\(\int \sec ^2(x) \tan ^2(x) \, dx\) [354]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 9, antiderivative size = 8 \[ \int \sec ^2(x) \tan ^2(x) \, dx=\frac {\tan ^3(x)}{3} \]

[Out]

1/3*tan(x)^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2687, 30} \[ \int \sec ^2(x) \tan ^2(x) \, dx=\frac {\tan ^3(x)}{3} \]

[In]

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

[Out]

Tan[x]^3/3

Rule 30

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

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int x^2 \, dx,x,\tan (x)\right ) \\ & = \frac {\tan ^3(x)}{3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \sec ^2(x) \tan ^2(x) \, dx=\frac {\tan ^3(x)}{3} \]

[In]

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

[Out]

Tan[x]^3/3

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method result size
derivativedivides \(\frac {\left (\tan ^{3}\left (x \right )\right )}{3}\) \(7\)
default \(\frac {\left (\tan ^{3}\left (x \right )\right )}{3}\) \(7\)
risch \(-\frac {2 i \left (3 \,{\mathrm e}^{4 i x}+1\right )}{3 \left ({\mathrm e}^{2 i x}+1\right )^{3}}\) \(22\)

[In]

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

[Out]

1/3*tan(x)^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 14 vs. \(2 (6) = 12\).

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.75 \[ \int \sec ^2(x) \tan ^2(x) \, dx=-\frac {{\left (\cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )}{3 \, \cos \left (x\right )^{3}} \]

[In]

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

[Out]

-1/3*(cos(x)^2 - 1)*sin(x)/cos(x)^3

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 17 vs. \(2 (5) = 10\).

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 2.12 \[ \int \sec ^2(x) \tan ^2(x) \, dx=- \frac {\sin {\left (x \right )}}{3 \cos {\left (x \right )}} + \frac {\sin {\left (x \right )}}{3 \cos ^{3}{\left (x \right )}} \]

[In]

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

[Out]

-sin(x)/(3*cos(x)) + sin(x)/(3*cos(x)**3)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \sec ^2(x) \tan ^2(x) \, dx=\frac {1}{3} \, \tan \left (x\right )^{3} \]

[In]

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

[Out]

1/3*tan(x)^3

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \sec ^2(x) \tan ^2(x) \, dx=\frac {1}{3} \, \tan \left (x\right )^{3} \]

[In]

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

[Out]

1/3*tan(x)^3

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \sec ^2(x) \tan ^2(x) \, dx=\frac {{\mathrm {tan}\left (x\right )}^3}{3} \]

[In]

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

[Out]

tan(x)^3/3