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

Optimal. Leaf size=26 \[ -\frac{i x}{2}+\frac{\cosh ^3(x)}{3}-\frac{1}{2} i \sinh (x) \cosh (x) \]

[Out]

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

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Rubi [A]  time = 0.0413358, antiderivative size = 26, normalized size of antiderivative = 1., 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 = {2682, 2635, 8} \[ -\frac{i x}{2}+\frac{\cosh ^3(x)}{3}-\frac{1}{2} i \sinh (x) \cosh (x) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^4/(I + Sinh[x]),x]

[Out]

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

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

Rubi steps

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

Mathematica [B]  time = 0.148906, size = 93, normalized size = 3.58 \[ \frac{\cosh ^5(x) \left (2 \sinh ^3(x)-i \sinh ^2(x)+5 \sinh (x)+\frac{6 i \sqrt{1-i \sinh (x)} \sin ^{-1}\left (\frac{\sqrt{1-i \sinh (x)}}{\sqrt{2}}\right )}{\sqrt{1+i \sinh (x)}}+2 i\right )}{6 (\sinh (x)-i)^2 (\sinh (x)+i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^4/(I + Sinh[x]),x]

[Out]

(Cosh[x]^5*(2*I + ((6*I)*ArcSin[Sqrt[1 - I*Sinh[x]]/Sqrt[2]]*Sqrt[1 - I*Sinh[x]])/Sqrt[1 + I*Sinh[x]] + 5*Sinh
[x] - I*Sinh[x]^2 + 2*Sinh[x]^3))/(6*(-I + Sinh[x])^2*(I + Sinh[x])^3)

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Maple [B]  time = 0.036, size = 126, normalized size = 4.9 \begin{align*} -{\frac{i}{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) +{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+{\frac{i}{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) -{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}-{{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}-{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.24569, size = 57, normalized size = 2.19 \begin{align*} -\frac{1}{48} \,{\left (6 i \, e^{\left (-x\right )} - 6 \, e^{\left (-2 \, x\right )} - 2\right )} e^{\left (3 \, x\right )} - \frac{1}{2} i \, x + \frac{1}{8} \, e^{\left (-x\right )} + \frac{1}{8} i \, e^{\left (-2 \, x\right )} + \frac{1}{24} \, e^{\left (-3 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(6*I*e^(-x) - 6*e^(-2*x) - 2)*e^(3*x) - 1/2*I*x + 1/8*e^(-x) + 1/8*I*e^(-2*x) + 1/24*e^(-3*x)

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Fricas [B]  time = 1.84849, size = 128, normalized size = 4.92 \begin{align*} \frac{1}{24} \,{\left (-12 i \, x e^{\left (3 \, x\right )} + e^{\left (6 \, x\right )} - 3 i \, e^{\left (5 \, x\right )} + 3 \, e^{\left (4 \, x\right )} + 3 \, e^{\left (2 \, x\right )} + 3 i \, e^{x} + 1\right )} e^{\left (-3 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(-12*I*x*e^(3*x) + e^(6*x) - 3*I*e^(5*x) + 3*e^(4*x) + 3*e^(2*x) + 3*I*e^x + 1)*e^(-3*x)

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Sympy [B]  time = 0.280378, size = 48, normalized size = 1.85 \begin{align*} - \frac{i x}{2} + \frac{e^{3 x}}{24} - \frac{i e^{2 x}}{8} + \frac{e^{x}}{8} + \frac{e^{- x}}{8} + \frac{i e^{- 2 x}}{8} + \frac{e^{- 3 x}}{24} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x/2 + exp(3*x)/24 - I*exp(2*x)/8 + exp(x)/8 + exp(-x)/8 + I*exp(-2*x)/8 + exp(-3*x)/24

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Giac [B]  time = 1.33892, size = 51, normalized size = 1.96 \begin{align*} \frac{1}{24} \,{\left (3 \, e^{\left (2 \, x\right )} + 3 i \, e^{x} + 1\right )} e^{\left (-3 \, x\right )} - \frac{1}{2} i \, x + \frac{1}{24} \, e^{\left (3 \, x\right )} - \frac{1}{8} i \, e^{\left (2 \, x\right )} + \frac{1}{8} \, e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*e^(2*x) + 3*I*e^x + 1)*e^(-3*x) - 1/2*I*x + 1/24*e^(3*x) - 1/8*I*e^(2*x) + 1/8*e^x