3.174 \(\int \log (a \sec ^2(x)) \, dx\)

Optimal. Leaf size=45 \[ x \log \left (a \sec ^2(x)\right )-i \text {Li}_2\left (-e^{2 i x}\right )-i x^2+2 x \log \left (1+e^{2 i x}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {2548, 12, 3719, 2190, 2279, 2391} \[ -i \text {PolyLog}\left (2,-e^{2 i x}\right )+x \log \left (a \sec ^2(x)\right )-i x^2+2 x \log \left (1+e^{2 i x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[a*Sec[x]^2],x]

[Out]

(-I)*x^2 + 2*x*Log[1 + E^((2*I)*x)] + x*Log[a*Sec[x]^2] - I*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 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 \left (a \sec ^2(x)\right ) \, dx &=x \log \left (a \sec ^2(x)\right )-\int 2 x \tan (x) \, dx\\ &=x \log \left (a \sec ^2(x)\right )-2 \int x \tan (x) \, dx\\ &=-i x^2+x \log \left (a \sec ^2(x)\right )+4 i \int \frac {e^{2 i x} x}{1+e^{2 i x}} \, dx\\ &=-i x^2+2 x \log \left (1+e^{2 i x}\right )+x \log \left (a \sec ^2(x)\right )-2 \int \log \left (1+e^{2 i x}\right ) \, dx\\ &=-i x^2+2 x \log \left (1+e^{2 i x}\right )+x \log \left (a \sec ^2(x)\right )+i \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i x}\right )\\ &=-i x^2+2 x \log \left (1+e^{2 i x}\right )+x \log \left (a \sec ^2(x)\right )-i \text {Li}_2\left (-e^{2 i x}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 43, normalized size = 0.96 \[ x \left (\log \left (a \sec ^2(x)\right )-i x+2 \log \left (1+e^{2 i x}\right )\right )-i \text {Li}_2\left (-e^{2 i x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[a*Sec[x]^2],x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.50, size = 102, normalized size = 2.27 \[ x \log \left (\frac {a}{\cos \relax (x)^{2}}\right ) + x \log \left (i \, \cos \relax (x) + \sin \relax (x) + 1\right ) + x \log \left (i \, \cos \relax (x) - \sin \relax (x) + 1\right ) + x \log \left (-i \, \cos \relax (x) + \sin \relax (x) + 1\right ) + x \log \left (-i \, \cos \relax (x) - \sin \relax (x) + 1\right ) + i \, {\rm Li}_2\left (i \, \cos \relax (x) + \sin \relax (x)\right ) - i \, {\rm Li}_2\left (i \, \cos \relax (x) - \sin \relax (x)\right ) - i \, {\rm Li}_2\left (-i \, \cos \relax (x) + \sin \relax (x)\right ) + i \, {\rm Li}_2\left (-i \, \cos \relax (x) - \sin \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*sec(x)^2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log \left (a \sec \relax (x)^{2}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*sec(x)^2),x, algorithm="giac")

[Out]

integrate(log(a*sec(x)^2), x)

________________________________________________________________________________________

maple [B]  time = 0.49, size = 118, normalized size = 2.62 \[ -i \ln \left (\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}\right ) \ln \left ({\mathrm e}^{i x}\right )-2 i \ln \left (-i {\mathrm e}^{i x}+1\right ) \ln \left ({\mathrm e}^{i x}\right )-2 i \ln \left (i {\mathrm e}^{i x}+1\right ) \ln \left ({\mathrm e}^{i x}\right )+i \ln \left ({\mathrm e}^{i x}\right )^{2}-2 i \dilog \left (-i {\mathrm e}^{i x}+1\right )-2 i \dilog \left (i {\mathrm e}^{i x}+1\right )-2 i \ln \relax (2) \ln \left ({\mathrm e}^{i x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(a*sec(x)^2),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.17, size = 61, normalized size = 1.36 \[ -i \, x^{2} + 2 i \, x \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right ) + 1\right ) + x \log \left (a \sec \relax (x)^{2}\right ) + x \log \left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right ) - i \, {\rm Li}_2\left (-e^{\left (2 i \, x\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a*sec(x)^2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.38, size = 39, normalized size = 0.87 \[ x\,\ln \left (\frac {a}{{\cos \relax (x)}^2}\right )-\mathrm {polylog}\left (2,-{\mathrm {e}}^{x\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}-x\,\left (x+\ln \left ({\mathrm {e}}^{x\,2{}\mathrm {i}}+1\right )\,2{}\mathrm {i}\right )\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(a/cos(x)^2),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log {\left (a \sec ^{2}{\relax (x )} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(a*sec(x)**2),x)

[Out]

Integral(log(a*sec(x)**2), x)

________________________________________________________________________________________