3.174 \(\int \frac{\cosh ^3(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=14 \[ \sinh (x)-2 i \log (\sinh (x)+i) \]

[Out]

(-2*I)*Log[I + Sinh[x]] + Sinh[x]

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Rubi [A]  time = 0.0377104, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2667, 43} \[ \sinh (x)-2 i \log (\sinh (x)+i) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^3/(I + Sinh[x])^2,x]

[Out]

(-2*I)*Log[I + Sinh[x]] + Sinh[x]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\cosh ^3(x)}{(i+\sinh (x))^2} \, dx &=-\operatorname{Subst}\left (\int \frac{i-x}{i+x} \, dx,x,\sinh (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (-1+\frac{2 i}{i+x}\right ) \, dx,x,\sinh (x)\right )\\ &=-2 i \log (i+\sinh (x))+\sinh (x)\\ \end{align*}

Mathematica [A]  time = 0.0124456, size = 14, normalized size = 1. \[ \sinh (x)-2 i \log (\sinh (x)+i) \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^3/(I + Sinh[x])^2,x]

[Out]

(-2*I)*Log[I + Sinh[x]] + Sinh[x]

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Maple [B]  time = 0.042, size = 53, normalized size = 3.8 \begin{align*} 2\,i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) - \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}+2\,i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) - \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}-4\,i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(I+sinh(x))^2,x)

[Out]

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

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Maxima [B]  time = 1.23432, size = 31, normalized size = 2.21 \begin{align*} -2 i \, x - \frac{1}{2} \, e^{\left (-x\right )} + \frac{1}{2} \, e^{x} - 4 i \, \log \left (e^{\left (-x\right )} - i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

-2*I*x - 1/2*e^(-x) + 1/2*e^x - 4*I*log(e^(-x) - I)

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Fricas [B]  time = 1.92183, size = 82, normalized size = 5.86 \begin{align*} \frac{1}{2} \,{\left (4 i \, x e^{x} - 8 i \, e^{x} \log \left (e^{x} + i\right ) + e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

1/2*(4*I*x*e^x - 8*I*e^x*log(e^x + I) + e^(2*x) - 1)*e^(-x)

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Sympy [B]  time = 9.33967, size = 26, normalized size = 1.86 \begin{align*} 2 i x + \frac{e^{x}}{2} - 4 i \log{\left (e^{x} + i \right )} - \frac{e^{- x}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**3/(I+sinh(x))**2,x)

[Out]

2*I*x + exp(x)/2 - 4*I*log(exp(x) + I) - exp(-x)/2

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Giac [A]  time = 1.31995, size = 28, normalized size = 2. \begin{align*} 2 i \, x - \frac{1}{2} \, e^{\left (-x\right )} + \frac{1}{2} \, e^{x} - 4 i \, \log \left (e^{x} + i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(I+sinh(x))^2,x, algorithm="giac")

[Out]

2*I*x - 1/2*e^(-x) + 1/2*e^x - 4*I*log(e^x + I)