+e^{2 x}} \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=\frac {\left (a+b \text {sech}^{-1}(c x)\right )^4}{4 b}-\left (a+b \text {sech}^{-1}(c x)\right )^3 \log \left (1+e^{2 \text {sech}^{-1}(c x)}\right )+(3 b) \operatorname {Subst}\left (\int (a+b x)^2 \log \left (1+e^{2 x}\right ) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=\frac {\left (a+b \text {sech}^{-1}(c x)\right )^4}{4 b}-\left (a+b \text {sech}^{-1}(c x)\right )^3 \log \left (1+e^{2 \text {sech}^{-1}(c x)}\right )-\frac {3}{2} b \left (a+b \text {sech}^{-1}(c x)\right )^2 \text {Li}_2\left (-e^{2 \text {sech}^{-1}(c x)}\right )+\left (3 b^2\right ) \operatorname {Subst}\left (\int (a+b x) \text {Li}_2\left (-e^{2 x}\right ) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=\frac {\left (a+b \text {sech}^{-1}(c x)\right )^4}{4 b}-\left (a+b \text {sech}^{-1}(c x)\right )^3 \log \left (1+e^{2 \text {sech}^{-1}(c x)}\right )-\frac {3}{2} b \left (a+b \text {sech}^{-1}(c x)\right )^2 \text {Li}_2\left (-e^{2 \text {sech}^{-1}(c x)}\right )+\frac {3}{2} b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \text {Li}_3\left (-e^{2 \text {sech}^{-1}(c x)}\right )-\frac {1}{2} \left (3 b^3\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (-e^{2 x}\right ) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=\frac {\left (a+b \text {sech}^{-1}(c x)\right )^4}{4 b}-\left (a+b \text {sech}^{-1}(c x)\right )^3 \log \left (1+e^{2 \text {sech}^{-1}(c x)}\right )-\frac {3}{2} b \left (a+b \text {sech}^{-1}(c x)\right )^2 \text {Li}_2\left (-e^{2 \text {sech}^{-1}(c x)}\right )+\frac {3}{2} b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \text {Li}_3\left (-e^{2 \text {sech}^{-1}(c x)}\right )-\frac {1}{4} \left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{2 \text {sech}^{-1}(c x)}\right )\\ &=\frac {\left (a+b \text {sech}^{-1}(c x)\right )^4}{4 b}-\left (a+b \text {sech}^{-1}(c x)\right )^3 \log \left (1+e^{2 \text {sech}^{-1}(c x)}\right )-\frac {3}{2} b \left (a+b \text {sech}^{-1}(c x)\right )^2 \text {Li}_2\left (-e^{2 \text {sech}^{-1}(c x)}\right )+\frac {3}{2} b^2 \left (a+b \text {sech}^{-1}(c x)\right ) \text {Li}_3\left (-e^{2 \text {sech}^{-1}(c x)}\right )-\frac {3}{4} b^3 \text {Li}_4\left (-e^{2 \text {sech}^{-1}(c x)}\right )\\ \end {align*}
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Mathematica [A] time = 0.25, size = 182, normalized size = 1.60 \[ \frac {1}{4} \left (4 a^3 \log (c x)-6 a^2 b \text {sech}^{-1}(c x)^2-12 a^2 b \text {sech}^{-1}(c x) \log \left (e^{-2 \text {sech}^{-1}(c x)}+1\right )+6 b^2 \text {Li}_3\left (-e^{-2 \text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )-4 a b^2 \text {sech}^{-1}(c x)^3-12 a b^2 \text {sech}^{-1}(c x)^2 \log \left (e^{-2 \text {sech}^{-1}(c x)}+1\right )+6 b \text {Li}_2\left (-e^{-2 \text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )^2+3 b^3 \text {Li}_4\left (-e^{-2 \text {sech}^{-1}(c x)}\right )-b^3 \text {sech}^{-1}(c x)^4-4 b^3 \text {sech}^{-1}(c x)^3 \log \left (e^{-2 \text {sech}^{-1}(c x)}+1\right )\right ) \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {arsech}\left (c x\right )^{3} + 3 \, a b^{2} \operatorname {arsech}\left (c x\right )^{2} + 3 \, a^{2} b \operatorname {arsech}\left (c x\right ) + a^{3}}{x}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {arsech}\left (c x\right ) + a\right )}^{3}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.15, size = 454, normalized size = 3.98 \[ a^{3} \ln \left (c x \right )+\frac {b^{3} \mathrm {arcsech}\left (c x \right )^{4}}{4}-b^{3} \mathrm {arcsech}\left (c x \right )^{3} \ln \left (1+\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )-\frac {3 b^{3} \mathrm {arcsech}\left (c x \right )^{2} \polylog \left (2, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )}{2}+\frac {3 b^{3} \mathrm {arcsech}\left (c x \right ) \polylog \left (3, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )}{2}-\frac {3 b^{3} \polylog \left (4, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )}{4}+a \,b^{2} \mathrm {arcsech}\left (c x \right )^{3}-3 a \,b^{2} \mathrm {arcsech}\left (c x \right )^{2} \ln \left (1+\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )-3 a \,b^{2} \mathrm {arcsech}\left (c x \right ) \polylog \left (2, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )+\frac {3 a \,b^{2} \polylog \left (3, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )}{2}+\frac {3 a^{2} b \mathrm {arcsech}\left (c x \right )^{2}}{2}-3 a^{2} b \,\mathrm {arcsech}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )-\frac {3 a^{2} b \polylog \left (2, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )}{2} \]
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 \[ a^{3} \log \relax (x) + \int \frac {b^{3} \log \left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )^{3}}{x} + \frac {3 \, a b^{2} \log \left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )^{2}}{x} + \frac {3 \, a^{2} b \log \left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}^3}{x} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {asech}{\left (c x \right )}\right )^{3}}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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