\(\int \frac {\sinh ^3(x)}{a+b \sinh (x)} \, dx\) [73]

Optimal result

Integrand size = 13, antiderivative size = 82 \[ \int \frac {\sinh ^3(x)}{a+b \sinh (x)} \, dx=\frac {\left (2 a^2-b^2\right ) x}{2 b^3}+\frac {2 a^3 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^3 \sqrt {a^2+b^2}}-\frac {a \cosh (x)}{b^2}+\frac {\cosh (x) \sinh (x)}{2 b} \] Output:

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

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \frac {\sinh ^3(x)}{a+b \sinh (x)} \, dx=\frac {4 a^2 x-2 b^2 x-\frac {8 a^3 \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-4 a b \cosh (x)+b^2 \sinh (2 x)}{4 b^3} \] Input:

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

Output:

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

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.61 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.29, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.923, Rules used = {3042, 26, 3272, 3042, 3502, 26, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3272

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

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 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(151\) vs. \(2(72)=144\).

Time = 0.27 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.85

method	result	size
default	\(-\frac {2 a^{3} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{3} \sqrt {a^{2}+b^{2}}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {-b -2 a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-2 a^{2}+b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{3}}-\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {-b +2 a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\left (2 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{3}}\)	\(152\)
risch	\(\frac {x \,a^{2}}{b^{3}}-\frac {x}{2 b}+\frac {{\mathrm e}^{2 x}}{8 b}-\frac {a \,{\mathrm e}^{x}}{2 b^{2}}-\frac {a \,{\mathrm e}^{-x}}{2 b^{2}}-\frac {{\mathrm e}^{-2 x}}{8 b}+\frac {a^{3} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}+b^{2}}+a^{2}+b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, b^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}+b^{2}}-a^{2}-b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, b^{3}}\)	\(159\)

Input:

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

Output:

-2*a^3/b^3/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/ 
2))+1/2/b/(tanh(1/2*x)-1)^2-1/2*(-b-2*a)/b^2/(tanh(1/2*x)-1)+1/2/b^3*(-2*a 
^2+b^2)*ln(tanh(1/2*x)-1)-1/2/b/(tanh(1/2*x)+1)^2-1/2*(-b+2*a)/b^2/(tanh(1 
/2*x)+1)+1/2*(2*a^2-b^2)/b^3*ln(tanh(1/2*x)+1)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 459 vs. \(2 (74) = 148\).

Time = 0.10 (sec) , antiderivative size = 459, normalized size of antiderivative = 5.60 \[ \int \frac {\sinh ^3(x)}{a+b \sinh (x)} \, dx=\frac {{\left (a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{4} + {\left (a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{4} - a^{2} b^{2} - b^{4} + 4 \, {\left (2 \, a^{4} + a^{2} b^{2} - b^{4}\right )} x \cosh \left (x\right )^{2} - 4 \, {\left (a^{3} b + a b^{3}\right )} \cosh \left (x\right )^{3} - 4 \, {\left (a^{3} b + a b^{3} - {\left (a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 2 \, {\left (3 \, {\left (a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (2 \, a^{4} + a^{2} b^{2} - b^{4}\right )} x - 6 \, {\left (a^{3} b + a b^{3}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )^{2} + 8 \, {\left (a^{3} \cosh \left (x\right )^{2} + 2 \, a^{3} \cosh \left (x\right ) \sinh \left (x\right ) + a^{3} \sinh \left (x\right )^{2}\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) - 4 \, {\left (a^{3} b + a b^{3}\right )} \cosh \left (x\right ) - 4 \, {\left (a^{3} b + a b^{3} - {\left (a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{3} - 2 \, {\left (2 \, a^{4} + a^{2} b^{2} - b^{4}\right )} x \cosh \left (x\right ) + 3 \, {\left (a^{3} b + a b^{3}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )}{8 \, {\left ({\left (a^{2} b^{3} + b^{5}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{2} b^{3} + b^{5}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{2} b^{3} + b^{5}\right )} \sinh \left (x\right )^{2}\right )}} \] Input:

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

Output:

1/8*((a^2*b^2 + b^4)*cosh(x)^4 + (a^2*b^2 + b^4)*sinh(x)^4 - a^2*b^2 - b^4 
 + 4*(2*a^4 + a^2*b^2 - b^4)*x*cosh(x)^2 - 4*(a^3*b + a*b^3)*cosh(x)^3 - 4 
*(a^3*b + a*b^3 - (a^2*b^2 + b^4)*cosh(x))*sinh(x)^3 + 2*(3*(a^2*b^2 + b^4 
)*cosh(x)^2 + 2*(2*a^4 + a^2*b^2 - b^4)*x - 6*(a^3*b + a*b^3)*cosh(x))*sin 
h(x)^2 + 8*(a^3*cosh(x)^2 + 2*a^3*cosh(x)*sinh(x) + a^3*sinh(x)^2)*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)) - 4*(a^3*b + a*b^3)*cosh(x) - 4*(a^3*b + a*b^3 - (a^2*b^2 + b^4)*c 
osh(x)^3 - 2*(2*a^4 + a^2*b^2 - b^4)*x*cosh(x) + 3*(a^3*b + a*b^3)*cosh(x) 
^2)*sinh(x))/((a^2*b^3 + b^5)*cosh(x)^2 + 2*(a^2*b^3 + b^5)*cosh(x)*sinh(x 
) + (a^2*b^3 + b^5)*sinh(x)^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sinh ^3(x)}{a+b \sinh (x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.44 \[ \int \frac {\sinh ^3(x)}{a+b \sinh (x)} \, dx=-\frac {a^{3} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{3}} - \frac {{\left (4 \, a e^{\left (-x\right )} - b\right )} e^{\left (2 \, x\right )}}{8 \, b^{2}} - \frac {4 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )}}{8 \, b^{2}} + \frac {{\left (2 \, a^{2} - b^{2}\right )} x}{2 \, b^{3}} \] Input:

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

Output:

-a^3*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^3) - 1/8*(4*a*e^(-x) - b)*e^(2*x)/b^2 - 1/8*(4*a*e^(- 
x) + b*e^(-2*x))/b^2 + 1/2*(2*a^2 - b^2)*x/b^3
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.43 \[ \int \frac {\sinh ^3(x)}{a+b \sinh (x)} \, dx=-\frac {a^{3} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{3}} + \frac {b e^{\left (2 \, x\right )} - 4 \, a e^{x}}{8 \, b^{2}} + \frac {{\left (2 \, a^{2} - b^{2}\right )} x}{2 \, b^{3}} - \frac {{\left (4 \, a b e^{x} + b^{2}\right )} e^{\left (-2 \, x\right )}}{8 \, b^{3}} \] Input:

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

Output:

-a^3*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^3) + 1/8*(b*e^(2*x) - 4*a*e^x)/b^2 + 1/2* 
(2*a^2 - b^2)*x/b^3 - 1/8*(4*a*b*e^x + b^2)*e^(-2*x)/b^3
 

Mupad [B] (verification not implemented)

Time = 1.92 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.94 \[ \int \frac {\sinh ^3(x)}{a+b \sinh (x)} \, dx=\frac {{\mathrm {e}}^{2\,x}}{8\,b}-\frac {{\mathrm {e}}^{-2\,x}}{8\,b}+\frac {x\,\left (2\,a^2-b^2\right )}{2\,b^3}-\frac {a\,{\mathrm {e}}^x}{2\,b^2}-\frac {a\,{\mathrm {e}}^{-x}}{2\,b^2}-\frac {a^3\,\ln \left (\frac {2\,a^3\,{\mathrm {e}}^x}{b^4}-\frac {2\,a^3\,\left (b-a\,{\mathrm {e}}^x\right )}{b^4\,\sqrt {a^2+b^2}}\right )}{b^3\,\sqrt {a^2+b^2}}+\frac {a^3\,\ln \left (\frac {2\,a^3\,{\mathrm {e}}^x}{b^4}+\frac {2\,a^3\,\left (b-a\,{\mathrm {e}}^x\right )}{b^4\,\sqrt {a^2+b^2}}\right )}{b^3\,\sqrt {a^2+b^2}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.11 \[ \int \frac {\sinh ^3(x)}{a+b \sinh (x)} \, dx=\frac {-16 e^{2 x} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{3} i +e^{4 x} a^{2} b^{2}+e^{4 x} b^{4}-4 e^{3 x} a^{3} b -4 e^{3 x} a \,b^{3}+8 e^{2 x} a^{4} x +4 e^{2 x} a^{2} b^{2} x -4 e^{2 x} b^{4} x -4 e^{x} a^{3} b -4 e^{x} a \,b^{3}-a^{2} b^{2}-b^{4}}{8 e^{2 x} b^{3} \left (a^{2}+b^{2}\right )} \] Input:

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

Output:

( - 16*e**(2*x)*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2)) 
*a**3*i + e**(4*x)*a**2*b**2 + e**(4*x)*b**4 - 4*e**(3*x)*a**3*b - 4*e**(3 
*x)*a*b**3 + 8*e**(2*x)*a**4*x + 4*e**(2*x)*a**2*b**2*x - 4*e**(2*x)*b**4* 
x - 4*e**x*a**3*b - 4*e**x*a*b**3 - a**2*b**2 - b**4)/(8*e**(2*x)*b**3*(a* 
*2 + b**2))