3.169 \(\int \log (a \tan ^n(x)) \, dx\)

Optimal. Leaf size=56 \[ x \log \left (a \tan ^n(x)\right )-\frac {1}{2} i n \text {Li}_2\left (-e^{2 i x}\right )+\frac {1}{2} i n \text {Li}_2\left (e^{2 i x}\right )+2 n x \tanh ^{-1}\left (e^{2 i x}\right ) \]

[Out]

2*n*x*arctanh(exp(2*I*x))+x*ln(a*tan(x)^n)-1/2*I*n*polylog(2,-exp(2*I*x))+1/2*I*n*polylog(2,exp(2*I*x))

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Rubi [A]  time = 0.05, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {2548, 12, 4419, 4183, 2279, 2391} \[ -\frac {1}{2} i n \text {PolyLog}\left (2,-e^{2 i x}\right )+\frac {1}{2} i n \text {PolyLog}\left (2,e^{2 i x}\right )+x \log \left (a \tan ^n(x)\right )+2 n x \tanh ^{-1}\left (e^{2 i x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[a*Tan[x]^n],x]

[Out]

2*n*x*ArcTanh[E^((2*I)*x)] + x*Log[a*Tan[x]^n] - (I/2)*n*PolyLog[2, -E^((2*I)*x)] + (I/2)*n*PolyLog[2, E^((2*I
)*x)]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 4419

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dist[
2^n, Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]

Rubi steps

\begin {align*} \int \log \left (a \tan ^n(x)\right ) \, dx &=x \log \left (a \tan ^n(x)\right )-\int n x \csc (x) \sec (x) \, dx\\ &=x \log \left (a \tan ^n(x)\right )-n \int x \csc (x) \sec (x) \, dx\\ &=x \log \left (a \tan ^n(x)\right )-(2 n) \int x \csc (2 x) \, dx\\ &=2 n x \tanh ^{-1}\left (e^{2 i x}\right )+x \log \left (a \tan ^n(x)\right )+n \int \log \left (1-e^{2 i x}\right ) \, dx-n \int \log \left (1+e^{2 i x}\right ) \, dx\\ &=2 n x \tanh ^{-1}\left (e^{2 i x}\right )+x \log \left (a \tan ^n(x)\right )-\frac {1}{2} (i n) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i x}\right )+\frac {1}{2} (i n) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i x}\right )\\ &=2 n x \tanh ^{-1}\left (e^{2 i x}\right )+x \log \left (a \tan ^n(x)\right )-\frac {1}{2} i n \text {Li}_2\left (-e^{2 i x}\right )+\frac {1}{2} i n \text {Li}_2\left (e^{2 i x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 81, normalized size = 1.45 \[ -\frac {1}{2} i \log (-i (-\tan (x)+i)) \log \left (a \tan ^n(x)\right )+\frac {1}{2} i \log (-i (\tan (x)+i)) \log \left (a \tan ^n(x)\right )-\frac {1}{2} i n \text {Li}_2(-i \tan (x))+\frac {1}{2} i n \text {Li}_2(i \tan (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[a*Tan[x]^n],x]

[Out]

(-1/2*I)*Log[(-I)*(I - Tan[x])]*Log[a*Tan[x]^n] + (I/2)*Log[a*Tan[x]^n]*Log[(-I)*(I + Tan[x])] - (I/2)*n*PolyL
og[2, (-I)*Tan[x]] + (I/2)*n*PolyLog[2, I*Tan[x]]

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fricas [B]  time = 0.52, size = 195, normalized size = 3.48 \[ -\frac {1}{2} \, n x \log \left (\frac {2 \, {\left (\tan \relax (x)^{2} + i \, \tan \relax (x)\right )}}{\tan \relax (x)^{2} + 1}\right ) - \frac {1}{2} \, n x \log \left (\frac {2 \, {\left (\tan \relax (x)^{2} - i \, \tan \relax (x)\right )}}{\tan \relax (x)^{2} + 1}\right ) + \frac {1}{2} \, n x \log \left (-\frac {2 \, {\left (i \, \tan \relax (x) - 1\right )}}{\tan \relax (x)^{2} + 1}\right ) + \frac {1}{2} \, n x \log \left (-\frac {2 \, {\left (-i \, \tan \relax (x) - 1\right )}}{\tan \relax (x)^{2} + 1}\right ) + n x \log \left (\tan \relax (x)\right ) - \frac {1}{4} i \, n {\rm Li}_2\left (-\frac {2 \, {\left (\tan \relax (x)^{2} + i \, \tan \relax (x)\right )}}{\tan \relax (x)^{2} + 1} + 1\right ) + \frac {1}{4} i \, n {\rm Li}_2\left (-\frac {2 \, {\left (\tan \relax (x)^{2} - i \, \tan \relax (x)\right )}}{\tan \relax (x)^{2} + 1} + 1\right ) + \frac {1}{4} i \, n {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \relax (x) - 1\right )}}{\tan \relax (x)^{2} + 1} + 1\right ) - \frac {1}{4} i \, n {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \relax (x) - 1\right )}}{\tan \relax (x)^{2} + 1} + 1\right ) + x \log \relax (a) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*tan(x)^n),x, algorithm="fricas")

[Out]

-1/2*n*x*log(2*(tan(x)^2 + I*tan(x))/(tan(x)^2 + 1)) - 1/2*n*x*log(2*(tan(x)^2 - I*tan(x))/(tan(x)^2 + 1)) + 1
/2*n*x*log(-2*(I*tan(x) - 1)/(tan(x)^2 + 1)) + 1/2*n*x*log(-2*(-I*tan(x) - 1)/(tan(x)^2 + 1)) + n*x*log(tan(x)
) - 1/4*I*n*dilog(-2*(tan(x)^2 + I*tan(x))/(tan(x)^2 + 1) + 1) + 1/4*I*n*dilog(-2*(tan(x)^2 - I*tan(x))/(tan(x
)^2 + 1) + 1) + 1/4*I*n*dilog(2*(I*tan(x) - 1)/(tan(x)^2 + 1) + 1) - 1/4*I*n*dilog(2*(-I*tan(x) - 1)/(tan(x)^2
+ 1) + 1) + x*log(a)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log \left (a \tan \relax (x)^{n}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*tan(x)^n),x, algorithm="giac")

[Out]

integrate(log(a*tan(x)^n), x)

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \ln \left (a \left (\tan ^{n}\relax (x )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(a*tan(x)^n),x)

[Out]

int(ln(a*tan(x)^n),x)

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maxima [A]  time = 1.51, size = 48, normalized size = 0.86 \[ -n x \log \left (\tan \relax (x)\right ) + \frac {1}{4} \, {\left (\pi \log \left (\tan \relax (x)^{2} + 1\right ) + 2 i \, {\rm Li}_2\left (i \, \tan \relax (x) + 1\right ) - 2 i \, {\rm Li}_2\left (-i \, \tan \relax (x) + 1\right )\right )} n + x \log \left (a \tan \relax (x)^{n}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*tan(x)^n),x, algorithm="maxima")

[Out]

-n*x*log(tan(x)) + 1/4*(pi*log(tan(x)^2 + 1) + 2*I*dilog(I*tan(x) + 1) - 2*I*dilog(-I*tan(x) + 1))*n + x*log(a
*tan(x)^n)

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mupad [B]  time = 0.07, size = 44, normalized size = 0.79 \[ \frac {n\,\mathrm {polylog}\left (2,{\mathrm {e}}^{x\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2}+x\,\ln \left (a\,{\mathrm {tan}\relax (x)}^n\right )-\frac {n\,\mathrm {polylog}\left (2,-{\mathrm {e}}^{x\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2}+2\,n\,x\,\mathrm {atanh}\left ({\mathrm {e}}^{x\,2{}\mathrm {i}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(a*tan(x)^n),x)

[Out]

(n*polylog(2, exp(x*2i))*1i)/2 + x*log(a*tan(x)^n) - (n*polylog(2, -exp(x*2i))*1i)/2 + 2*n*x*atanh(exp(x*2i))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log {\left (a \tan ^{n}{\relax (x )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(a*tan(x)**n),x)

[Out]

Integral(log(a*tan(x)**n), x)

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