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]

-1/8*I*arctan(sinh(x))-1/4*sech(x)^4-1/8*I*sech(x)*tanh(x)+1/4*I*sech(x)^3*tanh(x)

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Rubi [A]  time = 0.13, antiderivative size = 40, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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}^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*}

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Mathematica [A]  time = 0.06, size = 32, normalized size = 0.80 \[ \frac {1}{8} \left (-\frac {i}{\sinh (x)+i}+\frac {1}{(\sinh (x)-i)^2}-i \tan ^{-1}(\sinh (x))\right ) \]

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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fricas [B]  time = 0.41, size = 145, normalized size = 3.62 \[ \frac {{\left (e^{\left (6 \, x\right )} - 2 i \, e^{\left (5 \, x\right )} + e^{\left (4 \, x\right )} - 4 i \, e^{\left (3 \, x\right )} - e^{\left (2 \, x\right )} - 2 i \, e^{x} - 1\right )} \log \left (e^{x} + i\right ) - {\left (e^{\left (6 \, x\right )} - 2 i \, e^{\left (5 \, x\right )} + e^{\left (4 \, x\right )} - 4 i \, e^{\left (3 \, x\right )} - e^{\left (2 \, x\right )} - 2 i \, e^{x} - 1\right )} \log \left (e^{x} - i\right ) - 2 i \, e^{\left (5 \, x\right )} - 4 \, e^{\left (4 \, x\right )} + 20 i \, e^{\left (3 \, x\right )} + 4 \, e^{\left (2 \, x\right )} - 2 i \, e^{x}}{8 \, e^{\left (6 \, x\right )} - 16 i \, e^{\left (5 \, x\right )} + 8 \, e^{\left (4 \, x\right )} - 32 i \, e^{\left (3 \, x\right )} - 8 \, e^{\left (2 \, x\right )} - 16 i \, e^{x} - 8} \]

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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giac [B]  time = 0.13, size = 94, normalized size = 2.35 \[ -\frac {-i \, e^{\left (-x\right )} + i \, e^{x} - 6}{16 \, {\left (-i \, e^{\left (-x\right )} + i \, e^{x} - 2\right )}} + \frac {3 \, {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 12 i \, e^{\left (-x\right )} - 12 i \, e^{x} + 4}{32 \, {\left (e^{\left (-x\right )} - e^{x} + 2 i\right )}^{2}} + \frac {1}{16} \, \log \left (-e^{\left (-x\right )} + e^{x} + 2 i\right ) - \frac {1}{16} \, \log \left (-e^{\left (-x\right )} + e^{x} - 2 i\right ) \]

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)

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maple [B]  time = 0.20, size = 89, normalized size = 2.22 \[ \frac {i}{4 \tanh \left (\frac {x}{2}\right )+4 i}+\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+i\right )}{8}+\frac {i}{\left (\tanh \left (\frac {x}{2}\right )-i\right )^{3}}-\frac {i}{2 \left (\tanh \left (\frac {x}{2}\right )-i\right )}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{4}}+\frac {1}{\left (\tanh \left (\frac {x}{2}\right )-i\right )^{2}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-i\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.31, size = 92, normalized size = 2.30 \[ \frac {8 \, {\left (i \, e^{\left (-x\right )} + 2 \, e^{\left (-2 \, x\right )} - 10 i \, e^{\left (-3 \, x\right )} - 2 \, e^{\left (-4 \, x\right )} + i \, e^{\left (-5 \, x\right )}\right )}}{64 i \, e^{\left (-x\right )} - 32 \, e^{\left (-2 \, x\right )} + 128 i \, e^{\left (-3 \, x\right )} + 32 \, e^{\left (-4 \, x\right )} + 64 i \, e^{\left (-5 \, x\right )} + 32 \, e^{\left (-6 \, x\right )} - 32} - \frac {1}{8} \, \log \left (e^{\left (-x\right )} + i\right ) + \frac {1}{8} \, \log \left (e^{\left (-x\right )} - i\right ) \]

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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mupad [B]  time = 1.95, size = 122, normalized size = 3.05 \[ \frac {\ln \left (-\frac {1}{4}+\frac {{\mathrm {e}}^x\,1{}\mathrm {i}}{4}\right )}{8}-\frac {\ln \left (\frac {1}{4}+\frac {{\mathrm {e}}^x\,1{}\mathrm {i}}{4}\right )}{8}-\frac {1{}\mathrm {i}}{{\mathrm {e}}^{2\,x}\,3{}\mathrm {i}-{\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^x-\mathrm {i}}-\frac {1}{4\,\left ({\mathrm {e}}^{2\,x}-1+{\mathrm {e}}^x\,2{}\mathrm {i}\right )}-\frac {1}{2\,\left ({\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1-{\mathrm {e}}^{3\,x}\,4{}\mathrm {i}+{\mathrm {e}}^x\,4{}\mathrm {i}\right )}-\frac {1}{2\,\left (1-{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,2{}\mathrm {i}\right )}-\frac {1{}\mathrm {i}}{4\,\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^3*(1/sinh(x) + 1i)),x)

[Out]

log((exp(x)*1i)/4 - 1/4)/8 - log((exp(x)*1i)/4 + 1/4)/8 - 1i/(exp(2*x)*3i - exp(3*x) + 3*exp(x) - 1i) - 1/(4*(
exp(2*x) + exp(x)*2i - 1)) - 1/(2*(exp(4*x) - exp(3*x)*4i - 6*exp(2*x) + exp(x)*4i + 1)) - 1/(2*(exp(x)*2i - e
xp(2*x) + 1)) - 1i/(4*(exp(x) + 1i))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{3}{\relax (x )}}{\operatorname {csch}{\relax (x )} + i}\, dx \]

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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