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

Optimal. Leaf size=14 \[ x-\frac{2 \cosh (x)}{\sinh (x)+i} \]

[Out]

x - (2*Cosh[x])/(I + Sinh[x])

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Rubi [A]  time = 0.0336892, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2680, 8} \[ x-\frac{2 \cosh (x)}{\sinh (x)+i} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^2/(I + Sinh[x])^2,x]

[Out]

x - (2*Cosh[x])/(I + Sinh[x])

Rule 2680

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]

Rule 8

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

Rubi steps

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

Mathematica [B]  time = 0.0538803, size = 69, normalized size = 4.93 \[ \frac{2 \cosh ^3(x) \left (-1-\frac{\sqrt{1-i \sinh (x)} \sin ^{-1}\left (\frac{\sqrt{1-i \sinh (x)}}{\sqrt{2}}\right )}{\sqrt{1+i \sinh (x)}}\right )}{(\sinh (x)-i) (\sinh (x)+i)^2} \]

Antiderivative was successfully verified.

[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)

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Maple [B]  time = 0.04, size = 29, normalized size = 2.1 \begin{align*} \ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) -4\, \left ( \tanh \left ( x/2 \right ) +i \right ) ^{-1}-\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(tanh(1/2*x)+1)-4/(tanh(1/2*x)+I)-ln(tanh(1/2*x)-1)

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Maxima [A]  time = 1.21545, size = 16, normalized size = 1.14 \begin{align*} x + \frac{4 i}{e^{\left (-x\right )} - i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 4*I/(e^(-x) - I)

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Fricas [A]  time = 1.73517, size = 42, normalized size = 3. \begin{align*} \frac{x e^{x} + i \, x + 4 i}{e^{x} + i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*e^x + I*x + 4*I)/(e^x + I)

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Sympy [A]  time = 0.183728, size = 8, normalized size = 0.57 \begin{align*} x + \frac{4 i}{e^{x} + i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 4*I/(exp(x) + I)

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Giac [A]  time = 1.27628, size = 14, normalized size = 1. \begin{align*} x + \frac{4 i}{e^{x} + i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 4*I/(e^x + I)