Optimal. Leaf size=260 \[ -\frac{6 i a \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{6 i a \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{3 \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(a+b x)}\right )}{2 b^2}+\frac{6 i a \text{PolyLog}\left (3,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \text{PolyLog}\left (3,i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{a^2 \text{sech}^{-1}(a+b x)^3}{2 b^2}-\frac{3 \sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1) \text{sech}^{-1}(a+b x)^2}{2 b^2}-\frac{3 \text{sech}^{-1}(a+b x)^2}{2 b^2}+\frac{3 \text{sech}^{-1}(a+b x) \log \left (e^{2 \text{sech}^{-1}(a+b x)}+1\right )}{b^2}+\frac{6 a \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^3 \]
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Rubi [A] time = 0.257207, antiderivative size = 260, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 12, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.2, Rules used = {6321, 5468, 4190, 4180, 2531, 2282, 6589, 4184, 3718, 2190, 2279, 2391} \[ -\frac{6 i a \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{6 i a \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{3 \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(a+b x)}\right )}{2 b^2}+\frac{6 i a \text{PolyLog}\left (3,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \text{PolyLog}\left (3,i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{a^2 \text{sech}^{-1}(a+b x)^3}{2 b^2}-\frac{3 \sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1) \text{sech}^{-1}(a+b x)^2}{2 b^2}-\frac{3 \text{sech}^{-1}(a+b x)^2}{2 b^2}+\frac{3 \text{sech}^{-1}(a+b x) \log \left (e^{2 \text{sech}^{-1}(a+b x)}+1\right )}{b^2}+\frac{6 a \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^3 \]
Antiderivative was successfully verified.
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Rule 6321
Rule 5468
Rule 4190
Rule 4180
Rule 2531
Rule 2282
Rule 6589
Rule 4184
Rule 3718
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int x \text{sech}^{-1}(a+b x)^3 \, dx &=-\frac{\operatorname{Subst}\left (\int x^3 \text{sech}(x) (-a+\text{sech}(x)) \tanh (x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}\\ &=\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^3-\frac{3 \operatorname{Subst}\left (\int x^2 (-a+\text{sech}(x))^2 \, dx,x,\text{sech}^{-1}(a+b x)\right )}{2 b^2}\\ &=\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^3-\frac{3 \operatorname{Subst}\left (\int \left (a^2 x^2-2 a x^2 \text{sech}(x)+x^2 \text{sech}^2(x)\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac{a^2 \text{sech}^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^3-\frac{3 \operatorname{Subst}\left (\int x^2 \text{sech}^2(x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{2 b^2}+\frac{(3 a) \operatorname{Subst}\left (\int x^2 \text{sech}(x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{3 \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x) \text{sech}^{-1}(a+b x)^2}{2 b^2}-\frac{a^2 \text{sech}^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^3+\frac{6 a \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{3 \operatorname{Subst}\left (\int x \tanh (x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}-\frac{(6 i a) \operatorname{Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}+\frac{(6 i a) \operatorname{Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{3 \text{sech}^{-1}(a+b x)^2}{2 b^2}-\frac{3 \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x) \text{sech}^{-1}(a+b x)^2}{2 b^2}-\frac{a^2 \text{sech}^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^3+\frac{6 a \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \text{sech}^{-1}(a+b x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{6 i a \text{sech}^{-1}(a+b x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{6 \operatorname{Subst}\left (\int \frac{e^{2 x} x}{1+e^{2 x}} \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}+\frac{(6 i a) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}-\frac{(6 i a) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{3 \text{sech}^{-1}(a+b x)^2}{2 b^2}-\frac{3 \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x) \text{sech}^{-1}(a+b x)^2}{2 b^2}-\frac{a^2 \text{sech}^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^3+\frac{6 a \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{3 \text{sech}^{-1}(a+b x) \log \left (1+e^{2 \text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \text{sech}^{-1}(a+b x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{6 i a \text{sech}^{-1}(a+b x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{3 \operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^2}+\frac{(6 i a) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{(6 i a) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}\\ &=-\frac{3 \text{sech}^{-1}(a+b x)^2}{2 b^2}-\frac{3 \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x) \text{sech}^{-1}(a+b x)^2}{2 b^2}-\frac{a^2 \text{sech}^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^3+\frac{6 a \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{3 \text{sech}^{-1}(a+b x) \log \left (1+e^{2 \text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \text{sech}^{-1}(a+b x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{6 i a \text{sech}^{-1}(a+b x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{6 i a \text{Li}_3\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \text{Li}_3\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{3 \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \text{sech}^{-1}(a+b x)}\right )}{2 b^2}\\ &=-\frac{3 \text{sech}^{-1}(a+b x)^2}{2 b^2}-\frac{3 \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x) \text{sech}^{-1}(a+b x)^2}{2 b^2}-\frac{a^2 \text{sech}^{-1}(a+b x)^3}{2 b^2}+\frac{1}{2} x^2 \text{sech}^{-1}(a+b x)^3+\frac{6 a \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{3 \text{sech}^{-1}(a+b x) \log \left (1+e^{2 \text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \text{sech}^{-1}(a+b x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{6 i a \text{sech}^{-1}(a+b x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}+\frac{3 \text{Li}_2\left (-e^{2 \text{sech}^{-1}(a+b x)}\right )}{2 b^2}+\frac{6 i a \text{Li}_3\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}-\frac{6 i a \text{Li}_3\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b^2}\\ \end{align*}
Mathematica [A] time = 0.552299, size = 254, normalized size = 0.98 \[ \frac{-3 \text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}(a+b x)}\right )+6 i a \left (-2 \text{sech}^{-1}(a+b x) \left (\text{PolyLog}\left (2,-i e^{-\text{sech}^{-1}(a+b x)}\right )-\text{PolyLog}\left (2,i e^{-\text{sech}^{-1}(a+b x)}\right )\right )-2 \text{PolyLog}\left (3,-i e^{-\text{sech}^{-1}(a+b x)}\right )+2 \text{PolyLog}\left (3,i e^{-\text{sech}^{-1}(a+b x)}\right )+\text{sech}^{-1}(a+b x)^2 \left (-\left (\log \left (1-i e^{-\text{sech}^{-1}(a+b x)}\right )-\log \left (1+i e^{-\text{sech}^{-1}(a+b x)}\right )\right )\right )\right )+(a+b x)^2 \text{sech}^{-1}(a+b x)^3-2 a (a+b x) \text{sech}^{-1}(a+b x)^3-3 \sqrt{-\frac{a+b x-1}{a+b x+1}} (a+b x+1) \text{sech}^{-1}(a+b x)^2+3 \text{sech}^{-1}(a+b x) \left (\text{sech}^{-1}(a+b x)+2 \log \left (e^{-2 \text{sech}^{-1}(a+b x)}+1\right )\right )}{2 b^2} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.559, size = 0, normalized size = 0. \begin{align*} \int x \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*} \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 (x \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 x \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 x \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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