### 3.630 $$\int \frac{1}{(\text{sech}(x)+i \tanh (x))^2} \, dx$$

Optimal. Leaf size=20 $-x+\frac{2 i \cosh (x)}{1+i \sinh (x)}$

[Out]

-x + ((2*I)*Cosh[x])/(1 + I*Sinh[x])

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Rubi [A]  time = 0.0443195, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.273, Rules used = {4391, 2680, 8} $-x+\frac{2 i \cosh (x)}{1+i \sinh (x)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Sech[x] + I*Tanh[x])^(-2),x]

[Out]

-x + ((2*I)*Cosh[x])/(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 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{1}{(\text{sech}(x)+i \tanh (x))^2} \, dx &=\int \frac{\cosh ^2(x)}{(1+i \sinh (x))^2} \, dx\\ &=\frac{2 i \cosh (x)}{1+i \sinh (x)}-\int 1 \, dx\\ &=-x+\frac{2 i \cosh (x)}{1+i \sinh (x)}\\ \end{align*}

Mathematica [A]  time = 0.0272138, size = 31, normalized size = 1.55 $-x+\frac{4 \sinh \left (\frac{x}{2}\right )}{\cosh \left (\frac{x}{2}\right )+i \sinh \left (\frac{x}{2}\right )}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Sech[x] + I*Tanh[x])^(-2),x]

[Out]

-x + (4*Sinh[x/2])/(Cosh[x/2] + I*Sinh[x/2])

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Maple [A]  time = 0.051, size = 29, normalized size = 1.5 \begin{align*} -\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) +\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) +4\, \left ( \tanh \left ( x/2 \right ) -i \right ) ^{-1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sech(x)+I*tanh(x))^2,x)

[Out]

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

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Maxima [A]  time = 1.06524, size = 19, normalized size = 0.95 \begin{align*} -x + \frac{4 i}{e^{\left (-x\right )} + i} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)+I*tanh(x))^2,x, algorithm="maxima")

[Out]

-x + 4*I/(e^(-x) + I)

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Fricas [A]  time = 2.58902, size = 43, normalized size = 2.15 \begin{align*} -\frac{x e^{x} - i \, x - 4 i}{e^{x} - i} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)+I*tanh(x))^2,x, algorithm="fricas")

[Out]

-(x*e^x - I*x - 4*I)/(e^x - I)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (i \tanh{\left (x \right )} + \operatorname{sech}{\left (x \right )}\right )^{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)+I*tanh(x))**2,x)

[Out]

Integral((I*tanh(x) + sech(x))**(-2), x)

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Giac [A]  time = 1.14637, size = 16, normalized size = 0.8 \begin{align*} -x + \frac{4 i}{e^{x} - i} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sech(x)+I*tanh(x))^2,x, algorithm="giac")

[Out]

-x + 4*I/(e^x - I)