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

Optimal. Leaf size=40 \[ -\frac{1}{4} \text{sech}^4(x)-\frac{1}{8} i \tan ^{-1}(\sinh (x))+\frac{1}{4} i \tanh (x) \text{sech}^3(x)-\frac{1}{8} i \tanh (x) \text{sech}(x) \]

[Out]

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

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Rubi [A]  time = 0.130221, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {3872, 2835, 2606, 30, 2611, 3768, 3770} \[ -\frac{1}{4} \text{sech}^4(x)-\frac{1}{8} i \tan ^{-1}(\sinh (x))+\frac{1}{4} i \tanh (x) \text{sech}^3(x)-\frac{1}{8} i \tanh (x) \text{sech}(x) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(I + Csch[x]),x]

[Out]

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

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2835

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0586016, size = 32, normalized size = 0.8 \[ \frac{1}{8} \left (-\frac{i}{\sinh (x)+i}+\frac{1}{(\sinh (x)-i)^2}-i \tan ^{-1}(\sinh (x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^3/(I + Csch[x]),x]

[Out]

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

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Maple [B]  time = 0.043, size = 89, normalized size = 2.2 \begin{align*}{i \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-3}}-{{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}-{\frac{1}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-4}}+ \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-2}-{\frac{1}{8}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) }+{{\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{1}{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+csch(x)),x)

[Out]

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

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Maxima [B]  time = 1.05542, size = 124, normalized size = 3.1 \begin{align*} \frac{8 \,{\left (i \, e^{\left (-x\right )} + 2 \, e^{\left (-2 \, x\right )} - 10 i \, e^{\left (-3 \, x\right )} - 2 \, e^{\left (-4 \, x\right )} + 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{1}{8} \, \log \left (e^{\left (-x\right )} + i\right ) + \frac{1}{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+csch(x)),x, algorithm="maxima")

[Out]

8*(I*e^(-x) + 2*e^(-2*x) - 10*I*e^(-3*x) - 2*e^(-4*x) + 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) - 1/8*log(e^(-x) + I) + 1/8*log(e^(-x) - I)

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Fricas [B]  time = 1.69641, size = 431, normalized size = 10.78 \begin{align*} \frac{{\left (e^{\left (6 \, x\right )} - 2 i \, e^{\left (5 \, x\right )} + e^{\left (4 \, x\right )} - 4 i \, e^{\left (3 \, x\right )} - e^{\left (2 \, x\right )} - 2 i \, e^{x} - 1\right )} \log \left (e^{x} + i\right ) -{\left (e^{\left (6 \, x\right )} - 2 i \, e^{\left (5 \, x\right )} + e^{\left (4 \, x\right )} - 4 i \, e^{\left (3 \, x\right )} - e^{\left (2 \, x\right )} - 2 i \, e^{x} - 1\right )} \log \left (e^{x} - i\right ) - 2 i \, e^{\left (5 \, x\right )} - 4 \, e^{\left (4 \, x\right )} + 20 i \, e^{\left (3 \, x\right )} + 4 \, e^{\left (2 \, x\right )} - 2 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+csch(x)),x, algorithm="fricas")

[Out]

((e^(6*x) - 2*I*e^(5*x) + e^(4*x) - 4*I*e^(3*x) - e^(2*x) - 2*I*e^x - 1)*log(e^x + I) - (e^(6*x) - 2*I*e^(5*x)
+ e^(4*x) - 4*I*e^(3*x) - e^(2*x) - 2*I*e^x - 1)*log(e^x - I) - 2*I*e^(5*x) - 4*e^(4*x) + 20*I*e^(3*x) + 4*e^
(2*x) - 2*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 )}}{\operatorname{csch}{\left (x \right )} + i}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(x)**3/(csch(x) + I), x)

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Giac [B]  time = 1.16183, size = 127, normalized size = 3.18 \begin{align*} -\frac{-i \, e^{\left (-x\right )} + i \, e^{x} - 6}{16 \,{\left (-i \, e^{\left (-x\right )} + i \, e^{x} - 2\right )}} + \frac{3 \,{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 12 i \, e^{\left (-x\right )} - 12 i \, e^{x} + 4}{32 \,{\left (e^{\left (-x\right )} - e^{x} + 2 i\right )}^{2}} + \frac{1}{16} \, \log \left (-e^{\left (-x\right )} + e^{x} + 2 i\right ) - \frac{1}{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+csch(x)),x, algorithm="giac")

[Out]

-1/16*(-I*e^(-x) + I*e^x - 6)/(-I*e^(-x) + I*e^x - 2) + 1/32*(3*(e^(-x) - e^x)^2 + 12*I*e^(-x) - 12*I*e^x + 4)
/(e^(-x) - e^x + 2*I)^2 + 1/16*log(-e^(-x) + e^x + 2*I) - 1/16*log(-e^(-x) + e^x - 2*I)