3.40 \(\int \frac {\sinh ^4(x)}{i+\sinh (x)} \, dx\)

Optimal. Leaf size=46 \[ \frac {3 i x}{2}+\frac {4 \cosh ^3(x)}{3}-4 \cosh (x)-\frac {\sinh ^3(x) \cosh (x)}{\sinh (x)+i}-\frac {3}{2} i \sinh (x) \cosh (x) \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2767, 2748, 2635, 8, 2633} \[ \frac {3 i x}{2}+\frac {4 \cosh ^3(x)}{3}-4 \cosh (x)-\frac {\sinh ^3(x) \cosh (x)}{\sinh (x)+i}-\frac {3}{2} i \sinh (x) \cosh (x) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^4/(I + Sinh[x]),x]

[Out]

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

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 ^4(x)}{i+\sinh (x)} \, dx &=-\frac {\cosh (x) \sinh ^3(x)}{i+\sinh (x)}+\int \sinh ^2(x) (-3 i+4 \sinh (x)) \, dx\\ &=-\frac {\cosh (x) \sinh ^3(x)}{i+\sinh (x)}-3 i \int \sinh ^2(x) \, dx+4 \int \sinh ^3(x) \, dx\\ &=-\frac {3}{2} i \cosh (x) \sinh (x)-\frac {\cosh (x) \sinh ^3(x)}{i+\sinh (x)}+\frac {3}{2} i \int 1 \, dx-4 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh (x)\right )\\ &=\frac {3 i x}{2}-4 \cosh (x)+\frac {4 \cosh ^3(x)}{3}-\frac {3}{2} i \cosh (x) \sinh (x)-\frac {\cosh (x) \sinh ^3(x)}{i+\sinh (x)}\\ \end {align*}

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Mathematica [B]  time = 0.19, size = 134, normalized size = 2.91 \[ \frac {\cosh (x) \left (i \sinh ^{-1}(\sinh (x)) (\sinh (x)+i)+2 \sinh ^3(x) \sqrt {\cosh ^2(x)}-i \sinh ^2(x) \sqrt {\cosh ^2(x)}-\sinh (x) \left (7 \sqrt {\cosh ^2(x)}+16 \sin ^{-1}\left (\frac {\sqrt {1-i \sinh (x)}}{\sqrt {2}}\right )\right )-16 i \left (\sqrt {\cosh ^2(x)}+\sin ^{-1}\left (\frac {\sqrt {1-i \sinh (x)}}{\sqrt {2}}\right )\right )\right )}{6 (\sinh (x)+i) \sqrt {\cosh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^4/(I + Sinh[x]),x]

[Out]

(Cosh[x]*((-16*I)*(ArcSin[Sqrt[1 - I*Sinh[x]]/Sqrt[2]] + Sqrt[Cosh[x]^2]) - (16*ArcSin[Sqrt[1 - I*Sinh[x]]/Sqr
t[2]] + 7*Sqrt[Cosh[x]^2])*Sinh[x] - I*Sqrt[Cosh[x]^2]*Sinh[x]^2 + 2*Sqrt[Cosh[x]^2]*Sinh[x]^3 + I*ArcSinh[Sin
h[x]]*(I + Sinh[x])))/(6*Sqrt[Cosh[x]^2]*(I + Sinh[x]))

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fricas [A]  time = 0.62, size = 64, normalized size = 1.39 \[ \frac {{\left (36 i \, x - 21 i\right )} e^{\left (4 \, x\right )} - 3 \, {\left (12 \, x + 23\right )} e^{\left (3 \, x\right )} + e^{\left (7 \, x\right )} - 2 i \, e^{\left (6 \, x\right )} - 18 \, e^{\left (5 \, x\right )} - 18 i \, e^{\left (2 \, x\right )} - 2 \, e^{x} + i}{24 \, {\left (e^{\left (4 \, x\right )} + i \, e^{\left (3 \, x\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(I+sinh(x)),x, algorithm="fricas")

[Out]

1/24*((36*I*x - 21*I)*e^(4*x) - 3*(12*x + 23)*e^(3*x) + e^(7*x) - 2*I*e^(6*x) - 18*e^(5*x) - 18*I*e^(2*x) - 2*
e^x + I)/(e^(4*x) + I*e^(3*x))

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giac [A]  time = 0.20, size = 50, normalized size = 1.09 \[ \frac {3}{2} i \, x - \frac {{\left (69 \, e^{\left (3 \, x\right )} + 18 i \, e^{\left (2 \, x\right )} + 2 \, e^{x} - i\right )} e^{\left (-3 \, x\right )}}{24 \, {\left (e^{x} + i\right )}} + \frac {1}{24} \, e^{\left (3 \, x\right )} - \frac {1}{8} i \, e^{\left (2 \, x\right )} - \frac {7}{8} \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(I+sinh(x)),x, algorithm="giac")

[Out]

3/2*I*x - 1/24*(69*e^(3*x) + 18*I*e^(2*x) + 2*e^x - I)*e^(-3*x)/(e^x + I) + 1/24*e^(3*x) - 1/8*I*e^(2*x) - 7/8
*e^x

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maple [B]  time = 0.06, size = 138, normalized size = 3.00 \[ -\frac {3 i \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}+\frac {3}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {i}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {i}{2 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {1}{3 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}+\frac {3 i \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {i}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {i}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {1}{3 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2 i}{\tanh \left (\frac {x}{2}\right )+i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(I+sinh(x)),x)

[Out]

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

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maxima [A]  time = 0.32, size = 59, normalized size = 1.28 \[ \frac {3}{2} i \, x - \frac {4 \, e^{\left (-x\right )} - 36 i \, e^{\left (-2 \, x\right )} + 138 \, e^{\left (-3 \, x\right )} + 2 i}{16 \, {\left (-3 i \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-4 \, x\right )}\right )}} - \frac {7}{8} \, e^{\left (-x\right )} + \frac {1}{8} i \, e^{\left (-2 \, x\right )} + \frac {1}{24} \, e^{\left (-3 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(I+sinh(x)),x, algorithm="maxima")

[Out]

3/2*I*x - 1/16*(4*e^(-x) - 36*I*e^(-2*x) + 138*e^(-3*x) + 2*I)/(-3*I*e^(-3*x) + 3*e^(-4*x)) - 7/8*e^(-x) + 1/8
*I*e^(-2*x) + 1/24*e^(-3*x)

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mupad [B]  time = 0.49, size = 50, normalized size = 1.09 \[ \frac {x\,3{}\mathrm {i}}{2}-\frac {7\,{\mathrm {e}}^{-x}}{8}+\frac {{\mathrm {e}}^{-2\,x}\,1{}\mathrm {i}}{8}-\frac {{\mathrm {e}}^{2\,x}\,1{}\mathrm {i}}{8}+\frac {{\mathrm {e}}^{-3\,x}}{24}+\frac {{\mathrm {e}}^{3\,x}}{24}-\frac {7\,{\mathrm {e}}^x}{8}-\frac {2}{{\mathrm {e}}^x+1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(sinh(x) + 1i),x)

[Out]

(x*3i)/2 - (7*exp(-x))/8 + (exp(-2*x)*1i)/8 - (exp(2*x)*1i)/8 + exp(-3*x)/24 + exp(3*x)/24 - (7*exp(x))/8 - 2/
(exp(x) + 1i)

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sympy [A]  time = 0.20, size = 60, normalized size = 1.30 \[ \frac {3 i x}{2} + \frac {e^{3 x}}{24} - \frac {i e^{2 x}}{8} - \frac {7 e^{x}}{8} - \frac {7 e^{- x}}{8} + \frac {i e^{- 2 x}}{8} + \frac {e^{- 3 x}}{24} + \frac {2}{- e^{x} - i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**4/(I+sinh(x)),x)

[Out]

3*I*x/2 + exp(3*x)/24 - I*exp(2*x)/8 - 7*exp(x)/8 - 7*exp(-x)/8 + I*exp(-2*x)/8 + exp(-3*x)/24 + 2/(-exp(x) -
I)

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