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 = 0.03, antiderivative size = 14, normalized size of antiderivative = 1.00, 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 8

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

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*}

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Mathematica [B]  time = 0.05, 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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fricas [A]  time = 0.51, size = 16, normalized size = 1.14 \[ \frac {x e^{x} + i \, x + 4 i}{e^{x} + i} \]

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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giac [A]  time = 0.17, size = 10, normalized size = 0.71 \[ x + \frac {4 i}{e^{x} + i} \]

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)

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maple [B]  time = 0.06, size = 29, normalized size = 2.07 \[ -\frac {4}{\tanh \left (\frac {x}{2}\right )+i}-\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )+\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 12, normalized size = 0.86 \[ x + \frac {4 i}{e^{\left (-x\right )} - i} \]

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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mupad [B]  time = 0.65, size = 12, normalized size = 0.86 \[ x+\frac {4{}\mathrm {i}}{{\mathrm {e}}^x+1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^2/(sinh(x) + 1i)^2,x)

[Out]

x + 4i/(exp(x) + 1i)

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sympy [A]  time = 0.12, size = 8, normalized size = 0.57 \[ x + \frac {4}{- i e^{x} + 1} \]

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) + 1)

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