3.172 \(\int \frac{\cosh ^5(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=14 \[ -\frac{1}{3} (-\sinh (x)+i)^3 \]

[Out]

-(I - Sinh[x])^3/3

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Rubi [A]  time = 0.0335453, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2667, 32} \[ -\frac{1}{3} (-\sinh (x)+i)^3 \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^5/(I + Sinh[x])^2,x]

[Out]

-(I - Sinh[x])^3/3

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\cosh ^5(x)}{(i+\sinh (x))^2} \, dx &=\operatorname{Subst}\left (\int (i-x)^2 \, dx,x,\sinh (x)\right )\\ &=-\frac{1}{3} (i-\sinh (x))^3\\ \end{align*}

Mathematica [A]  time = 0.0193989, size = 18, normalized size = 1.29 \[ \frac{1}{6} \sinh (x) (-6 i \sinh (x)+\cosh (2 x)-7) \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^5/(I + Sinh[x])^2,x]

[Out]

((-7 + Cosh[2*x] - (6*I)*Sinh[x])*Sinh[x])/6

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Maple [B]  time = 0.047, size = 70, normalized size = 5. \begin{align*}{{\frac{1}{2}}-i \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{1+i \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+{1-i \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{{\frac{1}{2}}+i \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}-{\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)^5/(I+sinh(x))^2,x)

[Out]

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

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Maxima [B]  time = 1.2142, size = 53, normalized size = 3.79 \begin{align*} -\frac{1}{96} \,{\left (24 i \, e^{\left (-x\right )} + 60 \, e^{\left (-2 \, x\right )} - 4\right )} e^{\left (3 \, x\right )} + \frac{5}{8} \, e^{\left (-x\right )} - \frac{1}{4} 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)^5/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

-1/96*(24*I*e^(-x) + 60*e^(-2*x) - 4)*e^(3*x) + 5/8*e^(-x) - 1/4*I*e^(-2*x) - 1/24*e^(-3*x)

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Fricas [B]  time = 1.97653, size = 107, normalized size = 7.64 \begin{align*} \frac{1}{24} \,{\left (e^{\left (6 \, x\right )} - 6 i \, e^{\left (5 \, x\right )} - 15 \, e^{\left (4 \, x\right )} + 15 \, e^{\left (2 \, x\right )} - 6 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)^5/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

1/24*(e^(6*x) - 6*I*e^(5*x) - 15*e^(4*x) + 15*e^(2*x) - 6*I*e^x - 1)*e^(-3*x)

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Sympy [B]  time = 0.315745, size = 44, normalized size = 3.14 \begin{align*} \frac{e^{3 x}}{24} - \frac{i e^{2 x}}{4} - \frac{5 e^{x}}{8} + \frac{5 e^{- x}}{8} - \frac{i e^{- 2 x}}{4} - \frac{e^{- 3 x}}{24} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**5/(I+sinh(x))**2,x)

[Out]

exp(3*x)/24 - I*exp(2*x)/4 - 5*exp(x)/8 + 5*exp(-x)/8 - I*exp(-2*x)/4 - exp(-3*x)/24

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Giac [B]  time = 1.21791, size = 47, normalized size = 3.36 \begin{align*} \frac{1}{24} \,{\left (15 \, e^{\left (2 \, x\right )} - 6 i \, e^{x} - 1\right )} e^{\left (-3 \, x\right )} + \frac{1}{24} \, e^{\left (3 \, x\right )} - \frac{1}{4} i \, e^{\left (2 \, x\right )} - \frac{5}{8} \, e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^5/(I+sinh(x))^2,x, algorithm="giac")

[Out]

1/24*(15*e^(2*x) - 6*I*e^x - 1)*e^(-3*x) + 1/24*e^(3*x) - 1/4*I*e^(2*x) - 5/8*e^x