\(\int \frac {\tanh ^4(x)}{a+b \sinh (x)} \, dx\) [228]

Optimal result

Integrand size = 13, antiderivative size = 124 \[ \int \frac {\tanh ^4(x)}{a+b \sinh (x)} \, dx=-\frac {2 a^4 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {a^2 b \text {sech}(x)}{\left (a^2+b^2\right )^2}-\frac {b \text {sech}(x)}{a^2+b^2}+\frac {b \text {sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac {a^3 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac {a \tanh ^3(x)}{3 \left (a^2+b^2\right )} \]

[Out]

-2*a^4*arctanh((b-a*tanh(1/2*x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(5/2)-a^2*b*sech(x)/(a^2+b^2)^2-b*sech(x)/(a^2+b^2
)+1/3*b*sech(x)^3/(a^2+b^2)-a^3*tanh(x)/(a^2+b^2)^2-1/3*a*tanh(x)^3/(a^2+b^2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {2806, 2687, 30, 2686, 3852, 8, 2739, 632, 212} \[ \int \frac {\tanh ^4(x)}{a+b \sinh (x)} \, dx=-\frac {a \tanh ^3(x)}{3 \left (a^2+b^2\right )}+\frac {b \text {sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac {a^2 b \text {sech}(x)}{\left (a^2+b^2\right )^2}-\frac {b \text {sech}(x)}{a^2+b^2}-\frac {2 a^4 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {a^3 \tanh (x)}{\left (a^2+b^2\right )^2} \]

[In]

Int[Tanh[x]^4/(a + b*Sinh[x]),x]

[Out]

(-2*a^4*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (a^2*b*Sech[x])/(a^2 + b^2)^2 - (b*Sec
h[x])/(a^2 + b^2) + (b*Sech[x]^3)/(3*(a^2 + b^2)) - (a^3*Tanh[x])/(a^2 + b^2)^2 - (a*Tanh[x]^3)/(3*(a^2 + b^2)
)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2806

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[a/(a^2 - b^
2), Int[(g*Tan[e + f*x])^p/Sin[e + f*x]^2, x], x] + (-Dist[b*(g/(a^2 - b^2)), Int[(g*Tan[e + f*x])^(p - 1)/Cos
[e + f*x], x], x] - Dist[a^2*(g^2/(a^2 - b^2)), Int[(g*Tan[e + f*x])^(p - 2)/(a + b*Sin[e + f*x]), x], x]) /;
FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*p] && GtQ[p, 1]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {a \int \text {sech}^2(x) \tanh ^2(x) \, dx}{a^2+b^2}+\frac {a^2 \int \frac {\tanh ^2(x)}{a+b \sinh (x)} \, dx}{a^2+b^2}+\frac {b \int \text {sech}(x) \tanh ^3(x) \, dx}{a^2+b^2} \\ & = -\frac {a^3 \int \text {sech}^2(x) \, dx}{\left (a^2+b^2\right )^2}+\frac {a^4 \int \frac {1}{a+b \sinh (x)} \, dx}{\left (a^2+b^2\right )^2}+\frac {\left (a^2 b\right ) \int \text {sech}(x) \tanh (x) \, dx}{\left (a^2+b^2\right )^2}-\frac {(i a) \text {Subst}\left (\int x^2 \, dx,x,i \tanh (x)\right )}{a^2+b^2}+\frac {b \text {Subst}\left (\int \left (-1+x^2\right ) \, dx,x,\text {sech}(x)\right )}{a^2+b^2} \\ & = -\frac {b \text {sech}(x)}{a^2+b^2}+\frac {b \text {sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac {a \tanh ^3(x)}{3 \left (a^2+b^2\right )}-\frac {\left (i a^3\right ) \text {Subst}(\int 1 \, dx,x,-i \tanh (x))}{\left (a^2+b^2\right )^2}+\frac {\left (2 a^4\right ) \text {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{\left (a^2+b^2\right )^2}-\frac {\left (a^2 b\right ) \text {Subst}(\int 1 \, dx,x,\text {sech}(x))}{\left (a^2+b^2\right )^2} \\ & = -\frac {a^2 b \text {sech}(x)}{\left (a^2+b^2\right )^2}-\frac {b \text {sech}(x)}{a^2+b^2}+\frac {b \text {sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac {a^3 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac {a \tanh ^3(x)}{3 \left (a^2+b^2\right )}-\frac {\left (4 a^4\right ) \text {Subst}\left (\int \frac {1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac {x}{2}\right )\right )}{\left (a^2+b^2\right )^2} \\ & = -\frac {2 a^4 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {a^2 b \text {sech}(x)}{\left (a^2+b^2\right )^2}-\frac {b \text {sech}(x)}{a^2+b^2}+\frac {b \text {sech}^3(x)}{3 \left (a^2+b^2\right )}-\frac {a^3 \tanh (x)}{\left (a^2+b^2\right )^2}-\frac {a \tanh ^3(x)}{3 \left (a^2+b^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.87 \[ \int \frac {\tanh ^4(x)}{a+b \sinh (x)} \, dx=\frac {\frac {6 a^4 \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-3 b \left (2 a^2+b^2\right ) \text {sech}(x)+\left (a^2+b^2\right ) \text {sech}^3(x) (b+a \sinh (x))-a \left (4 a^2+b^2\right ) \tanh (x)}{3 \left (a^2+b^2\right )^2} \]

[In]

Integrate[Tanh[x]^4/(a + b*Sinh[x]),x]

[Out]

((6*a^4*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] - 3*b*(2*a^2 + b^2)*Sech[x] + (a^2 + b^2)
*Sech[x]^3*(b + a*Sinh[x]) - a*(4*a^2 + b^2)*Tanh[x])/(3*(a^2 + b^2)^2)

Maple [A] (verified)

Time = 1.91 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.38

method	result	size
default	\(\frac {32 a^{4} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (16 a^{4}+32 a^{2} b^{2}+16 b^{4}\right ) \sqrt {a^{2}+b^{2}}}+\frac {-2 a^{3} \tanh \left (\frac {x}{2}\right )^{5}-2 a^{2} b \tanh \left (\frac {x}{2}\right )^{4}+2 \left (-\frac {10}{3} a^{3}-\frac {4}{3} a \,b^{2}\right ) \tanh \left (\frac {x}{2}\right )^{3}+2 \left (-4 a^{2} b -2 b^{3}\right ) \tanh \left (\frac {x}{2}\right )^{2}-2 \tanh \left (\frac {x}{2}\right ) a^{3}-\frac {10 a^{2} b}{3}-\frac {4 b^{3}}{3}}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )^{3}}\)	\(171\)
risch	\(\frac {-4 a^{2} b \,{\mathrm e}^{5 x}-2 b^{3} {\mathrm e}^{5 x}+4 a^{3} {\mathrm e}^{4 x}+2 \,{\mathrm e}^{4 x} a \,b^{2}-\frac {16 a^{2} b \,{\mathrm e}^{3 x}}{3}-\frac {4 \,{\mathrm e}^{3 x} b^{3}}{3}+4 a^{3} {\mathrm e}^{2 x}-4 \,{\mathrm e}^{x} a^{2} b -2 b^{3} {\mathrm e}^{x}+\frac {8 a^{3}}{3}+\frac {2 a \,b^{2}}{3}}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (1+{\mathrm e}^{2 x}\right )^{3}}+\frac {a^{4} \ln \left ({\mathrm e}^{x}+\frac {\left (a^{2}+b^{2}\right )^{\frac {5}{2}} a -a^{6}-3 a^{4} b^{2}-3 a^{2} b^{4}-b^{6}}{b \left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}-\frac {a^{4} \ln \left ({\mathrm e}^{x}+\frac {\left (a^{2}+b^{2}\right )^{\frac {5}{2}} a +a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}{b \left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\)	\(254\)

[In]

int(tanh(x)^4/(a+b*sinh(x)),x,method=_RETURNVERBOSE)

[Out]

32*a^4/(16*a^4+32*a^2*b^2+16*b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))+2/(a^4+2*
a^2*b^2+b^4)*(-a^3*tanh(1/2*x)^5-a^2*b*tanh(1/2*x)^4+(-10/3*a^3-4/3*a*b^2)*tanh(1/2*x)^3+(-4*a^2*b-2*b^3)*tanh
(1/2*x)^2-tanh(1/2*x)*a^3-5/3*a^2*b-2/3*b^3)/(1+tanh(1/2*x)^2)^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1199 vs. \(2 (116) = 232\).

Time = 0.29 (sec) , antiderivative size = 1199, normalized size of antiderivative = 9.67 \[ \int \frac {\tanh ^4(x)}{a+b \sinh (x)} \, dx=\text {Too large to display} \]

[In]

integrate(tanh(x)^4/(a+b*sinh(x)),x, algorithm="fricas")

[Out]

-1/3*(6*(2*a^4*b + 3*a^2*b^3 + b^5)*cosh(x)^5 + 6*(2*a^4*b + 3*a^2*b^3 + b^5)*sinh(x)^5 - 8*a^5 - 10*a^3*b^2 -
2*a*b^4 - 6*(2*a^5 + 3*a^3*b^2 + a*b^4)*cosh(x)^4 - 6*(2*a^5 + 3*a^3*b^2 + a*b^4 - 5*(2*a^4*b + 3*a^2*b^3 + b
^5)*cosh(x))*sinh(x)^4 + 4*(4*a^4*b + 5*a^2*b^3 + b^5)*cosh(x)^3 + 4*(4*a^4*b + 5*a^2*b^3 + b^5 + 15*(2*a^4*b
+ 3*a^2*b^3 + b^5)*cosh(x)^2 - 6*(2*a^5 + 3*a^3*b^2 + a*b^4)*cosh(x))*sinh(x)^3 - 12*(a^5 + a^3*b^2)*cosh(x)^2
- 12*(a^5 + a^3*b^2 - 5*(2*a^4*b + 3*a^2*b^3 + b^5)*cosh(x)^3 + 3*(2*a^5 + 3*a^3*b^2 + a*b^4)*cosh(x)^2 - (4*
a^4*b + 5*a^2*b^3 + b^5)*cosh(x))*sinh(x)^2 - 3*(a^4*cosh(x)^6 + 6*a^4*cosh(x)*sinh(x)^5 + a^4*sinh(x)^6 + 3*a
^4*cosh(x)^4 + 3*a^4*cosh(x)^2 + 3*(5*a^4*cosh(x)^2 + a^4)*sinh(x)^4 + a^4 + 4*(5*a^4*cosh(x)^3 + 3*a^4*cosh(x
))*sinh(x)^3 + 3*(5*a^4*cosh(x)^4 + 6*a^4*cosh(x)^2 + a^4)*sinh(x)^2 + 6*(a^4*cosh(x)^5 + 2*a^4*cosh(x)^3 + a^
4*cosh(x))*sinh(x))*sqrt(a^2 + b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 + b^2 + 2*(b^2*
cosh(x) + a*b)*sinh(x) - 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(
x) + 2*(b*cosh(x) + a)*sinh(x) - b)) + 6*(2*a^4*b + 3*a^2*b^3 + b^5)*cosh(x) + 6*(2*a^4*b + 3*a^2*b^3 + b^5 +
5*(2*a^4*b + 3*a^2*b^3 + b^5)*cosh(x)^4 - 4*(2*a^5 + 3*a^3*b^2 + a*b^4)*cosh(x)^3 + 2*(4*a^4*b + 5*a^2*b^3 + b
^5)*cosh(x)^2 - 4*(a^5 + a^3*b^2)*cosh(x))*sinh(x))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^6 + 6*(a^6 +
3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)*sinh(x)^5 + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sinh(x)^6 + a^6 + 3*a^4*b
^2 + 3*a^2*b^4 + b^6 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^4 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6
+ 5*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x
)^3 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x))*sinh(x)^3 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)
^2 + 3*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + 5*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^4 + 6*(a^6 + 3*a^4*b
^2 + 3*a^2*b^4 + b^6)*cosh(x)^2)*sinh(x)^2 + 6*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^5 + 2*(a^6 + 3*a^4
*b^2 + 3*a^2*b^4 + b^6)*cosh(x)^3 + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*cosh(x))*sinh(x))

Sympy [F]

\[ \int \frac {\tanh ^4(x)}{a+b \sinh (x)} \, dx=\int \frac {\tanh ^{4}{\left (x \right )}}{a + b \sinh {\left (x \right )}}\, dx \]

[In]

integrate(tanh(x)**4/(a+b*sinh(x)),x)

[Out]

Integral(tanh(x)**4/(a + b*sinh(x)), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (116) = 232\).

Time = 0.31 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.94 \[ \int \frac {\tanh ^4(x)}{a+b \sinh (x)} \, dx=\frac {a^{4} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (6 \, a^{3} e^{\left (-2 \, x\right )} + 4 \, a^{3} + a b^{2} + 3 \, {\left (2 \, a^{2} b + b^{3}\right )} e^{\left (-x\right )} + 2 \, {\left (4 \, a^{2} b + b^{3}\right )} e^{\left (-3 \, x\right )} + 3 \, {\left (2 \, a^{3} + a b^{2}\right )} e^{\left (-4 \, x\right )} + 3 \, {\left (2 \, a^{2} b + b^{3}\right )} e^{\left (-5 \, x\right )}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + 3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-4 \, x\right )} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-6 \, x\right )}\right )}} \]

[In]

integrate(tanh(x)^4/(a+b*sinh(x)),x, algorithm="maxima")

[Out]

a^4*log((b*e^(-x) - a - sqrt(a^2 + b^2))/(b*e^(-x) - a + sqrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 +
b^2)) - 2/3*(6*a^3*e^(-2*x) + 4*a^3 + a*b^2 + 3*(2*a^2*b + b^3)*e^(-x) + 2*(4*a^2*b + b^3)*e^(-3*x) + 3*(2*a^
3 + a*b^2)*e^(-4*x) + 3*(2*a^2*b + b^3)*e^(-5*x))/(a^4 + 2*a^2*b^2 + b^4 + 3*(a^4 + 2*a^2*b^2 + b^4)*e^(-2*x)
+ 3*(a^4 + 2*a^2*b^2 + b^4)*e^(-4*x) + (a^4 + 2*a^2*b^2 + b^4)*e^(-6*x))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.59 \[ \int \frac {\tanh ^4(x)}{a+b \sinh (x)} \, dx=\frac {a^{4} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (6 \, a^{2} b e^{\left (5 \, x\right )} + 3 \, b^{3} e^{\left (5 \, x\right )} - 6 \, a^{3} e^{\left (4 \, x\right )} - 3 \, a b^{2} e^{\left (4 \, x\right )} + 8 \, a^{2} b e^{\left (3 \, x\right )} + 2 \, b^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} + 6 \, a^{2} b e^{x} + 3 \, b^{3} e^{x} - 4 \, a^{3} - a b^{2}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \]

[In]

integrate(tanh(x)^4/(a+b*sinh(x)),x, algorithm="giac")

[Out]

a^4*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 + b^4
)*sqrt(a^2 + b^2)) - 2/3*(6*a^2*b*e^(5*x) + 3*b^3*e^(5*x) - 6*a^3*e^(4*x) - 3*a*b^2*e^(4*x) + 8*a^2*b*e^(3*x)
+ 2*b^3*e^(3*x) - 6*a^3*e^(2*x) + 6*a^2*b*e^x + 3*b^3*e^x - 4*a^3 - a*b^2)/((a^4 + 2*a^2*b^2 + b^4)*(e^(2*x) +
1)^3)

Mupad [B] (verification not implemented)

Time = 2.24 (sec) , antiderivative size = 654, normalized size of antiderivative = 5.27 \[ \int \frac {\tanh ^4(x)}{a+b \sinh (x)} \, dx=\frac {\frac {2\,a\,\left (2\,a^2+b^2\right )}{{\left (a^2+b^2\right )}^2}-\frac {2\,b\,{\mathrm {e}}^x\,\left (2\,a^2+b^2\right )}{{\left (a^2+b^2\right )}^2}}{{\mathrm {e}}^{2\,x}+1}-\frac {\frac {4\,\left (a^3+a\,b^2\right )}{{\left (a^2+b^2\right )}^2}-\frac {8\,{\mathrm {e}}^x\,\left (a^2\,b+b^3\right )}{3\,{\left (a^2+b^2\right )}^2}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {\frac {8\,a}{3\,\left (a^2+b^2\right )}-\frac {8\,b\,{\mathrm {e}}^x}{3\,\left (a^2+b^2\right )}}{3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1}-\frac {2\,\mathrm {atan}\left (\left ({\mathrm {e}}^x\,\left (\frac {2\,a^4}{b^2\,\sqrt {a^8}\,{\left (a^2+b^2\right )}^2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {2\,\left (a^5\,\sqrt {a^8}+2\,a^3\,b^2\,\sqrt {a^8}+a\,b^4\,\sqrt {a^8}\right )}{a^3\,b^2\,\sqrt {-{\left (a^2+b^2\right )}^5}\,\left (a^4+2\,a^2\,b^2+b^4\right )\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}}\right )-\frac {2\,\left (b^5\,\sqrt {a^8}+2\,a^2\,b^3\,\sqrt {a^8}+a^4\,b\,\sqrt {a^8}\right )}{a^3\,b^2\,\sqrt {-{\left (a^2+b^2\right )}^5}\,\left (a^4+2\,a^2\,b^2+b^4\right )\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}}\right )\,\left (\frac {b^5\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}}{2}+\frac {a^4\,b\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}}{2}+a^2\,b^3\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}\right )\right )\,\sqrt {a^8}}{\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}} \]

[In]

int(tanh(x)^4/(a + b*sinh(x)),x)

[Out]

((2*a*(2*a^2 + b^2))/(a^2 + b^2)^2 - (2*b*exp(x)*(2*a^2 + b^2))/(a^2 + b^2)^2)/(exp(2*x) + 1) - ((4*(a*b^2 + a
^3))/(a^2 + b^2)^2 - (8*exp(x)*(a^2*b + b^3))/(3*(a^2 + b^2)^2))/(2*exp(2*x) + exp(4*x) + 1) + ((8*a)/(3*(a^2
+ b^2)) - (8*b*exp(x))/(3*(a^2 + b^2)))/(3*exp(2*x) + 3*exp(4*x) + exp(6*x) + 1) - (2*atan((exp(x)*((2*a^4)/(b
^2*(a^8)^(1/2)*(a^2 + b^2)^2*(a^4 + b^4 + 2*a^2*b^2)) + (2*(a^5*(a^8)^(1/2) + 2*a^3*b^2*(a^8)^(1/2) + a*b^4*(a
^8)^(1/2)))/(a^3*b^2*(-(a^2 + b^2)^5)^(1/2)*(a^4 + b^4 + 2*a^2*b^2)*(- a^10 - b^10 - 5*a^2*b^8 - 10*a^4*b^6 -
10*a^6*b^4 - 5*a^8*b^2)^(1/2))) - (2*(b^5*(a^8)^(1/2) + 2*a^2*b^3*(a^8)^(1/2) + a^4*b*(a^8)^(1/2)))/(a^3*b^2*(
-(a^2 + b^2)^5)^(1/2)*(a^4 + b^4 + 2*a^2*b^2)*(- a^10 - b^10 - 5*a^2*b^8 - 10*a^4*b^6 - 10*a^6*b^4 - 5*a^8*b^2
)^(1/2)))*((b^5*(- a^10 - b^10 - 5*a^2*b^8 - 10*a^4*b^6 - 10*a^6*b^4 - 5*a^8*b^2)^(1/2))/2 + (a^4*b*(- a^10 -
b^10 - 5*a^2*b^8 - 10*a^4*b^6 - 10*a^6*b^4 - 5*a^8*b^2)^(1/2))/2 + a^2*b^3*(- a^10 - b^10 - 5*a^2*b^8 - 10*a^4
*b^6 - 10*a^6*b^4 - 5*a^8*b^2)^(1/2)))*(a^8)^(1/2))/(- a^10 - b^10 - 5*a^2*b^8 - 10*a^4*b^6 - 10*a^6*b^4 - 5*a
^8*b^2)^(1/2)