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*ln(I+sinh(x))+sinh(x)

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Rubi [A]  time = 0.04, antiderivative size = 14, normalized size of antiderivative = 1.00, 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 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)}{(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*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 1.00 \[ \sinh (x)-2 i \log (\sinh (x)+i) \]

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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fricas [B]  time = 0.82, size = 26, normalized size = 1.86 \[ \frac {1}{2} \, {\left (4 i \, x e^{x} - 8 i \, e^{x} \log \left (e^{x} + i\right ) + e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )} \]

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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giac [B]  time = 0.16, size = 21, normalized size = 1.50 \[ 2 i \, x - \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} - 4 i \, \log \left (e^{x} + i\right ) \]

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)

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maple [B]  time = 0.07, size = 53, normalized size = 3.79 \[ 2 i \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )-\frac {1}{\tanh \left (\frac {x}{2}\right )-1}+2 i \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-\frac {1}{\tanh \left (\frac {x}{2}\right )+1}-4 i \ln \left (\tanh \left (\frac {x}{2}\right )+i\right ) \]

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 = 0.32, size = 23, normalized size = 1.64 \[ -2 i \, x - \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} - 4 i \, \log \left (e^{\left (-x\right )} - i\right ) \]

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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mupad [B]  time = 0.55, size = 24, normalized size = 1.71 \[ \frac {{\mathrm {e}}^x}{2}-\frac {{\mathrm {e}}^{-x}}{2}+x\,2{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,4{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(sinh(x) + 1i)^2,x)

[Out]

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

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sympy [B]  time = 0.18, size = 26, normalized size = 1.86 \[ 2 i x + \frac {e^{x}}{2} - 4 i \log {\left (e^{x} + i \right )} - \frac {e^{- x}}{2} \]

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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