\(\int \frac {\text {sech}^4(x)}{a+b \sinh (x)} \, dx\) [197]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 100 \[ \int \frac {\text {sech}^4(x)}{a+b \sinh (x)} \, dx=-\frac {2 b^4 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac {\text {sech}^3(x) (b+a \sinh (x))}{3 \left (a^2+b^2\right )}+\frac {\text {sech}(x) \left (3 b^3+a \left (2 a^2+5 b^2\right ) \sinh (x)\right )}{3 \left (a^2+b^2\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2775, 2945, 12, 2739, 632, 212} \[ \int \frac {\text {sech}^4(x)}{a+b \sinh (x)} \, dx=-\frac {2 b^4 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac {\text {sech}^3(x) (a \sinh (x)+b)}{3 \left (a^2+b^2\right )}+\frac {\text {sech}(x) \left (a \left (2 a^2+5 b^2\right ) \sinh (x)+3 b^3\right )}{3 \left (a^2+b^2\right )^2} \]

[In]

Int[Sech[x]^4/(a + b*Sinh[x]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 2775

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*Cos
[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b - a*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2945

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
 + (f_.)*(x_)]), x_Symbol] :> Simp[(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c -
b*d)*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
 + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.02 \[ \int \frac {\text {sech}^4(x)}{a+b \sinh (x)} \, dx=\frac {\frac {6 b^4 \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+3 b^3 \text {sech}(x)+\left (a^2+b^2\right ) \text {sech}^3(x) (b+a \sinh (x))+a \left (2 a^2+5 b^2\right ) \tanh (x)}{3 \left (a^2+b^2\right )^2} \]

[In]

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

[Out]

((6*b^4*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] + 3*b^3*Sech[x] + (a^2 + b^2)*Sech[x]^3*(
b + a*Sinh[x]) + a*(2*a^2 + 5*b^2)*Tanh[x])/(3*(a^2 + b^2)^2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(181\) vs. \(2(90)=180\).

Time = 47.08 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.82

method result size
default \(\frac {2 b^{4} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}+b^{2}}}-\frac {2 \left (\left (-a^{3}-2 a \,b^{2}\right ) \tanh \left (\frac {x}{2}\right )^{5}+\left (-a^{2} b -2 b^{3}\right ) \tanh \left (\frac {x}{2}\right )^{4}+\left (-\frac {2}{3} a^{3}-\frac {8}{3} a \,b^{2}\right ) \tanh \left (\frac {x}{2}\right )^{3}-2 b^{3} \tanh \left (\frac {x}{2}\right )^{2}+\left (-a^{3}-2 a \,b^{2}\right ) \tanh \left (\frac {x}{2}\right )-\frac {a^{2} b}{3}-\frac {4 b^{3}}{3}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )^{3}}\) \(182\)
risch \(-\frac {2 \left (-3 b^{3} {\mathrm e}^{5 x}+3 \,{\mathrm e}^{4 x} a \,b^{2}-4 a^{2} b \,{\mathrm e}^{3 x}-10 \,{\mathrm e}^{3 x} b^{3}+6 a^{3} {\mathrm e}^{2 x}+12 a \,{\mathrm e}^{2 x} b^{2}-3 b^{3} {\mathrm e}^{x}+2 a^{3}+5 a \,b^{2}\right )}{3 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (1+{\mathrm e}^{2 x}\right )^{3}}+\frac {b^{4} \ln \left ({\mathrm e}^{x}+\frac {\left (a^{2}+b^{2}\right )^{\frac {5}{2}} a -a^{6}-3 a^{4} b^{2}-3 a^{2} b^{4}-b^{6}}{b \left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}-\frac {b^{4} \ln \left ({\mathrm e}^{x}+\frac {\left (a^{2}+b^{2}\right )^{\frac {5}{2}} a +a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}{b \left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\) \(238\)

[In]

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

[Out]

2*b^4/(a^4+2*a^2*b^2+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^3-2*a*b^2)*tanh(1/2*x)^5+(-a^2*b-2*b^3)*tanh(1/2*x)^4+(-2/3*a^3-8/3*a*b^2)*tanh(1/2*x)^3-2*b^3*tanh(
1/2*x)^2+(-a^3-2*a*b^2)*tanh(1/2*x)-1/3*a^2*b-4/3*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. 1142 vs. \(2 (92) = 184\).

Time = 0.29 (sec) , antiderivative size = 1142, normalized size of antiderivative = 11.42 \[ \int \frac {\text {sech}^4(x)}{a+b \sinh (x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/3*(6*(a^2*b^3 + b^5)*cosh(x)^5 + 6*(a^2*b^3 + b^5)*sinh(x)^5 - 4*a^5 - 14*a^3*b^2 - 10*a*b^4 - 6*(a^3*b^2 +
a*b^4)*cosh(x)^4 - 6*(a^3*b^2 + a*b^4 - 5*(a^2*b^3 + b^5)*cosh(x))*sinh(x)^4 + 4*(2*a^4*b + 7*a^2*b^3 + 5*b^5)
*cosh(x)^3 + 4*(2*a^4*b + 7*a^2*b^3 + 5*b^5 + 15*(a^2*b^3 + b^5)*cosh(x)^2 - 6*(a^3*b^2 + a*b^4)*cosh(x))*sinh
(x)^3 - 12*(a^5 + 3*a^3*b^2 + 2*a*b^4)*cosh(x)^2 - 12*(a^5 + 3*a^3*b^2 + 2*a*b^4 - 5*(a^2*b^3 + b^5)*cosh(x)^3
 + 3*(a^3*b^2 + a*b^4)*cosh(x)^2 - (2*a^4*b + 7*a^2*b^3 + 5*b^5)*cosh(x))*sinh(x)^2 + 3*(b^4*cosh(x)^6 + 6*b^4
*cosh(x)*sinh(x)^5 + b^4*sinh(x)^6 + 3*b^4*cosh(x)^4 + 3*b^4*cosh(x)^2 + 3*(5*b^4*cosh(x)^2 + b^4)*sinh(x)^4 +
 b^4 + 4*(5*b^4*cosh(x)^3 + 3*b^4*cosh(x))*sinh(x)^3 + 3*(5*b^4*cosh(x)^4 + 6*b^4*cosh(x)^2 + b^4)*sinh(x)^2 +
 6*(b^4*cosh(x)^5 + 2*b^4*cosh(x)^3 + b^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*(a^2*b^3 + b^5)*cosh(x) + 6*(
a^2*b^3 + b^5 + 5*(a^2*b^3 + b^5)*cosh(x)^4 - 4*(a^3*b^2 + a*b^4)*cosh(x)^3 + 2*(2*a^4*b + 7*a^2*b^3 + 5*b^5)*
cosh(x)^2 - 4*(a^5 + 3*a^3*b^2 + 2*a*b^4)*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]

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

[In]

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

[Out]

Integral(sech(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. 230 vs. \(2 (92) = 184\).

Time = 0.29 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.30 \[ \int \frac {\text {sech}^4(x)}{a+b \sinh (x)} \, dx=\frac {b^{4} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (3 \, b^{3} e^{\left (-x\right )} + 3 \, a b^{2} e^{\left (-4 \, x\right )} + 3 \, b^{3} e^{\left (-5 \, x\right )} + 2 \, a^{3} + 5 \, a b^{2} + 6 \, {\left (a^{3} + 2 \, a b^{2}\right )} e^{\left (-2 \, x\right )} + 2 \, {\left (2 \, a^{2} b + 5 \, b^{3}\right )} e^{\left (-3 \, x\right )}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + 3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-2 \, x\right )} + 3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-4 \, x\right )} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} e^{\left (-6 \, x\right )}\right )}} \]

[In]

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

[Out]

b^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*(3*b^3*e^(-x) + 3*a*b^2*e^(-4*x) + 3*b^3*e^(-5*x) + 2*a^3 + 5*a*b^2 + 6*(a^3 + 2*a*b^2)*e^(-2*x)
+ 2*(2*a^2*b + 5*b^3)*e^(-3*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 = 0.28 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.80 \[ \int \frac {\text {sech}^4(x)}{a+b \sinh (x)} \, dx=\frac {b^{4} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (3 \, b^{3} e^{\left (5 \, x\right )} - 3 \, a b^{2} e^{\left (4 \, x\right )} + 4 \, a^{2} b e^{\left (3 \, x\right )} + 10 \, b^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} - 12 \, a b^{2} e^{\left (2 \, x\right )} + 3 \, b^{3} e^{x} - 2 \, a^{3} - 5 \, a b^{2}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \]

[In]

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

[Out]

b^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*(3*b^3*e^(5*x) - 3*a*b^2*e^(4*x) + 4*a^2*b*e^(3*x) + 10*b^3*e^(3*x) - 6*a^3*e^(2*x) -
 12*a*b^2*e^(2*x) + 3*b^3*e^x - 2*a^3 - 5*a*b^2)/((a^4 + 2*a^2*b^2 + b^4)*(e^(2*x) + 1)^3)

Mupad [B] (verification not implemented)

Time = 2.40 (sec) , antiderivative size = 634, normalized size of antiderivative = 6.34 \[ \int \frac {\text {sech}^4(x)}{a+b \sinh (x)} \, dx=\frac {\frac {2\,b^3\,{\mathrm {e}}^x}{{\left (a^2+b^2\right )}^2}-\frac {2\,a\,b^2}{{\left (a^2+b^2\right )}^2}}{{\mathrm {e}}^{2\,x}+1}-\frac {\frac {4\,\left (a^3+a\,b^2\right )}{{\left (a^2+b^2\right )}^2}-\frac {8\,{\mathrm {e}}^x\,\left (a^2\,b+b^3\right )}{3\,{\left (a^2+b^2\right )}^2}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {\frac {8\,a}{3\,\left (a^2+b^2\right )}-\frac {8\,b\,{\mathrm {e}}^x}{3\,\left (a^2+b^2\right )}}{3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1}-\frac {2\,\mathrm {atan}\left (\left ({\mathrm {e}}^x\,\left (\frac {2\,b^2}{\sqrt {b^8}\,{\left (a^2+b^2\right )}^2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {2\,a\,\left (a^5\,\sqrt {b^8}+2\,a^3\,b^2\,\sqrt {b^8}+a\,b^4\,\sqrt {b^8}\right )}{b^6\,\sqrt {-{\left (a^2+b^2\right )}^5}\,\left (a^4+2\,a^2\,b^2+b^4\right )\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}}\right )-\frac {2\,a\,\left (b^5\,\sqrt {b^8}+2\,a^2\,b^3\,\sqrt {b^8}+a^4\,b\,\sqrt {b^8}\right )}{b^6\,\sqrt {-{\left (a^2+b^2\right )}^5}\,\left (a^4+2\,a^2\,b^2+b^4\right )\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}}\right )\,\left (\frac {b^5\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}}{2}+\frac {a^4\,b\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}}{2}+a^2\,b^3\,\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}\right )\right )\,\sqrt {b^8}}{\sqrt {-a^{10}-5\,a^8\,b^2-10\,a^6\,b^4-10\,a^4\,b^6-5\,a^2\,b^8-b^{10}}} \]

[In]

int(1/(cosh(x)^4*(a + b*sinh(x))),x)

[Out]

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