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

Optimal. Leaf size=8 \[ \cosh (x)-i x \]

[Out]

-I*x+cosh(x)

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Rubi [A]  time = 0.03, antiderivative size = 8, 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 = {2682, 8} \[ \cosh (x)-i x \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^2/(I + Sinh[x]),x]

[Out]

(-I)*x + Cosh[x]

Rule 8

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

Rule 2682

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e
+ f*x])^(p - 1))/(b*f*(p - 1)), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rubi steps

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

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Mathematica [B]  time = 0.05, size = 34, normalized size = 4.25 \[ \cosh (x)+2 \sqrt {\cosh ^2(x)} \text {sech}(x) \sin ^{-1}\left (\frac {\sqrt {1-i \sinh (x)}}{\sqrt {2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^2/(I + Sinh[x]),x]

[Out]

Cosh[x] + 2*ArcSin[Sqrt[1 - I*Sinh[x]]/Sqrt[2]]*Sqrt[Cosh[x]^2]*Sech[x]

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fricas [B]  time = 0.61, size = 17, normalized size = 2.12 \[ \frac {1}{2} \, {\left (-2 i \, x e^{x} + e^{\left (2 \, x\right )} + 1\right )} e^{\left (-x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-2*I*x*e^x + e^(2*x) + 1)*e^(-x)

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giac [B]  time = 0.15, size = 14, normalized size = 1.75 \[ -i \, x + \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x + 1/2*e^(-x) + 1/2*e^x

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maple [B]  time = 0.05, size = 40, normalized size = 5.00 \[ i \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )-\frac {1}{\tanh \left (\frac {x}{2}\right )-1}-i \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )+\frac {1}{\tanh \left (\frac {x}{2}\right )+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.31, size = 14, normalized size = 1.75 \[ -i \, x + \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x + 1/2*e^(-x) + 1/2*e^x

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mupad [B]  time = 0.46, size = 7, normalized size = 0.88 \[ \mathrm {cosh}\relax (x)-x\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^2/(sinh(x) + 1i),x)

[Out]

cosh(x) - x*1i

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sympy [B]  time = 0.12, size = 14, normalized size = 1.75 \[ - i x + \frac {e^{x}}{2} + \frac {e^{- x}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x + exp(x)/2 + exp(-x)/2

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