3.48 \(\int \frac{\sinh ^4(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=58 \[ -\frac{7 x}{2}-\frac{16}{3} i \cosh (x)-\frac{\sinh ^3(x) \cosh (x)}{3 (\sinh (x)+i)^2}-\frac{8 \sinh ^2(x) \cosh (x)}{3 (\sinh (x)+i)}+\frac{7}{2} \sinh (x) \cosh (x) \]

[Out]

(-7*x)/2 - ((16*I)/3)*Cosh[x] + (7*Cosh[x]*Sinh[x])/2 - (Cosh[x]*Sinh[x]^3)/(3*(I + Sinh[x])^2) - (8*Cosh[x]*S
inh[x]^2)/(3*(I + Sinh[x]))

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Rubi [A]  time = 0.101936, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2765, 2977, 2734} \[ -\frac{7 x}{2}-\frac{16}{3} i \cosh (x)-\frac{\sinh ^3(x) \cosh (x)}{3 (\sinh (x)+i)^2}-\frac{8 \sinh ^2(x) \cosh (x)}{3 (\sinh (x)+i)}+\frac{7}{2} \sinh (x) \cosh (x) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^4/(I + Sinh[x])^2,x]

[Out]

(-7*x)/2 - ((16*I)/3)*Cosh[x] + (7*Cosh[x]*Sinh[x])/2 - (Cosh[x]*Sinh[x]^3)/(3*(I + Sinh[x])^2) - (8*Cosh[x]*S
inh[x]^2)/(3*(I + Sinh[x]))

Rule 2765

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

Rule 2977

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

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]

Rubi steps

\begin{align*} \int \frac{\sinh ^4(x)}{(i+\sinh (x))^2} \, dx &=-\frac{\cosh (x) \sinh ^3(x)}{3 (i+\sinh (x))^2}+\frac{1}{3} \int \frac{\sinh ^2(x) (-3 i+5 \sinh (x))}{i+\sinh (x)} \, dx\\ &=-\frac{\cosh (x) \sinh ^3(x)}{3 (i+\sinh (x))^2}-\frac{8 \cosh (x) \sinh ^2(x)}{3 (i+\sinh (x))}-\frac{1}{3} i \int (16+21 i \sinh (x)) \sinh (x) \, dx\\ &=-\frac{7 x}{2}-\frac{16}{3} i \cosh (x)+\frac{7}{2} \cosh (x) \sinh (x)-\frac{\cosh (x) \sinh ^3(x)}{3 (i+\sinh (x))^2}-\frac{8 \cosh (x) \sinh ^2(x)}{3 (i+\sinh (x))}\\ \end{align*}

Mathematica [B]  time = 0.185145, size = 147, normalized size = 2.53 \[ -\frac{\sinh ^3(x) \cosh (x)}{2 (1-i \sinh (x))^2}-\frac{i \sqrt{2} \sqrt{1+\frac{1}{2} (-1+i \sinh (x))} \cosh (x)}{\sqrt{1+i \sinh (x)}}-\frac{31 i \cosh (x)}{6 (1-i \sinh (x))}+\frac{5 i \cosh (x)}{6 (1-i \sinh (x))^2}-\frac{7 i \cosh (x) \sin ^{-1}\left (\frac{\sqrt{1-i \sinh (x)}}{\sqrt{2}}\right )}{\sqrt{1-i \sinh (x)} \sqrt{1+i \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^4/(I + Sinh[x])^2,x]

[Out]

(((5*I)/6)*Cosh[x])/(1 - I*Sinh[x])^2 - (((31*I)/6)*Cosh[x])/(1 - I*Sinh[x]) - (I*Sqrt[2]*Cosh[x]*Sqrt[1 + (-1
+ I*Sinh[x])/2])/Sqrt[1 + I*Sinh[x]] - ((7*I)*ArcSin[Sqrt[1 - I*Sinh[x]]/Sqrt[2]]*Cosh[x])/(Sqrt[1 - I*Sinh[x
]]*Sqrt[1 + I*Sinh[x]]) - (Cosh[x]*Sinh[x]^3)/(2*(1 - I*Sinh[x])^2)

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Maple [B]  time = 0.055, size = 116, normalized size = 2. \begin{align*}{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{2\,i \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}-{\frac{7}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{2\,i \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{\frac{7}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+{2\,i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}}+{\frac{4}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}+6\, \left ( \tanh \left ( x/2 \right ) +i \right ) ^{-1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(I+sinh(x))^2,x)

[Out]

1/2/(tanh(1/2*x)+1)-2*I/(tanh(1/2*x)+1)-1/2/(tanh(1/2*x)+1)^2-7/2*ln(tanh(1/2*x)+1)+1/2/(tanh(1/2*x)-1)+2*I/(t
anh(1/2*x)-1)+1/2/(tanh(1/2*x)-1)^2+7/2*ln(tanh(1/2*x)-1)+2*I/(tanh(1/2*x)+I)^2+4/3/(tanh(1/2*x)+I)^3+6/(tanh(
1/2*x)+I)

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Maxima [A]  time = 1.23467, size = 96, normalized size = 1.66 \begin{align*} -\frac{7}{2} \, x + \frac{30 \, e^{\left (-x\right )} + 478 i \, e^{\left (-2 \, x\right )} - 810 \, e^{\left (-3 \, x\right )} - 432 i \, e^{\left (-4 \, x\right )} + 6 i}{16 \,{\left (3 i \, e^{\left (-2 \, x\right )} - 9 \, e^{\left (-3 \, x\right )} - 9 i \, e^{\left (-4 \, x\right )} + 3 \, e^{\left (-5 \, x\right )}\right )}} - i \, e^{\left (-x\right )} - \frac{1}{8} \, e^{\left (-2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

-7/2*x + 1/16*(30*e^(-x) + 478*I*e^(-2*x) - 810*e^(-3*x) - 432*I*e^(-4*x) + 6*I)/(3*I*e^(-2*x) - 9*e^(-3*x) -
9*I*e^(-4*x) + 3*e^(-5*x)) - I*e^(-x) - 1/8*e^(-2*x)

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Fricas [B]  time = 2.03498, size = 275, normalized size = 4.74 \begin{align*} -\frac{21 \,{\left (4 \, x - 3\right )} e^{\left (5 \, x\right )} -{\left (-252 i \, x - 147 i\right )} e^{\left (4 \, x\right )} - 3 \,{\left (84 \, x + 127\right )} e^{\left (3 \, x\right )} -{\left (84 i \, x + 239 i\right )} e^{\left (2 \, x\right )} - 3 \, e^{\left (7 \, x\right )} + 15 i \, e^{\left (6 \, x\right )} + 15 \, e^{x} - 3 i}{24 \, e^{\left (5 \, x\right )} + 72 i \, e^{\left (4 \, x\right )} - 72 \, e^{\left (3 \, x\right )} - 24 i \, e^{\left (2 \, x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

-(21*(4*x - 3)*e^(5*x) - (-252*I*x - 147*I)*e^(4*x) - 3*(84*x + 127)*e^(3*x) - (84*I*x + 239*I)*e^(2*x) - 3*e^
(7*x) + 15*I*e^(6*x) + 15*e^x - 3*I)/(24*e^(5*x) + 72*I*e^(4*x) - 72*e^(3*x) - 24*I*e^(2*x))

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Sympy [A]  time = 0.531348, size = 68, normalized size = 1.17 \begin{align*} - \frac{7 x}{2} + \frac{- 8 i e^{2 x} + 14 e^{x} + \frac{22 i}{3}}{e^{3 x} + 3 i e^{2 x} - 3 e^{x} - i} + \frac{e^{2 x}}{8} - i e^{x} - i e^{- x} - \frac{e^{- 2 x}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**4/(I+sinh(x))**2,x)

[Out]

-7*x/2 + (-8*I*exp(2*x) + 14*exp(x) + 22*I/3)/(exp(3*x) + 3*I*exp(2*x) - 3*exp(x) - I) + exp(2*x)/8 - I*exp(x)
- I*exp(-x) - exp(-2*x)/8

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Giac [A]  time = 1.38319, size = 68, normalized size = 1.17 \begin{align*} -\frac{7}{2} \, x - \frac{{\left (216 i \, e^{\left (4 \, x\right )} - 405 \, e^{\left (3 \, x\right )} - 239 i \, e^{\left (2 \, x\right )} + 15 \, e^{x} - 3 i\right )} e^{\left (-2 \, x\right )}}{24 \,{\left (e^{x} + i\right )}^{3}} + \frac{1}{8} \, e^{\left (2 \, x\right )} - i \, e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(I+sinh(x))^2,x, algorithm="giac")

[Out]

-7/2*x - 1/24*(216*I*e^(4*x) - 405*e^(3*x) - 239*I*e^(2*x) + 15*e^x - 3*I)*e^(-2*x)/(e^x + I)^3 + 1/8*e^(2*x)
- I*e^x