3.89 \(\int \frac {\text {sech}^2(x)}{i+\text {csch}(x)} \, dx\)

Optimal. Leaf size=19 \[ -\frac {1}{3} \text {sech}^3(x)-\frac {1}{3} i \tanh ^3(x) \]

[Out]

-1/3*sech(x)^3-1/3*I*tanh(x)^3

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Rubi [A]  time = 0.11, antiderivative size = 19, 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 = {3872, 2839, 2606, 30, 2607} \[ -\frac {1}{3} \text {sech}^3(x)-\frac {1}{3} i \tanh ^3(x) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(I + Csch[x]),x]

[Out]

-Sech[x]^3/3 - (I/3)*Tanh[x]^3

Rule 30

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

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 2607

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

Rule 2839

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

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]

Rubi steps

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

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Mathematica [B]  time = 0.06, size = 64, normalized size = 3.37 \[ \frac {-2 i \sinh (x)+\cosh (x)+\cosh (2 x)+i \sinh (x) \cosh (x)-3}{6 \left (\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )\right ) \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^2/(I + Csch[x]),x]

[Out]

(-3 + Cosh[x] + Cosh[2*x] - (2*I)*Sinh[x] + I*Cosh[x]*Sinh[x])/(6*(Cosh[x/2] - I*Sinh[x/2])*(Cosh[x/2] + I*Sin
h[x/2])^3)

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fricas [B]  time = 0.39, size = 33, normalized size = 1.74 \[ \frac {6 i \, e^{\left (2 \, x\right )} + 4 \, e^{x} - 2 i}{3 \, e^{\left (4 \, x\right )} - 6 i \, e^{\left (3 \, x\right )} - 6 i \, e^{x} - 3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(I+csch(x)),x, algorithm="fricas")

[Out]

(6*I*e^(2*x) + 4*e^x - 2*I)/(3*e^(4*x) - 6*I*e^(3*x) - 6*I*e^x - 3)

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giac [B]  time = 0.14, size = 27, normalized size = 1.42 \[ -\frac {i}{2 \, {\left (i \, e^{x} - 1\right )}} + \frac {3 \, e^{\left (2 \, x\right )} - 1}{6 \, {\left (e^{x} - i\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(I+csch(x)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.20, size = 49, normalized size = 2.58 \[ -\frac {i}{2 \left (\tanh \left (\frac {x}{2}\right )+i\right )}+\frac {i}{2 \tanh \left (\frac {x}{2}\right )-2 i}-\frac {2 i}{3 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{3}}-\frac {1}{\left (\tanh \left (\frac {x}{2}\right )-i\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2/(I+csch(x)),x)

[Out]

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

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maxima [B]  time = 0.31, size = 81, normalized size = 4.26 \[ \frac {8 \, e^{\left (-x\right )}}{12 i \, e^{\left (-x\right )} + 12 i \, e^{\left (-3 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} - 6} - \frac {12 i \, e^{\left (-2 \, x\right )}}{12 i \, e^{\left (-x\right )} + 12 i \, e^{\left (-3 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} - 6} + \frac {4 i}{12 i \, e^{\left (-x\right )} + 12 i \, e^{\left (-3 \, x\right )} + 6 \, e^{\left (-4 \, x\right )} - 6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(I+csch(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.57, size = 31, normalized size = 1.63 \[ -\frac {\frac {2}{3}-2\,{\mathrm {e}}^{2\,x}+\frac {{\mathrm {e}}^x\,4{}\mathrm {i}}{3}}{\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,{\left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^2*(1/sinh(x) + 1i)),x)

[Out]

-((exp(x)*4i)/3 - 2*exp(2*x) + 2/3)/((exp(x) + 1i)*(exp(x)*1i + 1)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{2}{\relax (x )}}{\operatorname {csch}{\relax (x )} + i}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**2/(I+csch(x)),x)

[Out]

Integral(sech(x)**2/(csch(x) + I), x)

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