3.139 \(\int \frac {\tan (a+i \log (x))}{x} \, dx\)

Optimal. Leaf size=14 \[ i \log (\cos (a+i \log (x))) \]

[Out]

I*ln(cos(a+I*ln(x)))

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3475} \[ i \log (\cos (a+i \log (x))) \]

Antiderivative was successfully verified.

[In]

Int[Tan[a + I*Log[x]]/x,x]

[Out]

I*Log[Cos[a + I*Log[x]]]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\tan (a+i \log (x))}{x} \, dx &=\operatorname {Subst}(\int \tan (a+i x) \, dx,x,\log (x))\\ &=i \log (\cos (a+i \log (x)))\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 14, normalized size = 1.00 \[ i \log (\cos (a+i \log (x))) \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[a + I*Log[x]]/x,x]

[Out]

I*Log[Cos[a + I*Log[x]]]

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fricas [A]  time = 0.41, size = 16, normalized size = 1.14 \[ i \, \log \left (x^{2} + e^{\left (2 i \, a\right )}\right ) - i \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*log(x))/x,x, algorithm="fricas")

[Out]

I*log(x^2 + e^(2*I*a)) - I*log(x)

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giac [B]  time = 0.24, size = 43, normalized size = 3.07 \[ i \, \log \left (\frac {i \, {\left (x^{2} - 1\right )} \tan \relax (a)}{x^{2} + 1} - 1\right ) - \frac {1}{2} i \, \log \left (-\frac {{\left (x^{2} - 1\right )}^{2}}{{\left (x^{2} + 1\right )}^{2}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*log(x))/x,x, algorithm="giac")

[Out]

I*log(I*(x^2 - 1)*tan(a)/(x^2 + 1) - 1) - 1/2*I*log(-(x^2 - 1)^2/(x^2 + 1)^2 + 1)

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maple [A]  time = 0.00, size = 17, normalized size = 1.21 \[ -\frac {i \ln \left (1+\tan ^{2}\left (a +i \ln \relax (x )\right )\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(a+I*ln(x))/x,x)

[Out]

-1/2*I*ln(1+tan(a+I*ln(x))^2)

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maxima [A]  time = 0.33, size = 10, normalized size = 0.71 \[ -i \, \log \left (\sec \left (a + i \, \log \relax (x)\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*log(x))/x,x, algorithm="maxima")

[Out]

-I*log(sec(a + I*log(x)))

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mupad [B]  time = 3.73, size = 16, normalized size = 1.14 \[ -\frac {\ln \left ({\mathrm {tan}\left (a+\ln \relax (x)\,1{}\mathrm {i}\right )}^2+1\right )\,1{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(a + log(x)*1i)/x,x)

[Out]

-(log(tan(a + log(x)*1i)^2 + 1)*1i)/2

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sympy [A]  time = 0.27, size = 17, normalized size = 1.21 \[ - i \log {\relax (x )} + i \log {\left (x^{2} + e^{2 i a} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(a+I*ln(x))/x,x)

[Out]

-I*log(x) + I*log(x**2 + exp(2*I*a))

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