3.181 \(\int \frac {\cosh ^3(x)}{(1+i \sinh (x))^3} \, dx\)

Optimal. Leaf size=28 \[ \frac {2 i}{1+i \sinh (x)}+i \log (-\sinh (x)+i) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2667, 43} \[ \frac {2 i}{1+i \sinh (x)}+i \log (-\sinh (x)+i) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^3/(1 + I*Sinh[x])^3,x]

[Out]

I*Log[I - Sinh[x]] + (2*I)/(1 + I*Sinh[x])

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])

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])

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 45, normalized size = 1.61 \[ \frac {2 i \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )+\log (\cosh (x))+\sinh (x) \left (-2 \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )+i \log (\cosh (x))\right )+2}{\sinh (x)-i} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^3/(1 + I*Sinh[x])^3,x]

[Out]

(2 + (2*I)*ArcTan[Tanh[x/2]] + Log[Cosh[x]] + (-2*ArcTan[Tanh[x/2]] + I*Log[Cosh[x]])*Sinh[x])/(-I + Sinh[x])

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fricas [B]  time = 0.48, size = 49, normalized size = 1.75 \[ \frac {-i \, x e^{\left (2 \, x\right )} - 2 \, {\left (x - 2\right )} e^{x} + {\left (2 i \, e^{\left (2 \, x\right )} + 4 \, e^{x} - 2 i\right )} \log \left (e^{x} - i\right ) + i \, x}{e^{\left (2 \, x\right )} - 2 i \, e^{x} - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(1+I*sinh(x))^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 27, normalized size = 0.96 \[ \frac {4 \, e^{x}}{{\left (e^{x} - i\right )}^{2}} - i \, \log \left (i \, e^{x}\right ) + 2 i \, \log \left (-i \, e^{x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(1+I*sinh(x))^3,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.08, size = 56, normalized size = 2.00 \[ -i \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )-i \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )+2 i \ln \left (\tanh \left (\frac {x}{2}\right )-i\right )-\frac {4 i}{\left (\tanh \left (\frac {x}{2}\right )-i\right )^{2}}-\frac {4}{\tanh \left (\frac {x}{2}\right )-i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(1+I*sinh(x))^3,x)

[Out]

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

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maxima [A]  time = 0.35, size = 33, normalized size = 1.18 \[ i \, x - \frac {4 \, e^{\left (-x\right )}}{2 i \, e^{\left (-x\right )} + e^{\left (-2 \, x\right )} - 1} + 2 i \, \log \left (e^{\left (-x\right )} + i\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(1+I*sinh(x))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.19, size = 41, normalized size = 1.46 \[ -x\,1{}\mathrm {i}+\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,2{}\mathrm {i}-\frac {4{}\mathrm {i}}{1-{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,2{}\mathrm {i}}+\frac {4}{{\mathrm {e}}^x-\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(sinh(x)*1i + 1)^3,x)

[Out]

log(exp(x) - 1i)*2i - x*1i - 4i/(exp(x)*2i - exp(2*x) + 1) + 4/(exp(x) - 1i)

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sympy [A]  time = 0.17, size = 31, normalized size = 1.11 \[ x \left (-2 + i\right ) + 2 \log {\left (e^{x} - i \right )} - \frac {4 e^{x}}{- e^{2 x} + 2 i e^{x} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**3/(1+I*sinh(x))**3,x)

[Out]

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

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