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

Optimal. Leaf size=38 \[ -\frac {3 i x}{8}+\frac {\cosh ^5(x)}{5}-\frac {1}{4} i \sinh (x) \cosh ^3(x)-\frac {3}{8} i \sinh (x) \cosh (x) \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2682, 2635, 8} \[ -\frac {3 i x}{8}+\frac {\cosh ^5(x)}{5}-\frac {1}{4} i \sinh (x) \cosh ^3(x)-\frac {3}{8} i \sinh (x) \cosh (x) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^6/(I + Sinh[x]),x]

[Out]

((-3*I)/8)*x + Cosh[x]^5/5 - ((3*I)/8)*Cosh[x]*Sinh[x] - (I/4)*Cosh[x]^3*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 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)} \, dx &=\frac {\cosh ^5(x)}{5}-i \int \cosh ^4(x) \, dx\\ &=\frac {\cosh ^5(x)}{5}-\frac {1}{4} i \cosh ^3(x) \sinh (x)-\frac {3}{4} i \int \cosh ^2(x) \, dx\\ &=\frac {\cosh ^5(x)}{5}-\frac {3}{8} i \cosh (x) \sinh (x)-\frac {1}{4} i \cosh ^3(x) \sinh (x)-\frac {3}{8} i \int 1 \, dx\\ &=-\frac {3 i x}{8}+\frac {\cosh ^5(x)}{5}-\frac {3}{8} i \cosh (x) \sinh (x)-\frac {1}{4} i \cosh ^3(x) \sinh (x)\\ \end {align*}

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Mathematica [B]  time = 0.24, size = 131, normalized size = 3.45 \[ -\frac {i \cosh ^7(x) \left (8 \sinh ^5(x)-2 i \sinh ^4(x)+26 \sinh ^3(x)-9 i \sinh ^2(x)+33 \sinh (x)+\frac {30 i \sqrt {1-i \sinh (x)} \sin ^{-1}\left (\frac {\sqrt {1-i \sinh (x)}}{\sqrt {2}}\right )}{\sqrt {1+i \sinh (x)}}+8 i\right )}{40 \left (\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )\right )^8 \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^6/(I + Sinh[x]),x]

[Out]

((-1/40*I)*Cosh[x]^7*(8*I + ((30*I)*ArcSin[Sqrt[1 - I*Sinh[x]]/Sqrt[2]]*Sqrt[1 - I*Sinh[x]])/Sqrt[1 + I*Sinh[x
]] + 33*Sinh[x] - (9*I)*Sinh[x]^2 + 26*Sinh[x]^3 - (2*I)*Sinh[x]^4 + 8*Sinh[x]^5))/((Cosh[x/2] - I*Sinh[x/2])^
8*(Cosh[x/2] + I*Sinh[x/2])^6)

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fricas [B]  time = 0.68, size = 67, normalized size = 1.76 \[ \frac {1}{320} \, {\left (-120 i \, x e^{\left (5 \, x\right )} + 2 \, e^{\left (10 \, x\right )} - 5 i \, e^{\left (9 \, x\right )} + 10 \, e^{\left (8 \, x\right )} - 40 i \, e^{\left (7 \, x\right )} + 20 \, e^{\left (6 \, x\right )} + 20 \, e^{\left (4 \, x\right )} + 40 i \, e^{\left (3 \, x\right )} + 10 \, e^{\left (2 \, x\right )} + 5 i \, e^{x} + 2\right )} e^{\left (-5 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/320*(-120*I*x*e^(5*x) + 2*e^(10*x) - 5*I*e^(9*x) + 10*e^(8*x) - 40*I*e^(7*x) + 20*e^(6*x) + 20*e^(4*x) + 40*
I*e^(3*x) + 10*e^(2*x) + 5*I*e^x + 2)*e^(-5*x)

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giac [B]  time = 0.23, size = 62, normalized size = 1.63 \[ \frac {1}{320} \, {\left (20 \, e^{\left (4 \, x\right )} + 40 i \, e^{\left (3 \, x\right )} + 10 \, e^{\left (2 \, x\right )} + 5 i \, e^{x} + 2\right )} e^{\left (-5 \, x\right )} - \frac {3}{8} i \, x + \frac {1}{160} \, e^{\left (5 \, x\right )} - \frac {1}{64} i \, e^{\left (4 \, x\right )} + \frac {1}{32} \, e^{\left (3 \, x\right )} - \frac {1}{8} i \, e^{\left (2 \, x\right )} + \frac {1}{16} \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/320*(20*e^(4*x) + 40*I*e^(3*x) + 10*e^(2*x) + 5*I*e^x + 2)*e^(-5*x) - 3/8*I*x + 1/160*e^(5*x) - 1/64*I*e^(4*
x) + 1/32*e^(3*x) - 1/8*I*e^(2*x) + 1/16*e^x

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maple [B]  time = 0.06, size = 210, normalized size = 5.53 \[ -\frac {i}{4 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}-\frac {i}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {3}{8 \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {3 i \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8}-\frac {5}{8 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {7 i}{8 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {3}{4 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {5 i}{8 \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {1}{5 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{5}}+\frac {7 i}{8 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {3 i \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8}+\frac {3}{4 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {i}{4 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {3}{8 \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {i}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {5}{8 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {5 i}{8 \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {1}{5 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I/(tanh(1/2*x)-1)^4-1/2/(tanh(1/2*x)-1)^4-1/2*I/(tanh(1/2*x)+1)^3-3/8/(tanh(1/2*x)-1)-3/8*I*ln(tanh(1/2*x
)+1)-5/8/(tanh(1/2*x)-1)^2-7/8*I/(tanh(1/2*x)-1)^2-3/4/(tanh(1/2*x)-1)^3-5/8*I/(tanh(1/2*x)+1)-1/5/(tanh(1/2*x
)-1)^5+7/8*I/(tanh(1/2*x)+1)^2-1/2/(tanh(1/2*x)+1)^4+3/8*I*ln(tanh(1/2*x)-1)+3/4/(tanh(1/2*x)+1)^3+1/4*I/(tanh
(1/2*x)+1)^4+3/8/(tanh(1/2*x)+1)-1/2*I/(tanh(1/2*x)-1)^3-5/8/(tanh(1/2*x)+1)^2-5/8*I/(tanh(1/2*x)-1)+1/5/(tanh
(1/2*x)+1)^5

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maxima [B]  time = 0.32, size = 66, normalized size = 1.74 \[ -\frac {1}{320} \, {\left (5 i \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 40 i \, e^{\left (-3 \, x\right )} - 20 \, e^{\left (-4 \, x\right )} - 2\right )} e^{\left (5 \, x\right )} - \frac {3}{8} i \, x + \frac {1}{16} \, e^{\left (-x\right )} + \frac {1}{8} i \, e^{\left (-2 \, x\right )} + \frac {1}{32} \, e^{\left (-3 \, x\right )} + \frac {1}{64} i \, e^{\left (-4 \, x\right )} + \frac {1}{160} \, e^{\left (-5 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/320*(5*I*e^(-x) - 10*e^(-2*x) + 40*I*e^(-3*x) - 20*e^(-4*x) - 2)*e^(5*x) - 3/8*I*x + 1/16*e^(-x) + 1/8*I*e^
(-2*x) + 1/32*e^(-3*x) + 1/64*I*e^(-4*x) + 1/160*e^(-5*x)

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mupad [B]  time = 0.62, size = 67, normalized size = 1.76 \[ \frac {{\mathrm {e}}^{-x}}{16}+\frac {{\mathrm {e}}^{-3\,x}}{32}+\frac {{\mathrm {e}}^{3\,x}}{32}+\frac {{\mathrm {e}}^{-5\,x}}{160}+\frac {{\mathrm {e}}^{5\,x}}{160}+\frac {{\mathrm {e}}^x}{16}-\frac {x\,3{}\mathrm {i}}{8}+\frac {{\mathrm {e}}^{-2\,x}\,1{}\mathrm {i}}{8}-\frac {{\mathrm {e}}^{2\,x}\,1{}\mathrm {i}}{8}+\frac {{\mathrm {e}}^{-4\,x}\,1{}\mathrm {i}}{64}-\frac {{\mathrm {e}}^{4\,x}\,1{}\mathrm {i}}{64} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^6/(sinh(x) + 1i),x)

[Out]

exp(-x)/16 - (x*3i)/8 + (exp(-2*x)*1i)/8 - (exp(2*x)*1i)/8 + exp(-3*x)/32 + exp(3*x)/32 + (exp(-4*x)*1i)/64 -
(exp(4*x)*1i)/64 + exp(-5*x)/160 + exp(5*x)/160 + exp(x)/16

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sympy [B]  time = 0.23, size = 82, normalized size = 2.16 \[ - \frac {3 i x}{8} + \frac {e^{5 x}}{160} - \frac {i e^{4 x}}{64} + \frac {e^{3 x}}{32} - \frac {i e^{2 x}}{8} + \frac {e^{x}}{16} + \frac {e^{- x}}{16} + \frac {i e^{- 2 x}}{8} + \frac {e^{- 3 x}}{32} + \frac {i e^{- 4 x}}{64} + \frac {e^{- 5 x}}{160} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*I*x/8 + exp(5*x)/160 - I*exp(4*x)/64 + exp(3*x)/32 - I*exp(2*x)/8 + exp(x)/16 + exp(-x)/16 + I*exp(-2*x)/8
+ exp(-3*x)/32 + I*exp(-4*x)/64 + exp(-5*x)/160

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