3.168 \(\int \frac{\text{sech}^3(x)}{i+\sinh (x)} \, dx\)

Optimal. Leaf size=52 \[ \frac{i}{8 (-\sinh (x)+i)}-\frac{i}{4 (\sinh (x)+i)}+\frac{1}{8 (\sinh (x)+i)^2}-\frac{3}{8} i \tan ^{-1}(\sinh (x)) \]

[Out]

((-3*I)/8)*ArcTan[Sinh[x]] + (I/8)/(I - Sinh[x]) + 1/(8*(I + Sinh[x])^2) - (I/4)/(I + Sinh[x])

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Rubi [A]  time = 0.0525301, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2667, 44, 203} \[ \frac{i}{8 (-\sinh (x)+i)}-\frac{i}{4 (\sinh (x)+i)}+\frac{1}{8 (\sinh (x)+i)^2}-\frac{3}{8} i \tan ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(I + Sinh[x]),x]

[Out]

((-3*I)/8)*ArcTan[Sinh[x]] + (I/8)/(I - Sinh[x]) + 1/(8*(I + Sinh[x])^2) - (I/4)/(I + Sinh[x])

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])

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\text{sech}^3(x)}{i+\sinh (x)} \, dx &=\operatorname{Subst}\left (\int \frac{1}{(i-x)^2 (i+x)^3} \, dx,x,\sinh (x)\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{i}{8 (-i+x)^2}-\frac{1}{4 (i+x)^3}+\frac{i}{4 (i+x)^2}-\frac{3 i}{8 \left (1+x^2\right )}\right ) \, dx,x,\sinh (x)\right )\\ &=\frac{i}{8 (i-\sinh (x))}+\frac{1}{8 (i+\sinh (x))^2}-\frac{i}{4 (i+\sinh (x))}-\frac{3}{8} i \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac{3}{8} i \tan ^{-1}(\sinh (x))+\frac{i}{8 (i-\sinh (x))}+\frac{1}{8 (i+\sinh (x))^2}-\frac{i}{4 (i+\sinh (x))}\\ \end{align*}

Mathematica [A]  time = 0.0414301, size = 61, normalized size = 1.17 \[ -\frac{i \text{sech}^2(x) \left (3 \sinh ^3(x) \tan ^{-1}(\sinh (x))+\sinh ^2(x) \left (3+3 i \tan ^{-1}(\sinh (x))\right )+3 \sinh (x) \left (\tan ^{-1}(\sinh (x))+i\right )+3 i \tan ^{-1}(\sinh (x))+2\right )}{8 (\sinh (x)+i)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^3/(I + Sinh[x]),x]

[Out]

((-I/8)*Sech[x]^2*(2 + (3*I)*ArcTan[Sinh[x]] + 3*(I + ArcTan[Sinh[x]])*Sinh[x] + (3 + (3*I)*ArcTan[Sinh[x]])*S
inh[x]^2 + 3*ArcTan[Sinh[x]]*Sinh[x]^3))/(I + Sinh[x])

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Maple [B]  time = 0.036, size = 91, normalized size = 1.8 \begin{align*}{{\frac{i}{4}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}-{\frac{1}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-2}}-{\frac{3}{8}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) }-{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-4}}+{i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}}-{i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}+{\frac{3}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}}+{\frac{3}{8}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I/(tanh(1/2*x)-I)-1/4/(tanh(1/2*x)-I)^2-3/8*ln(tanh(1/2*x)-I)-1/2/(tanh(1/2*x)+I)^4+I/(tanh(1/2*x)+I)-I/(t
anh(1/2*x)+I)^3+3/2/(tanh(1/2*x)+I)^2+3/8*ln(tanh(1/2*x)+I)

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Maxima [B]  time = 1.18619, size = 124, normalized size = 2.38 \begin{align*} \frac{8 \,{\left (3 i \, e^{\left (-x\right )} - 6 \, e^{\left (-2 \, x\right )} + 2 i \, e^{\left (-3 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} + 3 i \, e^{\left (-5 \, x\right )}\right )}}{-64 i \, e^{\left (-x\right )} - 32 \, e^{\left (-2 \, x\right )} - 128 i \, e^{\left (-3 \, x\right )} + 32 \, e^{\left (-4 \, x\right )} - 64 i \, e^{\left (-5 \, x\right )} + 32 \, e^{\left (-6 \, x\right )} - 32} - \frac{3}{8} \, \log \left (e^{\left (-x\right )} + i\right ) + \frac{3}{8} \, \log \left (e^{\left (-x\right )} - i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*(3*I*e^(-x) - 6*e^(-2*x) + 2*I*e^(-3*x) + 6*e^(-4*x) + 3*I*e^(-5*x))/(-64*I*e^(-x) - 32*e^(-2*x) - 128*I*e^(
-3*x) + 32*e^(-4*x) - 64*I*e^(-5*x) + 32*e^(-6*x) - 32) - 3/8*log(e^(-x) + I) + 3/8*log(e^(-x) - I)

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Fricas [B]  time = 1.75851, size = 451, normalized size = 8.67 \begin{align*} \frac{{\left (3 \, e^{\left (6 \, x\right )} + 6 i \, e^{\left (5 \, x\right )} + 3 \, e^{\left (4 \, x\right )} + 12 i \, e^{\left (3 \, x\right )} - 3 \, e^{\left (2 \, x\right )} + 6 i \, e^{x} - 3\right )} \log \left (e^{x} + i\right ) -{\left (3 \, e^{\left (6 \, x\right )} + 6 i \, e^{\left (5 \, x\right )} + 3 \, e^{\left (4 \, x\right )} + 12 i \, e^{\left (3 \, x\right )} - 3 \, e^{\left (2 \, x\right )} + 6 i \, e^{x} - 3\right )} \log \left (e^{x} - i\right ) - 6 i \, e^{\left (5 \, x\right )} + 12 \, e^{\left (4 \, x\right )} - 4 i \, e^{\left (3 \, x\right )} - 12 \, e^{\left (2 \, x\right )} - 6 i \, e^{x}}{8 \, e^{\left (6 \, x\right )} + 16 i \, e^{\left (5 \, x\right )} + 8 \, e^{\left (4 \, x\right )} + 32 i \, e^{\left (3 \, x\right )} - 8 \, e^{\left (2 \, x\right )} + 16 i \, e^{x} - 8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*e^(6*x) + 6*I*e^(5*x) + 3*e^(4*x) + 12*I*e^(3*x) - 3*e^(2*x) + 6*I*e^x - 3)*log(e^x + I) - (3*e^(6*x) + 6*
I*e^(5*x) + 3*e^(4*x) + 12*I*e^(3*x) - 3*e^(2*x) + 6*I*e^x - 3)*log(e^x - I) - 6*I*e^(5*x) + 12*e^(4*x) - 4*I*
e^(3*x) - 12*e^(2*x) - 6*I*e^x)/(8*e^(6*x) + 16*I*e^(5*x) + 8*e^(4*x) + 32*I*e^(3*x) - 8*e^(2*x) + 16*I*e^x -
8)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{3}{\left (x \right )}}{\sinh{\left (x \right )} + i}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(x)**3/(sinh(x) + I), x)

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Giac [B]  time = 1.32803, size = 124, normalized size = 2.38 \begin{align*} \frac{3 \, e^{\left (-x\right )} - 3 \, e^{x} + 10 i}{16 \,{\left (e^{\left (-x\right )} - e^{x} + 2 i\right )}} - \frac{9 \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} - 52 i \, e^{\left (-x\right )} + 52 i \, e^{x} - 84}{32 \,{\left (e^{\left (-x\right )} - e^{x} - 2 i\right )}^{2}} + \frac{3}{16} \, \log \left (-e^{\left (-x\right )} + e^{x} + 2 i\right ) - \frac{3}{16} \, \log \left (-e^{\left (-x\right )} + e^{x} - 2 i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*e^(-x) - 3*e^x + 10*I)/(e^(-x) - e^x + 2*I) - 1/32*(9*(e^(-x) - e^x)^2 - 52*I*e^(-x) + 52*I*e^x - 84)/
(e^(-x) - e^x - 2*I)^2 + 3/16*log(-e^(-x) + e^x + 2*I) - 3/16*log(-e^(-x) + e^x - 2*I)