### 3.639 $$\int \frac{1}{\text{sech}(x)-i \tanh (x)} \, dx$$

Optimal. Leaf size=11 $i \log (\sinh (x)+i)$

[Out]

I*Log[I + Sinh[x]]

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Rubi [A]  time = 0.0301713, antiderivative size = 11, 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 = {3159, 2667, 31} $i \log (\sinh (x)+i)$

Antiderivative was successfully veriﬁed.

[In]

Int[(Sech[x] - I*Tanh[x])^(-1),x]

[Out]

I*Log[I + Sinh[x]]

Rule 3159

Int[((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Int[Cos[d + e*x
]/(b + a*Cos[d + e*x] + c*Sin[d + e*x]), x] /; FreeQ[{a, b, c, d, e}, x]

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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0173788, size = 17, normalized size = 1.55 $2 \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+i \log (\cosh (x))$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Sech[x] - I*Tanh[x])^(-1),x]

[Out]

2*ArcTan[Tanh[x/2]] + I*Log[Cosh[x]]

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Maple [B]  time = 0.045, size = 33, normalized size = 3. \begin{align*} -i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) -i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) +2\,i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.04957, size = 20, normalized size = 1.82 \begin{align*} i \, x + 2 i \, \log \left (i \, e^{\left (-x\right )} + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*x + 2*I*log(I*e^(-x) + 1)

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Fricas [A]  time = 2.05302, size = 34, normalized size = 3.09 \begin{align*} -i \, x + 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)),x, algorithm="fricas")

[Out]

-I*x + 2*I*log(e^x + I)

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Sympy [B]  time = 0.663409, size = 22, normalized size = 2. \begin{align*} i x + i \log{\left (- i \tanh{\left (x \right )} + \operatorname{sech}{\left (x \right )} \right )} - i \log{\left (\tanh{\left (x \right )} + 1 \right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*x + I*log(-I*tanh(x) + sech(x)) - I*log(tanh(x) + 1)

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Giac [A]  time = 1.17548, size = 18, normalized size = 1.64 \begin{align*} -i \, x + 2 i \, \log \left (-i \, e^{x} + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x + 2*I*log(-I*e^x + 1)