Optimal. Leaf size=140 \[ \frac{6 i b^2 \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )}{c}-\frac{6 i b^2 \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )}{c}-\frac{6 i b^3 \text{PolyLog}\left (3,-i e^{\text{sech}^{-1}(c x)}\right )}{c}+\frac{6 i b^3 \text{PolyLog}\left (3,i e^{\text{sech}^{-1}(c x)}\right )}{c}+x \left (a+b \text{sech}^{-1}(c x)\right )^3-\frac{6 b \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )^2}{c} \]
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Rubi [A] time = 0.111417, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6279, 5451, 4180, 2531, 2282, 6589} \[ \frac{6 i b^2 \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )}{c}-\frac{6 i b^2 \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )}{c}-\frac{6 i b^3 \text{PolyLog}\left (3,-i e^{\text{sech}^{-1}(c x)}\right )}{c}+\frac{6 i b^3 \text{PolyLog}\left (3,i e^{\text{sech}^{-1}(c x)}\right )}{c}+x \left (a+b \text{sech}^{-1}(c x)\right )^3-\frac{6 b \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )^2}{c} \]
Antiderivative was successfully verified.
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Rule 6279
Rule 5451
Rule 4180
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \left (a+b \text{sech}^{-1}(c x)\right )^3 \, dx &=-\frac{\operatorname{Subst}\left (\int (a+b x)^3 \text{sech}(x) \tanh (x) \, dx,x,\text{sech}^{-1}(c x)\right )}{c}\\ &=x \left (a+b \text{sech}^{-1}(c x)\right )^3-\frac{(3 b) \operatorname{Subst}\left (\int (a+b x)^2 \text{sech}(x) \, dx,x,\text{sech}^{-1}(c x)\right )}{c}\\ &=x \left (a+b \text{sech}^{-1}(c x)\right )^3-\frac{6 b \left (a+b \text{sech}^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right )}{c}+\frac{\left (6 i b^2\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\text{sech}^{-1}(c x)\right )}{c}-\frac{\left (6 i b^2\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\text{sech}^{-1}(c x)\right )}{c}\\ &=x \left (a+b \text{sech}^{-1}(c x)\right )^3-\frac{6 b \left (a+b \text{sech}^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right )}{c}+\frac{6 i b^2 \left (a+b \text{sech}^{-1}(c x)\right ) \text{Li}_2\left (-i e^{\text{sech}^{-1}(c x)}\right )}{c}-\frac{6 i b^2 \left (a+b \text{sech}^{-1}(c x)\right ) \text{Li}_2\left (i e^{\text{sech}^{-1}(c x)}\right )}{c}-\frac{\left (6 i b^3\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^x\right ) \, dx,x,\text{sech}^{-1}(c x)\right )}{c}+\frac{\left (6 i b^3\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^x\right ) \, dx,x,\text{sech}^{-1}(c x)\right )}{c}\\ &=x \left (a+b \text{sech}^{-1}(c x)\right )^3-\frac{6 b \left (a+b \text{sech}^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right )}{c}+\frac{6 i b^2 \left (a+b \text{sech}^{-1}(c x)\right ) \text{Li}_2\left (-i e^{\text{sech}^{-1}(c x)}\right )}{c}-\frac{6 i b^2 \left (a+b \text{sech}^{-1}(c x)\right ) \text{Li}_2\left (i e^{\text{sech}^{-1}(c x)}\right )}{c}-\frac{\left (6 i b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{\text{sech}^{-1}(c x)}\right )}{c}+\frac{\left (6 i b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{\text{sech}^{-1}(c x)}\right )}{c}\\ &=x \left (a+b \text{sech}^{-1}(c x)\right )^3-\frac{6 b \left (a+b \text{sech}^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(c x)}\right )}{c}+\frac{6 i b^2 \left (a+b \text{sech}^{-1}(c x)\right ) \text{Li}_2\left (-i e^{\text{sech}^{-1}(c x)}\right )}{c}-\frac{6 i b^2 \left (a+b \text{sech}^{-1}(c x)\right ) \text{Li}_2\left (i e^{\text{sech}^{-1}(c x)}\right )}{c}-\frac{6 i b^3 \text{Li}_3\left (-i e^{\text{sech}^{-1}(c x)}\right )}{c}+\frac{6 i b^3 \text{Li}_3\left (i e^{\text{sech}^{-1}(c x)}\right )}{c}\\ \end{align*}
Mathematica [B] time = 0.408751, size = 282, normalized size = 2.01 \[ \frac{3 i a b^2 \left (2 \text{PolyLog}\left (2,-i e^{-\text{sech}^{-1}(c x)}\right )-2 \text{PolyLog}\left (2,i e^{-\text{sech}^{-1}(c x)}\right )+\text{sech}^{-1}(c x) \left (-i c x \text{sech}^{-1}(c x)+2 \log \left (1-i e^{-\text{sech}^{-1}(c x)}\right )-2 \log \left (1+i e^{-\text{sech}^{-1}(c x)}\right )\right )\right )}{c}+\frac{b^3 \left (c x \text{sech}^{-1}(c x)^3-3 i \left (-2 \text{sech}^{-1}(c x) \left (\text{PolyLog}\left (2,-i e^{-\text{sech}^{-1}(c x)}\right )-\text{PolyLog}\left (2,i e^{-\text{sech}^{-1}(c x)}\right )\right )-2 \left (\text{PolyLog}\left (3,-i e^{-\text{sech}^{-1}(c x)}\right )-\text{PolyLog}\left (3,i e^{-\text{sech}^{-1}(c x)}\right )\right )+\text{sech}^{-1}(c x)^2 \left (-\left (\log \left (1-i e^{-\text{sech}^{-1}(c x)}\right )-\log \left (1+i e^{-\text{sech}^{-1}(c x)}\right )\right )\right )\right )\right )}{c}-\frac{3 a^2 b \tan ^{-1}\left (\frac{c x \sqrt{\frac{1-c x}{c x+1}}}{c x-1}\right )}{c}+3 a^2 b x \text{sech}^{-1}(c x)+a^3 x \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.329, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\rm arcsech} \left (cx\right ) \right ) ^{3}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (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\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{asech}{\left (c x \right )}\right )^{3}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arsech}\left (c x\right ) + a\right )}^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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