\(\int \sec ^2(x) \tan ^4(x) \, dx\) [86]

Optimal result

Integrand size = 9, antiderivative size = 8 \[ \int \sec ^2(x) \tan ^4(x) \, dx=\frac {\tan ^5(x)}{5} \]

[Out]

1/5*tan(x)^5

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} \[ \int \sec ^2(x) \tan ^4(x) \, dx=\frac {\tan ^5(x)}{5} \]

[In]

Int[Sec[x]^2*Tan[x]^4,x]

[Out]

Tan[x]^5/5

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^4 \, dx,x,\tan (x)\right ) \\ & = \frac {\tan ^5(x)}{5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \sec ^2(x) \tan ^4(x) \, dx=\frac {\tan ^5(x)}{5} \]

[In]

Integrate[Sec[x]^2*Tan[x]^4,x]

[Out]

Tan[x]^5/5

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\left (\tan ^{5}\left (x \right )\right )}{5}\)	\(7\)
default	\(\frac {\left (\tan ^{5}\left (x \right )\right )}{5}\)	\(7\)
risch	\(\frac {2 i \left (5 \,{\mathrm e}^{8 i x}+10 \,{\mathrm e}^{4 i x}+1\right )}{5 \left ({\mathrm e}^{2 i x}+1\right )^{5}}\)	\(29\)

[In]

int(sec(x)^2*tan(x)^4,x,method=_RETURNVERBOSE)

[Out]

1/5*tan(x)^5

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 20 vs. \(2 (6) = 12\).

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 2.50 \[ \int \sec ^2(x) \tan ^4(x) \, dx=\frac {{\left (\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1\right )} \sin \left (x\right )}{5 \, \cos \left (x\right )^{5}} \]

[In]

integrate(sec(x)^2*tan(x)^4,x, algorithm="fricas")

[Out]

1/5*(cos(x)^4 - 2*cos(x)^2 + 1)*sin(x)/cos(x)^5

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (5) = 10\).

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 3.62 \[ \int \sec ^2(x) \tan ^4(x) \, dx=\frac {\sin {\left (x \right )}}{5 \cos {\left (x \right )}} - \frac {2 \sin {\left (x \right )}}{5 \cos ^{3}{\left (x \right )}} + \frac {\sin {\left (x \right )}}{5 \cos ^{5}{\left (x \right )}} \]

[In]

integrate(sec(x)**2*tan(x)**4,x)

[Out]

sin(x)/(5*cos(x)) - 2*sin(x)/(5*cos(x)**3) + sin(x)/(5*cos(x)**5)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \sec ^2(x) \tan ^4(x) \, dx=\frac {1}{5} \, \tan \left (x\right )^{5} \]

[In]

integrate(sec(x)^2*tan(x)^4,x, algorithm="maxima")

[Out]

1/5*tan(x)^5

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \sec ^2(x) \tan ^4(x) \, dx=\frac {1}{5} \, \tan \left (x\right )^{5} \]

[In]

integrate(sec(x)^2*tan(x)^4,x, algorithm="giac")

[Out]

1/5*tan(x)^5

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \sec ^2(x) \tan ^4(x) \, dx=\frac {{\mathrm {tan}\left (x\right )}^5}{5} \]

[In]

int(tan(x)^4/cos(x)^2,x)

[Out]

tan(x)^5/5