3.171 \(\int \frac {\cosh ^6(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=40 \[ -\frac {5 x}{8}-\frac {5}{12} i \cosh ^3(x)+\frac {\cosh ^5(x)}{4 (\sinh (x)+i)}-\frac {5}{8} \sinh (x) \cosh (x) \]

[Out]

-5/8*x-5/12*I*cosh(x)^3-5/8*cosh(x)*sinh(x)+1/4*cosh(x)^5/(I+sinh(x))

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2679, 2682, 2635, 8} \[ -\frac {5 x}{8}-\frac {5}{12} i \cosh ^3(x)+\frac {\cosh ^5(x)}{4 (\sinh (x)+i)}-\frac {5}{8} \sinh (x) \cosh (x) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^6/(I + Sinh[x])^2,x]

[Out]

(-5*x)/8 - ((5*I)/12)*Cosh[x]^3 - (5*Cosh[x]*Sinh[x])/8 + Cosh[x]^5/(4*(I + Sinh[x]))

Rule 8

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

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 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 ^6(x)}{(i+\sinh (x))^2} \, dx &=\frac {\cosh ^5(x)}{4 (i+\sinh (x))}-\frac {5}{4} i \int \frac {\cosh ^4(x)}{i+\sinh (x)} \, dx\\ &=-\frac {5}{12} i \cosh ^3(x)+\frac {\cosh ^5(x)}{4 (i+\sinh (x))}-\frac {5}{4} \int \cosh ^2(x) \, dx\\ &=-\frac {5}{12} i \cosh ^3(x)-\frac {5}{8} \cosh (x) \sinh (x)+\frac {\cosh ^5(x)}{4 (i+\sinh (x))}-\frac {5 \int 1 \, dx}{8}\\ &=-\frac {5 x}{8}-\frac {5}{12} i \cosh ^3(x)-\frac {5}{8} \cosh (x) \sinh (x)+\frac {\cosh ^5(x)}{4 (i+\sinh (x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.19, size = 121, normalized size = 3.02 \[ -\frac {i \cosh ^7(x) \left (6 \sinh ^4(x)-10 i \sinh ^3(x)+7 \sinh ^2(x)-25 i \sinh (x)+\frac {30 \sqrt {1-i \sinh (x)} \sin ^{-1}\left (\frac {\sqrt {1-i \sinh (x)}}{\sqrt {2}}\right )}{\sqrt {1+i \sinh (x)}}+16\right )}{24 \left (\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )\right )^8 \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^6/(I + Sinh[x])^2,x]

[Out]

((-1/24*I)*Cosh[x]^7*(16 + (30*ArcSin[Sqrt[1 - I*Sinh[x]]/Sqrt[2]]*Sqrt[1 - I*Sinh[x]])/Sqrt[1 + I*Sinh[x]] -
(25*I)*Sinh[x] + 7*Sinh[x]^2 - (10*I)*Sinh[x]^3 + 6*Sinh[x]^4))/((Cosh[x/2] - I*Sinh[x/2])^8*(Cosh[x/2] + I*Si
nh[x/2])^6)

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 55, normalized size = 1.38 \[ -\frac {1}{192} \, {\left (120 \, x e^{\left (4 \, x\right )} - 3 \, e^{\left (8 \, x\right )} + 16 i \, e^{\left (7 \, x\right )} + 24 \, e^{\left (6 \, x\right )} + 48 i \, e^{\left (5 \, x\right )} + 48 i \, e^{\left (3 \, x\right )} - 24 \, e^{\left (2 \, x\right )} + 16 i \, e^{x} + 3\right )} e^{\left (-4 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^6/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

-1/192*(120*x*e^(4*x) - 3*e^(8*x) + 16*I*e^(7*x) + 24*e^(6*x) + 48*I*e^(5*x) + 48*I*e^(3*x) - 24*e^(2*x) + 16*
I*e^x + 3)*e^(-4*x)

________________________________________________________________________________________

giac [A]  time = 0.18, size = 50, normalized size = 1.25 \[ -\frac {1}{192} \, {\left (48 i \, e^{\left (3 \, x\right )} - 24 \, e^{\left (2 \, x\right )} + 16 i \, e^{x} + 3\right )} e^{\left (-4 \, x\right )} - \frac {5}{8} \, x + \frac {1}{64} \, e^{\left (4 \, x\right )} - \frac {1}{12} i \, e^{\left (3 \, x\right )} - \frac {1}{8} \, e^{\left (2 \, x\right )} - \frac {1}{4} i \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^6/(I+sinh(x))^2,x, algorithm="giac")

[Out]

-1/192*(48*I*e^(3*x) - 24*e^(2*x) + 16*I*e^x + 3)*e^(-4*x) - 5/8*x + 1/64*e^(4*x) - 1/12*I*e^(3*x) - 1/8*e^(2*
x) - 1/4*I*e^x

________________________________________________________________________________________

maple [B]  time = 0.07, size = 166, normalized size = 4.15 \[ \frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {i}{\tanh \left (\frac {x}{2}\right )+1}-\frac {1}{8 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {i}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3}{8 \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {2 i}{3 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {5 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8}+\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {i}{\left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{8 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {2 i}{3 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {3}{8 \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {i}{\tanh \left (\frac {x}{2}\right )-1}-\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}-\frac {5 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^6/(I+sinh(x))^2,x)

[Out]

1/2/(tanh(1/2*x)-1)^3-I/(tanh(1/2*x)+1)-1/8/(tanh(1/2*x)-1)^2+I/(tanh(1/2*x)+1)^2-3/8/(tanh(1/2*x)-1)-2/3*I/(t
anh(1/2*x)+1)^3+1/4/(tanh(1/2*x)-1)^4+5/8*ln(tanh(1/2*x)-1)+1/2/(tanh(1/2*x)+1)^3+I/(tanh(1/2*x)-1)^2+1/8/(tan
h(1/2*x)+1)^2+2/3*I/(tanh(1/2*x)-1)^3-3/8/(tanh(1/2*x)+1)+I/(tanh(1/2*x)-1)-1/4/(tanh(1/2*x)+1)^4-5/8*ln(tanh(
1/2*x)+1)

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 54, normalized size = 1.35 \[ -\frac {1}{192} \, {\left (16 i \, e^{\left (-x\right )} + 24 \, e^{\left (-2 \, x\right )} + 48 i \, e^{\left (-3 \, x\right )} - 3\right )} e^{\left (4 \, x\right )} - \frac {5}{8} \, x - \frac {1}{4} i \, e^{\left (-x\right )} + \frac {1}{8} \, e^{\left (-2 \, x\right )} - \frac {1}{12} i \, e^{\left (-3 \, x\right )} - \frac {1}{64} \, e^{\left (-4 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^6/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

-1/192*(16*I*e^(-x) + 24*e^(-2*x) + 48*I*e^(-3*x) - 3)*e^(4*x) - 5/8*x - 1/4*I*e^(-x) + 1/8*e^(-2*x) - 1/12*I*
e^(-3*x) - 1/64*e^(-4*x)

________________________________________________________________________________________

mupad [B]  time = 0.14, size = 54, normalized size = 1.35 \[ \frac {{\mathrm {e}}^{-2\,x}}{8}-\frac {{\mathrm {e}}^{-x}\,1{}\mathrm {i}}{4}-\frac {5\,x}{8}-\frac {{\mathrm {e}}^{2\,x}}{8}-\frac {{\mathrm {e}}^{-3\,x}\,1{}\mathrm {i}}{12}-\frac {{\mathrm {e}}^{3\,x}\,1{}\mathrm {i}}{12}-\frac {{\mathrm {e}}^{-4\,x}}{64}+\frac {{\mathrm {e}}^{4\,x}}{64}-\frac {{\mathrm {e}}^x\,1{}\mathrm {i}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^6/(sinh(x) + 1i)^2,x)

[Out]

exp(-2*x)/8 - (exp(-x)*1i)/4 - (5*x)/8 - exp(2*x)/8 - (exp(-3*x)*1i)/12 - (exp(3*x)*1i)/12 - exp(-4*x)/64 + ex
p(4*x)/64 - (exp(x)*1i)/4

________________________________________________________________________________________

sympy [A]  time = 0.24, size = 65, normalized size = 1.62 \[ - \frac {5 x}{8} + \frac {e^{4 x}}{64} - \frac {i e^{3 x}}{12} - \frac {e^{2 x}}{8} - \frac {i e^{x}}{4} - \frac {i e^{- x}}{4} + \frac {e^{- 2 x}}{8} - \frac {i e^{- 3 x}}{12} - \frac {e^{- 4 x}}{64} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**6/(I+sinh(x))**2,x)

[Out]

-5*x/8 + exp(4*x)/64 - I*exp(3*x)/12 - exp(2*x)/8 - I*exp(x)/4 - I*exp(-x)/4 + exp(-2*x)/8 - I*exp(-3*x)/12 -
exp(-4*x)/64

________________________________________________________________________________________