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)+1/2*sinh(x)^2-1/3*I*sinh(x)^3+1/4*sinh(x)^4

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Rubi [A]  time = 0.04, antiderivative size = 33, normalized size of antiderivative = 1.00, 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 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])

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])

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*}

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Mathematica [A]  time = 0.02, 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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fricas [B]  time = 0.52, size = 48, normalized size = 1.45 \[ \frac {1}{192} \, {\left (3 \, e^{\left (8 \, x\right )} - 8 i \, e^{\left (7 \, x\right )} + 12 \, e^{\left (6 \, x\right )} - 72 i \, e^{\left (5 \, x\right )} + 72 i \, e^{\left (3 \, x\right )} + 12 \, e^{\left (2 \, x\right )} + 8 i \, e^{x} + 3\right )} e^{\left (-4 \, x\right )} \]

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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giac [B]  time = 0.36, size = 47, normalized size = 1.42 \[ -\frac {1}{192} \, {\left (-72 i \, e^{\left (3 \, x\right )} - 12 \, e^{\left (2 \, x\right )} - 8 i \, e^{x} - 3\right )} e^{\left (-4 \, x\right )} + \frac {1}{64} \, e^{\left (4 \, x\right )} - \frac {1}{24} i \, e^{\left (3 \, x\right )} + \frac {1}{16} \, e^{\left (2 \, x\right )} - \frac {3}{8} i \, e^{x} \]

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

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maple [B]  time = 0.06, size = 94, normalized size = 2.85 \[ \frac {\frac {5}{8}+\frac {i}{2}}{\left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {\frac {1}{2}+\frac {i}{3}}{\left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}+\frac {\frac {3}{8}+i}{\tanh \left (\frac {x}{2}\right )-1}+\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {\frac {5}{8}-\frac {i}{2}}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {-\frac {1}{2}+\frac {i}{3}}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {-\frac {3}{8}+i}{\tanh \left (\frac {x}{2}\right )+1}+\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}} \]

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 = 0.31, size = 51, normalized size = 1.55 \[ -\frac {1}{192} \, {\left (8 i \, e^{\left (-x\right )} - 12 \, e^{\left (-2 \, x\right )} + 72 i \, e^{\left (-3 \, x\right )} - 3\right )} e^{\left (4 \, x\right )} + \frac {3}{8} i \, e^{\left (-x\right )} + \frac {1}{16} \, e^{\left (-2 \, x\right )} + \frac {1}{24} i \, e^{\left (-3 \, x\right )} + \frac {1}{64} \, e^{\left (-4 \, x\right )} \]

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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mupad [B]  time = 0.56, size = 51, normalized size = 1.55 \[ \frac {{\mathrm {e}}^{-x}\,3{}\mathrm {i}}{8}+\frac {{\mathrm {e}}^{-2\,x}}{16}+\frac {{\mathrm {e}}^{2\,x}}{16}+\frac {{\mathrm {e}}^{-3\,x}\,1{}\mathrm {i}}{24}-\frac {{\mathrm {e}}^{3\,x}\,1{}\mathrm {i}}{24}+\frac {{\mathrm {e}}^{-4\,x}}{64}+\frac {{\mathrm {e}}^{4\,x}}{64}-\frac {{\mathrm {e}}^x\,3{}\mathrm {i}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^5/(sinh(x) + 1i),x)

[Out]

(exp(-x)*3i)/8 + exp(-2*x)/16 + exp(2*x)/16 + (exp(-3*x)*1i)/24 - (exp(3*x)*1i)/24 + exp(-4*x)/64 + exp(4*x)/6
4 - (exp(x)*3i)/8

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sympy [B]  time = 0.20, size = 63, normalized size = 1.91 \[ \frac {e^{4 x}}{64} - \frac {i e^{3 x}}{24} + \frac {e^{2 x}}{16} - \frac {3 i e^{x}}{8} + \frac {3 i e^{- x}}{8} + \frac {e^{- 2 x}}{16} + \frac {i e^{- 3 x}}{24} + \frac {e^{- 4 x}}{64} \]

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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