3.173 \(\int \log (a \sec (x)) \, dx\)

Optimal. Leaf size=46 \[ x \log (a \sec (x))-\frac {1}{2} i \text {Li}_2\left (-e^{2 i x}\right )-\frac {i x^2}{2}+x \log \left (1+e^{2 i x}\right ) \]

[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))

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Rubi [A]  time = 0.05, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {2548, 3719, 2190, 2279, 2391} \[ -\frac {1}{2} i \text {PolyLog}\left (2,-e^{2 i x}\right )+x \log (a \sec (x))-\frac {i x^2}{2}+x \log \left (1+e^{2 i x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 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 3719

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*} \int \log (a \sec (x)) \, dx &=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 \operatorname {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*}

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Mathematica [A]  time = 0.01, size = 46, normalized size = 1.00 \[ x \log (a \sec (x))-\frac {1}{2} i \text {Li}_2\left (-e^{2 i x}\right )-\frac {i x^2}{2}+x \log \left (1+e^{2 i x}\right ) \]

Antiderivative was successfully verified.

[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)]

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fricas [B]  time = 0.53, size = 106, normalized size = 2.30 \[ x \log \left (\frac {a}{\cos \relax (x)}\right ) + \frac {1}{2} \, x \log \left (i \, \cos \relax (x) + \sin \relax (x) + 1\right ) + \frac {1}{2} \, x \log \left (i \, \cos \relax (x) - \sin \relax (x) + 1\right ) + \frac {1}{2} \, x \log \left (-i \, \cos \relax (x) + \sin \relax (x) + 1\right ) + \frac {1}{2} \, x \log \left (-i \, \cos \relax (x) - \sin \relax (x) + 1\right ) + \frac {1}{2} i \, {\rm Li}_2\left (i \, \cos \relax (x) + \sin \relax (x)\right ) - \frac {1}{2} i \, {\rm Li}_2\left (i \, \cos \relax (x) - \sin \relax (x)\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-i \, \cos \relax (x) + \sin \relax (x)\right ) + \frac {1}{2} i \, {\rm Li}_2\left (-i \, \cos \relax (x) - \sin \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(a*sec(x)), x)

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maple [B]  time = 0.54, size = 118, normalized size = 2.57 \[ -i \ln \left (\frac {a \,{\mathrm e}^{i x}}{{\mathrm e}^{2 i x}+1}\right ) \ln \left ({\mathrm e}^{i x}\right )-i \ln \left (-i {\mathrm e}^{i x}+1\right ) \ln \left ({\mathrm e}^{i x}\right )-i \ln \left (i {\mathrm e}^{i x}+1\right ) \ln \left ({\mathrm e}^{i x}\right )+\frac {i \ln \left ({\mathrm e}^{i x}\right )^{2}}{2}-i \dilog \left (-i {\mathrm e}^{i x}+1\right )-i \dilog \left (i {\mathrm e}^{i x}+1\right )-i \ln \relax (2) \ln \left ({\mathrm e}^{i x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.27, size = 60, normalized size = 1.30 \[ -\frac {1}{2} i \, x^{2} + i \, x \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right ) + 1\right ) + \frac {1}{2} \, x \log \left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right ) + x \log \left (a \sec \relax (x)\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-e^{\left (2 i \, x\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 0.09, size = 39, normalized size = 0.85 \[ x\,\ln \left (\frac {a}{\cos \relax (x)}\right )-\frac {\mathrm {polylog}\left (2,-{\mathrm {e}}^{x\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2}-\frac {x\,\left (x+\ln \left ({\mathrm {e}}^{x\,2{}\mathrm {i}}+1\right )\,2{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(a*sec(x)), x)

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