\(\int \frac {\cosh ^2(x)}{(a+b \sinh (x))^2} \, dx\) [202]

Optimal result

Integrand size = 13, antiderivative size = 62 \[ \int \frac {\cosh ^2(x)}{(a+b \sinh (x))^2} \, dx=\frac {x}{b^2}+\frac {2 a \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2}}-\frac {\cosh (x)}{b (a+b \sinh (x))} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.29 (sec) , antiderivative size = 502, normalized size of antiderivative = 8.10 \[ \int \frac {\cosh ^2(x)}{(a+b \sinh (x))^2} \, dx=\frac {\cosh (x) \left (2 a \sqrt {a-i b} \sqrt {a+i b} \text {arctanh}\left (\frac {\sqrt {-\frac {b (i+\sinh (x))}{a-i b}}}{\sqrt {-\frac {b (-i+\sinh (x))}{a+i b}}}\right ) \sqrt {1+i \sinh (x)} (a+b \sinh (x))-2 a (a-i b) \text {arctanh}\left (\frac {\sqrt {a-i b} \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}}{\sqrt {a+i b} \sqrt {-\frac {b (-i+\sinh (x))}{a+i b}}}\right ) \sqrt {1+i \sinh (x)} (a+b \sinh (x))+\sqrt {a+i b} \sqrt {-\frac {b (-i+\sinh (x))}{a+i b}} \left (2 \sqrt [4]{-1} a \sqrt {b} (i a+b) \arcsin \left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {a-i b} \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}}{\sqrt {b}}\right )+2 \sqrt [4]{-1} b^{3/2} (i a+b) \arcsin \left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {a-i b} \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}}{\sqrt {b}}\right ) \sinh (x)-\sqrt {a-i b} \left (a^2+b^2\right ) \sqrt {1+i \sinh (x)} \sqrt {-\frac {b (i+\sinh (x))}{a-i b}}\right )\right )}{(a-i b)^{3/2} (a+i b)^{3/2} b \sqrt {1+i \sinh (x)} \sqrt {-\frac {b (-i+\sinh (x))}{a+i b}} \sqrt {-\frac {b (i+\sinh (x))}{a-i b}} (a+b \sinh (x))} \] Input:

Integrate[Cosh[x]^2/(a + b*Sinh[x])^2,x]
 

Output:

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

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.40 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.29, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {3042, 3172, 26, 3042, 26, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (i x)^2}{(a-i b \sin (i x))^2}dx\)

\(\Big \downarrow \) 3172

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

\(\Big \downarrow \) 26

\(\displaystyle \frac {\int \frac {\sinh (x)}{a+b \sinh (x)}dx}{b}-\frac {\cosh (x)}{b (a+b \sinh (x))}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 219

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

Input:

Int[Cosh[x]^2/(a + b*Sinh[x])^2,x]
 

Output:

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

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 3172

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x 
])^(m + 1)/(b*f*(m + 1))), x] + Simp[g^2*((p - 1)/(b*(m + 1)))   Int[(g*Cos 
[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; Fre 
eQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && I 
ntegersQ[2*m, 2*p]
 

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]
 

Maple [A] (verified)

Time = 2.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.63

method	result	size
default	\(\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b^{2}}+\frac {\frac {2 \left (\frac {b^{2} \tanh \left (\frac {x}{2}\right )}{a}+b \right )}{\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a}-\frac {2 a \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}}{b^{2}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b^{2}}\)	\(101\)
risch	\(\frac {x}{b^{2}}+\frac {2 \,{\mathrm e}^{x} a -2 b}{b^{2} \left (b \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} a -b \right )}+\frac {a \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}+b^{2}}+a^{2}+b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, b^{2}}-\frac {a \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}+b^{2}}-a^{2}-b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, b^{2}}\)	\(140\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (58) = 116\).

Time = 0.11 (sec) , antiderivative size = 362, normalized size of antiderivative = 5.84 \[ \int \frac {\cosh ^2(x)}{(a+b \sinh (x))^2} \, dx=-\frac {{\left (a^{2} b + b^{3}\right )} x \cosh \left (x\right )^{2} + {\left (a^{2} b + b^{3}\right )} x \sinh \left (x\right )^{2} - 2 \, a^{2} b - 2 \, b^{3} + {\left (a b \cosh \left (x\right )^{2} + a b \sinh \left (x\right )^{2} + 2 \, a^{2} \cosh \left (x\right ) - a b + 2 \, {\left (a b \cosh \left (x\right ) + a^{2}\right )} \sinh \left (x\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) - {\left (a^{2} b + b^{3}\right )} x + 2 \, {\left (a^{3} + a b^{2} + {\left (a^{3} + a b^{2}\right )} x\right )} \cosh \left (x\right ) + 2 \, {\left (a^{3} + a b^{2} + {\left (a^{2} b + b^{3}\right )} x \cosh \left (x\right ) + {\left (a^{3} + a b^{2}\right )} x\right )} \sinh \left (x\right )}{a^{2} b^{3} + b^{5} - {\left (a^{2} b^{3} + b^{5}\right )} \cosh \left (x\right )^{2} - {\left (a^{2} b^{3} + b^{5}\right )} \sinh \left (x\right )^{2} - 2 \, {\left (a^{3} b^{2} + a b^{4}\right )} \cosh \left (x\right ) - 2 \, {\left (a^{3} b^{2} + a b^{4} + {\left (a^{2} b^{3} + b^{5}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )} \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cosh ^2(x)}{(a+b \sinh (x))^2} \, dx=\text {Timed out} \] Input:

integrate(cosh(x)**2/(a+b*sinh(x))**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.61 \[ \int \frac {\cosh ^2(x)}{(a+b \sinh (x))^2} \, dx=-\frac {2 \, {\left (a e^{\left (-x\right )} + b\right )}}{2 \, a b^{2} e^{\left (-x\right )} - b^{3} e^{\left (-2 \, x\right )} + b^{3}} - \frac {a \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b^{2}} + \frac {x}{b^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.56 \[ \int \frac {\cosh ^2(x)}{(a+b \sinh (x))^2} \, dx=-\frac {a \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} b^{2}} + \frac {x}{b^{2}} + \frac {2 \, {\left (a e^{x} - b\right )}}{{\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b\right )} b^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 1.54 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.13 \[ \int \frac {\cosh ^2(x)}{(a+b \sinh (x))^2} \, dx=\frac {x}{b^2}-\frac {\frac {2}{b}-\frac {2\,a\,{\mathrm {e}}^x}{b^2}}{2\,a\,{\mathrm {e}}^x-b+b\,{\mathrm {e}}^{2\,x}}-\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^x}{b^3}-\frac {2\,a\,\left (b-a\,{\mathrm {e}}^x\right )}{b^3\,\sqrt {a^2+b^2}}\right )}{b^2\,\sqrt {a^2+b^2}}+\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^x}{b^3}+\frac {2\,a\,\left (b-a\,{\mathrm {e}}^x\right )}{b^3\,\sqrt {a^2+b^2}}\right )}{b^2\,\sqrt {a^2+b^2}} \] Input:

int(cosh(x)^2/(a + b*sinh(x))^2,x)
                                                                                    
                                                                                    
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.31 \[ \int \frac {\cosh ^2(x)}{(a+b \sinh (x))^2} \, dx=\frac {-2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) \sinh \left (x \right ) a b i -2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{2} i -\cosh \left (x \right ) a^{2} b -\cosh \left (x \right ) b^{3}+\sinh \left (x \right ) a^{2} b x +\sinh \left (x \right ) b^{3} x +a^{3} x +a \,b^{2} x}{b^{2} \left (\sinh \left (x \right ) a^{2} b +\sinh \left (x \right ) b^{3}+a^{3}+a \,b^{2}\right )} \] Input:

int(cosh(x)^2/(a+b*sinh(x))^2,x)
 

Output:

( - 2*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*sinh(x)*a 
*b*i - 2*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*a**2*i 
 - cosh(x)*a**2*b - cosh(x)*b**3 + sinh(x)*a**2*b*x + sinh(x)*b**3*x + a** 
3*x + a*b**2*x)/(b**2*(sinh(x)*a**2*b + sinh(x)*b**3 + a**3 + a*b**2))