\(\int \csc (x) \log (\tan (x)) \sec (x) \, dx\) [749]

Optimal result

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

[Out]

1/2*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 = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2700, 29, 6818} \[ \int \csc (x) \log (\tan (x)) \sec (x) \, dx=\frac {1}{2} \log ^2(\tan (x)) \]

[In]

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

[Out]

Log[Tan[x]]^2/2

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2700

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

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}{2} \log ^2(\tan (x)) \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

Log[Tan[x]]^2/2

Maple [A] (verified)

Time = 1.24 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.33 \[ \int \csc (x) \log (\tan (x)) \sec (x) \, dx=\frac {1}{2} \, \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right )^{2} \]

[In]

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

[Out]

1/2*log(sin(x)/cos(x))^2

Sympy [F]

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

[In]

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

[Out]

Integral(log(tan(x))*csc(x)*sec(x), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \csc (x) \log (\tan (x)) \sec (x) \, dx=\frac {1}{2} \, \log \left (\tan \left (x\right )\right )^{2} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \csc (x) \log (\tan (x)) \sec (x) \, dx=\frac {1}{2} \, \log \left (\tan \left (x\right )\right )^{2} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 29.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 3.00 \[ \int \csc (x) \log (\tan (x)) \sec (x) \, dx=\frac {{\ln \left (-\frac {{\mathrm {e}}^{x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}}{{\mathrm {e}}^{x\,2{}\mathrm {i}}+1}\right )}^2}{2} \]

[In]

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

[Out]

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