\(\int \frac {1}{(a+b \sinh (c+d x))^2} \, dx\) [102]

Optimal result

Integrand size = 12, antiderivative size = 79 \[ \int \frac {1}{(a+b \sinh (c+d x))^2} \, dx=-\frac {2 a \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}-\frac {b \cosh (c+d x)}{\left (a^2+b^2\right ) d (a+b \sinh (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(a+b \sinh (c+d x))^2} \, dx=-\frac {\frac {2 a \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}+\frac {b \cosh (c+d x)}{\left (a^2+b^2\right ) (a+b \sinh (c+d x))}}{d} \] Input:

Integrate[(a + b*Sinh[c + d*x])^(-2),x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3143, 25, 27, 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.

\(\displaystyle \int \frac {1}{(a+b \sinh (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3143

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

Int[(a + b*Sinh[c + d*x])^(-2),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 3143

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos 
[c + d*x]*((a + b*Sin[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 - b^2))), x] + Simp 
[1/((n + 1)*(a^2 - b^2))   Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1) 
- b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.49

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (a^{2}+b^{2}\right )}-\frac {b}{a^{2}+b^{2}}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a}+\frac {2 a \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}}{d}\)	\(118\)
default	\(\frac {-\frac {2 \left (-\frac {b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (a^{2}+b^{2}\right )}-\frac {b}{a^{2}+b^{2}}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a}+\frac {2 a \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}}{d}\)	\(118\)
risch	\(\frac {2 a \,{\mathrm e}^{d x +c}-2 b}{d \left (a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}-b \right )}+\frac {a \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d}-\frac {a \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d}\)	\(181\)

Input:

int(1/(a+b*sinh(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-2*(-b^2/a/(a^2+b^2)*tanh(1/2*d*x+1/2*c)-b/(a^2+b^2))/(tanh(1/2*d*x+1 
/2*c)^2*a-2*b*tanh(1/2*d*x+1/2*c)-a)+2*a/(a^2+b^2)^(3/2)*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. 423 vs. \(2 (76) = 152\).

Time = 0.09 (sec) , antiderivative size = 423, normalized size of antiderivative = 5.35 \[ \int \frac {1}{(a+b \sinh (c+d x))^2} \, dx=-\frac {2 \, a^{2} b + 2 \, b^{3} - {\left (a b \cosh \left (d x + c\right )^{2} + a b \sinh \left (d x + c\right )^{2} + 2 \, a^{2} \cosh \left (d x + c\right ) - a b + 2 \, {\left (a b \cosh \left (d x + c\right ) + a^{2}\right )} \sinh \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) - 2 \, {\left (a^{3} + a b^{2}\right )} \cosh \left (d x + c\right ) - 2 \, {\left (a^{3} + a b^{2}\right )} \sinh \left (d x + c\right )}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right )^{2} + {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \sinh \left (d x + c\right )^{2} + 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d \cosh \left (d x + c\right ) - {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d + 2 \, {\left ({\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right ) + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )} \sinh \left (d x + c\right )} \] Input:

integrate(1/(a+b*sinh(d*x+c))^2,x, algorithm="fricas")
 

Output:

-(2*a^2*b + 2*b^3 - (a*b*cosh(d*x + c)^2 + a*b*sinh(d*x + c)^2 + 2*a^2*cos 
h(d*x + c) - a*b + 2*(a*b*cosh(d*x + c) + a^2)*sinh(d*x + c))*sqrt(a^2 + b 
^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*cosh(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)) 
 - 2*(a^3 + a*b^2)*cosh(d*x + c) - 2*(a^3 + a*b^2)*sinh(d*x + c))/((a^4*b 
+ 2*a^2*b^3 + b^5)*d*cosh(d*x + c)^2 + (a^4*b + 2*a^2*b^3 + b^5)*d*sinh(d* 
x + c)^2 + 2*(a^5 + 2*a^3*b^2 + a*b^4)*d*cosh(d*x + c) - (a^4*b + 2*a^2*b^ 
3 + b^5)*d + 2*((a^4*b + 2*a^2*b^3 + b^5)*d*cosh(d*x + c) + (a^5 + 2*a^3*b 
^2 + a*b^4)*d)*sinh(d*x + c))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2082 vs. \(2 (68) = 136\).

Time = 81.16 (sec) , antiderivative size = 2082, normalized size of antiderivative = 26.35 \[ \int \frac {1}{(a+b \sinh (c+d x))^2} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*sinh(d*x+c))**2,x)
 

Output:

Piecewise((zoo*x/sinh(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((-tanh(c/2 
+ d*x/2)/(2*d) - 1/(2*d*tanh(c/2 + d*x/2)))/b**2, Eq(a, 0)), (-6*b*tanh(c/ 
2 + d*x/2)**2/(3*b**3*d*tanh(c/2 + d*x/2)**3 - 9*b**3*d*tanh(c/2 + d*x/2) 
- 9*b**2*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**2 + 3*b**2*d*sqrt(-b**2)) + 4*b/ 
(3*b**3*d*tanh(c/2 + d*x/2)**3 - 9*b**3*d*tanh(c/2 + d*x/2) - 9*b**2*d*sqr 
t(-b**2)*tanh(c/2 + d*x/2)**2 + 3*b**2*d*sqrt(-b**2)) + 6*sqrt(-b**2)*tanh 
(c/2 + d*x/2)/(3*b**3*d*tanh(c/2 + d*x/2)**3 - 9*b**3*d*tanh(c/2 + d*x/2) 
- 9*b**2*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**2 + 3*b**2*d*sqrt(-b**2)), Eq(a, 
 -sqrt(-b**2))), (-6*b*tanh(c/2 + d*x/2)**2/(3*b**3*d*tanh(c/2 + d*x/2)**3 
 - 9*b**3*d*tanh(c/2 + d*x/2) + 9*b**2*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**2 
- 3*b**2*d*sqrt(-b**2)) + 4*b/(3*b**3*d*tanh(c/2 + d*x/2)**3 - 9*b**3*d*ta 
nh(c/2 + d*x/2) + 9*b**2*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**2 - 3*b**2*d*sqr 
t(-b**2)) - 6*sqrt(-b**2)*tanh(c/2 + d*x/2)/(3*b**3*d*tanh(c/2 + d*x/2)**3 
 - 9*b**3*d*tanh(c/2 + d*x/2) + 9*b**2*d*sqrt(-b**2)*tanh(c/2 + d*x/2)**2 
- 3*b**2*d*sqrt(-b**2)), Eq(a, sqrt(-b**2))), (x/(a + b*sinh(c))**2, Eq(d, 
 0)), (-a**3*log(tanh(c/2 + d*x/2) - b/a - sqrt(a**2 + b**2)/a)*tanh(c/2 + 
 d*x/2)**2/(a**4*d*sqrt(a**2 + b**2)*tanh(c/2 + d*x/2)**2 - a**4*d*sqrt(a* 
*2 + b**2) - 2*a**3*b*d*sqrt(a**2 + b**2)*tanh(c/2 + d*x/2) + a**2*b**2*d* 
sqrt(a**2 + b**2)*tanh(c/2 + d*x/2)**2 - a**2*b**2*d*sqrt(a**2 + b**2) - 2 
*a*b**3*d*sqrt(a**2 + b**2)*tanh(c/2 + d*x/2)) + a**3*log(tanh(c/2 + d*...
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.75 \[ \int \frac {1}{(a+b \sinh (c+d x))^2} \, dx=\frac {a \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}} d} - \frac {2 \, {\left (a e^{\left (-d x - c\right )} + b\right )}}{{\left (a^{2} b + b^{3} + 2 \, {\left (a^{3} + a b^{2}\right )} e^{\left (-d x - c\right )} - {\left (a^{2} b + b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} \] Input:

integrate(1/(a+b*sinh(d*x+c))^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.51 \[ \int \frac {1}{(a+b \sinh (c+d x))^2} \, dx=\frac {\frac {a \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (a e^{\left (d x + c\right )} - b\right )}}{{\left (a^{2} + b^{2}\right )} {\left (b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - b\right )}}}{d} \] Input:

integrate(1/(a+b*sinh(d*x+c))^2,x, algorithm="giac")
 

Output:

(a*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)))/(a^2 + b^2)^(3/2) + 2*(a*e^(d*x + c) - b)/((a^ 
2 + b^2)*(b*e^(2*d*x + 2*c) + 2*a*e^(d*x + c) - b)))/d
 

Mupad [B] (verification not implemented)

Time = 1.97 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.53 \[ \int \frac {1}{(a+b \sinh (c+d x))^2} \, dx=\frac {a\,\ln \left (\frac {2\,a\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b\,{\left (a^2+b^2\right )}^{3/2}}-\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{b\,\left (a^2+b^2\right )}\right )}{d\,{\left (a^2+b^2\right )}^{3/2}}-\frac {a\,\ln \left (-\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{b\,\left (a^2+b^2\right )}-\frac {2\,a\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b\,{\left (a^2+b^2\right )}^{3/2}}\right )}{d\,{\left (a^2+b^2\right )}^{3/2}}-\frac {\frac {2\,b^2}{d\,\left (a^2\,b+b^3\right )}-\frac {2\,a\,b\,{\mathrm {e}}^{c+d\,x}}{d\,\left (a^2\,b+b^3\right )}}{2\,a\,{\mathrm {e}}^{c+d\,x}-b+b\,{\mathrm {e}}^{2\,c+2\,d\,x}} \] Input:

int(1/(a + b*sinh(c + d*x))^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 291, normalized size of antiderivative = 3.68 \[ \int \frac {1}{(a+b \sinh (c+d x))^2} \, dx=\frac {2 e^{2 d x +2 c} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a b i +4 e^{d x +c} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{2} i -2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a b i -e^{2 d x +2 c} a^{2} b -e^{2 d x +2 c} b^{3}-a^{2} b -b^{3}}{d \left (e^{2 d x +2 c} a^{4} b +2 e^{2 d x +2 c} a^{2} b^{3}+e^{2 d x +2 c} b^{5}+2 e^{d x +c} a^{5}+4 e^{d x +c} a^{3} b^{2}+2 e^{d x +c} a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}\right )} \] Input:

int(1/(a+b*sinh(d*x+c))^2,x)
 

Output:

(2*e**(2*c + 2*d*x)*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sqrt(a 
**2 + b**2))*a*b*i + 4*e**(c + d*x)*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b 
*i + a*i)/sqrt(a**2 + b**2))*a**2*i - 2*sqrt(a**2 + b**2)*atan((e**(c + d* 
x)*b*i + a*i)/sqrt(a**2 + b**2))*a*b*i - e**(2*c + 2*d*x)*a**2*b - e**(2*c 
 + 2*d*x)*b**3 - a**2*b - b**3)/(d*(e**(2*c + 2*d*x)*a**4*b + 2*e**(2*c + 
2*d*x)*a**2*b**3 + e**(2*c + 2*d*x)*b**5 + 2*e**(c + d*x)*a**5 + 4*e**(c + 
 d*x)*a**3*b**2 + 2*e**(c + d*x)*a*b**4 - a**4*b - 2*a**2*b**3 - b**5))