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

Optimal. Leaf size=45 $-i \text{PolyLog}\left (2,-e^{2 i x}\right )+x \log \left (a \sec ^2(x)\right )-i x^2+2 x \log \left (1+e^{2 i x}\right )$

[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)]

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Rubi [A]  time = 0.0543116, antiderivative size = 45, normalized size of antiderivative = 1., 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 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 12

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

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]

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 = 0.02009, size = 43, normalized size = 0.96 $x \left (\log \left (a \sec ^2(x)\right )-i x+2 \log \left (1+e^{2 i x}\right )\right )-i \text{PolyLog}\left (2,-e^{2 i x}\right )$

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)]

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Maple [B]  time = 0.105, size = 118, normalized size = 2.6 \begin{align*} i \left ( \ln \left ({{\rm e}^{ix}} \right ) \right ) ^{2}-i\ln \left ({{\rm e}^{ix}} \right ) \ln \left ({\frac{a{{\rm e}^{2\,ix}}}{ \left ( 1+{{\rm e}^{2\,ix}} \right ) ^{2}}} \right ) -2\,i\ln \left ({{\rm e}^{ix}} \right ) \ln \left ( 1+i{{\rm e}^{ix}} \right ) -2\,i\ln \left ({{\rm e}^{ix}} \right ) \ln \left ( 1-i{{\rm e}^{ix}} \right ) -2\,i\ln \left ({{\rm e}^{ix}} \right ) \ln \left ( 2 \right ) -2\,i{\it dilog} \left ( 1+i{{\rm e}^{ix}} \right ) -2\,i{\it dilog} \left ( 1-i{{\rm e}^{ix}} \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.93225, size = 82, normalized size = 1.82 \begin{align*} -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 \left (x\right )^{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 ) \end{align*}

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))

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Fricas [B]  time = 2.30184, size = 355, normalized size = 7.89 \begin{align*} x \log \left (\frac{a}{\cos \left (x\right )^{2}}\right ) + x \log \left (i \, \cos \left (x\right ) + \sin \left (x\right ) + 1\right ) + x \log \left (i \, \cos \left (x\right ) - \sin \left (x\right ) + 1\right ) + x \log \left (-i \, \cos \left (x\right ) + \sin \left (x\right ) + 1\right ) + x \log \left (-i \, \cos \left (x\right ) - \sin \left (x\right ) + 1\right ) + i \,{\rm Li}_2\left (i \, \cos \left (x\right ) + \sin \left (x\right )\right ) - i \,{\rm Li}_2\left (i \, \cos \left (x\right ) - \sin \left (x\right )\right ) - i \,{\rm Li}_2\left (-i \, \cos \left (x\right ) + \sin \left (x\right )\right ) + i \,{\rm Li}_2\left (-i \, \cos \left (x\right ) - \sin \left (x\right )\right ) \end{align*}

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))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log{\left (a \sec ^{2}{\left (x \right )} \right )}\, dx \end{align*}

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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log \left (a \sec \left (x\right )^{2}\right )\,{d x} \end{align*}

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)