\(\int \sec ^3(x) \tan ^5(x) \, dx\) [93]

Optimal result

Integrand size = 9, antiderivative size = 25 \[ \int \sec ^3(x) \tan ^5(x) \, dx=\frac {\sec ^3(x)}{3}-\frac {2 \sec ^5(x)}{5}+\frac {\sec ^7(x)}{7} \]

[Out]

1/3*sec(x)^3-2/5*sec(x)^5+1/7*sec(x)^7

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, 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, 276} \[ \int \sec ^3(x) \tan ^5(x) \, dx=\frac {\sec ^7(x)}{7}-\frac {2 \sec ^5(x)}{5}+\frac {\sec ^3(x)}{3} \]

[In]

Int[Sec[x]^3*Tan[x]^5,x]

[Out]

Sec[x]^3/3 - (2*Sec[x]^5)/5 + Sec[x]^7/7

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && 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 x^2 \left (-1+x^2\right )^2 \, dx,x,\sec (x)\right ) \\ & = \text {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sec (x)\right ) \\ & = \frac {\sec ^3(x)}{3}-\frac {2 \sec ^5(x)}{5}+\frac {\sec ^7(x)}{7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \sec ^3(x) \tan ^5(x) \, dx=\frac {\sec ^3(x)}{3}-\frac {2 \sec ^5(x)}{5}+\frac {\sec ^7(x)}{7} \]

[In]

Integrate[Sec[x]^3*Tan[x]^5,x]

[Out]

Sec[x]^3/3 - (2*Sec[x]^5)/5 + Sec[x]^7/7

Maple [A] (verified)

Time = 2.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {\left (\sec ^{3}\left (x \right )\right )}{3}-\frac {2 \left (\sec ^{5}\left (x \right )\right )}{5}+\frac {\left (\sec ^{7}\left (x \right )\right )}{7}\)	\(20\)
default	\(\frac {\left (\sec ^{3}\left (x \right )\right )}{3}-\frac {2 \left (\sec ^{5}\left (x \right )\right )}{5}+\frac {\left (\sec ^{7}\left (x \right )\right )}{7}\)	\(20\)
risch	\(\frac {\frac {8 \,{\mathrm e}^{11 i x}}{3}-\frac {32 \,{\mathrm e}^{9 i x}}{15}+\frac {304 \,{\mathrm e}^{7 i x}}{35}-\frac {32 \,{\mathrm e}^{5 i x}}{15}+\frac {8 \,{\mathrm e}^{3 i x}}{3}}{\left ({\mathrm e}^{2 i x}+1\right )^{7}}\)	\(48\)

[In]

int(sec(x)^3*tan(x)^5,x,method=_RETURNVERBOSE)

[Out]

1/3*sec(x)^3-2/5*sec(x)^5+1/7*sec(x)^7

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \sec ^3(x) \tan ^5(x) \, dx=\frac {35 \, \cos \left (x\right )^{4} - 42 \, \cos \left (x\right )^{2} + 15}{105 \, \cos \left (x\right )^{7}} \]

[In]

integrate(sec(x)^3*tan(x)^5,x, algorithm="fricas")

[Out]

1/105*(35*cos(x)^4 - 42*cos(x)^2 + 15)/cos(x)^7

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \sec ^3(x) \tan ^5(x) \, dx=- \frac {- 35 \cos ^{4}{\left (x \right )} + 42 \cos ^{2}{\left (x \right )} - 15}{105 \cos ^{7}{\left (x \right )}} \]

[In]

integrate(sec(x)**3*tan(x)**5,x)

[Out]

-(-35*cos(x)**4 + 42*cos(x)**2 - 15)/(105*cos(x)**7)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \sec ^3(x) \tan ^5(x) \, dx=\frac {35 \, \cos \left (x\right )^{4} - 42 \, \cos \left (x\right )^{2} + 15}{105 \, \cos \left (x\right )^{7}} \]

[In]

integrate(sec(x)^3*tan(x)^5,x, algorithm="maxima")

[Out]

1/105*(35*cos(x)^4 - 42*cos(x)^2 + 15)/cos(x)^7

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \sec ^3(x) \tan ^5(x) \, dx=\frac {35 \, \cos \left (x\right )^{4} - 42 \, \cos \left (x\right )^{2} + 15}{105 \, \cos \left (x\right )^{7}} \]

[In]

integrate(sec(x)^3*tan(x)^5,x, algorithm="giac")

[Out]

1/105*(35*cos(x)^4 - 42*cos(x)^2 + 15)/cos(x)^7

Mupad [B] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \sec ^3(x) \tan ^5(x) \, dx=\frac {\frac {{\cos \left (x\right )}^4}{3}-\frac {2\,{\cos \left (x\right )}^2}{5}+\frac {1}{7}}{{\cos \left (x\right )}^7} \]

[In]

int(tan(x)^5/cos(x)^3,x)

[Out]

(cos(x)^4/3 - (2*cos(x)^2)/5 + 1/7)/cos(x)^7