\(\int \log (a \sec (x)) \, dx\) [173]

Optimal result

Integrand size = 5, antiderivative size = 46 \[ \int \log (a \sec (x)) \, dx=-\frac {i x^2}{2}+x \log \left (1+e^{2 i x}\right )+x \log (a \sec (x))-\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {2628, 3800, 2221, 2317, 2438} \[ \int \log (a \sec (x)) \, dx=x \log (a \sec (x))-\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i x}\right )-\frac {i x^2}{2}+x \log \left (1+e^{2 i x}\right ) \]

[In]

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

[Out]

(-1/2*I)*x^2 + x*Log[1 + E^((2*I)*x)] + x*Log[a*Sec[x]] - (I/2)*PolyLog[2, -E^((2*I)*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 (a \sec (x))-\int x \tan (x) \, dx \\ & = -\frac {i x^2}{2}+x \log (a \sec (x))+2 i \int \frac {e^{2 i x} x}{1+e^{2 i x}} \, dx \\ & = -\frac {i x^2}{2}+x \log \left (1+e^{2 i x}\right )+x \log (a \sec (x))-\int \log \left (1+e^{2 i x}\right ) \, dx \\ & = -\frac {i x^2}{2}+x \log \left (1+e^{2 i x}\right )+x \log (a \sec (x))+\frac {1}{2} i \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i x}\right ) \\ & = -\frac {i x^2}{2}+x \log \left (1+e^{2 i x}\right )+x \log (a \sec (x))-\frac {1}{2} i \text {Li}_2\left (-e^{2 i x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \log (a \sec (x)) \, dx=-\frac {i x^2}{2}+x \log \left (1+e^{2 i x}\right )+x \log (a \sec (x))-\frac {1}{2} i \operatorname {PolyLog}\left (2,-e^{2 i x}\right ) \]

[In]

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

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (36 ) = 72\).

Time = 1.32 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.35

method	result	size
default	\(-i \left (\ln \left (2\right ) \ln \left ({\mathrm e}^{i x}\right )+\ln \left ({\mathrm e}^{i x}\right ) \ln \left (\frac {a \,{\mathrm e}^{i x}}{1+{\mathrm e}^{2 i x}}\right )+\ln \left ({\mathrm e}^{i x}\right ) \ln \left (1+i {\mathrm e}^{i x}\right )+\ln \left ({\mathrm e}^{i x}\right ) \ln \left (1-i {\mathrm e}^{i x}\right )+\operatorname {dilog}\left (1+i {\mathrm e}^{i x}\right )+\operatorname {dilog}\left (1-i {\mathrm e}^{i x}\right )-\frac {\ln \left ({\mathrm e}^{i x}\right )^{2}}{2}\right )\)	\(108\)
risch	\(x \ln \left ({\mathrm e}^{i x}\right )-\frac {i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{i x}}{1+{\mathrm e}^{2 i x}}\right )^{3} x}{2}-\frac {i x^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{i x}\right ) \operatorname {csgn}\left (\frac {i}{1+{\mathrm e}^{2 i x}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{i x}}{1+{\mathrm e}^{2 i x}}\right ) x}{2}-i \ln \left ({\mathrm e}^{i x}\right ) \ln \left (1-i {\mathrm e}^{i x}\right )+\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{i x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{i x}}{1+{\mathrm e}^{2 i x}}\right )^{2} x}{2}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{i x}}{1+{\mathrm e}^{2 i x}}\right ) \operatorname {csgn}\left (\frac {i a \,{\mathrm e}^{i x}}{1+{\mathrm e}^{2 i x}}\right ) \operatorname {csgn}\left (i a \right ) x}{2}+x \ln \left (2\right )+\ln \left (a \right ) x -i \operatorname {dilog}\left (1+i {\mathrm e}^{i x}\right )+\frac {i \pi \,\operatorname {csgn}\left (\frac {i}{1+{\mathrm e}^{2 i x}}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{i x}}{1+{\mathrm e}^{2 i x}}\right )^{2} x}{2}+\frac {i \pi \operatorname {csgn}\left (\frac {i a \,{\mathrm e}^{i x}}{1+{\mathrm e}^{2 i x}}\right )^{2} \operatorname {csgn}\left (i a \right ) x}{2}+i \ln \left ({\mathrm e}^{i x}\right ) \ln \left (1+{\mathrm e}^{2 i x}\right )-i \ln \left ({\mathrm e}^{i x}\right ) \ln \left (1+i {\mathrm e}^{i x}\right )-i \operatorname {dilog}\left (1-i {\mathrm e}^{i x}\right )-\frac {i \pi \operatorname {csgn}\left (\frac {i a \,{\mathrm e}^{i x}}{1+{\mathrm e}^{2 i x}}\right )^{3} x}{2}+\frac {i \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{i x}}{1+{\mathrm e}^{2 i x}}\right ) \operatorname {csgn}\left (\frac {i a \,{\mathrm e}^{i x}}{1+{\mathrm e}^{2 i x}}\right )^{2} x}{2}\)	\(399\)

[In]

int(ln(a*sec(x)),x,method=_RETURNVERBOSE)

[Out]

-I*(ln(2)*ln(exp(I*x))+ln(exp(I*x))*ln(a*exp(I*x)/(exp(I*x)^2+1))+ln(exp(I*x))*ln(1+I*exp(I*x))+ln(exp(I*x))*l
n(1-I*exp(I*x))+dilog(1+I*exp(I*x))+dilog(1-I*exp(I*x))-1/2*ln(exp(I*x))^2)

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. 106 vs. \(2 (31) = 62\).

Time = 0.32 (sec) , antiderivative size = 106, normalized size of antiderivative = 2.30 \[ \int \log (a \sec (x)) \, dx=x \log \left (\frac {a}{\cos \left (x\right )}\right ) + \frac {1}{2} \, x \log \left (i \, \cos \left (x\right ) + \sin \left (x\right ) + 1\right ) + \frac {1}{2} \, x \log \left (i \, \cos \left (x\right ) - \sin \left (x\right ) + 1\right ) + \frac {1}{2} \, x \log \left (-i \, \cos \left (x\right ) + \sin \left (x\right ) + 1\right ) + \frac {1}{2} \, x \log \left (-i \, \cos \left (x\right ) - \sin \left (x\right ) + 1\right ) + \frac {1}{2} i \, {\rm Li}_2\left (i \, \cos \left (x\right ) + \sin \left (x\right )\right ) - \frac {1}{2} i \, {\rm Li}_2\left (i \, \cos \left (x\right ) - \sin \left (x\right )\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-i \, \cos \left (x\right ) + \sin \left (x\right )\right ) + \frac {1}{2} i \, {\rm Li}_2\left (-i \, \cos \left (x\right ) - \sin \left (x\right )\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(log(a*sec(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.30 \[ \int \log (a \sec (x)) \, dx=-\frac {1}{2} i \, x^{2} + i \, x \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right ) + 1\right ) + \frac {1}{2} \, x \log \left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right ) + x \log \left (a \sec \left (x\right )\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-e^{\left (2 i \, x\right )}\right ) \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(log(a*sec(x)), x)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \log (a \sec (x)) \, dx=x\,\ln \left (\frac {a}{\cos \left (x\right )}\right )-\frac {\mathrm {polylog}\left (2,-{\mathrm {e}}^{x\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2}-\frac {x\,\left (x+\ln \left ({\mathrm {e}}^{x\,2{}\mathrm {i}}+1\right )\,2{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \]

[In]

int(log(a/cos(x)),x)

[Out]

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