\(\int \frac {\cosh ^2(x)}{(i+\sinh (x))^2} \, dx\) [175]

Optimal result

Integrand size = 13, antiderivative size = 14 \[ \int \frac {\cosh ^2(x)}{(i+\sinh (x))^2} \, dx=x-\frac {2 \cosh (x)}{i+\sinh (x)} \]

[Out]

x-2*cosh(x)/(I+sinh(x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2759, 8} \[ \int \frac {\cosh ^2(x)}{(i+\sinh (x))^2} \, dx=x-\frac {2 \cosh (x)}{\sinh (x)+i} \]

[In]

Int[Cosh[x]^2/(I + Sinh[x])^2,x]

[Out]

x - (2*Cosh[x])/(I + Sinh[x])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2759

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[2*g*(g
*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(2*m +
p + 1))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \cosh (x)}{i+\sinh (x)}+\int 1 \, dx \\ & = x-\frac {2 \cosh (x)}{i+\sinh (x)} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(69\) vs. \(2(14)=28\).

Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 4.93 \[ \int \frac {\cosh ^2(x)}{(i+\sinh (x))^2} \, dx=\frac {2 \cosh ^3(x) \left (-1-\frac {\arcsin \left (\frac {\sqrt {1-i \sinh (x)}}{\sqrt {2}}\right ) \sqrt {1-i \sinh (x)}}{\sqrt {1+i \sinh (x)}}\right )}{(-i+\sinh (x)) (i+\sinh (x))^2} \]

[In]

Integrate[Cosh[x]^2/(I + Sinh[x])^2,x]

[Out]

(2*Cosh[x]^3*(-1 - (ArcSin[Sqrt[1 - I*Sinh[x]]/Sqrt[2]]*Sqrt[1 - I*Sinh[x]])/Sqrt[1 + I*Sinh[x]]))/((-I + Sinh
[x])*(I + Sinh[x])^2)

Maple [A] (verified)

Time = 13.10 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
risch	\(x +\frac {4 i}{{\mathrm e}^{x}+i}\)	\(13\)
default	\(-\frac {4}{\tanh \left (\frac {x}{2}\right )+i}-\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )+\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )\)	\(29\)

[In]

int(cosh(x)^2/(I+sinh(x))^2,x,method=_RETURNVERBOSE)

[Out]

x+4*I/(exp(x)+I)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {\cosh ^2(x)}{(i+\sinh (x))^2} \, dx=\frac {x e^{x} + i \, x + 4 i}{e^{x} + i} \]

[In]

integrate(cosh(x)^2/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

(x*e^x + I*x + 4*I)/(e^x + I)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.57 \[ \int \frac {\cosh ^2(x)}{(i+\sinh (x))^2} \, dx=x + \frac {4 i}{e^{x} + i} \]

[In]

integrate(cosh(x)**2/(I+sinh(x))**2,x)

[Out]

x + 4*I/(exp(x) + I)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {\cosh ^2(x)}{(i+\sinh (x))^2} \, dx=x + \frac {4 i}{e^{\left (-x\right )} - i} \]

[In]

integrate(cosh(x)^2/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

x + 4*I/(e^(-x) - I)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {\cosh ^2(x)}{(i+\sinh (x))^2} \, dx=x + \frac {4 i}{e^{x} + i} \]

[In]

integrate(cosh(x)^2/(I+sinh(x))^2,x, algorithm="giac")

[Out]

x + 4*I/(e^x + I)

Mupad [B] (verification not implemented)

Time = 1.32 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {\cosh ^2(x)}{(i+\sinh (x))^2} \, dx=x+\frac {4{}\mathrm {i}}{{\mathrm {e}}^x+1{}\mathrm {i}} \]

[In]

int(cosh(x)^2/(sinh(x) + 1i)^2,x)

[Out]

x + 4i/(exp(x) + 1i)