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

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

[Out]

I*ln(I+sinh(x))

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Rubi [A]  time = 0.03, antiderivative size = 11, normalized size of antiderivative = 1.00, 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 31

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

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.44, size = 11, normalized size = 1.00 \[ -i \, x + 2 i \, \log \left (e^{x} + i\right ) \]

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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giac [A]  time = 0.12, size = 13, normalized size = 1.18 \[ -i \, x + 2 i \, \log \left (-i \, e^{x} + 1\right ) \]

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)

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maple [B]  time = 0.23, size = 33, normalized size = 3.00 \[ -i \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )-i \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )+2 i \ln \left (\tanh \left (\frac {x}{2}\right )+i\right ) \]

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 = 0.32, size = 15, normalized size = 1.36 \[ i \, x + 2 i \, \log \left (i \, e^{\left (-x\right )} + 1\right ) \]

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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mupad [B]  time = 0.13, size = 14, normalized size = 1.27 \[ -x\,1{}\mathrm {i}+\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,2{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(tanh(x)*1i - 1/cosh(x)),x)

[Out]

log(exp(x) + 1i)*2i - x*1i

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sympy [B]  time = 0.32, size = 22, normalized size = 2.00 \[ i x + i \log {\left (- i \tanh {\relax (x )} + \operatorname {sech}{\relax (x )} \right )} - i \log {\left (\tanh {\relax (x )} + 1 \right )} \]

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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