\(\int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx\) [72]

Optimal result

Integrand size = 13, antiderivative size = 108 \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=-\frac {a \left (2 a^2-b^2\right ) x}{2 b^4}-\frac {2 a^4 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^4 \sqrt {a^2+b^2}}-\frac {\left (2-\frac {3 a^2}{b^2}\right ) \cosh (x)}{3 b}-\frac {a \cosh (x) \sinh (x)}{2 b^2}+\frac {\cosh (x) \sinh ^2(x)}{3 b} \]

[Out]

-1/2*a*(2*a^2-b^2)*x/b^4-1/3*(2-3*a^2/b^2)*cosh(x)/b-1/2*a*cosh(x)*sinh(x)/b^2+1/3*cosh(x)*sinh(x)^2/b-2*a^4*a
rctanh((b-a*tanh(1/2*x))/(a^2+b^2)^(1/2))/b^4/(a^2+b^2)^(1/2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {2872, 3128, 3102, 2814, 2739, 632, 212} \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=-\frac {\left (2-\frac {3 a^2}{b^2}\right ) \cosh (x)}{3 b}-\frac {a x \left (2 a^2-b^2\right )}{2 b^4}-\frac {2 a^4 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^4 \sqrt {a^2+b^2}}-\frac {a \sinh (x) \cosh (x)}{2 b^2}+\frac {\sinh ^2(x) \cosh (x)}{3 b} \]

[In]

Int[Sinh[x]^4/(a + b*Sinh[x]),x]

[Out]

-1/2*(a*(2*a^2 - b^2)*x)/b^4 - (2*a^4*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(b^4*Sqrt[a^2 + b^2]) - ((2
- (3*a^2)/b^2)*Cosh[x])/(3*b) - (a*Cosh[x]*Sinh[x])/(2*b^2) + (Cosh[x]*Sinh[x]^2)/(3*b)

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 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2872

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Dist[1/
(d*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a
*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n
- 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m]
|| (EqQ[a, 0] && NeQ[c, 0])))

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3128

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e
+ f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n +
2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rubi steps \begin{align*} \text {integral}& = \frac {\cosh (x) \sinh ^2(x)}{3 b}-\frac {\int \frac {\sinh (x) \left (2 a+2 b \sinh (x)+3 a \sinh ^2(x)\right )}{a+b \sinh (x)} \, dx}{3 b} \\ & = -\frac {a \cosh (x) \sinh (x)}{2 b^2}+\frac {\cosh (x) \sinh ^2(x)}{3 b}-\frac {\int \frac {-3 a^2+a b \sinh (x)-2 \left (3 a^2-2 b^2\right ) \sinh ^2(x)}{a+b \sinh (x)} \, dx}{6 b^2} \\ & = \frac {\left (3 a^2-2 b^2\right ) \cosh (x)}{3 b^3}-\frac {a \cosh (x) \sinh (x)}{2 b^2}+\frac {\cosh (x) \sinh ^2(x)}{3 b}-\frac {i \int \frac {3 i a^2 b-3 i a \left (2 a^2-b^2\right ) \sinh (x)}{a+b \sinh (x)} \, dx}{6 b^3} \\ & = -\frac {a \left (2 a^2-b^2\right ) x}{2 b^4}+\frac {\left (3 a^2-2 b^2\right ) \cosh (x)}{3 b^3}-\frac {a \cosh (x) \sinh (x)}{2 b^2}+\frac {\cosh (x) \sinh ^2(x)}{3 b}+\frac {a^4 \int \frac {1}{a+b \sinh (x)} \, dx}{b^4} \\ & = -\frac {a \left (2 a^2-b^2\right ) x}{2 b^4}+\frac {\left (3 a^2-2 b^2\right ) \cosh (x)}{3 b^3}-\frac {a \cosh (x) \sinh (x)}{2 b^2}+\frac {\cosh (x) \sinh ^2(x)}{3 b}+\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 )}{b^4} \\ & = -\frac {a \left (2 a^2-b^2\right ) x}{2 b^4}+\frac {\left (3 a^2-2 b^2\right ) \cosh (x)}{3 b^3}-\frac {a \cosh (x) \sinh (x)}{2 b^2}+\frac {\cosh (x) \sinh ^2(x)}{3 b}-\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 )}{b^4} \\ & = -\frac {a \left (2 a^2-b^2\right ) x}{2 b^4}-\frac {2 a^4 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^4 \sqrt {a^2+b^2}}+\frac {\left (3 a^2-2 b^2\right ) \cosh (x)}{3 b^3}-\frac {a \cosh (x) \sinh (x)}{2 b^2}+\frac {\cosh (x) \sinh ^2(x)}{3 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.64 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.97 \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=\frac {3 b \left (4 a^2-3 b^2\right ) \cosh (x)+b^3 \cosh (3 x)+3 a \left (-4 a^2 x+2 b^2 x+\frac {8 a^3 \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-b^2 \sinh (2 x)\right )}{12 b^4} \]

[In]

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

[Out]

(3*b*(4*a^2 - 3*b^2)*Cosh[x] + b^3*Cosh[3*x] + 3*a*(-4*a^2*x + 2*b^2*x + (8*a^3*ArcTan[(b - a*Tanh[x/2])/Sqrt[
-a^2 - b^2]])/Sqrt[-a^2 - b^2] - b^2*Sinh[2*x]))/(12*b^4)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(200\) vs. \(2(94)=188\).

Time = 0.67 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.86

method	result	size
risch	\(-\frac {a^{3} x}{b^{4}}+\frac {a x}{2 b^{2}}+\frac {{\mathrm e}^{3 x}}{24 b}-\frac {a \,{\mathrm e}^{2 x}}{8 b^{2}}+\frac {{\mathrm e}^{x} a^{2}}{2 b^{3}}-\frac {3 \,{\mathrm e}^{x}}{8 b}+\frac {{\mathrm e}^{-x} a^{2}}{2 b^{3}}-\frac {3 \,{\mathrm e}^{-x}}{8 b}+\frac {a \,{\mathrm e}^{-2 x}}{8 b^{2}}+\frac {{\mathrm e}^{-3 x}}{24 b}+\frac {a^{4} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}+b^{2}}-a^{2}-b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, b^{4}}-\frac {a^{4} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}+b^{2}}+a^{2}+b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, b^{4}}\)	\(201\)
default	\(-\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a +b}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {2 a^{2}+a b -b^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a \left (2 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{4}}+\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {b -a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {-2 a^{2}+a b +b^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a \left (2 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{4}}+\frac {2 a^{4} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{4} \sqrt {a^{2}+b^{2}}}\)	\(202\)

[In]

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

[Out]

-a^3*x/b^4+1/2*a*x/b^2+1/24/b*exp(x)^3-1/8*a/b^2*exp(x)^2+1/2/b^3*exp(x)*a^2-3/8/b*exp(x)+1/2/b^3/exp(x)*a^2-3
/8/b/exp(x)+1/8*a/b^2/exp(x)^2+1/24/b/exp(x)^3+1/(a^2+b^2)^(1/2)*a^4/b^4*ln(exp(x)+(a*(a^2+b^2)^(1/2)-a^2-b^2)
/(a^2+b^2)^(1/2)/b)-1/(a^2+b^2)^(1/2)*a^4/b^4*ln(exp(x)+(a*(a^2+b^2)^(1/2)+a^2+b^2)/(a^2+b^2)^(1/2)/b)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 799 vs. \(2 (96) = 192\).

Time = 0.30 (sec) , antiderivative size = 799, normalized size of antiderivative = 7.40 \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/24*((a^2*b^3 + b^5)*cosh(x)^6 + (a^2*b^3 + b^5)*sinh(x)^6 - 3*(a^3*b^2 + a*b^4)*cosh(x)^5 - 3*(a^3*b^2 + a*b
^4 - 2*(a^2*b^3 + b^5)*cosh(x))*sinh(x)^5 + a^2*b^3 + b^5 - 12*(2*a^5 + a^3*b^2 - a*b^4)*x*cosh(x)^3 + 3*(4*a^
4*b + a^2*b^3 - 3*b^5)*cosh(x)^4 + 3*(4*a^4*b + a^2*b^3 - 3*b^5 + 5*(a^2*b^3 + b^5)*cosh(x)^2 - 5*(a^3*b^2 + a
*b^4)*cosh(x))*sinh(x)^4 + 2*(10*(a^2*b^3 + b^5)*cosh(x)^3 - 15*(a^3*b^2 + a*b^4)*cosh(x)^2 - 6*(2*a^5 + a^3*b
^2 - a*b^4)*x + 6*(4*a^4*b + a^2*b^3 - 3*b^5)*cosh(x))*sinh(x)^3 + 3*(4*a^4*b + a^2*b^3 - 3*b^5)*cosh(x)^2 + 3
*(4*a^4*b + a^2*b^3 - 3*b^5 + 5*(a^2*b^3 + b^5)*cosh(x)^4 - 10*(a^3*b^2 + a*b^4)*cosh(x)^3 - 12*(2*a^5 + a^3*b
^2 - a*b^4)*x*cosh(x) + 6*(4*a^4*b + a^2*b^3 - 3*b^5)*cosh(x)^2)*sinh(x)^2 + 24*(a^4*cosh(x)^3 + 3*a^4*cosh(x)
^2*sinh(x) + 3*a^4*cosh(x)*sinh(x)^2 + a^4*sinh(x)^3)*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*c
osh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) - b)) + 3*(a^3*b^2 + a*b^4)*cosh(x) + 3*(2*(a
^2*b^3 + b^5)*cosh(x)^5 + a^3*b^2 + a*b^4 - 5*(a^3*b^2 + a*b^4)*cosh(x)^4 - 12*(2*a^5 + a^3*b^2 - a*b^4)*x*cos
h(x)^2 + 4*(4*a^4*b + a^2*b^3 - 3*b^5)*cosh(x)^3 + 2*(4*a^4*b + a^2*b^3 - 3*b^5)*cosh(x))*sinh(x))/((a^2*b^4 +
b^6)*cosh(x)^3 + 3*(a^2*b^4 + b^6)*cosh(x)^2*sinh(x) + 3*(a^2*b^4 + b^6)*cosh(x)*sinh(x)^2 + (a^2*b^4 + b^6)*
sinh(x)^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.46 \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=\frac {a^{4} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}} - \frac {{\left (3 \, a b e^{\left (-x\right )} - b^{2} - 3 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )} e^{\left (-2 \, x\right )}\right )} e^{\left (3 \, x\right )}}{24 \, b^{3}} + \frac {3 \, a b e^{\left (-2 \, x\right )} + b^{2} e^{\left (-3 \, x\right )} + 3 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )} e^{\left (-x\right )}}{24 \, b^{3}} - \frac {{\left (2 \, a^{3} - a b^{2}\right )} x}{2 \, b^{4}} \]

[In]

integrate(sinh(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)))/(sqrt(a^2 + b^2)*b^4) - 1/24*(3*a*b
*e^(-x) - b^2 - 3*(4*a^2 - 3*b^2)*e^(-2*x))*e^(3*x)/b^3 + 1/24*(3*a*b*e^(-2*x) + b^2*e^(-3*x) + 3*(4*a^2 - 3*b
^2)*e^(-x))/b^3 - 1/2*(2*a^3 - a*b^2)*x/b^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.44 \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=\frac {a^{4} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}} + \frac {b^{2} e^{\left (3 \, x\right )} - 3 \, a b e^{\left (2 \, x\right )} + 12 \, a^{2} e^{x} - 9 \, b^{2} e^{x}}{24 \, b^{3}} - \frac {{\left (2 \, a^{3} - a b^{2}\right )} x}{2 \, b^{4}} + \frac {{\left (3 \, a b^{2} e^{x} + b^{3} + 3 \, {\left (4 \, a^{2} b - 3 \, b^{3}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-3 \, x\right )}}{24 \, b^{4}} \]

[In]

integrate(sinh(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)))/(sqrt(a^2 + b^2)*b^4) +
1/24*(b^2*e^(3*x) - 3*a*b*e^(2*x) + 12*a^2*e^x - 9*b^2*e^x)/b^3 - 1/2*(2*a^3 - a*b^2)*x/b^4 + 1/24*(3*a*b^2*e
^x + b^3 + 3*(4*a^2*b - 3*b^3)*e^(2*x))*e^(-3*x)/b^4

Mupad [B] (verification not implemented)

Time = 1.54 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.84 \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=\frac {{\mathrm {e}}^{-3\,x}}{24\,b}+\frac {{\mathrm {e}}^{3\,x}}{24\,b}+\frac {x\,\left (a\,b^2-2\,a^3\right )}{2\,b^4}+\frac {{\mathrm {e}}^x\,\left (4\,a^2-3\,b^2\right )}{8\,b^3}+\frac {a\,{\mathrm {e}}^{-2\,x}}{8\,b^2}-\frac {a\,{\mathrm {e}}^{2\,x}}{8\,b^2}+\frac {{\mathrm {e}}^{-x}\,\left (4\,a^2-3\,b^2\right )}{8\,b^3}-\frac {a^4\,\ln \left (-\frac {2\,a^4\,{\mathrm {e}}^x}{b^5}-\frac {2\,a^4\,\left (b-a\,{\mathrm {e}}^x\right )}{b^5\,\sqrt {a^2+b^2}}\right )}{b^4\,\sqrt {a^2+b^2}}+\frac {a^4\,\ln \left (\frac {2\,a^4\,\left (b-a\,{\mathrm {e}}^x\right )}{b^5\,\sqrt {a^2+b^2}}-\frac {2\,a^4\,{\mathrm {e}}^x}{b^5}\right )}{b^4\,\sqrt {a^2+b^2}} \]

[In]

int(sinh(x)^4/(a + b*sinh(x)),x)

[Out]

exp(-3*x)/(24*b) + exp(3*x)/(24*b) + (x*(a*b^2 - 2*a^3))/(2*b^4) + (exp(x)*(4*a^2 - 3*b^2))/(8*b^3) + (a*exp(-
2*x))/(8*b^2) - (a*exp(2*x))/(8*b^2) + (exp(-x)*(4*a^2 - 3*b^2))/(8*b^3) - (a^4*log(- (2*a^4*exp(x))/b^5 - (2*
a^4*(b - a*exp(x)))/(b^5*(a^2 + b^2)^(1/2))))/(b^4*(a^2 + b^2)^(1/2)) + (a^4*log((2*a^4*(b - a*exp(x)))/(b^5*(
a^2 + b^2)^(1/2)) - (2*a^4*exp(x))/b^5))/(b^4*(a^2 + b^2)^(1/2))