\(\int 2 x \sec ^2(x) \tan (x) \, dx\) [886]

Optimal result

Integrand size = 9, antiderivative size = 11 \[ \int 2 x \sec ^2(x) \tan (x) \, dx=x \sec ^2(x)-\tan (x) \]

[Out]

x*sec(x)^2-tan(x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {12, 3842, 3852, 8} \[ \int 2 x \sec ^2(x) \tan (x) \, dx=x \sec ^2(x)-\tan (x) \]

[In]

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

[Out]

x*Sec[x]^2 - Tan[x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 3842

Int[(x_)^(m_.)*Sec[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tan[(a_.) + (b_.)*(x_)^(n_.)]^(q_.), x_Symbol] :> Simp[x^(m
- n + 1)*(Sec[a + b*x^n]^p/(b*n*p)), x] - Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Sec[a + b*x^n]^p, x], x] /;
FreeQ[{a, b, p}, x] && IntegerQ[n] && GeQ[m, n] && EqQ[q, 1]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = 2 \int x \sec ^2(x) \tan (x) \, dx \\ & = x \sec ^2(x)-\int \sec ^2(x) \, dx \\ & = x \sec ^2(x)+\text {Subst}(\int 1 \, dx,x,-\tan (x)) \\ & = x \sec ^2(x)-\tan (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.64 \[ \int 2 x \sec ^2(x) \tan (x) \, dx=2 \left (\frac {1}{2} x \sec ^2(x)-\frac {\tan (x)}{2}\right ) \]

[In]

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

[Out]

2*((x*Sec[x]^2)/2 - Tan[x]/2)

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {x}{\cos \left (x \right )^{2}}-\tan \left (x \right )\)	\(12\)
risch	\(\frac {-2 i {\mathrm e}^{2 i x}+4 \,{\mathrm e}^{2 i x} x -2 i}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}\)	\(31\)

[In]

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

[Out]

x/cos(x)^2-tan(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.36 \[ \int 2 x \sec ^2(x) \tan (x) \, dx=-\frac {\cos \left (x\right ) \sin \left (x\right ) - x}{\cos \left (x\right )^{2}} \]

[In]

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

[Out]

-(cos(x)*sin(x) - x)/cos(x)^2

Sympy [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int 2 x \sec ^2(x) \tan (x) \, dx=x \sec ^{2}{\left (x \right )} - \tan {\left (x \right )} \]

[In]

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

[Out]

x*sec(x)**2 - tan(x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (11) = 22\).

Time = 0.21 (sec) , antiderivative size = 133, normalized size of antiderivative = 12.09 \[ \int 2 x \sec ^2(x) \tan (x) \, dx=\frac {2 \, {\left (4 \, x \cos \left (2 \, x\right )^{2} + 4 \, x \sin \left (2 \, x\right )^{2} + {\left (2 \, x \cos \left (2 \, x\right ) + \sin \left (2 \, x\right )\right )} \cos \left (4 \, x\right ) + 2 \, x \cos \left (2 \, x\right ) + {\left (2 \, x \sin \left (2 \, x\right ) - \cos \left (2 \, x\right ) - 1\right )} \sin \left (4 \, x\right ) - \sin \left (2 \, x\right )\right )}}{2 \, {\left (2 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) + 1} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (11) = 22\).

Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 4.73 \[ \int 2 x \sec ^2(x) \tan (x) \, dx=\frac {x \tan \left (\frac {1}{2} \, x\right )^{4} + 2 \, x \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, \tan \left (\frac {1}{2} \, x\right )^{3} + x - 2 \, \tan \left (\frac {1}{2} \, x\right )}{\tan \left (\frac {1}{2} \, x\right )^{4} - 2 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 1} \]

[In]

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

[Out]

(x*tan(1/2*x)^4 + 2*x*tan(1/2*x)^2 + 2*tan(1/2*x)^3 + x - 2*tan(1/2*x))/(tan(1/2*x)^4 - 2*tan(1/2*x)^2 + 1)

Mupad [B] (verification not implemented)

Time = 26.48 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.45 \[ \int 2 x \sec ^2(x) \tan (x) \, dx=\frac {2\,x-\sin \left (2\,x\right )}{2\,{\cos \left (x\right )}^2} \]

[In]

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

[Out]

(2*x - sin(2*x))/(2*cos(x)^2)