\(\int x \tan ^2(x) \, dx\) [130]

Optimal result

Integrand size = 6, antiderivative size = 15 \[ \int x \tan ^2(x) \, dx=-\frac {x^2}{2}+\log (\cos (x))+x \tan (x) \]

[Out]

-1/2*x^2+ln(cos(x))+x*tan(x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3801, 3556, 30} \[ \int x \tan ^2(x) \, dx=-\frac {x^2}{2}+x \tan (x)+\log (\cos (x)) \]

[In]

Int[x*Tan[x]^2,x]

[Out]

-1/2*x^2 + Log[Cos[x]] + x*Tan[x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3556

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

Rule 3801

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(c + d*x)^m*((b*Tan[e
+ f*x])^(n - 1)/(f*(n - 1))), x] + (-Dist[b*d*(m/(f*(n - 1))), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = x \tan (x)-\int x \, dx-\int \tan (x) \, dx \\ & = -\frac {x^2}{2}+\log (\cos (x))+x \tan (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int x \tan ^2(x) \, dx=-\frac {x^2}{2}+\log (\cos (x))+x \tan (x) \]

[In]

Integrate[x*Tan[x]^2,x]

[Out]

-1/2*x^2 + Log[Cos[x]] + x*Tan[x]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33

method	result	size
norman	\(x \tan \left (x \right )-\frac {x^{2}}{2}-\frac {\ln \left (1+\tan ^{2}\left (x \right )\right )}{2}\)	\(20\)
parallelrisch	\(x \tan \left (x \right )-\frac {x^{2}}{2}-\frac {\ln \left (1+\tan ^{2}\left (x \right )\right )}{2}\)	\(20\)
risch	\(-\frac {x^{2}}{2}-2 i x +\frac {2 i x}{{\mathrm e}^{2 i x}+1}+\ln \left ({\mathrm e}^{2 i x}+1\right )\)	\(32\)

[In]

int(x*tan(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.40 \[ \int x \tan ^2(x) \, dx=-\frac {1}{2} \, x^{2} + x \tan \left (x\right ) + \frac {1}{2} \, \log \left (\frac {1}{\tan \left (x\right )^{2} + 1}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int x \tan ^2(x) \, dx=- \frac {x^{2}}{2} + x \tan {\left (x \right )} - \frac {\log {\left (\tan ^{2}{\left (x \right )} + 1 \right )}}{2} \]

[In]

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

[Out]

-x**2/2 + x*tan(x) - log(tan(x)**2 + 1)/2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (13) = 26\).

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 7.13 \[ \int x \tan ^2(x) \, dx=-\frac {x^{2} \cos \left (2 \, x\right )^{2} + x^{2} \sin \left (2 \, x\right )^{2} + 2 \, x^{2} \cos \left (2 \, x\right ) + x^{2} - {\left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right )} \log \left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right ) - 4 \, x \sin \left (2 \, x\right )}{2 \, {\left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right )}} \]

[In]

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

[Out]

-1/2*(x^2*cos(2*x)^2 + x^2*sin(2*x)^2 + 2*x^2*cos(2*x) + x^2 - (cos(2*x)^2 + sin(2*x)^2 + 2*cos(2*x) + 1)*log(
cos(2*x)^2 + sin(2*x)^2 + 2*cos(2*x) + 1) - 4*x*sin(2*x))/(cos(2*x)^2 + sin(2*x)^2 + 2*cos(2*x) + 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.53 \[ \int x \tan ^2(x) \, dx=-\frac {1}{2} \, x^{2} + x \tan \left (x\right ) + \frac {1}{2} \, \log \left (\frac {4}{\tan \left (x\right )^{2} + 1}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int x \tan ^2(x) \, dx=\ln \left (\cos \left (x\right )\right )+x\,\mathrm {tan}\left (x\right )-\frac {x^2}{2} \]

[In]

int(x*tan(x)^2,x)

[Out]

log(cos(x)) + x*tan(x) - x^2/2