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

Optimal result

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

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

Mathematica [A] (verified)

Time = 1.54 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.97 \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=\frac {3 b \left (4 a^2-3 b^2\right ) \cosh (x)+b^3 \cosh (3 x)+3 a \left (-4 a^2 x+2 b^2 x+\frac {8 a^3 \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-b^2 \sinh (2 x)\right )}{12 b^4} \] Input:

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

Output:

(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*S 
inh[2*x]))/(12*b^4)
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.81 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.28, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.231, Rules used = {3042, 3272, 26, 3042, 26, 3528, 25, 3042, 3502, 27, 3042, 3214, 3042, 3139, 1083, 219}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (i x)^4}{a-i b \sin (i x)}dx\)

\(\Big \downarrow \) 3272

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sinh ^2(x) \cosh (x)}{3 b}-\frac {\int -\frac {i \sin (i x) \left (-3 a \sin (i x)^2-2 i b \sin (i x)+2 a\right )}{a-i b \sin (i x)}dx}{3 b}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\sinh ^2(x) \cosh (x)}{3 b}+\frac {i \int \frac {\sin (i x) \left (-3 a \sin (i x)^2-2 i b \sin (i x)+2 a\right )}{a-i b \sin (i x)}dx}{3 b}\)

\(\Big \downarrow \) 3528

\(\displaystyle \frac {\sinh ^2(x) \cosh (x)}{3 b}+\frac {i \left (\frac {i \int -\frac {3 a^2-b \sinh (x) a+2 \left (3 a^2-2 b^2\right ) \sinh ^2(x)}{a+b \sinh (x)}dx}{2 b}+\frac {3 i a \sinh (x) \cosh (x)}{2 b}\right )}{3 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sinh ^2(x) \cosh (x)}{3 b}+\frac {i \left (\frac {3 i a \sinh (x) \cosh (x)}{2 b}-\frac {i \int \frac {3 a^2-b \sinh (x) a+2 \left (3 a^2-2 b^2\right ) \sinh ^2(x)}{a+b \sinh (x)}dx}{2 b}\right )}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sinh ^2(x) \cosh (x)}{3 b}+\frac {i \left (\frac {3 i a \sinh (x) \cosh (x)}{2 b}-\frac {i \int \frac {3 a^2+i b \sin (i x) a-2 \left (3 a^2-2 b^2\right ) \sin (i x)^2}{a-i b \sin (i x)}dx}{2 b}\right )}{3 b}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\sinh ^2(x) \cosh (x)}{3 b}+\frac {i \left (\frac {3 i a \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {2 \left (3 a^2-2 b^2\right ) \cosh (x)}{b}+\frac {i \int -\frac {3 i \left (a^2 b-a \left (2 a^2-b^2\right ) \sinh (x)\right )}{a+b \sinh (x)}dx}{b}\right )}{2 b}\right )}{3 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sinh ^2(x) \cosh (x)}{3 b}+\frac {i \left (\frac {3 i a \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {3 \int \frac {a^2 b-a \left (2 a^2-b^2\right ) \sinh (x)}{a+b \sinh (x)}dx}{b}+\frac {2 \left (3 a^2-2 b^2\right ) \cosh (x)}{b}\right )}{2 b}\right )}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sinh ^2(x) \cosh (x)}{3 b}+\frac {i \left (\frac {3 i a \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {2 \left (3 a^2-2 b^2\right ) \cosh (x)}{b}+\frac {3 \int \frac {b a^2+i \left (2 a^2-b^2\right ) \sin (i x) a}{a-i b \sin (i x)}dx}{b}\right )}{2 b}\right )}{3 b}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\sinh ^2(x) \cosh (x)}{3 b}+\frac {i \left (\frac {3 i a \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {3 \left (\frac {2 a^4 \int \frac {1}{a+b \sinh (x)}dx}{b}-\frac {a x \left (2 a^2-b^2\right )}{b}\right )}{b}+\frac {2 \left (3 a^2-2 b^2\right ) \cosh (x)}{b}\right )}{2 b}\right )}{3 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sinh ^2(x) \cosh (x)}{3 b}+\frac {i \left (\frac {3 i a \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {2 \left (3 a^2-2 b^2\right ) \cosh (x)}{b}+\frac {3 \left (-\frac {a x \left (2 a^2-b^2\right )}{b}+\frac {2 a^4 \int \frac {1}{a-i b \sin (i x)}dx}{b}\right )}{b}\right )}{2 b}\right )}{3 b}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {\sinh ^2(x) \cosh (x)}{3 b}+\frac {i \left (\frac {3 i a \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {3 \left (\frac {4 a^4 \int \frac {1}{-a \tanh ^2\left (\frac {x}{2}\right )+2 b \tanh \left (\frac {x}{2}\right )+a}d\tanh \left (\frac {x}{2}\right )}{b}-\frac {a x \left (2 a^2-b^2\right )}{b}\right )}{b}+\frac {2 \left (3 a^2-2 b^2\right ) \cosh (x)}{b}\right )}{2 b}\right )}{3 b}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\sinh ^2(x) \cosh (x)}{3 b}+\frac {i \left (\frac {3 i a \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {3 \left (-\frac {8 a^4 \int \frac {1}{4 \left (a^2+b^2\right )-\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )^2}d\left (2 b-2 a \tanh \left (\frac {x}{2}\right )\right )}{b}-\frac {a x \left (2 a^2-b^2\right )}{b}\right )}{b}+\frac {2 \left (3 a^2-2 b^2\right ) \cosh (x)}{b}\right )}{2 b}\right )}{3 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sinh ^2(x) \cosh (x)}{3 b}+\frac {i \left (\frac {3 i a \sinh (x) \cosh (x)}{2 b}-\frac {i \left (\frac {2 \left (3 a^2-2 b^2\right ) \cosh (x)}{b}+\frac {3 \left (-\frac {a x \left (2 a^2-b^2\right )}{b}-\frac {4 a^4 \text {arctanh}\left (\frac {2 b-2 a \tanh \left (\frac {x}{2}\right )}{2 \sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2}}\right )}{b}\right )}{2 b}\right )}{3 b}\)

Input:

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

Output:

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

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 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1083

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

rule 3042

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

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[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 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(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 3272

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-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] + Simp[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))*Si 
n[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 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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 3528

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] + Simp[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])))
 

Maple [B] (verified)

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

Time = 0.40 (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	\(-\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a +b}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {2 a^{2}+a b -b^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a \left (2 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{4}}+\frac {2 a^{4} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{4} \sqrt {a^{2}+b^{2}}}+\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {-a +b}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {-2 a^{2}+a b +b^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a \left (2 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{4}}\)	\(202\)

Input:

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

Output:

-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 = 0.10 (sec) , antiderivative size = 799, normalized size of antiderivative = 7.40 \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=\text {Too large to display} \] Input:

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

Output:

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*s 
inh(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*co 
sh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*co 
sh(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*cosh(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. \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.46 \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=\frac {a^{4} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}} - \frac {{\left (3 \, a b e^{\left (-x\right )} - b^{2} - 3 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )} e^{\left (-2 \, x\right )}\right )} e^{\left (3 \, x\right )}}{24 \, b^{3}} + \frac {3 \, a b e^{\left (-2 \, x\right )} + b^{2} e^{\left (-3 \, x\right )} + 3 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )} e^{\left (-x\right )}}{24 \, b^{3}} - \frac {{\left (2 \, a^{3} - a b^{2}\right )} x}{2 \, b^{4}} \] Input:

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

Output:

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)

Time = 0.13 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.44 \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=\frac {a^{4} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{4}} + \frac {b^{2} e^{\left (3 \, x\right )} - 3 \, a b e^{\left (2 \, x\right )} + 12 \, a^{2} e^{x} - 9 \, b^{2} e^{x}}{24 \, b^{3}} - \frac {{\left (2 \, a^{3} - a b^{2}\right )} x}{2 \, b^{4}} + \frac {{\left (3 \, a b^{2} e^{x} + b^{3} + 3 \, {\left (4 \, a^{2} b - 3 \, b^{3}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-3 \, x\right )}}{24 \, b^{4}} \] Input:

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

Output:

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) + 1 
2*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 = 2.11 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.84 \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=\frac {{\mathrm {e}}^{-3\,x}}{24\,b}+\frac {{\mathrm {e}}^{3\,x}}{24\,b}+\frac {x\,\left (a\,b^2-2\,a^3\right )}{2\,b^4}+\frac {{\mathrm {e}}^x\,\left (4\,a^2-3\,b^2\right )}{8\,b^3}+\frac {a\,{\mathrm {e}}^{-2\,x}}{8\,b^2}-\frac {a\,{\mathrm {e}}^{2\,x}}{8\,b^2}+\frac {{\mathrm {e}}^{-x}\,\left (4\,a^2-3\,b^2\right )}{8\,b^3}-\frac {a^4\,\ln \left (-\frac {2\,a^4\,{\mathrm {e}}^x}{b^5}-\frac {2\,a^4\,\left (b-a\,{\mathrm {e}}^x\right )}{b^5\,\sqrt {a^2+b^2}}\right )}{b^4\,\sqrt {a^2+b^2}}+\frac {a^4\,\ln \left (\frac {2\,a^4\,\left (b-a\,{\mathrm {e}}^x\right )}{b^5\,\sqrt {a^2+b^2}}-\frac {2\,a^4\,{\mathrm {e}}^x}{b^5}\right )}{b^4\,\sqrt {a^2+b^2}} \] Input:

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

Output:

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))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.25 \[ \int \frac {\sinh ^4(x)}{a+b \sinh (x)} \, dx=\frac {48 e^{3 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{4} i +e^{6 x} a^{2} b^{3}+e^{6 x} b^{5}-3 e^{5 x} a^{3} b^{2}-3 e^{5 x} a \,b^{4}+12 e^{4 x} a^{4} b +3 e^{4 x} a^{2} b^{3}-9 e^{4 x} b^{5}-24 e^{3 x} a^{5} x -12 e^{3 x} a^{3} b^{2} x +12 e^{3 x} a \,b^{4} x +12 e^{2 x} a^{4} b +3 e^{2 x} a^{2} b^{3}-9 e^{2 x} b^{5}+3 e^{x} a^{3} b^{2}+3 e^{x} a \,b^{4}+a^{2} b^{3}+b^{5}}{24 e^{3 x} b^{4} \left (a^{2}+b^{2}\right )} \] Input:

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

Output:

(48*e**(3*x)*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*a* 
*4*i + e**(6*x)*a**2*b**3 + e**(6*x)*b**5 - 3*e**(5*x)*a**3*b**2 - 3*e**(5 
*x)*a*b**4 + 12*e**(4*x)*a**4*b + 3*e**(4*x)*a**2*b**3 - 9*e**(4*x)*b**5 - 
 24*e**(3*x)*a**5*x - 12*e**(3*x)*a**3*b**2*x + 12*e**(3*x)*a*b**4*x + 12* 
e**(2*x)*a**4*b + 3*e**(2*x)*a**2*b**3 - 9*e**(2*x)*b**5 + 3*e**x*a**3*b** 
2 + 3*e**x*a*b**4 + a**2*b**3 + b**5)/(24*e**(3*x)*b**4*(a**2 + b**2))