\(\int \frac {\tanh ^3(x)}{a+b \sinh (x)} \, dx\) [229]

Optimal result

Integrand size = 13, antiderivative size = 88 \[ \int \frac {\tanh ^3(x)}{a+b \sinh (x)} \, dx=\frac {b \left (3 a^2+b^2\right ) \arctan (\sinh (x))}{2 \left (a^2+b^2\right )^2}+\frac {a^3 \log (\cosh (x))}{\left (a^2+b^2\right )^2}-\frac {a^3 \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac {\text {sech}^2(x) (a-b \sinh (x))}{2 \left (a^2+b^2\right )} \] Output:

1/2*b*(3*a^2+b^2)*arctan(sinh(x))/(a^2+b^2)^2+a^3*ln(cosh(x))/(a^2+b^2)^2- 
a^3*ln(a+b*sinh(x))/(a^2+b^2)^2+sech(x)^2*(a-b*sinh(x))/(2*a^2+2*b^2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.74 \[ \int \frac {\tanh ^3(x)}{a+b \sinh (x)} \, dx=-\frac {b \arctan (\sinh (x))}{2 \left (a^2+b^2\right )}+\frac {\left (a^3-i \left (2 a^2 b+b^3\right )\right ) \log (i-\sinh (x))}{2 \left (a^2+b^2\right )^2}+\frac {\left (a^3+i \left (2 a^2 b+b^3\right )\right ) \log (i+\sinh (x))}{2 \left (a^2+b^2\right )^2}-\frac {a^3 \log (a+b \sinh (x))}{\left (a^2+b^2\right )^2}+\frac {a \text {sech}^2(x)}{2 \left (a^2+b^2\right )}-\frac {b \text {sech}(x) \tanh (x)}{2 \left (a^2+b^2\right )} \] Input:

Integrate[Tanh[x]^3/(a + b*Sinh[x]),x]
 

Output:

-1/2*(b*ArcTan[Sinh[x]])/(a^2 + b^2) + ((a^3 - I*(2*a^2*b + b^3))*Log[I - 
Sinh[x]])/(2*(a^2 + b^2)^2) + ((a^3 + I*(2*a^2*b + b^3))*Log[I + Sinh[x]]) 
/(2*(a^2 + b^2)^2) - (a^3*Log[a + b*Sinh[x]])/(a^2 + b^2)^2 + (a*Sech[x]^2 
)/(2*(a^2 + b^2)) - (b*Sech[x]*Tanh[x])/(2*(a^2 + b^2))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.39, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3042, 26, 3200, 601, 25, 27, 657, 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.

\(\displaystyle \int \frac {\tanh ^3(x)}{a+b \sinh (x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i \tan (i x)^3}{a-i b \sin (i x)}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {\tan (i x)^3}{a-i b \sin (i x)}dx\)

\(\Big \downarrow \) 3200

\(\displaystyle \int \frac {b^3 \sinh ^3(x)}{\left (b^2 \sinh ^2(x)+b^2\right )^2 (a+b \sinh (x))}d(b \sinh (x))\)

\(\Big \downarrow \) 601

\(\displaystyle \frac {b^2 (a-b \sinh (x))}{2 \left (a^2+b^2\right ) \left (b^2 \sinh ^2(x)+b^2\right )}-\frac {\int -\frac {b^2 \left (a b^2+\left (2 a^2+b^2\right ) \sinh (x) b\right )}{\left (a^2+b^2\right ) (a+b \sinh (x)) \left (\sinh ^2(x) b^2+b^2\right )}d(b \sinh (x))}{2 b^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {b^2 \left (a b^2+\left (2 a^2+b^2\right ) \sinh (x) b\right )}{\left (a^2+b^2\right ) (a+b \sinh (x)) \left (\sinh ^2(x) b^2+b^2\right )}d(b \sinh (x))}{2 b^2}+\frac {b^2 (a-b \sinh (x))}{2 \left (a^2+b^2\right ) \left (b^2 \sinh ^2(x)+b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a b^2+\left (2 a^2+b^2\right ) \sinh (x) b}{(a+b \sinh (x)) \left (\sinh ^2(x) b^2+b^2\right )}d(b \sinh (x))}{2 \left (a^2+b^2\right )}+\frac {b^2 (a-b \sinh (x))}{2 \left (a^2+b^2\right ) \left (b^2 \sinh ^2(x)+b^2\right )}\)

\(\Big \downarrow \) 657

\(\displaystyle \frac {\int \left (\frac {b^4+3 a^2 b^2+2 a^3 \sinh (x) b}{\left (a^2+b^2\right ) \left (\sinh ^2(x) b^2+b^2\right )}-\frac {2 a^3}{\left (a^2+b^2\right ) (a+b \sinh (x))}\right )d(b \sinh (x))}{2 \left (a^2+b^2\right )}+\frac {b^2 (a-b \sinh (x))}{2 \left (a^2+b^2\right ) \left (b^2 \sinh ^2(x)+b^2\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^2 (a-b \sinh (x))}{2 \left (a^2+b^2\right ) \left (b^2 \sinh ^2(x)+b^2\right )}+\frac {\frac {b \left (3 a^2+b^2\right ) \arctan (\sinh (x))}{a^2+b^2}+\frac {a^3 \log \left (b^2 \sinh ^2(x)+b^2\right )}{a^2+b^2}-\frac {2 a^3 \log (a+b \sinh (x))}{a^2+b^2}}{2 \left (a^2+b^2\right )}\)

Input:

Int[Tanh[x]^3/(a + b*Sinh[x]),x]
 

Output:

((b*(3*a^2 + b^2)*ArcTan[Sinh[x]])/(a^2 + b^2) - (2*a^3*Log[a + b*Sinh[x]] 
)/(a^2 + b^2) + (a^3*Log[b^2 + b^2*Sinh[x]^2])/(a^2 + b^2))/(2*(a^2 + b^2) 
) + (b^2*(a - b*Sinh[x]))/(2*(a^2 + b^2)*(b^2 + b^2*Sinh[x]^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 601

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c 
+ d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* 
(2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] 
 && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
 

rule 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3200

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p 
_.), x_Symbol] :> Simp[1/f   Subst[Int[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1) 
/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b 
^2, 0] && IntegerQ[(p + 1)/2]
 

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.89

method	result	size
default	\(-\frac {8 a^{3} \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a \right )}{8 a^{4}+16 a^{2} b^{2}+8 b^{4}}+\frac {\frac {2 \left (\left (\frac {1}{2} a^{2} b +\frac {1}{2} b^{3}\right ) \tanh \left (\frac {x}{2}\right )^{3}+\left (-a^{3}-a \,b^{2}\right ) \tanh \left (\frac {x}{2}\right )^{2}+\left (-\frac {1}{2} a^{2} b -\frac {1}{2} b^{3}\right ) \tanh \left (\frac {x}{2}\right )\right )}{\left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )^{2}}+a^{3} \ln \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )+\left (3 a^{2} b +b^{3}\right ) \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a^{4}+2 a^{2} b^{2}+b^{4}}\)	\(166\)
risch	\(\frac {{\mathrm e}^{x} \left (-b \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} a +b \right )}{\left ({\mathrm e}^{2 x}+1\right )^{2} \left (a^{2}+b^{2}\right )}+\frac {3 i \ln \left ({\mathrm e}^{x}+i\right ) a^{2} b}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {i \ln \left ({\mathrm e}^{x}+i\right ) b^{3}}{2 a^{4}+4 a^{2} b^{2}+2 b^{4}}+\frac {\ln \left ({\mathrm e}^{x}+i\right ) a^{3}}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {3 i \ln \left ({\mathrm e}^{x}-i\right ) a^{2} b}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {i \ln \left ({\mathrm e}^{x}-i\right ) b^{3}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {\ln \left ({\mathrm e}^{x}-i\right ) a^{3}}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {a^{3} \ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}-1\right )}{a^{4}+2 a^{2} b^{2}+b^{4}}\)	\(245\)

Input:

int(tanh(x)^3/(a+b*sinh(x)),x,method=_RETURNVERBOSE)
 

Output:

-8*a^3/(8*a^4+16*a^2*b^2+8*b^4)*ln(tanh(1/2*x)^2*a-2*b*tanh(1/2*x)-a)+2/(a 
^4+2*a^2*b^2+b^4)*(((1/2*a^2*b+1/2*b^3)*tanh(1/2*x)^3+(-a^3-a*b^2)*tanh(1/ 
2*x)^2+(-1/2*a^2*b-1/2*b^3)*tanh(1/2*x))/(1+tanh(1/2*x)^2)^2+1/2*a^3*ln(1+ 
tanh(1/2*x)^2)+1/2*(3*a^2*b+b^3)*arctan(tanh(1/2*x)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 655 vs. \(2 (85) = 170\).

Time = 0.11 (sec) , antiderivative size = 655, normalized size of antiderivative = 7.44 \[ \int \frac {\tanh ^3(x)}{a+b \sinh (x)} \, dx =\text {Too large to display} \] Input:

integrate(tanh(x)^3/(a+b*sinh(x)),x, algorithm="fricas")
 

Output:

-((a^2*b + b^3)*cosh(x)^3 + (a^2*b + b^3)*sinh(x)^3 - 2*(a^3 + a*b^2)*cosh 
(x)^2 - (2*a^3 + 2*a*b^2 - 3*(a^2*b + b^3)*cosh(x))*sinh(x)^2 - ((3*a^2*b 
+ b^3)*cosh(x)^4 + 4*(3*a^2*b + b^3)*cosh(x)*sinh(x)^3 + (3*a^2*b + b^3)*s 
inh(x)^4 + 3*a^2*b + b^3 + 2*(3*a^2*b + b^3)*cosh(x)^2 + 2*(3*a^2*b + b^3 
+ 3*(3*a^2*b + b^3)*cosh(x)^2)*sinh(x)^2 + 4*((3*a^2*b + b^3)*cosh(x)^3 + 
(3*a^2*b + b^3)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) - (a^2*b + b^3 
)*cosh(x) + (a^3*cosh(x)^4 + 4*a^3*cosh(x)*sinh(x)^3 + a^3*sinh(x)^4 + 2*a 
^3*cosh(x)^2 + a^3 + 2*(3*a^3*cosh(x)^2 + a^3)*sinh(x)^2 + 4*(a^3*cosh(x)^ 
3 + a^3*cosh(x))*sinh(x))*log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x))) - (a^ 
3*cosh(x)^4 + 4*a^3*cosh(x)*sinh(x)^3 + a^3*sinh(x)^4 + 2*a^3*cosh(x)^2 + 
a^3 + 2*(3*a^3*cosh(x)^2 + a^3)*sinh(x)^2 + 4*(a^3*cosh(x)^3 + a^3*cosh(x) 
)*sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) - (a^2*b + b^3 - 3*(a^2*b + 
b^3)*cosh(x)^2 + 4*(a^3 + a*b^2)*cosh(x))*sinh(x))/((a^4 + 2*a^2*b^2 + b^4 
)*cosh(x)^4 + 4*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)*sinh(x)^3 + (a^4 + 2*a^2*b 
^2 + b^4)*sinh(x)^4 + a^4 + 2*a^2*b^2 + b^4 + 2*(a^4 + 2*a^2*b^2 + b^4)*co 
sh(x)^2 + 2*(a^4 + 2*a^2*b^2 + b^4 + 3*(a^4 + 2*a^2*b^2 + b^4)*cosh(x)^2)* 
sinh(x)^2 + 4*((a^4 + 2*a^2*b^2 + b^4)*cosh(x)^3 + (a^4 + 2*a^2*b^2 + b^4) 
*cosh(x))*sinh(x))
 

Sympy [F]

\[ \int \frac {\tanh ^3(x)}{a+b \sinh (x)} \, dx=\int \frac {\tanh ^{3}{\left (x \right )}}{a + b \sinh {\left (x \right )}}\, dx \] Input:

integrate(tanh(x)**3/(a+b*sinh(x)),x)
 

Output:

Integral(tanh(x)**3/(a + b*sinh(x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.82 \[ \int \frac {\tanh ^3(x)}{a+b \sinh (x)} \, dx=-\frac {a^{3} \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {a^{3} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (3 \, a^{2} b + b^{3}\right )} \arctan \left (e^{\left (-x\right )}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {b e^{\left (-x\right )} - 2 \, a e^{\left (-2 \, x\right )} - b e^{\left (-3 \, x\right )}}{a^{2} + b^{2} + 2 \, {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, x\right )} + {\left (a^{2} + b^{2}\right )} e^{\left (-4 \, x\right )}} \] Input:

integrate(tanh(x)^3/(a+b*sinh(x)),x, algorithm="maxima")
 

Output:

-a^3*log(-2*a*e^(-x) + b*e^(-2*x) - b)/(a^4 + 2*a^2*b^2 + b^4) + a^3*log(e 
^(-2*x) + 1)/(a^4 + 2*a^2*b^2 + b^4) - (3*a^2*b + b^3)*arctan(e^(-x))/(a^4 
 + 2*a^2*b^2 + b^4) - (b*e^(-x) - 2*a*e^(-2*x) - b*e^(-3*x))/(a^2 + b^2 + 
2*(a^2 + b^2)*e^(-2*x) + (a^2 + b^2)*e^(-4*x))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (85) = 170\).

Time = 0.14 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.40 \[ \int \frac {\tanh ^3(x)}{a+b \sinh (x)} \, dx=-\frac {a^{3} b \log \left ({\left | -b {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} + \frac {a^{3} \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} + \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} {\left (3 \, a^{2} b + b^{3}\right )}}{4 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}} - \frac {a^{3} {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} - 2 \, a^{2} b {\left (e^{\left (-x\right )} - e^{x}\right )} - 2 \, b^{3} {\left (e^{\left (-x\right )} - e^{x}\right )} - 4 \, a b^{2}}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}} \] Input:

integrate(tanh(x)^3/(a+b*sinh(x)),x, algorithm="giac")
 

Output:

-a^3*b*log(abs(-b*(e^(-x) - e^x) + 2*a))/(a^4*b + 2*a^2*b^3 + b^5) + 1/2*a 
^3*log((e^(-x) - e^x)^2 + 4)/(a^4 + 2*a^2*b^2 + b^4) + 1/4*(pi + 2*arctan( 
1/2*(e^(2*x) - 1)*e^(-x)))*(3*a^2*b + b^3)/(a^4 + 2*a^2*b^2 + b^4) - 1/2*( 
a^3*(e^(-x) - e^x)^2 - 2*a^2*b*(e^(-x) - e^x) - 2*b^3*(e^(-x) - e^x) - 4*a 
*b^2)/((a^4 + 2*a^2*b^2 + b^4)*((e^(-x) - e^x)^2 + 4))
 

Mupad [B] (verification not implemented)

Time = 3.24 (sec) , antiderivative size = 291, normalized size of antiderivative = 3.31 \[ \int \frac {\tanh ^3(x)}{a+b \sinh (x)} \, dx=\frac {\frac {2\,\left (a^3+a\,b^2\right )}{{\left (a^2+b^2\right )}^2}-\frac {{\mathrm {e}}^x\,\left (a^2\,b+b^3\right )}{{\left (a^2+b^2\right )}^2}}{{\mathrm {e}}^{2\,x}+1}-\frac {\frac {2\,a}{a^2+b^2}-\frac {2\,b\,{\mathrm {e}}^x}{a^2+b^2}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,\left (2\,a+b\,1{}\mathrm {i}\right )}{2\,\left (a^2+a\,b\,2{}\mathrm {i}-b^2\right )}-\frac {a^3\,\ln \left (b^7\,{\mathrm {e}}^{2\,x}-16\,a^6\,b-b^7-6\,a^2\,b^5-9\,a^4\,b^3+32\,a^7\,{\mathrm {e}}^x+6\,a^2\,b^5\,{\mathrm {e}}^{2\,x}+9\,a^4\,b^3\,{\mathrm {e}}^{2\,x}+2\,a\,b^6\,{\mathrm {e}}^x+16\,a^6\,b\,{\mathrm {e}}^{2\,x}+12\,a^3\,b^4\,{\mathrm {e}}^x+18\,a^5\,b^2\,{\mathrm {e}}^x\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,\left (b+a\,2{}\mathrm {i}\right )}{2\,\left (a^2\,1{}\mathrm {i}+2\,a\,b-b^2\,1{}\mathrm {i}\right )} \] Input:

int(tanh(x)^3/(a + b*sinh(x)),x)
 

Output:

((2*(a*b^2 + a^3))/(a^2 + b^2)^2 - (exp(x)*(a^2*b + b^3))/(a^2 + b^2)^2)/( 
exp(2*x) + 1) - ((2*a)/(a^2 + b^2) - (2*b*exp(x))/(a^2 + b^2))/(2*exp(2*x) 
 + exp(4*x) + 1) + (log(exp(x)*1i + 1)*(2*a + b*1i))/(2*(a*b*2i + a^2 - b^ 
2)) - (a^3*log(b^7*exp(2*x) - 16*a^6*b - b^7 - 6*a^2*b^5 - 9*a^4*b^3 + 32* 
a^7*exp(x) + 6*a^2*b^5*exp(2*x) + 9*a^4*b^3*exp(2*x) + 2*a*b^6*exp(x) + 16 
*a^6*b*exp(2*x) + 12*a^3*b^4*exp(x) + 18*a^5*b^2*exp(x)))/(a^4 + b^4 + 2*a 
^2*b^2) + (log(exp(x) + 1i)*(a*2i + b))/(2*(2*a*b + a^2*1i - b^2*1i))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 352, normalized size of antiderivative = 4.00 \[ \int \frac {\tanh ^3(x)}{a+b \sinh (x)} \, dx=\frac {3 e^{4 x} \mathit {atan} \left (e^{x}\right ) a^{2} b +e^{4 x} \mathit {atan} \left (e^{x}\right ) b^{3}+6 e^{2 x} \mathit {atan} \left (e^{x}\right ) a^{2} b +2 e^{2 x} \mathit {atan} \left (e^{x}\right ) b^{3}+3 \mathit {atan} \left (e^{x}\right ) a^{2} b +\mathit {atan} \left (e^{x}\right ) b^{3}+e^{4 x} \mathrm {log}\left (e^{2 x}+1\right ) a^{3}-e^{4 x} \mathrm {log}\left (e^{2 x} b +2 e^{x} a -b \right ) a^{3}-e^{4 x} a^{3}-e^{4 x} a \,b^{2}-e^{3 x} a^{2} b -e^{3 x} b^{3}+2 e^{2 x} \mathrm {log}\left (e^{2 x}+1\right ) a^{3}-2 e^{2 x} \mathrm {log}\left (e^{2 x} b +2 e^{x} a -b \right ) a^{3}+e^{x} a^{2} b +e^{x} b^{3}+\mathrm {log}\left (e^{2 x}+1\right ) a^{3}-\mathrm {log}\left (e^{2 x} b +2 e^{x} a -b \right ) a^{3}-a^{3}-a \,b^{2}}{e^{4 x} a^{4}+2 e^{4 x} a^{2} b^{2}+e^{4 x} b^{4}+2 e^{2 x} a^{4}+4 e^{2 x} a^{2} b^{2}+2 e^{2 x} b^{4}+a^{4}+2 a^{2} b^{2}+b^{4}} \] Input:

int(tanh(x)^3/(a+b*sinh(x)),x)
 

Output:

(3*e**(4*x)*atan(e**x)*a**2*b + e**(4*x)*atan(e**x)*b**3 + 6*e**(2*x)*atan 
(e**x)*a**2*b + 2*e**(2*x)*atan(e**x)*b**3 + 3*atan(e**x)*a**2*b + atan(e* 
*x)*b**3 + e**(4*x)*log(e**(2*x) + 1)*a**3 - e**(4*x)*log(e**(2*x)*b + 2*e 
**x*a - b)*a**3 - e**(4*x)*a**3 - e**(4*x)*a*b**2 - e**(3*x)*a**2*b - e**( 
3*x)*b**3 + 2*e**(2*x)*log(e**(2*x) + 1)*a**3 - 2*e**(2*x)*log(e**(2*x)*b 
+ 2*e**x*a - b)*a**3 + e**x*a**2*b + e**x*b**3 + log(e**(2*x) + 1)*a**3 - 
log(e**(2*x)*b + 2*e**x*a - b)*a**3 - a**3 - a*b**2)/(e**(4*x)*a**4 + 2*e* 
*(4*x)*a**2*b**2 + e**(4*x)*b**4 + 2*e**(2*x)*a**4 + 4*e**(2*x)*a**2*b**2 
+ 2*e**(2*x)*b**4 + a**4 + 2*a**2*b**2 + b**4)