3.173 \(\int \frac {\cosh ^4(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=30 \[ -\frac {3 x}{2}-\frac {3}{2} i \cosh (x)+\frac {\cosh ^3(x)}{2 (\sinh (x)+i)} \]

[Out]

-3/2*x-3/2*I*cosh(x)+1/2*cosh(x)^3/(I+sinh(x))

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Rubi [A]  time = 0.07, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2679, 2682, 8} \[ -\frac {3 x}{2}-\frac {3}{2} i \cosh (x)+\frac {\cosh ^3(x)}{2 (\sinh (x)+i)} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^4/(I + Sinh[x])^2,x]

[Out]

(-3*x)/2 - ((3*I)/2)*Cosh[x] + Cosh[x]^3/(2*(I + Sinh[x]))

Rule 8

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

Rule 2679

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

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 ^4(x)}{(i+\sinh (x))^2} \, dx &=\frac {\cosh ^3(x)}{2 (i+\sinh (x))}-\frac {3}{2} i \int \frac {\cosh ^2(x)}{i+\sinh (x)} \, dx\\ &=-\frac {3}{2} i \cosh (x)+\frac {\cosh ^3(x)}{2 (i+\sinh (x))}-\frac {3 \int 1 \, dx}{2}\\ &=-\frac {3 x}{2}-\frac {3}{2} i \cosh (x)+\frac {\cosh ^3(x)}{2 (i+\sinh (x))}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 46, normalized size = 1.53 \[ \frac {1}{2} (\sinh (x)-4 i) \cosh (x)-3 i \sqrt {\cosh ^2(x)} \text {sech}(x) \sin ^{-1}\left (\frac {\sqrt {1-i \sinh (x)}}{\sqrt {2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^4/(I + Sinh[x])^2,x]

[Out]

(-3*I)*ArcSin[Sqrt[1 - I*Sinh[x]]/Sqrt[2]]*Sqrt[Cosh[x]^2]*Sech[x] + (Cosh[x]*(-4*I + Sinh[x]))/2

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fricas [A]  time = 0.46, size = 31, normalized size = 1.03 \[ -\frac {1}{8} \, {\left (12 \, x e^{\left (2 \, x\right )} - e^{\left (4 \, x\right )} + 8 i \, e^{\left (3 \, x\right )} + 8 i \, e^{x} + 1\right )} e^{\left (-2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

-1/8*(12*x*e^(2*x) - e^(4*x) + 8*I*e^(3*x) + 8*I*e^x + 1)*e^(-2*x)

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giac [A]  time = 0.49, size = 26, normalized size = 0.87 \[ -\frac {1}{8} \, {\left (8 i \, e^{x} + 1\right )} e^{\left (-2 \, x\right )} - \frac {3}{2} \, x + \frac {1}{8} \, e^{\left (2 \, x\right )} - i \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(I+sinh(x))^2,x, algorithm="giac")

[Out]

-1/8*(8*I*e^x + 1)*e^(-2*x) - 3/2*x + 1/8*e^(2*x) - I*e^x

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maple [B]  time = 0.07, size = 82, normalized size = 2.73 \[ \frac {1}{2 \tanh \left (\frac {x}{2}\right )-2}+\frac {2 i}{\tanh \left (\frac {x}{2}\right )-1}+\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}+\frac {1}{2 \tanh \left (\frac {x}{2}\right )+2}-\frac {2 i}{\tanh \left (\frac {x}{2}\right )+1}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^4/(I+sinh(x))^2,x)

[Out]

1/2/(tanh(1/2*x)-1)+2*I/(tanh(1/2*x)-1)+1/2/(tanh(1/2*x)-1)^2+3/2*ln(tanh(1/2*x)-1)+1/2/(tanh(1/2*x)+1)-2*I/(t
anh(1/2*x)+1)-1/2/(tanh(1/2*x)+1)^2-3/2*ln(tanh(1/2*x)+1)

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maxima [A]  time = 0.33, size = 30, normalized size = 1.00 \[ -\frac {1}{8} \, {\left (8 i \, e^{\left (-x\right )} - 1\right )} e^{\left (2 \, x\right )} - \frac {3}{2} \, x - i \, e^{\left (-x\right )} - \frac {1}{8} \, e^{\left (-2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

-1/8*(8*I*e^(-x) - 1)*e^(2*x) - 3/2*x - I*e^(-x) - 1/8*e^(-2*x)

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mupad [B]  time = 0.48, size = 28, normalized size = 0.93 \[ \frac {{\mathrm {e}}^{2\,x}}{8}-{\mathrm {e}}^{-x}\,1{}\mathrm {i}-\frac {{\mathrm {e}}^{-2\,x}}{8}-\frac {3\,x}{2}-{\mathrm {e}}^x\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^4/(sinh(x) + 1i)^2,x)

[Out]

exp(2*x)/8 - exp(-x)*1i - exp(-2*x)/8 - (3*x)/2 - exp(x)*1i

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sympy [A]  time = 0.16, size = 29, normalized size = 0.97 \[ - \frac {3 x}{2} + \frac {e^{2 x}}{8} - i e^{x} - i e^{- x} - \frac {e^{- 2 x}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**4/(I+sinh(x))**2,x)

[Out]

-3*x/2 + exp(2*x)/8 - I*exp(x) - I*exp(-x) - exp(-2*x)/8

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