Integrand size = 13, antiderivative size = 178 \[ \int \frac {\coth ^5(x)}{a+b \text {sech}(x)} \, dx=\frac {\log (\cosh (x))}{a}+\frac {\left (8 a^2+21 a b+15 b^2\right ) \log (1-\text {sech}(x))}{16 (a+b)^3}+\frac {\left (8 a^2-21 a b+15 b^2\right ) \log (1+\text {sech}(x))}{16 (a-b)^3}-\frac {b^6 \log (a+b \text {sech}(x))}{a \left (a^2-b^2\right )^3}-\frac {1}{16 (a+b) (1-\text {sech}(x))^2}-\frac {5 a+7 b}{16 (a+b)^2 (1-\text {sech}(x))}-\frac {1}{16 (a-b) (1+\text {sech}(x))^2}-\frac {5 a-7 b}{16 (a-b)^2 (1+\text {sech}(x))} \] Output:
ln(cosh(x))/a+1/16*(8*a^2+21*a*b+15*b^2)*ln(1-sech(x))/(a+b)^3+1/16*(8*a^2 -21*a*b+15*b^2)*ln(1+sech(x))/(a-b)^3-b^6*ln(a+b*sech(x))/a/(a^2-b^2)^3-1/ 16/(a+b)/(1-sech(x))^2-1/16*(5*a+7*b)/(a+b)^2/(1-sech(x))-1/16/(a-b)/(1+se ch(x))^2-1/16*(5*a-7*b)/(a-b)^2/(1+sech(x))
Time = 0.69 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.92 \[ \int \frac {\coth ^5(x)}{a+b \text {sech}(x)} \, dx=\frac {1}{16} \left (\frac {16 \log (\cosh (x))}{a}+\frac {\left (8 a^2+21 a b+15 b^2\right ) \log (1-\text {sech}(x))}{(a+b)^3}+\frac {\left (8 a^2-21 a b+15 b^2\right ) \log (1+\text {sech}(x))}{(a-b)^3}-\frac {16 b^6 \log (a+b \text {sech}(x))}{a (a-b)^3 (a+b)^3}-\frac {1}{(a+b) (-1+\text {sech}(x))^2}+\frac {5 a+7 b}{(a+b)^2 (-1+\text {sech}(x))}-\frac {1}{(a-b) (1+\text {sech}(x))^2}+\frac {-5 a+7 b}{(a-b)^2 (1+\text {sech}(x))}\right ) \] Input:
Integrate[Coth[x]^5/(a + b*Sech[x]),x]
Output:
((16*Log[Cosh[x]])/a + ((8*a^2 + 21*a*b + 15*b^2)*Log[1 - Sech[x]])/(a + b )^3 + ((8*a^2 - 21*a*b + 15*b^2)*Log[1 + Sech[x]])/(a - b)^3 - (16*b^6*Log [a + b*Sech[x]])/(a*(a - b)^3*(a + b)^3) - 1/((a + b)*(-1 + Sech[x])^2) + (5*a + 7*b)/((a + b)^2*(-1 + Sech[x])) - 1/((a - b)*(1 + Sech[x])^2) + (-5 *a + 7*b)/((a - b)^2*(1 + Sech[x])))/16
Time = 0.55 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.19, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 26, 4373, 615, 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.
| \(\displaystyle \int \frac {\coth ^5(x)}{a+b \text {sech}(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {i}{\cot \left (\frac {\pi }{2}+i x\right )^5 \left (a+b \csc \left (\frac {\pi }{2}+i x\right )\right )}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -i \int \frac {1}{\cot \left (i x+\frac {\pi }{2}\right )^5 \left (a+b \csc \left (i x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4373 |
| \(\displaystyle -b^6 \int \frac {\cosh (x)}{b (a+b \text {sech}(x)) \left (b^2-b^2 \text {sech}^2(x)\right )^3}d(b \text {sech}(x))\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle -b^6 \int \left (\frac {7 b-5 a}{16 (a-b)^2 b^5 (\text {sech}(x) b+b)^2}+\frac {\cosh (x)}{a b^7}+\frac {8 a^2+21 b a+15 b^2}{16 b^6 (a+b)^3 (b-b \text {sech}(x))}+\frac {1}{a (a-b)^3 (a+b)^3 (a+b \text {sech}(x))}+\frac {8 a^2-21 b a+15 b^2}{16 b^6 (b-a)^3 (\text {sech}(x) b+b)}+\frac {5 a+7 b}{16 b^5 (a+b)^2 (b-b \text {sech}(x))^2}+\frac {1}{8 b^4 (a+b) (b-b \text {sech}(x))^3}+\frac {1}{8 b^4 (b-a) (\text {sech}(x) b+b)^3}\right )d(b \text {sech}(x))\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -b^6 \left (\frac {\log (a+b \text {sech}(x))}{a \left (a^2-b^2\right )^3}-\frac {\left (8 a^2+21 a b+15 b^2\right ) \log (b-b \text {sech}(x))}{16 b^6 (a+b)^3}-\frac {\left (8 a^2-21 a b+15 b^2\right ) \log (b \text {sech}(x)+b)}{16 b^6 (a-b)^3}+\frac {\log (b \text {sech}(x))}{a b^6}+\frac {5 a-7 b}{16 b^5 (a-b)^2 (b \text {sech}(x)+b)}+\frac {5 a+7 b}{16 b^5 (a+b)^2 (b-b \text {sech}(x))}+\frac {1}{16 b^4 (a+b) (b-b \text {sech}(x))^2}+\frac {1}{16 b^4 (a-b) (b \text {sech}(x)+b)^2}\right )\) |
Input:
Int[Coth[x]^5/(a + b*Sech[x]),x]
Output:
-(b^6*(Log[b*Sech[x]]/(a*b^6) - ((8*a^2 + 21*a*b + 15*b^2)*Log[b - b*Sech[ x]])/(16*b^6*(a + b)^3) + Log[a + b*Sech[x]]/(a*(a^2 - b^2)^3) - ((8*a^2 - 21*a*b + 15*b^2)*Log[b + b*Sech[x]])/(16*(a - b)^3*b^6) + 1/(16*b^4*(a + b)*(b - b*Sech[x])^2) + (5*a + 7*b)/(16*b^5*(a + b)^2*(b - b*Sech[x])) + 1 /(16*(a - b)*b^4*(b + b*Sech[x])^2) + (5*a - 7*b)/(16*(a - b)^2*b^5*(b + b *Sech[x]))))
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1)) Subst[Int[(b^2 - x^ 2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 1.74 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.91
| method | result | size |
| default | \(-\frac {\left (a \tanh \left (\frac {x}{2}\right )^{2}-b \tanh \left (\frac {x}{2}\right )^{2}+6 a -8 b \right )^{2}}{64 \left (a -b \right )^{3}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}-\frac {b^{6} \ln \left (a \tanh \left (\frac {x}{2}\right )^{2}-b \tanh \left (\frac {x}{2}\right )^{2}+a +b \right )}{\left (a -b \right )^{3} \left (a +b \right )^{3} a}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a}-\frac {1}{64 \left (a +b \right ) \tanh \left (\frac {x}{2}\right )^{4}}-\frac {6 a +8 b}{32 \left (a +b \right )^{2} \tanh \left (\frac {x}{2}\right )^{2}}+\frac {\left (16 a^{2}+42 a b +30 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{16 \left (a +b \right )^{3}}\) | \(162\) |
| risch | \(\frac {x}{a}-\frac {x \,a^{2}}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}-\frac {21 x a b}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {15 x \,b^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {x \,a^{2}}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}+\frac {21 x a b}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {15 x \,b^{2}}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {2 x \,b^{6}}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) a}-\frac {\left (-5 a^{2} b \,{\mathrm e}^{6 x}+9 b^{3} {\mathrm e}^{6 x}+16 a^{3} {\mathrm e}^{5 x}-24 a \,b^{2} {\mathrm e}^{5 x}-3 a^{2} b \,{\mathrm e}^{4 x}-b^{3} {\mathrm e}^{4 x}-16 a^{3} {\mathrm e}^{3 x}+32 a \,b^{2} {\mathrm e}^{3 x}-3 a^{2} b \,{\mathrm e}^{2 x}-b^{3} {\mathrm e}^{2 x}+16 a^{3} {\mathrm e}^{x}-24 \,{\mathrm e}^{x} b^{2} a -5 a^{2} b +9 b^{3}\right ) {\mathrm e}^{x}}{4 \left (a^{2}-b^{2}\right )^{2} \left ({\mathrm e}^{2 x}-1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{x}-1\right ) a^{2}}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {21 \ln \left ({\mathrm e}^{x}-1\right ) a b}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {15 \ln \left ({\mathrm e}^{x}-1\right ) b^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {\ln \left (1+{\mathrm e}^{x}\right ) a^{2}}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}-\frac {21 \ln \left (1+{\mathrm e}^{x}\right ) a b}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {15 \ln \left (1+{\mathrm e}^{x}\right ) b^{2}}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {b^{6} \ln \left ({\mathrm e}^{2 x}+\frac {2 b \,{\mathrm e}^{x}}{a}+1\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) a}\) | \(592\) |
Input:
int(coth(x)^5/(a+b*sech(x)),x,method=_RETURNVERBOSE)
Output:
-1/64*(a*tanh(1/2*x)^2-b*tanh(1/2*x)^2+6*a-8*b)^2/(a-b)^3-1/a*ln(tanh(1/2* x)-1)-1/(a-b)^3*b^6/(a+b)^3/a*ln(a*tanh(1/2*x)^2-b*tanh(1/2*x)^2+a+b)-1/a* ln(tanh(1/2*x)+1)-1/64/(a+b)/tanh(1/2*x)^4-1/32*(6*a+8*b)/(a+b)^2/tanh(1/2 *x)^2+1/16/(a+b)^3*(16*a^2+42*a*b+30*b^2)*ln(tanh(1/2*x))
Leaf count of result is larger than twice the leaf count of optimal. 5181 vs. \(2 (162) = 324\).
Time = 0.20 (sec) , antiderivative size = 5181, normalized size of antiderivative = 29.11 \[ \int \frac {\coth ^5(x)}{a+b \text {sech}(x)} \, dx=\text {Too large to display} \] Input:
integrate(coth(x)^5/(a+b*sech(x)),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\coth ^5(x)}{a+b \text {sech}(x)} \, dx=\int \frac {\coth ^{5}{\left (x \right )}}{a + b \operatorname {sech}{\left (x \right )}}\, dx \] Input:
integrate(coth(x)**5/(a+b*sech(x)),x)
Output:
Integral(coth(x)**5/(a + b*sech(x)), x)
Leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (162) = 324\).
Time = 0.05 (sec) , antiderivative size = 366, normalized size of antiderivative = 2.06 \[ \int \frac {\coth ^5(x)}{a+b \text {sech}(x)} \, dx=-\frac {b^{6} \log \left (2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a\right )}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}} + \frac {{\left (8 \, a^{2} - 21 \, a b + 15 \, b^{2}\right )} \log \left (e^{\left (-x\right )} + 1\right )}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac {{\left (8 \, a^{2} + 21 \, a b + 15 \, b^{2}\right )} \log \left (e^{\left (-x\right )} - 1\right )}{8 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} + \frac {{\left (5 \, a^{2} b - 9 \, b^{3}\right )} e^{\left (-x\right )} - 8 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} e^{\left (-2 \, x\right )} + {\left (3 \, a^{2} b + b^{3}\right )} e^{\left (-3 \, x\right )} + 16 \, {\left (a^{3} - 2 \, a b^{2}\right )} e^{\left (-4 \, x\right )} + {\left (3 \, a^{2} b + b^{3}\right )} e^{\left (-5 \, x\right )} - 8 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} e^{\left (-6 \, x\right )} + {\left (5 \, a^{2} b - 9 \, b^{3}\right )} e^{\left (-7 \, x\right )}}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} - 4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-2 \, x\right )} + 6 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-4 \, x\right )} - 4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-6 \, x\right )} + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-8 \, x\right )}\right )}} + \frac {x}{a} \] Input:
integrate(coth(x)^5/(a+b*sech(x)),x, algorithm="maxima")
Output:
-b^6*log(2*b*e^(-x) + a*e^(-2*x) + a)/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 ) + 1/8*(8*a^2 - 21*a*b + 15*b^2)*log(e^(-x) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + 1/8*(8*a^2 + 21*a*b + 15*b^2)*log(e^(-x) - 1)/(a^3 + 3*a^2*b + 3 *a*b^2 + b^3) + 1/4*((5*a^2*b - 9*b^3)*e^(-x) - 8*(2*a^3 - 3*a*b^2)*e^(-2* x) + (3*a^2*b + b^3)*e^(-3*x) + 16*(a^3 - 2*a*b^2)*e^(-4*x) + (3*a^2*b + b ^3)*e^(-5*x) - 8*(2*a^3 - 3*a*b^2)*e^(-6*x) + (5*a^2*b - 9*b^3)*e^(-7*x))/ (a^4 - 2*a^2*b^2 + b^4 - 4*(a^4 - 2*a^2*b^2 + b^4)*e^(-2*x) + 6*(a^4 - 2*a ^2*b^2 + b^4)*e^(-4*x) - 4*(a^4 - 2*a^2*b^2 + b^4)*e^(-6*x) + (a^4 - 2*a^2 *b^2 + b^4)*e^(-8*x)) + x/a
Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (162) = 324\).
Time = 0.12 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.13 \[ \int \frac {\coth ^5(x)}{a+b \text {sech}(x)} \, dx=-\frac {b^{6} \log \left ({\left | a {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b \right |}\right )}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}} + \frac {{\left (8 \, a^{2} - 21 \, a b + 15 \, b^{2}\right )} \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{16 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac {{\left (8 \, a^{2} + 21 \, a b + 15 \, b^{2}\right )} \log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {3 \, a^{5} {\left (e^{\left (-x\right )} + e^{x}\right )}^{4} - 9 \, a^{3} b^{2} {\left (e^{\left (-x\right )} + e^{x}\right )}^{4} + 9 \, a b^{4} {\left (e^{\left (-x\right )} + e^{x}\right )}^{4} - 5 \, a^{4} b {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} + 14 \, a^{2} b^{3} {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - 9 \, b^{5} {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - 8 \, a^{5} {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 32 \, a^{3} b^{2} {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 48 \, a b^{4} {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 12 \, a^{4} b {\left (e^{\left (-x\right )} + e^{x}\right )} - 40 \, a^{2} b^{3} {\left (e^{\left (-x\right )} + e^{x}\right )} + 28 \, b^{5} {\left (e^{\left (-x\right )} + e^{x}\right )} - 16 \, a^{3} b^{2} + 64 \, a b^{4}}{4 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left ({\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4\right )}^{2}} \] Input:
integrate(coth(x)^5/(a+b*sech(x)),x, algorithm="giac")
Output:
-b^6*log(abs(a*(e^(-x) + e^x) + 2*b))/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 ) + 1/16*(8*a^2 - 21*a*b + 15*b^2)*log(e^(-x) + e^x + 2)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + 1/16*(8*a^2 + 21*a*b + 15*b^2)*log(e^(-x) + e^x - 2)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 1/4*(3*a^5*(e^(-x) + e^x)^4 - 9*a^3*b^2*(e^( -x) + e^x)^4 + 9*a*b^4*(e^(-x) + e^x)^4 - 5*a^4*b*(e^(-x) + e^x)^3 + 14*a^ 2*b^3*(e^(-x) + e^x)^3 - 9*b^5*(e^(-x) + e^x)^3 - 8*a^5*(e^(-x) + e^x)^2 + 32*a^3*b^2*(e^(-x) + e^x)^2 - 48*a*b^4*(e^(-x) + e^x)^2 + 12*a^4*b*(e^(-x ) + e^x) - 40*a^2*b^3*(e^(-x) + e^x) + 28*b^5*(e^(-x) + e^x) - 16*a^3*b^2 + 64*a*b^4)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*((e^(-x) + e^x)^2 - 4)^2)
Time = 3.55 (sec) , antiderivative size = 623, normalized size of antiderivative = 3.50 \[ \int \frac {\coth ^5(x)}{a+b \text {sech}(x)} \, dx=\frac {\ln \left ({\mathrm {e}}^x-1\right )\,\left (8\,a^2+21\,a\,b+15\,b^2\right )}{8\,a^3+24\,a^2\,b+24\,a\,b^2+8\,b^3}-\frac {\frac {2\,\left (4\,a^4-5\,a^2\,b^2\right )}{a\,{\left (a^2-b^2\right )}^2}-\frac {{\mathrm {e}}^x\,\left (9\,a^2\,b-13\,b^3\right )}{2\,{\left (a^2-b^2\right )}^2}}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}-\frac {\frac {2\,\left (2\,a^6-5\,a^4\,b^2+3\,a^2\,b^4\right )}{a\,{\left (a^2-b^2\right )}^3}-\frac {{\mathrm {e}}^x\,\left (5\,a^4\,b-14\,a^2\,b^3+9\,b^5\right )}{4\,{\left (a^2-b^2\right )}^3}}{{\mathrm {e}}^{2\,x}-1}-\frac {\frac {8\,\left (a^4-a^2\,b^2\right )}{a\,{\left (a^2-b^2\right )}^2}-\frac {6\,{\mathrm {e}}^x\,\left (a^2\,b-b^3\right )}{{\left (a^2-b^2\right )}^2}}{3\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}-1}-\frac {x}{a}-\frac {\frac {4\,a}{a^2-b^2}-\frac {4\,b\,{\mathrm {e}}^x}{a^2-b^2}}{6\,{\mathrm {e}}^{4\,x}-4\,{\mathrm {e}}^{2\,x}-4\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{8\,x}+1}+\frac {\ln \left ({\mathrm {e}}^x+1\right )\,\left (8\,a^2-21\,a\,b+15\,b^2\right )}{8\,a^3-24\,a^2\,b+24\,a\,b^2-8\,b^3}+\frac {b^6\,\ln \left (64\,a^{13}\,{\mathrm {e}}^{2\,x}+64\,a\,b^{12}+64\,a^{13}+159\,a^3\,b^{10}+492\,a^5\,b^8-1214\,a^7\,b^6+1020\,a^9\,b^4-393\,a^{11}\,b^2+128\,b^{13}\,{\mathrm {e}}^x+159\,a^3\,b^{10}\,{\mathrm {e}}^{2\,x}+492\,a^5\,b^8\,{\mathrm {e}}^{2\,x}-1214\,a^7\,b^6\,{\mathrm {e}}^{2\,x}+1020\,a^9\,b^4\,{\mathrm {e}}^{2\,x}-393\,a^{11}\,b^2\,{\mathrm {e}}^{2\,x}+128\,a^{12}\,b\,{\mathrm {e}}^x+64\,a\,b^{12}\,{\mathrm {e}}^{2\,x}+318\,a^2\,b^{11}\,{\mathrm {e}}^x+984\,a^4\,b^9\,{\mathrm {e}}^x-2428\,a^6\,b^7\,{\mathrm {e}}^x+2040\,a^8\,b^5\,{\mathrm {e}}^x-786\,a^{10}\,b^3\,{\mathrm {e}}^x\right )}{-a^7+3\,a^5\,b^2-3\,a^3\,b^4+a\,b^6} \] Input:
int(coth(x)^5/(a + b/cosh(x)),x)
Output:
(log(exp(x) - 1)*(21*a*b + 8*a^2 + 15*b^2))/(24*a*b^2 + 24*a^2*b + 8*a^3 + 8*b^3) - ((2*(4*a^4 - 5*a^2*b^2))/(a*(a^2 - b^2)^2) - (exp(x)*(9*a^2*b - 13*b^3))/(2*(a^2 - b^2)^2))/(exp(4*x) - 2*exp(2*x) + 1) - ((2*(2*a^6 + 3*a ^2*b^4 - 5*a^4*b^2))/(a*(a^2 - b^2)^3) - (exp(x)*(5*a^4*b + 9*b^5 - 14*a^2 *b^3))/(4*(a^2 - b^2)^3))/(exp(2*x) - 1) - ((8*(a^4 - a^2*b^2))/(a*(a^2 - b^2)^2) - (6*exp(x)*(a^2*b - b^3))/(a^2 - b^2)^2)/(3*exp(2*x) - 3*exp(4*x) + exp(6*x) - 1) - x/a - ((4*a)/(a^2 - b^2) - (4*b*exp(x))/(a^2 - b^2))/(6 *exp(4*x) - 4*exp(2*x) - 4*exp(6*x) + exp(8*x) + 1) + (log(exp(x) + 1)*(8* a^2 - 21*a*b + 15*b^2))/(24*a*b^2 - 24*a^2*b + 8*a^3 - 8*b^3) + (b^6*log(6 4*a^13*exp(2*x) + 64*a*b^12 + 64*a^13 + 159*a^3*b^10 + 492*a^5*b^8 - 1214* a^7*b^6 + 1020*a^9*b^4 - 393*a^11*b^2 + 128*b^13*exp(x) + 159*a^3*b^10*exp (2*x) + 492*a^5*b^8*exp(2*x) - 1214*a^7*b^6*exp(2*x) + 1020*a^9*b^4*exp(2* x) - 393*a^11*b^2*exp(2*x) + 128*a^12*b*exp(x) + 64*a*b^12*exp(2*x) + 318* a^2*b^11*exp(x) + 984*a^4*b^9*exp(x) - 2428*a^6*b^7*exp(x) + 2040*a^8*b^5* exp(x) - 786*a^10*b^3*exp(x)))/(a*b^6 - a^7 - 3*a^3*b^4 + 3*a^5*b^2)
\[ \int \frac {\coth ^5(x)}{a+b \text {sech}(x)} \, dx=\int \frac {\coth \left (x \right )^{5}}{a +b \,\mathrm {sech}\left (x \right )}d x \] Input:
int(coth(x)^5/(a+b*sech(x)),x)
Output:
int(coth(x)^5/(a+b*sech(x)),x)