3.41 \(\int \frac {\sinh ^3(x)}{i+\sinh (x)} \, dx\)

Optimal. Leaf size=36 \[ -\frac {3 x}{2}-2 i \cosh (x)-\frac {\sinh ^2(x) \cosh (x)}{\sinh (x)+i}+\frac {3}{2} \sinh (x) \cosh (x) \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2767, 2734} \[ -\frac {3 x}{2}-2 i \cosh (x)-\frac {\sinh ^2(x) \cosh (x)}{\sinh (x)+i}+\frac {3}{2} \sinh (x) \cosh (x) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^3/(I + Sinh[x]),x]

[Out]

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

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2767

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n - 1))/(a*f*(a + b*Sin[e + f*x])), x] - Dist[d/(a*b), Int[(c +
d*Sin[e + f*x])^(n - 2)*Simp[b*d*(n - 1) - a*c*n + (b*c*(n - 1) - a*d*n)*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 1] && (IntegerQ[2
*n] || EqQ[c, 0])

Rubi steps

\begin {align*} \int \frac {\sinh ^3(x)}{i+\sinh (x)} \, dx &=-\frac {\cosh (x) \sinh ^2(x)}{i+\sinh (x)}+\int \sinh (x) (-2 i+3 \sinh (x)) \, dx\\ &=-\frac {3 x}{2}-2 i \cosh (x)+\frac {3}{2} \cosh (x) \sinh (x)-\frac {\cosh (x) \sinh ^2(x)}{i+\sinh (x)}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 41, normalized size = 1.14 \[ \frac {1}{2} \cosh (x) \left (-\frac {3 \sinh ^{-1}(\sinh (x))}{\sqrt {\cosh ^2(x)}}+\frac {\sinh ^2(x)-i \sinh (x)+4}{\sinh (x)+i}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^3/(I + Sinh[x]),x]

[Out]

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

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fricas [B]  time = 0.61, size = 57, normalized size = 1.58 \[ -\frac {4 \, {\left (3 \, x - 1\right )} e^{\left (3 \, x\right )} - {\left (-12 i \, x - 20 i\right )} e^{\left (2 \, x\right )} - e^{\left (5 \, x\right )} + 3 i \, e^{\left (4 \, x\right )} - 3 \, e^{x} + i}{8 \, e^{\left (3 \, x\right )} + 8 i \, e^{\left (2 \, x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(I+sinh(x)),x, algorithm="fricas")

[Out]

-(4*(3*x - 1)*e^(3*x) - (-12*I*x - 20*I)*e^(2*x) - e^(5*x) + 3*I*e^(4*x) - 3*e^x + I)/(8*e^(3*x) + 8*I*e^(2*x)
)

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giac [A]  time = 0.25, size = 38, normalized size = 1.06 \[ -\frac {3}{2} \, x - \frac {{\left (20 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} + i\right )} e^{\left (-2 \, x\right )}}{8 \, {\left (e^{x} + i\right )}} + \frac {1}{8} \, e^{\left (2 \, x\right )} - \frac {1}{2} i \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(I+sinh(x)),x, algorithm="giac")

[Out]

-3/2*x - 1/8*(20*I*e^(2*x) - 3*e^x + I)*e^(-2*x)/(e^x + I) + 1/8*e^(2*x) - 1/2*I*e^x

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maple [B]  time = 0.05, size = 93, normalized size = 2.58 \[ \frac {1}{2 \tanh \left (\frac {x}{2}\right )-2}+\frac {i}{\tanh \left (\frac {x}{2}\right )-1}+\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}+\frac {1}{2 \tanh \left (\frac {x}{2}\right )+2}-\frac {i}{\tanh \left (\frac {x}{2}\right )+1}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2}+\frac {2}{\tanh \left (\frac {x}{2}\right )+i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(I+sinh(x)),x)

[Out]

1/2/(tanh(1/2*x)-1)+I/(tanh(1/2*x)-1)+1/2/(tanh(1/2*x)-1)^2+3/2*ln(tanh(1/2*x)-1)+1/2/(tanh(1/2*x)+1)-I/(tanh(
1/2*x)+1)-1/2/(tanh(1/2*x)+1)^2-3/2*ln(tanh(1/2*x)+1)+2/(tanh(1/2*x)+I)

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maxima [A]  time = 0.44, size = 45, normalized size = 1.25 \[ -\frac {3}{2} \, x - \frac {3 \, e^{\left (-x\right )} + 20 i \, e^{\left (-2 \, x\right )} + i}{8 \, {\left (-i \, e^{\left (-2 \, x\right )} + e^{\left (-3 \, x\right )}\right )}} - \frac {1}{2} i \, e^{\left (-x\right )} - \frac {1}{8} \, e^{\left (-2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(I+sinh(x)),x, algorithm="maxima")

[Out]

-3/2*x - 1/8*(3*e^(-x) + 20*I*e^(-2*x) + I)/(-I*e^(-2*x) + e^(-3*x)) - 1/2*I*e^(-x) - 1/8*e^(-2*x)

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mupad [B]  time = 0.43, size = 38, normalized size = 1.06 \[ \frac {{\mathrm {e}}^{2\,x}}{8}-\frac {{\mathrm {e}}^{-x}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{-2\,x}}{8}-\frac {3\,x}{2}-\frac {{\mathrm {e}}^x\,1{}\mathrm {i}}{2}-\frac {2{}\mathrm {i}}{{\mathrm {e}}^x+1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(sinh(x) + 1i),x)

[Out]

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

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sympy [A]  time = 0.17, size = 41, normalized size = 1.14 \[ - \frac {3 x}{2} + \frac {e^{2 x}}{8} - \frac {i e^{x}}{2} - \frac {i e^{- x}}{2} - \frac {e^{- 2 x}}{8} + \frac {2}{i e^{x} - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**3/(I+sinh(x)),x)

[Out]

-3*x/2 + exp(2*x)/8 - I*exp(x)/2 - I*exp(-x)/2 - exp(-2*x)/8 + 2/(I*exp(x) - 1)

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