\(\int \frac {\cosh ^5(x)}{(i+\sinh (x))^2} \, dx\) [172]

Optimal result

Integrand size = 13, antiderivative size = 14 \[ \int \frac {\cosh ^5(x)}{(i+\sinh (x))^2} \, dx=-\frac {1}{3} (i-\sinh (x))^3 \]

[Out]

-1/3*(I-sinh(x))^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2746, 32} \[ \int \frac {\cosh ^5(x)}{(i+\sinh (x))^2} \, dx=-\frac {1}{3} (-\sinh (x)+i)^3 \]

[In]

Int[Cosh[x]^5/(I + Sinh[x])^2,x]

[Out]

-1/3*(I - Sinh[x])^3

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]

Rule 2746

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*} \text {integral}& = \text {Subst}\left (\int (i-x)^2 \, dx,x,\sinh (x)\right ) \\ & = -\frac {1}{3} (i-\sinh (x))^3 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.29 \[ \int \frac {\cosh ^5(x)}{(i+\sinh (x))^2} \, dx=\frac {1}{6} (-7+\cosh (2 x)-6 i \sinh (x)) \sinh (x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86

\[-\frac {\left (i-\sinh \left (x \right )\right )^{3}}{3}\]

[In]

int(cosh(x)^5/(I+sinh(x))^2,x)

[Out]

-1/3*(I-sinh(x))^3

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (8) = 16\).

Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.43 \[ \int \frac {\cosh ^5(x)}{(i+\sinh (x))^2} \, dx=\frac {1}{24} \, {\left (e^{\left (6 \, x\right )} - 6 i \, e^{\left (5 \, x\right )} - 15 \, e^{\left (4 \, x\right )} + 15 \, e^{\left (2 \, x\right )} - 6 i \, e^{x} - 1\right )} e^{\left (-3 \, x\right )} \]

[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)

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (8) = 16\).

Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 3.14 \[ \int \frac {\cosh ^5(x)}{(i+\sinh (x))^2} \, dx=\frac {e^{3 x}}{24} - \frac {i e^{2 x}}{4} - \frac {5 e^{x}}{8} + \frac {5 e^{- x}}{8} - \frac {i e^{- 2 x}}{4} - \frac {e^{- 3 x}}{24} \]

[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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (8) = 16\).

Time = 0.20 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.79 \[ \int \frac {\cosh ^5(x)}{(i+\sinh (x))^2} \, dx=-\frac {1}{24} \, {\left (6 i \, e^{\left (-x\right )} + 15 \, e^{\left (-2 \, x\right )} - 1\right )} e^{\left (3 \, x\right )} + \frac {5}{8} \, e^{\left (-x\right )} - \frac {1}{4} i \, e^{\left (-2 \, x\right )} - \frac {1}{24} \, e^{\left (-3 \, x\right )} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (8) = 16\).

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.50 \[ \int \frac {\cosh ^5(x)}{(i+\sinh (x))^2} \, dx=\frac {1}{24} \, {\left (15 \, e^{\left (2 \, x\right )} - 6 i \, e^{x} - 1\right )} e^{\left (-3 \, x\right )} + \frac {1}{24} \, e^{\left (3 \, x\right )} - \frac {1}{4} i \, e^{\left (2 \, x\right )} - \frac {5}{8} \, e^{x} \]

[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

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.64 \[ \int \frac {\cosh ^5(x)}{(i+\sinh (x))^2} \, dx=\frac {5\,{\mathrm {e}}^{-x}}{8}-\frac {{\mathrm {e}}^{-3\,x}}{24}+\frac {{\mathrm {e}}^{3\,x}}{24}-\frac {5\,{\mathrm {e}}^x}{8}-\frac {{\mathrm {e}}^{-2\,x}\,1{}\mathrm {i}}{4}-\frac {{\mathrm {e}}^{2\,x}\,1{}\mathrm {i}}{4} \]

[In]

int(cosh(x)^5/(sinh(x) + 1i)^2,x)

[Out]

(5*exp(-x))/8 - (exp(-2*x)*1i)/4 - (exp(2*x)*1i)/4 - exp(-3*x)/24 + exp(3*x)/24 - (5*exp(x))/8