Optimal. Leaf size=91 \[ -\frac {i \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}+\frac {i \text {Li}_2\left (i e^{a+b x}\right )}{2 b^2}-\frac {\text {sech}(a+b x)}{2 b^2}+\frac {x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {x \tanh (a+b x) \text {sech}(a+b x)}{2 b} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.10, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5455, 4180, 2279, 2391, 4185} \[ -\frac {i \text {PolyLog}\left (2,-i e^{a+b x}\right )}{2 b^2}+\frac {i \text {PolyLog}\left (2,i e^{a+b x}\right )}{2 b^2}-\frac {\text {sech}(a+b x)}{2 b^2}+\frac {x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {x \tanh (a+b x) \text {sech}(a+b x)}{2 b} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 2279
Rule 2391
Rule 4180
Rule 4185
Rule 5455
Rubi steps
\begin {align*} \int x \text {sech}(a+b x) \tanh ^2(a+b x) \, dx &=\int x \text {sech}(a+b x) \, dx-\int x \text {sech}^3(a+b x) \, dx\\ &=\frac {2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\text {sech}(a+b x)}{2 b^2}-\frac {x \text {sech}(a+b x) \tanh (a+b x)}{2 b}-\frac {1}{2} \int x \text {sech}(a+b x) \, dx-\frac {i \int \log \left (1-i e^{a+b x}\right ) \, dx}{b}+\frac {i \int \log \left (1+i e^{a+b x}\right ) \, dx}{b}\\ &=\frac {x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\text {sech}(a+b x)}{2 b^2}-\frac {x \text {sech}(a+b x) \tanh (a+b x)}{2 b}-\frac {i \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^2}+\frac {i \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{a+b x}\right )}{b^2}+\frac {i \int \log \left (1-i e^{a+b x}\right ) \, dx}{2 b}-\frac {i \int \log \left (1+i e^{a+b x}\right ) \, dx}{2 b}\\ &=\frac {x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {i \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}+\frac {i \text {Li}_2\left (i e^{a+b x}\right )}{b^2}-\frac {\text {sech}(a+b x)}{2 b^2}-\frac {x \text {sech}(a+b x) \tanh (a+b x)}{2 b}+\frac {i \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{a+b x}\right )}{2 b^2}-\frac {i \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{a+b x}\right )}{2 b^2}\\ &=\frac {x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {i \text {Li}_2\left (-i e^{a+b x}\right )}{2 b^2}+\frac {i \text {Li}_2\left (i e^{a+b x}\right )}{2 b^2}-\frac {\text {sech}(a+b x)}{2 b^2}-\frac {x \text {sech}(a+b x) \tanh (a+b x)}{2 b}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.78, size = 93, normalized size = 1.02 \[ -\frac {i \text {Li}_2(-i (\cosh (a+b x)+\sinh (a+b x)))-i \text {Li}_2(i (\cosh (a+b x)+\sinh (a+b x)))+\text {sech}(a+b x)+b x \tanh (a+b x) \text {sech}(a+b x)-2 b x \tan ^{-1}(\sinh (a+b x)+\cosh (a+b x))}{2 b^2} \]
Warning: Unable to verify antiderivative.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.99, size = 1046, normalized size = 11.49 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {sech}\left (b x + a\right )^{3} \sinh \left (b x + a\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [B] time = 0.68, size = 178, normalized size = 1.96 \[ -\frac {{\mathrm e}^{b x +a} \left (b x \,{\mathrm e}^{2 b x +2 a}-b x +{\mathrm e}^{2 b x +2 a}+1\right )}{b^{2} \left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}-\frac {i \ln \left (1+i {\mathrm e}^{b x +a}\right ) x}{2 b}-\frac {i \ln \left (1+i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}+\frac {i \ln \left (1-i {\mathrm e}^{b x +a}\right ) x}{2 b}+\frac {i \ln \left (1-i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}-\frac {i \dilog \left (1+i {\mathrm e}^{b x +a}\right )}{2 b^{2}}+\frac {i \dilog \left (1-i {\mathrm e}^{b x +a}\right )}{2 b^{2}}-\frac {a \arctan \left ({\mathrm e}^{b x +a}\right )}{b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (b x e^{\left (3 \, a\right )} + e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )} - {\left (b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + 2 \, \int \frac {x e^{\left (b x + a\right )}}{2 \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{{\mathrm {cosh}\left (a+b\,x\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sinh ^{2}{\left (a + b x \right )} \operatorname {sech}^{3}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________