### 3.643 $$\int \frac{1}{(\text{sech}(x)-i \tanh (x))^5} \, dx$$

Optimal. Leaf size=40 $\frac{4 i}{1-i \sinh (x)}-\frac{2 i}{(1-i \sinh (x))^2}+i \log (\sinh (x)+i)$

[Out]

I*Log[I + Sinh[x]] - (2*I)/(1 - I*Sinh[x])^2 + (4*I)/(1 - I*Sinh[x])

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Rubi [A]  time = 0.0551645, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.273, Rules used = {4391, 2667, 43} $\frac{4 i}{1-i \sinh (x)}-\frac{2 i}{(1-i \sinh (x))^2}+i \log (\sinh (x)+i)$

Antiderivative was successfully veriﬁed.

[In]

Int[(Sech[x] - I*Tanh[x])^(-5),x]

[Out]

I*Log[I + Sinh[x]] - (2*I)/(1 - I*Sinh[x])^2 + (4*I)/(1 - I*Sinh[x])

Rule 4391

Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a*Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{1}{(\text{sech}(x)-i \tanh (x))^5} \, dx &=\int \frac{\cosh ^5(x)}{(1-i \sinh (x))^5} \, dx\\ &=i \operatorname{Subst}\left (\int \frac{(1-x)^2}{(1+x)^3} \, dx,x,-i \sinh (x)\right )\\ &=i \operatorname{Subst}\left (\int \left (\frac{4}{(1+x)^3}-\frac{4}{(1+x)^2}+\frac{1}{1+x}\right ) \, dx,x,-i \sinh (x)\right )\\ &=i \log (i+\sinh (x))-\frac{2 i}{(1-i \sinh (x))^2}+\frac{4 i}{1-i \sinh (x)}\\ \end{align*}

Mathematica [A]  time = 0.0983803, size = 45, normalized size = 1.12 $2 \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+i \log (\cosh (x))+\frac{4 \sinh (x)+2 i}{\left (\cosh \left (\frac{x}{2}\right )-i \sinh \left (\frac{x}{2}\right )\right )^4}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Sech[x] - I*Tanh[x])^(-5),x]

[Out]

2*ArcTan[Tanh[x/2]] + I*Log[Cosh[x]] + (2*I + 4*Sinh[x])/(Cosh[x/2] - I*Sinh[x/2])^4

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Maple [A]  time = 0.098, size = 68, normalized size = 1.7 \begin{align*} -i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) -i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) +{8\,i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}}+2\,i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) -{8\,i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-4}}+16\, \left ( \tanh \left ( x/2 \right ) +i \right ) ^{-3} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sech(x)-I*tanh(x))^5,x)

[Out]

-I*ln(tanh(1/2*x)+1)-I*ln(tanh(1/2*x)-1)+8*I/(tanh(1/2*x)+I)^2+2*I*ln(tanh(1/2*x)+I)-8*I/(tanh(1/2*x)+I)^4+16/
(tanh(1/2*x)+I)^3

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Maxima [B]  time = 1.08159, size = 81, normalized size = 2.02 \begin{align*} i \, x - \frac{8 \, e^{\left (-x\right )} + 8 i \, e^{\left (-2 \, x\right )} - 8 \, e^{\left (-3 \, x\right )}}{4 i \, e^{\left (-x\right )} - 6 \, e^{\left (-2 \, x\right )} - 4 i \, e^{\left (-3 \, x\right )} + e^{\left (-4 \, x\right )} + 1} + 2 i \, \log \left (e^{\left (-x\right )} - i\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)-I*tanh(x))^5,x, algorithm="maxima")

[Out]

I*x - (8*e^(-x) + 8*I*e^(-2*x) - 8*e^(-3*x))/(4*I*e^(-x) - 6*e^(-2*x) - 4*I*e^(-3*x) + e^(-4*x) + 1) + 2*I*log
(e^(-x) - I)

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Fricas [B]  time = 2.16657, size = 273, normalized size = 6.82 \begin{align*} \frac{-i \, x e^{\left (4 \, x\right )} + 4 \,{\left (x - 2\right )} e^{\left (3 \, x\right )} +{\left (6 i \, x - 8 i\right )} e^{\left (2 \, x\right )} - 4 \,{\left (x - 2\right )} e^{x} +{\left (2 i \, e^{\left (4 \, x\right )} - 8 \, e^{\left (3 \, x\right )} - 12 i \, e^{\left (2 \, x\right )} + 8 \, e^{x} + 2 i\right )} \log \left (e^{x} + i\right ) - i \, x}{e^{\left (4 \, x\right )} + 4 i \, e^{\left (3 \, x\right )} - 6 \, e^{\left (2 \, x\right )} - 4 i \, e^{x} + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)-I*tanh(x))^5,x, algorithm="fricas")

[Out]

(-I*x*e^(4*x) + 4*(x - 2)*e^(3*x) + (6*I*x - 8*I)*e^(2*x) - 4*(x - 2)*e^x + (2*I*e^(4*x) - 8*e^(3*x) - 12*I*e^
(2*x) + 8*e^x + 2*I)*log(e^x + I) - I*x)/(e^(4*x) + 4*I*e^(3*x) - 6*e^(2*x) - 4*I*e^x + 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)-I*tanh(x))**5,x)

[Out]

Timed out

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Giac [A]  time = 1.15358, size = 54, normalized size = 1.35 \begin{align*} -\frac{8 \, e^{\left (3 \, x\right )} + 8 i \, e^{\left (2 \, x\right )} - 8 \, e^{x}}{{\left (e^{x} + i\right )}^{4}} - i \, \log \left (-i \, e^{x}\right ) + 2 i \, \log \left (e^{x} + i\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)-I*tanh(x))^5,x, algorithm="giac")

[Out]

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