Optimal. Leaf size=136 \[ \frac{6 i \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{PolyLog}\left (3,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{6 i \text{PolyLog}\left (3,i e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{(a+b x) \text{sech}^{-1}(a+b x)^3}{b}-\frac{6 \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b} \]
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Rubi [A] time = 0.104783, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.875, Rules used = {6315, 6279, 5418, 4180, 2531, 2282, 6589} \[ \frac{6 i \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{PolyLog}\left (3,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{6 i \text{PolyLog}\left (3,i e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{(a+b x) \text{sech}^{-1}(a+b x)^3}{b}-\frac{6 \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 6315
Rule 6279
Rule 5418
Rule 4180
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \text{sech}^{-1}(a+b x)^3 \, dx &=\frac{\operatorname{Subst}\left (\int \text{sech}^{-1}(x)^3 \, dx,x,a+b x\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int x^3 \text{sech}(x) \tanh (x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}\\ &=\frac{(a+b x) \text{sech}^{-1}(a+b x)^3}{b}-\frac{3 \operatorname{Subst}\left (\int x^2 \text{sech}(x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}\\ &=\frac{(a+b x) \text{sech}^{-1}(a+b x)^3}{b}-\frac{6 \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{(6 i) \operatorname{Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}-\frac{(6 i) \operatorname{Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}\\ &=\frac{(a+b x) \text{sech}^{-1}(a+b x)^3}{b}-\frac{6 \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{6 i \text{sech}^{-1}(a+b x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{sech}^{-1}(a+b x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{(6 i) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}+\frac{(6 i) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}\\ &=\frac{(a+b x) \text{sech}^{-1}(a+b x)^3}{b}-\frac{6 \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{6 i \text{sech}^{-1}(a+b x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{sech}^{-1}(a+b x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{\text{sech}^{-1}(a+b x)}\right )}{b}\\ &=\frac{(a+b x) \text{sech}^{-1}(a+b x)^3}{b}-\frac{6 \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{6 i \text{sech}^{-1}(a+b x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{sech}^{-1}(a+b x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{Li}_3\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{6 i \text{Li}_3\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b}\\ \end{align*}
Mathematica [A] time = 0.140201, size = 153, normalized size = 1.12 \[ -\frac{-(a+b x) \text{sech}^{-1}(a+b x)^3+3 i \left (-2 \text{sech}^{-1}(a+b x) \left (\text{PolyLog}\left (2,-i e^{-\text{sech}^{-1}(a+b x)}\right )-\text{PolyLog}\left (2,i e^{-\text{sech}^{-1}(a+b x)}\right )\right )-2 \left (\text{PolyLog}\left (3,-i e^{-\text{sech}^{-1}(a+b x)}\right )-\text{PolyLog}\left (3,i e^{-\text{sech}^{-1}(a+b x)}\right )\right )+\text{sech}^{-1}(a+b x)^2 \left (-\left (\log \left (1-i e^{-\text{sech}^{-1}(a+b x)}\right )-\log \left (1+i e^{-\text{sech}^{-1}(a+b x)}\right )\right )\right )\right )}{b} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.353, size = 0, normalized size = 0. \begin{align*} \int \left ({\rm arcsech} \left (bx+a\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*} x \log \left (\sqrt{b x + a + 1} \sqrt{-b x - a + 1} b x + \sqrt{b x + a + 1} \sqrt{-b x - a + 1} a + b x + a\right )^{3} - \int \frac{8 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt{b x + a + 1} \sqrt{-b x - a + 1} \log \left (b x + a\right )^{3} + 8 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right )^{3} + 3 \,{\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} +{\left (a^{2} b - b\right )} x + 2 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right ) +{\left ({\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt{b x + a + 1} \log \left (b x + a\right ) +{\left (2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} +{\left (2 \, a^{2} b - b\right )} x +{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right )\right )} \sqrt{b x + a + 1}\right )} \sqrt{-b x - a + 1}\right )} \log \left (\sqrt{b x + a + 1} \sqrt{-b x - a + 1} b x + \sqrt{b x + a + 1} \sqrt{-b x - a + 1} a + b x + a\right )^{2} - 12 \,{\left ({\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt{b x + a + 1} \sqrt{-b x - a + 1} \log \left (b x + a\right )^{2} +{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right )^{2}\right )} \log \left (\sqrt{b x + a + 1} \sqrt{-b x - a + 1} b x + \sqrt{b x + a + 1} \sqrt{-b x - a + 1} a + b x + a\right )}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt{b x + a + 1} \sqrt{-b x - a + 1} +{\left (3 \, a^{2} b - b\right )} x - a}\,{d x} \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 (\operatorname{arsech}\left (b x + a\right )^{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 \operatorname{asech}^{3}{\left (a + b x \right )}\, 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 \operatorname{arsech}\left (b x + a\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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