3.49 \(\int \frac{\sinh ^3(x)}{(i+\sinh (x))^2} \, dx\)

Optimal. Leaf size=44 \[ -2 i x+\frac{4 \cosh (x)}{3}-\frac{\sinh ^2(x) \cosh (x)}{3 (\sinh (x)+i)^2}+\frac{2 i \cosh (x)}{\sinh (x)+i} \]

[Out]

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

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Rubi [A]  time = 0.129489, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2765, 2968, 3023, 12, 2735, 2648} \[ -2 i x+\frac{4 \cosh (x)}{3}-\frac{\sinh ^2(x) \cosh (x)}{3 (\sinh (x)+i)^2}+\frac{2 i \cosh (x)}{\sinh (x)+i} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^3/(I + Sinh[x])^2,x]

[Out]

(-2*I)*x + (4*Cosh[x])/3 - (Cosh[x]*Sinh[x]^2)/(3*(I + Sinh[x])^2) + ((2*I)*Cosh[x])/(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 2968

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2735

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\sinh ^3(x)}{(i+\sinh (x))^2} \, dx &=-\frac{\cosh (x) \sinh ^2(x)}{3 (i+\sinh (x))^2}+\frac{1}{3} \int \frac{\sinh (x) (-2 i+4 \sinh (x))}{i+\sinh (x)} \, dx\\ &=-\frac{\cosh (x) \sinh ^2(x)}{3 (i+\sinh (x))^2}-\frac{1}{3} i \int \frac{2 \sinh (x)+4 i \sinh ^2(x)}{i+\sinh (x)} \, dx\\ &=\frac{4 \cosh (x)}{3}-\frac{\cosh (x) \sinh ^2(x)}{3 (i+\sinh (x))^2}+\frac{1}{3} \int -\frac{6 i \sinh (x)}{i+\sinh (x)} \, dx\\ &=\frac{4 \cosh (x)}{3}-\frac{\cosh (x) \sinh ^2(x)}{3 (i+\sinh (x))^2}-2 i \int \frac{\sinh (x)}{i+\sinh (x)} \, dx\\ &=-2 i x+\frac{4 \cosh (x)}{3}-\frac{\cosh (x) \sinh ^2(x)}{3 (i+\sinh (x))^2}-2 \int \frac{1}{i+\sinh (x)} \, dx\\ &=-2 i x+\frac{4 \cosh (x)}{3}-\frac{\cosh (x) \sinh ^2(x)}{3 (i+\sinh (x))^2}+\frac{2 i \cosh (x)}{i+\sinh (x)}\\ \end{align*}

Mathematica [A]  time = 0.120159, size = 45, normalized size = 1.02 \[ \frac{1}{3} \cosh (x) \left (\frac{3 \sinh ^2(x)+14 i \sinh (x)-10}{(\sinh (x)+i)^2}-\frac{6 i \sinh ^{-1}(\sinh (x))}{\sqrt{\cosh ^2(x)}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^3/(I + Sinh[x])^2,x]

[Out]

(Cosh[x]*(((-6*I)*ArcSinh[Sinh[x]])/Sqrt[Cosh[x]^2] + (-10 + (14*I)*Sinh[x] + 3*Sinh[x]^2)/(I + Sinh[x])^2))/3

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Maple [B]  time = 0.05, size = 75, normalized size = 1.7 \begin{align*} -2\,i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) + \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}+2\,i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) - \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}+{{\frac{4\,i}{3}} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}+{4\,i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}}-2\, \left ( \tanh \left ( x/2 \right ) +i \right ) ^{-2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3/(I+sinh(x))^2,x)

[Out]

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

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Maxima [A]  time = 1.31374, size = 80, normalized size = 1.82 \begin{align*} -2 i \, x - \frac{164 \, e^{\left (-x\right )} + 276 i \, e^{\left (-2 \, x\right )} - 156 \, e^{\left (-3 \, x\right )} - 12 i}{8 \,{\left (3 i \, e^{\left (-x\right )} - 9 \, e^{\left (-2 \, x\right )} - 9 i \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-4 \, x\right )}\right )}} + \frac{1}{2} \, e^{\left (-x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

-2*I*x - 1/8*(164*e^(-x) + 276*I*e^(-2*x) - 156*e^(-3*x) - 12*I)/(3*I*e^(-x) - 9*e^(-2*x) - 9*I*e^(-3*x) + 3*e
^(-4*x)) + 1/2*e^(-x)

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Fricas [B]  time = 2.06003, size = 217, normalized size = 4.93 \begin{align*} \frac{{\left (-12 i \, x + 9 i\right )} e^{\left (4 \, x\right )} + 6 \,{\left (6 \, x + 5\right )} e^{\left (3 \, x\right )} +{\left (36 i \, x + 66 i\right )} e^{\left (2 \, x\right )} -{\left (12 \, x + 41\right )} e^{x} + 3 \, e^{\left (5 \, x\right )} - 3 i}{6 \, e^{\left (4 \, x\right )} + 18 i \, e^{\left (3 \, x\right )} - 18 \, e^{\left (2 \, x\right )} - 6 i \, e^{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(I+sinh(x))^2,x, algorithm="fricas")

[Out]

((-12*I*x + 9*I)*e^(4*x) + 6*(6*x + 5)*e^(3*x) + (36*I*x + 66*I)*e^(2*x) - (12*x + 41)*e^x + 3*e^(5*x) - 3*I)/
(6*e^(4*x) + 18*I*e^(3*x) - 18*e^(2*x) - 6*I*e^x)

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Sympy [A]  time = 0.401296, size = 53, normalized size = 1.2 \begin{align*} - 2 i x + \frac{6 e^{2 x} + 10 i e^{x} - \frac{16}{3}}{e^{3 x} + 3 i e^{2 x} - 3 e^{x} - i} + \frac{e^{x}}{2} + \frac{e^{- x}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**3/(I+sinh(x))**2,x)

[Out]

-2*I*x + (6*exp(2*x) + 10*I*exp(x) - 16/3)/(exp(3*x) + 3*I*exp(2*x) - 3*exp(x) - I) + exp(x)/2 + exp(-x)/2

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Giac [A]  time = 1.38429, size = 51, normalized size = 1.16 \begin{align*} -2 i \, x + \frac{{\left (39 \, e^{\left (3 \, x\right )} + 69 i \, e^{\left (2 \, x\right )} - 41 \, e^{x} - 3 i\right )} e^{\left (-x\right )}}{6 \,{\left (e^{x} + i\right )}^{3}} + \frac{1}{2} \, e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3/(I+sinh(x))^2,x, algorithm="giac")

[Out]

-2*I*x + 1/6*(39*e^(3*x) + 69*I*e^(2*x) - 41*e^x - 3*I)*e^(-x)/(e^x + I)^3 + 1/2*e^x