\(\int \log (a \sec ^n(x)) \, dx\) [175]

Optimal result

Integrand size = 7, antiderivative size = 51 \[ \int \log \left (a \sec ^n(x)\right ) \, dx=-\frac {1}{2} i n x^2+n x \log \left (1+e^{2 i x}\right )+x \log \left (a \sec ^n(x)\right )-\frac {1}{2} i n \operatorname {PolyLog}\left (2,-e^{2 i x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {2628, 12, 3800, 2221, 2317, 2438} \[ \int \log \left (a \sec ^n(x)\right ) \, dx=x \log \left (a \sec ^n(x)\right )-\frac {1}{2} i n \operatorname {PolyLog}\left (2,-e^{2 i x}\right )-\frac {1}{2} i n x^2+n x \log \left (1+e^{2 i x}\right ) \]

[In]

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

[Out]

(-1/2*I)*n*x^2 + n*x*Log[1 + E^((2*I)*x)] + x*Log[a*Sec[x]^n] - (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 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

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 2438

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 2628

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

Rule 3800

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \log \left (a \sec ^n(x)\right ) \, dx=-\frac {1}{2} i n x^2+n x \log \left (1+e^{2 i x}\right )+x \log \left (a \sec ^n(x)\right )-\frac {1}{2} i n \operatorname {PolyLog}\left (2,-e^{2 i x}\right ) \]

[In]

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

[Out]

(-1/2*I)*n*x^2 + n*x*Log[1 + E^((2*I)*x)] + x*Log[a*Sec[x]^n] - (I/2)*n*PolyLog[2, -E^((2*I)*x)]

Maple [F]

\[\int \ln \left (a \left (\sec ^{n}\left (x \right )\right )\right )d x\]

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (36) = 72\).

Time = 0.34 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.29 \[ \int \log \left (a \sec ^n(x)\right ) \, dx=n x \log \left (\frac {1}{\cos \left (x\right )}\right ) + \frac {1}{2} \, n x \log \left (i \, \cos \left (x\right ) + \sin \left (x\right ) + 1\right ) + \frac {1}{2} \, n x \log \left (i \, \cos \left (x\right ) - \sin \left (x\right ) + 1\right ) + \frac {1}{2} \, n x \log \left (-i \, \cos \left (x\right ) + \sin \left (x\right ) + 1\right ) + \frac {1}{2} \, n x \log \left (-i \, \cos \left (x\right ) - \sin \left (x\right ) + 1\right ) + \frac {1}{2} i \, n {\rm Li}_2\left (i \, \cos \left (x\right ) + \sin \left (x\right )\right ) - \frac {1}{2} i \, n {\rm Li}_2\left (i \, \cos \left (x\right ) - \sin \left (x\right )\right ) - \frac {1}{2} i \, n {\rm Li}_2\left (-i \, \cos \left (x\right ) + \sin \left (x\right )\right ) + \frac {1}{2} i \, n {\rm Li}_2\left (-i \, \cos \left (x\right ) - \sin \left (x\right )\right ) + x \log \left (a\right ) \]

[In]

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

[Out]

n*x*log(1/cos(x)) + 1/2*n*x*log(I*cos(x) + sin(x) + 1) + 1/2*n*x*log(I*cos(x) - sin(x) + 1) + 1/2*n*x*log(-I*c
os(x) + sin(x) + 1) + 1/2*n*x*log(-I*cos(x) - sin(x) + 1) + 1/2*I*n*dilog(I*cos(x) + sin(x)) - 1/2*I*n*dilog(I
*cos(x) - sin(x)) - 1/2*I*n*dilog(-I*cos(x) + sin(x)) + 1/2*I*n*dilog(-I*cos(x) - sin(x)) + x*log(a)

Sympy [F]

\[ \int \log \left (a \sec ^n(x)\right ) \, dx=\int \log {\left (a \sec ^{n}{\left (x \right )} \right )}\, dx \]

[In]

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

[Out]

Integral(log(a*sec(x)**n), x)

Maxima [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.27 \[ \int \log \left (a \sec ^n(x)\right ) \, dx=\frac {1}{2} \, {\left (-i \, x^{2} + 2 i \, x \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right ) + 1\right ) + x \log \left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right ) - i \, {\rm Li}_2\left (-e^{\left (2 i \, x\right )}\right )\right )} n + x \log \left (a \sec \left (x\right )^{n}\right ) \]

[In]

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

[Out]

1/2*(-I*x^2 + 2*I*x*arctan2(sin(2*x), cos(2*x) + 1) + x*log(cos(2*x)^2 + sin(2*x)^2 + 2*cos(2*x) + 1) - I*dilo
g(-e^(2*I*x)))*n + x*log(a*sec(x)^n)

Giac [F]

\[ \int \log \left (a \sec ^n(x)\right ) \, dx=\int { \log \left (a \sec \left (x\right )^{n}\right ) \,d x } \]

[In]

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

[Out]

integrate(log(a*sec(x)^n), x)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.84 \[ \int \log \left (a \sec ^n(x)\right ) \, dx=x\,\ln \left (a\,{\left (\frac {1}{\cos \left (x\right )}\right )}^n\right )-\frac {n\,\mathrm {polylog}\left (2,-{\mathrm {e}}^{x\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2}-\frac {n\,x\,\left (x+\ln \left ({\mathrm {e}}^{x\,2{}\mathrm {i}}+1\right )\,2{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \]

[In]

int(log(a*(1/cos(x))^n),x)

[Out]

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