\(\int \sec (x) \tan ^5(x) \, dx\) [92]

Optimal result

Integrand size = 7, antiderivative size = 19 \[ \int \sec (x) \tan ^5(x) \, dx=\sec (x)-\frac {2 \sec ^3(x)}{3}+\frac {\sec ^5(x)}{5} \]

[Out]

sec(x)-2/3*sec(x)^3+1/5*sec(x)^5

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2686, 200} \[ \int \sec (x) \tan ^5(x) \, dx=\frac {\sec ^5(x)}{5}-\frac {2 \sec ^3(x)}{3}+\sec (x) \]

[In]

Int[Sec[x]*Tan[x]^5,x]

[Out]

Sec[x] - (2*Sec[x]^3)/3 + Sec[x]^5/5

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \sec (x) \tan ^5(x) \, dx=\sec (x)-\frac {2 \sec ^3(x)}{3}+\frac {\sec ^5(x)}{5} \]

[In]

Integrate[Sec[x]*Tan[x]^5,x]

[Out]

Sec[x] - (2*Sec[x]^3)/3 + Sec[x]^5/5

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\sec \left (x \right )-\frac {2 \left (\sec ^{3}\left (x \right )\right )}{3}+\frac {\left (\sec ^{5}\left (x \right )\right )}{5}\)	\(16\)
default	\(\sec \left (x \right )-\frac {2 \left (\sec ^{3}\left (x \right )\right )}{3}+\frac {\left (\sec ^{5}\left (x \right )\right )}{5}\)	\(16\)
risch	\(\frac {2 \,{\mathrm e}^{9 i x}+\frac {8 \,{\mathrm e}^{7 i x}}{3}+\frac {116 \,{\mathrm e}^{5 i x}}{15}+\frac {8 \,{\mathrm e}^{3 i x}}{3}+2 \,{\mathrm e}^{i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{5}}\)	\(48\)

[In]

int(sec(x)*tan(x)^5,x,method=_RETURNVERBOSE)

[Out]

sec(x)-2/3*sec(x)^3+1/5*sec(x)^5

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \sec (x) \tan ^5(x) \, dx=\frac {15 \, \cos \left (x\right )^{4} - 10 \, \cos \left (x\right )^{2} + 3}{15 \, \cos \left (x\right )^{5}} \]

[In]

integrate(sec(x)*tan(x)^5,x, algorithm="fricas")

[Out]

1/15*(15*cos(x)^4 - 10*cos(x)^2 + 3)/cos(x)^5

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \sec (x) \tan ^5(x) \, dx=- \frac {- 15 \cos ^{4}{\left (x \right )} + 10 \cos ^{2}{\left (x \right )} - 3}{15 \cos ^{5}{\left (x \right )}} \]

[In]

integrate(sec(x)*tan(x)**5,x)

[Out]

-(-15*cos(x)**4 + 10*cos(x)**2 - 3)/(15*cos(x)**5)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \sec (x) \tan ^5(x) \, dx=\frac {15 \, \cos \left (x\right )^{4} - 10 \, \cos \left (x\right )^{2} + 3}{15 \, \cos \left (x\right )^{5}} \]

[In]

integrate(sec(x)*tan(x)^5,x, algorithm="maxima")

[Out]

1/15*(15*cos(x)^4 - 10*cos(x)^2 + 3)/cos(x)^5

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \sec (x) \tan ^5(x) \, dx=\frac {15 \, \cos \left (x\right )^{4} - 10 \, \cos \left (x\right )^{2} + 3}{15 \, \cos \left (x\right )^{5}} \]

[In]

integrate(sec(x)*tan(x)^5,x, algorithm="giac")

[Out]

1/15*(15*cos(x)^4 - 10*cos(x)^2 + 3)/cos(x)^5

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \sec (x) \tan ^5(x) \, dx=\frac {{\cos \left (x\right )}^4-\frac {2\,{\cos \left (x\right )}^2}{3}+\frac {1}{5}}{{\cos \left (x\right )}^5} \]

[In]

int(tan(x)^5/cos(x),x)

[Out]

(cos(x)^4 - (2*cos(x)^2)/3 + 1/5)/cos(x)^5