3.177 \(\int \log (a \csc ^2(x)) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.06, 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, 3717, 2190, 2279, 2391} \[ -i \text {PolyLog}\left (2,e^{2 i x}\right )+x \log \left (a \csc ^2(x)\right )-i x^2+2 x \log \left (1-e^{2 i x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-I)*x^2 + 2*x*Log[1 - E^((2*I)*x)] + x*Log[a*Csc[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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rubi steps

\begin {align*} \int \log \left (a \csc ^2(x)\right ) \, dx &=x \log \left (a \csc ^2(x)\right )-\int -2 x \cot (x) \, dx\\ &=x \log \left (a \csc ^2(x)\right )+2 \int x \cot (x) \, dx\\ &=-i x^2+x \log \left (a \csc ^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 \csc ^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 \csc ^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 \csc ^2(x)\right )-i \text {Li}_2\left (e^{2 i x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 42, normalized size = 0.93 \[ x \log \left (a \csc ^2(x)\right )-i \left (x^2+\text {Li}_2\left (e^{2 i x}\right )\right )+2 x \log \left (1-e^{2 i x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.54, size = 107, normalized size = 2.38 \[ x \log \left (-\frac {a}{\cos \relax (x)^{2} - 1}\right ) + x \log \left (\cos \relax (x) + i \, \sin \relax (x) + 1\right ) + x \log \left (\cos \relax (x) - i \, \sin \relax (x) + 1\right ) + x \log \left (-\cos \relax (x) + i \, \sin \relax (x) + 1\right ) + x \log \left (-\cos \relax (x) - i \, \sin \relax (x) + 1\right ) - i \, {\rm Li}_2\left (\cos \relax (x) + i \, \sin \relax (x)\right ) + i \, {\rm Li}_2\left (\cos \relax (x) - i \, \sin \relax (x)\right ) + i \, {\rm Li}_2\left (-\cos \relax (x) + i \, \sin \relax (x)\right ) - i \, {\rm Li}_2\left (-\cos \relax (x) - i \, \sin \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(a*csc(x)^2), x)

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maple [B]  time = 0.55, size = 88, normalized size = 1.96 \[ -i \ln \left (-\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2}}\right ) \ln \left ({\mathrm e}^{i x}\right )-2 i \ln \left ({\mathrm e}^{i x}+1\right ) \ln \left ({\mathrm e}^{i x}\right )+i \ln \left ({\mathrm e}^{i x}\right )^{2}-2 i \dilog \left ({\mathrm e}^{i x}+1\right )+2 i \dilog \left ({\mathrm e}^{i x}\right )-2 i \ln \relax (2) \ln \left ({\mathrm e}^{i x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 2.31, size = 87, normalized size = 1.93 \[ -i \, x^{2} + 2 i \, x \arctan \left (\sin \relax (x), \cos \relax (x) + 1\right ) - 2 i \, x \arctan \left (\sin \relax (x), -\cos \relax (x) + 1\right ) + x \log \left (a \csc \relax (x)^{2}\right ) + x \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \cos \relax (x) + 1\right ) + x \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \cos \relax (x) + 1\right ) - 2 i \, {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) - 2 i \, {\rm Li}_2\left (e^{\left (i \, x\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x^2 + 2*I*x*arctan2(sin(x), cos(x) + 1) - 2*I*x*arctan2(sin(x), -cos(x) + 1) + x*log(a*csc(x)^2) + x*log(co
s(x)^2 + sin(x)^2 + 2*cos(x) + 1) + x*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) - 2*I*dilog(-e^(I*x)) - 2*I*dilo
g(e^(I*x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \ln \left (\frac {a}{{\sin \relax (x)}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(log(a/sin(x)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(a*csc(x)**2), x)

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