Integrand size = 10, antiderivative size = 260 \[ \int x \text {sech}^{-1}(a+b x)^3 \, dx=-\frac {3 \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {3 \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {a^2 \text {sech}^{-1}(a+b x)^3}{2 b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^3+\frac {6 a \text {sech}^{-1}(a+b x)^2 \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {3 \text {sech}^{-1}(a+b x) \log \left (1+e^{2 \text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {3 \operatorname {PolyLog}\left (2,-e^{2 \text {sech}^{-1}(a+b x)}\right )}{2 b^2}+\frac {6 i a \operatorname {PolyLog}\left (3,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \operatorname {PolyLog}\left (3,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2} \]
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Time = 0.20 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.200, Rules used = {6456, 5576, 4275, 4265, 2611, 2320, 6724, 4269, 3799, 2221, 2317, 2438} \[ \int x \text {sech}^{-1}(a+b x)^3 \, dx=-\frac {a^2 \text {sech}^{-1}(a+b x)^3}{2 b^2}+\frac {6 a \text {sech}^{-1}(a+b x)^2 \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {3 \operatorname {PolyLog}\left (2,-e^{2 \text {sech}^{-1}(a+b x)}\right )}{2 b^2}+\frac {6 i a \operatorname {PolyLog}\left (3,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \operatorname {PolyLog}\left (3,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {3 \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)^2}{2 b^2}-\frac {3 \text {sech}^{-1}(a+b x)^2}{2 b^2}+\frac {3 \text {sech}^{-1}(a+b x) \log \left (e^{2 \text {sech}^{-1}(a+b x)}+1\right )}{b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^3 \]
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Rule 2221
Rule 2317
Rule 2320
Rule 2438
Rule 2611
Rule 3799
Rule 4265
Rule 4269
Rule 4275
Rule 5576
Rule 6456
Rule 6724
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {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 \text {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 \text {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 \text {Subst}\left (\int x^2 \text {sech}^2(x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{2 b^2}+\frac {(3 a) \text {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 \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {3 \text {Subst}\left (\int x \tanh (x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^2}-\frac {(6 i a) \text {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) \text {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 \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 \text {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) \text {Subst}\left (\int \operatorname {PolyLog}\left (2,-i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^2}-\frac {(6 i a) \text {Subst}\left (\int \operatorname {PolyLog}\left (2,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 \arctan \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) \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {3 \text {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) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {(6 i a) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(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 \arctan \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) \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \operatorname {PolyLog}\left (3,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \operatorname {PolyLog}\left (3,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {3 \text {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 \arctan \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) \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {6 i a \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {3 \operatorname {PolyLog}\left (2,-e^{2 \text {sech}^{-1}(a+b x)}\right )}{2 b^2}+\frac {6 i a \operatorname {PolyLog}\left (3,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {6 i a \operatorname {PolyLog}\left (3,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2} \\ \end{align*}
Time = 0.65 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.98 \[ \int x \text {sech}^{-1}(a+b x)^3 \, dx=\frac {-3 \sqrt {-\frac {-1+a+b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)^2-2 a (a+b x) \text {sech}^{-1}(a+b x)^3+(a+b x)^2 \text {sech}^{-1}(a+b x)^3+3 \text {sech}^{-1}(a+b x) \left (\text {sech}^{-1}(a+b x)+2 \log \left (1+e^{-2 \text {sech}^{-1}(a+b x)}\right )\right )-3 \operatorname {PolyLog}\left (2,-e^{-2 \text {sech}^{-1}(a+b x)}\right )+6 i a \left (-\text {sech}^{-1}(a+b x)^2 \left (\log \left (1-i e^{-\text {sech}^{-1}(a+b x)}\right )-\log \left (1+i e^{-\text {sech}^{-1}(a+b x)}\right )\right )-2 \text {sech}^{-1}(a+b x) \left (\operatorname {PolyLog}\left (2,-i e^{-\text {sech}^{-1}(a+b x)}\right )-\operatorname {PolyLog}\left (2,i e^{-\text {sech}^{-1}(a+b x)}\right )\right )-2 \operatorname {PolyLog}\left (3,-i e^{-\text {sech}^{-1}(a+b x)}\right )+2 \operatorname {PolyLog}\left (3,i e^{-\text {sech}^{-1}(a+b x)}\right )\right )}{2 b^2} \]
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\[\int x \operatorname {arcsech}\left (b x +a \right )^{3}d x\]
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\[ \int x \text {sech}^{-1}(a+b x)^3 \, dx=\int { x \operatorname {arsech}\left (b x + a\right )^{3} \,d x } \]
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\[ \int x \text {sech}^{-1}(a+b x)^3 \, dx=\int x \operatorname {asech}^{3}{\left (a + b x \right )}\, dx \]
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\[ \int x \text {sech}^{-1}(a+b x)^3 \, dx=\int { x \operatorname {arsech}\left (b x + a\right )^{3} \,d x } \]
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\[ \int x \text {sech}^{-1}(a+b x)^3 \, dx=\int { x \operatorname {arsech}\left (b x + a\right )^{3} \,d x } \]
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Timed out. \[ \int x \text {sech}^{-1}(a+b x)^3 \, dx=\int x\,{\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}^3 \,d x \]
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