\(\int \sec ^4(x) \tan ^3(x) \, dx\) [271]

Optimal result

Integrand size = 9, antiderivative size = 17 \[ \int \sec ^4(x) \tan ^3(x) \, dx=-\frac {1}{4} \sec ^4(x)+\frac {\sec ^6(x)}{6} \]

[Out]

-1/4*sec(x)^4+1/6*sec(x)^6

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} \[ \int \sec ^4(x) \tan ^3(x) \, dx=\frac {\sec ^6(x)}{6}-\frac {\sec ^4(x)}{4} \]

[In]

Int[Sec[x]^4*Tan[x]^3,x]

[Out]

-1/4*Sec[x]^4 + Sec[x]^6/6

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^3 \left (-1+x^2\right ) \, dx,x,\sec (x)\right ) \\ & = \text {Subst}\left (\int \left (-x^3+x^5\right ) \, dx,x,\sec (x)\right ) \\ & = -\frac {1}{4} \sec ^4(x)+\frac {\sec ^6(x)}{6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \sec ^4(x) \tan ^3(x) \, dx=-\frac {1}{4} \sec ^4(x)+\frac {\sec ^6(x)}{6} \]

[In]

Integrate[Sec[x]^4*Tan[x]^3,x]

[Out]

-1/4*Sec[x]^4 + Sec[x]^6/6

Maple [A] (verified)

Time = 2.79 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(-\frac {\left (\sec ^{4}\left (x \right )\right )}{4}+\frac {\left (\sec ^{6}\left (x \right )\right )}{6}\)	\(14\)
default	\(-\frac {\left (\sec ^{4}\left (x \right )\right )}{4}+\frac {\left (\sec ^{6}\left (x \right )\right )}{6}\)	\(14\)
risch	\(-\frac {4 \left (3 \,{\mathrm e}^{8 i x}-2 \,{\mathrm e}^{6 i x}+3 \,{\mathrm e}^{4 i x}\right )}{3 \left ({\mathrm e}^{2 i x}+1\right )^{6}}\)	\(34\)

[In]

int(sec(x)^4*tan(x)^3,x,method=_RETURNVERBOSE)

[Out]

-1/4*sec(x)^4+1/6*sec(x)^6

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \sec ^4(x) \tan ^3(x) \, dx=-\frac {3 \, \cos \left (x\right )^{2} - 2}{12 \, \cos \left (x\right )^{6}} \]

[In]

integrate(sec(x)^4*tan(x)^3,x, algorithm="fricas")

[Out]

-1/12*(3*cos(x)^2 - 2)/cos(x)^6

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \sec ^4(x) \tan ^3(x) \, dx=\frac {2 - 3 \cos ^{2}{\left (x \right )}}{12 \cos ^{6}{\left (x \right )}} \]

[In]

integrate(sec(x)**4*tan(x)**3,x)

[Out]

(2 - 3*cos(x)**2)/(12*cos(x)**6)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 30 vs. \(2 (13) = 26\).

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.76 \[ \int \sec ^4(x) \tan ^3(x) \, dx=-\frac {3 \, \sin \left (x\right )^{2} - 1}{12 \, {\left (\sin \left (x\right )^{6} - 3 \, \sin \left (x\right )^{4} + 3 \, \sin \left (x\right )^{2} - 1\right )}} \]

[In]

integrate(sec(x)^4*tan(x)^3,x, algorithm="maxima")

[Out]

-1/12*(3*sin(x)^2 - 1)/(sin(x)^6 - 3*sin(x)^4 + 3*sin(x)^2 - 1)

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \sec ^4(x) \tan ^3(x) \, dx=-\frac {3 \, \cos \left (x\right )^{2} - 2}{12 \, \cos \left (x\right )^{6}} \]

[In]

integrate(sec(x)^4*tan(x)^3,x, algorithm="giac")

[Out]

-1/12*(3*cos(x)^2 - 2)/cos(x)^6

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \sec ^4(x) \tan ^3(x) \, dx=\frac {{\mathrm {tan}\left (x\right )}^4\,\left (2\,{\mathrm {tan}\left (x\right )}^2+3\right )}{12} \]

[In]

int(tan(x)^3/cos(x)^4,x)

[Out]

(tan(x)^4*(2*tan(x)^2 + 3))/12