\(\int \sec ^3(x) \tan ^3(x) \, dx\) [256]

Optimal result

Integrand size = 9, antiderivative size = 17 \[ \int \sec ^3(x) \tan ^3(x) \, dx=-\frac {1}{3} \sec ^3(x)+\frac {\sec ^5(x)}{5} \]

[Out]

-1/3*sec(x)^3+1/5*sec(x)^5

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2686, 14} \[ \int \sec ^3(x) \tan ^3(x) \, dx=\frac {\sec ^5(x)}{5}-\frac {\sec ^3(x)}{3} \]

[In]

Int[Sec[x]^3*Tan[x]^3,x]

[Out]

-1/3*Sec[x]^3 + Sec[x]^5/5

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2686

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

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int x^2 \left (-1+x^2\right ) \, dx,x,\sec (x)\right ) \\ & = \text {Subst}\left (\int \left (-x^2+x^4\right ) \, dx,x,\sec (x)\right ) \\ & = -\frac {1}{3} \sec ^3(x)+\frac {\sec ^5(x)}{5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \sec ^3(x) \tan ^3(x) \, dx=-\frac {1}{3} \sec ^3(x)+\frac {\sec ^5(x)}{5} \]

[In]

Integrate[Sec[x]^3*Tan[x]^3,x]

[Out]

-1/3*Sec[x]^3 + Sec[x]^5/5

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(-\frac {\left (\sec ^{3}\left (x \right )\right )}{3}+\frac {\left (\sec ^{5}\left (x \right )\right )}{5}\)	\(14\)
default	\(-\frac {\left (\sec ^{3}\left (x \right )\right )}{3}+\frac {\left (\sec ^{5}\left (x \right )\right )}{5}\)	\(14\)
risch	\(-\frac {8 \left (5 \,{\mathrm e}^{7 i x}-2 \,{\mathrm e}^{5 i x}+5 \,{\mathrm e}^{3 i x}\right )}{15 \left ({\mathrm e}^{2 i x}+1\right )^{5}}\)	\(34\)

[In]

int(sec(x)^3*tan(x)^3,x,method=_RETURNVERBOSE)

[Out]

-1/3*sec(x)^3+1/5*sec(x)^5

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \sec ^3(x) \tan ^3(x) \, dx=-\frac {5 \, \cos \left (x\right )^{2} - 3}{15 \, \cos \left (x\right )^{5}} \]

[In]

integrate(sec(x)^3*tan(x)^3,x, algorithm="fricas")

[Out]

-1/15*(5*cos(x)^2 - 3)/cos(x)^5

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \sec ^3(x) \tan ^3(x) \, dx=\frac {3 - 5 \cos ^{2}{\left (x \right )}}{15 \cos ^{5}{\left (x \right )}} \]

[In]

integrate(sec(x)**3*tan(x)**3,x)

[Out]

(3 - 5*cos(x)**2)/(15*cos(x)**5)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \sec ^3(x) \tan ^3(x) \, dx=-\frac {5 \, \cos \left (x\right )^{2} - 3}{15 \, \cos \left (x\right )^{5}} \]

[In]

integrate(sec(x)^3*tan(x)^3,x, algorithm="maxima")

[Out]

-1/15*(5*cos(x)^2 - 3)/cos(x)^5

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \sec ^3(x) \tan ^3(x) \, dx=-\frac {5 \, \cos \left (x\right )^{2} - 3}{15 \, \cos \left (x\right )^{5}} \]

[In]

integrate(sec(x)^3*tan(x)^3,x, algorithm="giac")

[Out]

-1/15*(5*cos(x)^2 - 3)/cos(x)^5

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \sec ^3(x) \tan ^3(x) \, dx=\frac {1}{5\,{\cos \left (x\right )}^5}-\frac {1}{3\,{\cos \left (x\right )}^3} \]

[In]

int(tan(x)^3/cos(x)^3,x)

[Out]

1/(5*cos(x)^5) - 1/(3*cos(x)^3)