\(\int \csc (2 x) \log (\tan (x)) \, dx\) [750]

Optimal result

Integrand size = 8, antiderivative size = 9 \[ \int \csc (2 x) \log (\tan (x)) \, dx=\frac {1}{4} \log ^2(\tan (x)) \]

[Out]

1/4*ln(tan(x))^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3855, 6818} \[ \int \csc (2 x) \log (\tan (x)) \, dx=\frac {1}{4} \log ^2(\tan (x)) \]

[In]

Int[Csc[2*x]*Log[Tan[x]],x]

[Out]

Log[Tan[x]]^2/4

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \log ^2(\tan (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \csc (2 x) \log (\tan (x)) \, dx=\frac {1}{4} \log ^2(\tan (x)) \]

[In]

Integrate[Csc[2*x]*Log[Tan[x]],x]

[Out]

Log[Tan[x]]^2/4

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {\ln \left (\tan \left (x \right )\right )^{2}}{4}\)	\(8\)
default	\(\frac {\ln \left (\tan \left (x \right )\right )^{2}}{4}\)	\(8\)
norman	\(\frac {\ln \left (\tan \left (x \right )\right )^{2}}{4}\)	\(8\)
parallelrisch	\(\frac {\ln \left (\tan \left (x \right )\right )^{2}}{4}\)	\(8\)
risch	\(\text {Expression too large to display}\)	\(764\)

[In]

int(csc(2*x)*ln(tan(x)),x,method=_RETURNVERBOSE)

[Out]

1/4*ln(tan(x))^2

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \csc (2 x) \log (\tan (x)) \, dx=\frac {1}{4} \, \log \left (\tan \left (x\right )\right )^{2} \]

[In]

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

[Out]

1/4*log(tan(x))^2

Sympy [F(-1)]

Timed out. \[ \int \csc (2 x) \log (\tan (x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (7) = 14\).

Time = 0.31 (sec) , antiderivative size = 265, normalized size of antiderivative = 29.44 \[ \int \csc (2 x) \log (\tan (x)) \, dx=\frac {1}{4} \, {\left (\pi - 2 \, \arctan \left (\sin \left (x\right ), \cos \left (x\right ) + 1\right ) - 2 \, \arctan \left (\sin \left (x\right ), \cos \left (x\right ) - 1\right )\right )} \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right ) + 1\right ) + \frac {1}{4} \, \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right ) + 1\right )^{2} - \frac {1}{4} \, {\left (\pi - 2 \, \arctan \left (\sin \left (x\right ), \cos \left (x\right ) - 1\right )\right )} \arctan \left (\sin \left (x\right ), \cos \left (x\right ) + 1\right ) + \frac {1}{4} \, \arctan \left (\sin \left (x\right ), \cos \left (x\right ) + 1\right )^{2} - \frac {1}{4} \, \pi \arctan \left (\sin \left (x\right ), \cos \left (x\right ) - 1\right ) + \frac {1}{4} \, \arctan \left (\sin \left (x\right ), \cos \left (x\right ) - 1\right )^{2} + \frac {1}{8} \, {\left (\log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) + \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right )\right )} \log \left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right ) - \frac {1}{16} \, \log \left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right )^{2} - \frac {1}{16} \, \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right )^{2} - \frac {1}{8} \, \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right ) - \frac {1}{16} \, \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right )^{2} - \frac {1}{2} \, \log \left (\cot \left (2 \, x\right ) + \csc \left (2 \, x\right )\right ) \log \left (\tan \left (x\right )\right ) \]

[In]

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

[Out]

1/4*(pi - 2*arctan2(sin(x), cos(x) + 1) - 2*arctan2(sin(x), cos(x) - 1))*arctan2(sin(2*x), cos(2*x) + 1) + 1/4
*arctan2(sin(2*x), cos(2*x) + 1)^2 - 1/4*(pi - 2*arctan2(sin(x), cos(x) - 1))*arctan2(sin(x), cos(x) + 1) + 1/
4*arctan2(sin(x), cos(x) + 1)^2 - 1/4*pi*arctan2(sin(x), cos(x) - 1) + 1/4*arctan2(sin(x), cos(x) - 1)^2 + 1/8
*(log(cos(x)^2 + sin(x)^2 + 2*cos(x) + 1) + log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1))*log(cos(2*x)^2 + sin(2*x)
^2 + 2*cos(2*x) + 1) - 1/16*log(cos(2*x)^2 + sin(2*x)^2 + 2*cos(2*x) + 1)^2 - 1/16*log(cos(x)^2 + sin(x)^2 + 2
*cos(x) + 1)^2 - 1/8*log(cos(x)^2 + sin(x)^2 + 2*cos(x) + 1)*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) - 1/16*lo
g(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1)^2 - 1/2*log(cot(2*x) + csc(2*x))*log(tan(x))

Giac [F]

\[ \int \csc (2 x) \log (\tan (x)) \, dx=\int { \csc \left (2 \, x\right ) \log \left (\tan \left (x\right )\right ) \,d x } \]

[In]

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

[Out]

integrate(csc(2*x)*log(tan(x)), x)

Mupad [B] (verification not implemented)

Time = 27.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 3.00 \[ \int \csc (2 x) \log (\tan (x)) \, dx=\frac {{\ln \left (-\frac {{\mathrm {e}}^{x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}}{{\mathrm {e}}^{x\,2{}\mathrm {i}}+1}\right )}^2}{4} \]

[In]

int(log(tan(x))/sin(2*x),x)

[Out]

log(-(exp(x*2i)*1i - 1i)/(exp(x*2i) + 1))^2/4