Integrand size = 16, antiderivative size = 95 \[ \int x \text {csch}(a+b x) \text {sech}^3(a+b x) \, dx=\frac {x}{2 b}-\frac {2 x \text {arctanh}\left (e^{2 a+2 b x}\right )}{b}-\frac {\operatorname {PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}+\frac {\operatorname {PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}-\frac {\tanh (a+b x)}{2 b^2}-\frac {x \tanh ^2(a+b x)}{2 b} \] Output:
1/2*x/b-2*x*arctanh(exp(2*b*x+2*a))/b-1/2*polylog(2,-exp(2*b*x+2*a))/b^2+1 /2*polylog(2,exp(2*b*x+2*a))/b^2-1/2*tanh(b*x+a)/b^2-1/2*x*tanh(b*x+a)^2/b
Time = 0.36 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.92 \[ \int x \text {csch}(a+b x) \text {sech}^3(a+b x) \, dx=\frac {2 b x \log \left (1-e^{-2 (a+b x)}\right )-2 b x \log \left (1+e^{-2 (a+b x)}\right )+\operatorname {PolyLog}\left (2,-e^{-2 (a+b x)}\right )-\operatorname {PolyLog}\left (2,e^{-2 (a+b x)}\right )+b x \text {sech}^2(a+b x)-\tanh (a+b x)}{2 b^2} \] Input:
Integrate[x*Csch[a + b*x]*Sech[a + b*x]^3,x]
Output:
(2*b*x*Log[1 - E^(-2*(a + b*x))] - 2*b*x*Log[1 + E^(-2*(a + b*x))] + PolyL og[2, -E^(-2*(a + b*x))] - PolyLog[2, E^(-2*(a + b*x))] + b*x*Sech[a + b*x ]^2 - Tanh[a + b*x])/(2*b^2)
Time = 0.56 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5985, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int x \text {csch}(a+b x) \text {sech}^3(a+b x) \, dx\) |
| \(\Big \downarrow \) 5985 |
| \(\displaystyle -\int \left (\frac {\log (\tanh (a+b x))}{b}-\frac {\tanh ^2(a+b x)}{2 b}\right )dx-\frac {x \tanh ^2(a+b x)}{2 b}+\frac {x \log (\tanh (a+b x))}{b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 x \text {arctanh}\left (e^{2 a+2 b x}\right )}{b}-\frac {\operatorname {PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}+\frac {\operatorname {PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}-\frac {\tanh (a+b x)}{2 b^2}-\frac {x \tanh ^2(a+b x)}{2 b}+\frac {x}{2 b}\) |
Input:
Int[x*Csch[a + b*x]*Sech[a + b*x]^3,x]
Output:
x/(2*b) - (2*x*ArcTanh[E^(2*a + 2*b*x)])/b - PolyLog[2, -E^(2*a + 2*b*x)]/ (2*b^2) + PolyLog[2, E^(2*a + 2*b*x)]/(2*b^2) - Tanh[a + b*x]/(2*b^2) - (x *Tanh[a + b*x]^2)/(2*b)
Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> With[{u = IntHide[Csch[a + b*x]^n*Sech[a + b*x]^p, x]}, Simp[(c + d*x)^m u, x] - Simp[d*m Int[(c + d*x)^(m - 1)*u, x], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n , p]
Leaf count of result is larger than twice the leaf count of optimal. \(165\) vs. \(2(82)=164\).
Time = 4.49 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.75
| method | result | size |
| risch | \(\frac {2 b x \,{\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 b x +2 a}+1}{b^{2} \left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) x}{b}+\frac {\operatorname {polylog}\left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {x \ln \left ({\mathrm e}^{2 b x +2 a}+1\right )}{b}-\frac {\operatorname {polylog}\left (2, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{2}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x}{b}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a}{b^{2}}+\frac {\operatorname {polylog}\left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {a \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{2}}\) | \(166\) |
Input:
int(x*csch(b*x+a)*sech(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
(2*b*x*exp(2*b*x+2*a)+exp(2*b*x+2*a)+1)/b^2/(exp(2*b*x+2*a)+1)^2+1/b*ln(ex p(b*x+a)+1)*x+polylog(2,-exp(b*x+a))/b^2-x*ln(exp(2*b*x+2*a)+1)/b-1/2*poly log(2,-exp(2*b*x+2*a))/b^2+1/b*ln(1-exp(b*x+a))*x+1/b^2*ln(1-exp(b*x+a))*a +polylog(2,exp(b*x+a))/b^2-1/b^2*a*ln(exp(b*x+a)-1)
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 1543, normalized size of antiderivative = 16.24 \[ \int x \text {csch}(a+b x) \text {sech}^3(a+b x) \, dx=\text {Too large to display} \] Input:
integrate(x*csch(b*x+a)*sech(b*x+a)^3,x, algorithm="fricas")
Output:
((2*b*x + 1)*cosh(b*x + a)^2 + 2*(2*b*x + 1)*cosh(b*x + a)*sinh(b*x + a) + (2*b*x + 1)*sinh(b*x + a)^2 + (cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2* cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*d ilog(cosh(b*x + a) + sinh(b*x + a)) - (cosh(b*x + a)^4 + 4*cosh(b*x + a)*s inh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a) ^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - (cosh(b*x + a)^4 + 4*cosh (b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*si nh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + cosh(b*x + a))*si nh(b*x + a) + 1)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + (cosh(b*x + a )^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + cosh( b*x + a))*sinh(b*x + a) + 1)*dilog(-cosh(b*x + a) - sinh(b*x + a)) + (b*x* cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sinh(b*x + a)^ 4 + 2*b*x*cosh(b*x + a)^2 + 2*(3*b*x*cosh(b*x + a)^2 + b*x)*sinh(b*x + a)^ 2 + b*x + 4*(b*x*cosh(b*x + a)^3 + b*x*cosh(b*x + a))*sinh(b*x + a))*log(c osh(b*x + a) + sinh(b*x + a) + 1) + (a*cosh(b*x + a)^4 + 4*a*cosh(b*x + a) *sinh(b*x + a)^3 + a*sinh(b*x + a)^4 + 2*a*cosh(b*x + a)^2 + 2*(3*a*cosh(b *x + a)^2 + a)*sinh(b*x + a)^2 + 4*(a*cosh(b*x + a)^3 + a*cosh(b*x + a)...
\[ \int x \text {csch}(a+b x) \text {sech}^3(a+b x) \, dx=\int x \operatorname {csch}{\left (a + b x \right )} \operatorname {sech}^{3}{\left (a + b x \right )}\, dx \] Input:
integrate(x*csch(b*x+a)*sech(b*x+a)**3,x)
Output:
Integral(x*csch(a + b*x)*sech(a + b*x)**3, x)
Time = 0.06 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.49 \[ \int x \text {csch}(a+b x) \text {sech}^3(a+b x) \, dx=\frac {{\left (2 \, b x e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + 1}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} - \frac {2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )}{2 \, b^{2}} + \frac {b x \log \left (e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (b x + a\right )}\right )}{b^{2}} + \frac {b x \log \left (-e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (b x + a\right )}\right )}{b^{2}} \] Input:
integrate(x*csch(b*x+a)*sech(b*x+a)^3,x, algorithm="maxima")
Output:
((2*b*x*e^(2*a) + e^(2*a))*e^(2*b*x) + 1)/(b^2*e^(4*b*x + 4*a) + 2*b^2*e^( 2*b*x + 2*a) + b^2) - 1/2*(2*b*x*log(e^(2*b*x + 2*a) + 1) + dilog(-e^(2*b* x + 2*a)))/b^2 + (b*x*log(e^(b*x + a) + 1) + dilog(-e^(b*x + a)))/b^2 + (b *x*log(-e^(b*x + a) + 1) + dilog(e^(b*x + a)))/b^2
\[ \int x \text {csch}(a+b x) \text {sech}^3(a+b x) \, dx=\int { x \operatorname {csch}\left (b x + a\right ) \operatorname {sech}\left (b x + a\right )^{3} \,d x } \] Input:
integrate(x*csch(b*x+a)*sech(b*x+a)^3,x, algorithm="giac")
Output:
integrate(x*csch(b*x + a)*sech(b*x + a)^3, x)
Timed out. \[ \int x \text {csch}(a+b x) \text {sech}^3(a+b x) \, dx=\int \frac {x}{{\mathrm {cosh}\left (a+b\,x\right )}^3\,\mathrm {sinh}\left (a+b\,x\right )} \,d x \] Input:
int(x/(cosh(a + b*x)^3*sinh(a + b*x)),x)
Output:
int(x/(cosh(a + b*x)^3*sinh(a + b*x)), x)
\[ \int x \text {csch}(a+b x) \text {sech}^3(a+b x) \, dx=\frac {16 e^{4 b x +6 a} \left (\int \frac {e^{2 b x} x}{e^{8 b x +8 a}+2 e^{6 b x +6 a}-2 e^{2 b x +2 a}-1}d x \right ) b^{2}-2 e^{4 b x +4 a} \mathrm {log}\left (e^{2 b x +2 a}+1\right )+4 e^{4 b x +4 a} b x -e^{4 b x +4 a}+32 e^{2 b x +4 a} \left (\int \frac {e^{2 b x} x}{e^{8 b x +8 a}+2 e^{6 b x +6 a}-2 e^{2 b x +2 a}-1}d x \right ) b^{2}-4 e^{2 b x +2 a} \mathrm {log}\left (e^{2 b x +2 a}+1\right )+8 e^{2 b x +2 a} b x +16 e^{2 a} \left (\int \frac {e^{2 b x} x}{e^{8 b x +8 a}+2 e^{6 b x +6 a}-2 e^{2 b x +2 a}-1}d x \right ) b^{2}-2 \,\mathrm {log}\left (e^{2 b x +2 a}+1\right )+1}{b^{2} \left (e^{4 b x +4 a}+2 e^{2 b x +2 a}+1\right )} \] Input:
int(x*csch(b*x+a)*sech(b*x+a)^3,x)
Output:
(16*e**(6*a + 4*b*x)*int((e**(2*b*x)*x)/(e**(8*a + 8*b*x) + 2*e**(6*a + 6* b*x) - 2*e**(2*a + 2*b*x) - 1),x)*b**2 - 2*e**(4*a + 4*b*x)*log(e**(2*a + 2*b*x) + 1) + 4*e**(4*a + 4*b*x)*b*x - e**(4*a + 4*b*x) + 32*e**(4*a + 2*b *x)*int((e**(2*b*x)*x)/(e**(8*a + 8*b*x) + 2*e**(6*a + 6*b*x) - 2*e**(2*a + 2*b*x) - 1),x)*b**2 - 4*e**(2*a + 2*b*x)*log(e**(2*a + 2*b*x) + 1) + 8*e **(2*a + 2*b*x)*b*x + 16*e**(2*a)*int((e**(2*b*x)*x)/(e**(8*a + 8*b*x) + 2 *e**(6*a + 6*b*x) - 2*e**(2*a + 2*b*x) - 1),x)*b**2 - 2*log(e**(2*a + 2*b* x) + 1) + 1)/(b**2*(e**(4*a + 4*b*x) + 2*e**(2*a + 2*b*x) + 1))