\(\int \frac {\sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx\) [236]

Optimal result

Integrand size = 21, antiderivative size = 107 \[ \int \frac {\sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\left (2 a^2-b^2\right ) x}{2 b^3}+\frac {2 a^3 \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b^3 \sqrt {a^2+b^2} d}-\frac {a \cosh (c+d x)}{b^2 d}+\frac {\cosh (c+d x) \sinh (c+d x)}{2 b d} \] Output:

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

Mathematica [A] (verified)

Time = 2.85 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.94 \[ \int \frac {\sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {-2 \left (-2 a^2+b^2\right ) (c+d x)-\frac {8 a^3 \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-4 a b \cosh (c+d x)+b^2 \sinh (2 (c+d x))}{4 b^3 d} \] Input:

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

Output:

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

Rubi [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.66 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.16, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 26, 3272, 3042, 3502, 26, 3042, 3214, 3042, 3139, 1083, 217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3272

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

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

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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[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 [A] (verified)

Time = 0.28 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.79

method	result	size
derivativedivides	\(\frac {\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-b -2 a}{2 b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-2 a^{2}+b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}-\frac {1}{2 b \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {-b +2 a}{2 b^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\left (2 a^{2}-b^{2}\right ) \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 b^{3}}-\frac {2 a^{3} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{3} \sqrt {a^{2}+b^{2}}}}{d}\)	\(191\)
default	\(\frac {\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-b -2 a}{2 b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-2 a^{2}+b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}-\frac {1}{2 b \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {-b +2 a}{2 b^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\left (2 a^{2}-b^{2}\right ) \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 b^{3}}-\frac {2 a^{3} \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{b^{3} \sqrt {a^{2}+b^{2}}}}{d}\)	\(191\)
risch	\(\frac {x \,a^{2}}{b^{3}}-\frac {x}{2 b}+\frac {{\mathrm e}^{2 d x +2 c}}{8 b d}-\frac {a \,{\mathrm e}^{d x +c}}{2 b^{2} d}-\frac {a \,{\mathrm e}^{-d x -c}}{2 b^{2} d}-\frac {{\mathrm e}^{-2 d x -2 c}}{8 b d}+\frac {a^{3} \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}+a^{2}+b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d \,b^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}-a^{2}-b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d \,b^{3}}\)	\(204\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 601 vs. \(2 (100) = 200\).

Time = 0.10 (sec) , antiderivative size = 601, normalized size of antiderivative = 5.62 \[ \int \frac {\sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

1/8*(4*(2*a^4 + a^2*b^2 - b^4)*d*x*cosh(d*x + c)^2 + (a^2*b^2 + b^4)*cosh( 
d*x + c)^4 + (a^2*b^2 + b^4)*sinh(d*x + c)^4 - a^2*b^2 - b^4 - 4*(a^3*b + 
a*b^3)*cosh(d*x + c)^3 - 4*(a^3*b + a*b^3 - (a^2*b^2 + b^4)*cosh(d*x + c)) 
*sinh(d*x + c)^3 + 2*(2*(2*a^4 + a^2*b^2 - b^4)*d*x + 3*(a^2*b^2 + b^4)*co 
sh(d*x + c)^2 - 6*(a^3*b + a*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 + 8*(a^3* 
cosh(d*x + c)^2 + 2*a^3*cosh(d*x + c)*sinh(d*x + c) + a^3*sinh(d*x + c)^2) 
*sqrt(a^2 + b^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*co 
sh(d*x + c) + 2*a^2 + b^2 + 2*(b^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2* 
sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d*x + c) + a))/(b*cosh(d*x + c)^ 
2 + b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d 
*x + c) - b)) - 4*(a^3*b + a*b^3)*cosh(d*x + c) - 4*(a^3*b + a*b^3 - 2*(2* 
a^4 + a^2*b^2 - b^4)*d*x*cosh(d*x + c) - (a^2*b^2 + b^4)*cosh(d*x + c)^3 + 
 3*(a^3*b + a*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/((a^2*b^3 + b^5)*d*cosh 
(d*x + c)^2 + 2*(a^2*b^3 + b^5)*d*cosh(d*x + c)*sinh(d*x + c) + (a^2*b^3 + 
 b^5)*d*sinh(d*x + c)^2)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.53 \[ \int \frac {\sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {a^{3} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{3} d} - \frac {{\left (4 \, a e^{\left (-d x - c\right )} - b\right )} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, b^{2} d} + \frac {{\left (2 \, a^{2} - b^{2}\right )} {\left (d x + c\right )}}{2 \, b^{3} d} - \frac {4 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, b^{2} d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.41 \[ \int \frac {\sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {8 \, a^{3} \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{3}} - \frac {4 \, {\left (2 \, a^{2} - b^{2}\right )} {\left (d x + c\right )}}{b^{3}} - \frac {b e^{\left (2 \, d x + 2 \, c\right )} - 4 \, a e^{\left (d x + c\right )}}{b^{2}} + \frac {{\left (4 \, a b e^{\left (d x + c\right )} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{b^{3}}}{8 \, d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.98 \[ \int \frac {\sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {x\,\left (2\,a^2-b^2\right )}{2\,b^3}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,b\,d}+\frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,b\,d}-\frac {a\,{\mathrm {e}}^{-c-d\,x}}{2\,b^2\,d}-\frac {a\,{\mathrm {e}}^{c+d\,x}}{2\,b^2\,d}-\frac {a^3\,\ln \left (\frac {2\,a^3\,{\mathrm {e}}^{c+d\,x}}{b^4}-\frac {2\,a^3\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^4\,\sqrt {a^2+b^2}}\right )}{b^3\,d\,\sqrt {a^2+b^2}}+\frac {a^3\,\ln \left (\frac {2\,a^3\,{\mathrm {e}}^{c+d\,x}}{b^4}+\frac {2\,a^3\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^4\,\sqrt {a^2+b^2}}\right )}{b^3\,d\,\sqrt {a^2+b^2}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.21 \[ \int \frac {\sinh ^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {-16 e^{2 d x +2 c} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{3} i +e^{4 d x +4 c} a^{2} b^{2}+e^{4 d x +4 c} b^{4}-4 e^{3 d x +3 c} a^{3} b -4 e^{3 d x +3 c} a \,b^{3}+8 e^{2 d x +2 c} a^{4} d x +4 e^{2 d x +2 c} a^{2} b^{2} d x -4 e^{2 d x +2 c} b^{4} d x -4 e^{d x +c} a^{3} b -4 e^{d x +c} a \,b^{3}-a^{2} b^{2}-b^{4}}{8 e^{2 d x +2 c} b^{3} d \left (a^{2}+b^{2}\right )} \] Input:

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

Output:

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