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

Optimal. Leaf size=33 \[ \frac{\sinh ^4(x)}{4}-\frac{1}{3} i \sinh ^3(x)+\frac{\sinh ^2(x)}{2}-i \sinh (x) \]

[Out]

(-I)*Sinh[x] + Sinh[x]^2/2 - (I/3)*Sinh[x]^3 + Sinh[x]^4/4

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Rubi [A]  time = 0.0391253, antiderivative size = 33, normalized size of antiderivative = 1., 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} \[ \frac{\sinh ^4(x)}{4}-\frac{1}{3} i \sinh ^3(x)+\frac{\sinh ^2(x)}{2}-i \sinh (x) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^5/(I + Sinh[x]),x]

[Out]

(-I)*Sinh[x] + Sinh[x]^2/2 - (I/3)*Sinh[x]^3 + Sinh[x]^4/4

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 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])

Rubi steps

\begin{align*} \int \frac{\cosh ^5(x)}{i+\sinh (x)} \, dx &=\operatorname{Subst}\left (\int (i-x)^2 (i+x) \, dx,x,\sinh (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-i+x-i x^2+x^3\right ) \, dx,x,\sinh (x)\right )\\ &=-i \sinh (x)+\frac{\sinh ^2(x)}{2}-\frac{1}{3} i \sinh ^3(x)+\frac{\sinh ^4(x)}{4}\\ \end{align*}

Mathematica [A]  time = 0.0200937, size = 28, normalized size = 0.85 \[ \frac{1}{12} \sinh (x) \left (3 \sinh ^3(x)-4 i \sinh ^2(x)+6 \sinh (x)-12 i\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^5/(I + Sinh[x]),x]

[Out]

(Sinh[x]*(-12*I + 6*Sinh[x] - (4*I)*Sinh[x]^2 + 3*Sinh[x]^3))/12

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Maple [B]  time = 0.043, size = 94, normalized size = 2.9 \begin{align*}{{\frac{5}{8}}-{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}-{{\frac{1}{2}}-{\frac{i}{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}-{{\frac{3}{8}}-i \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{1}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-4}}+{{\frac{5}{8}}+{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{{\frac{1}{2}}+{\frac{i}{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}+{{\frac{3}{8}}+i \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{1}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^5/(I+sinh(x)),x)

[Out]

(5/8-1/2*I)/(tanh(1/2*x)+1)^2+(-1/2+1/3*I)/(tanh(1/2*x)+1)^3+(-3/8+I)/(tanh(1/2*x)+1)+1/4/(tanh(1/2*x)+1)^4+(5
/8+1/2*I)/(tanh(1/2*x)-1)^2+(1/2+1/3*I)/(tanh(1/2*x)-1)^3+(3/8+I)/(tanh(1/2*x)-1)+1/4/(tanh(1/2*x)-1)^4

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Maxima [B]  time = 1.11092, size = 69, normalized size = 2.09 \begin{align*} -\frac{1}{192} \,{\left (8 i \, e^{\left (-x\right )} - 12 \, e^{\left (-2 \, x\right )} + 72 i \, e^{\left (-3 \, x\right )} - 3\right )} e^{\left (4 \, x\right )} + \frac{3}{8} i \, e^{\left (-x\right )} + \frac{1}{16} \, e^{\left (-2 \, x\right )} + \frac{1}{24} i \, e^{\left (-3 \, x\right )} + \frac{1}{64} \, e^{\left (-4 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(8*I*e^(-x) - 12*e^(-2*x) + 72*I*e^(-3*x) - 3)*e^(4*x) + 3/8*I*e^(-x) + 1/16*e^(-2*x) + 1/24*I*e^(-3*x)
+ 1/64*e^(-4*x)

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Fricas [B]  time = 1.80942, size = 151, normalized size = 4.58 \begin{align*} \frac{1}{192} \,{\left (3 \, e^{\left (8 \, x\right )} - 8 i \, e^{\left (7 \, x\right )} + 12 \, e^{\left (6 \, x\right )} - 72 i \, e^{\left (5 \, x\right )} + 72 i \, e^{\left (3 \, x\right )} + 12 \, e^{\left (2 \, x\right )} + 8 i \, e^{x} + 3\right )} e^{\left (-4 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*e^(8*x) - 8*I*e^(7*x) + 12*e^(6*x) - 72*I*e^(5*x) + 72*I*e^(3*x) + 12*e^(2*x) + 8*I*e^x + 3)*e^(-4*x)

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Sympy [B]  time = 0.370981, size = 63, normalized size = 1.91 \begin{align*} \frac{e^{4 x}}{64} - \frac{i e^{3 x}}{24} + \frac{e^{2 x}}{16} - \frac{3 i e^{x}}{8} + \frac{3 i e^{- x}}{8} + \frac{e^{- 2 x}}{16} + \frac{i e^{- 3 x}}{24} + \frac{e^{- 4 x}}{64} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(4*x)/64 - I*exp(3*x)/24 + exp(2*x)/16 - 3*I*exp(x)/8 + 3*I*exp(-x)/8 + exp(-2*x)/16 + I*exp(-3*x)/24 + exp
(-4*x)/64

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Giac [B]  time = 1.2244, size = 63, normalized size = 1.91 \begin{align*} -\frac{1}{192} \,{\left (-72 i \, e^{\left (3 \, x\right )} - 12 \, e^{\left (2 \, x\right )} - 8 i \, e^{x} - 3\right )} e^{\left (-4 \, x\right )} + \frac{1}{64} \, e^{\left (4 \, x\right )} - \frac{1}{24} i \, e^{\left (3 \, x\right )} + \frac{1}{16} \, e^{\left (2 \, x\right )} - \frac{3}{8} i \, e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(-72*I*e^(3*x) - 12*e^(2*x) - 8*I*e^x - 3)*e^(-4*x) + 1/64*e^(4*x) - 1/24*I*e^(3*x) + 1/16*e^(2*x) - 3/
8*I*e^x