\(\int \sec ^{\frac {13}{2}}(x) \sin ^5(x) \, dx\) [359]

Optimal result

Integrand size = 11, antiderivative size = 31 \[ \int \sec ^{\frac {13}{2}}(x) \sin ^5(x) \, dx=\frac {2}{3} \sec ^{\frac {3}{2}}(x)-\frac {4}{7} \sec ^{\frac {7}{2}}(x)+\frac {2}{11} \sec ^{\frac {11}{2}}(x) \]

[Out]

2/3*sec(x)^(3/2)-4/7*sec(x)^(7/2)+2/11*sec(x)^(11/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2702, 276} \[ \int \sec ^{\frac {13}{2}}(x) \sin ^5(x) \, dx=\frac {2}{11} \sec ^{\frac {11}{2}}(x)-\frac {4}{7} \sec ^{\frac {7}{2}}(x)+\frac {2}{3} \sec ^{\frac {3}{2}}(x) \]

[In]

Int[Sec[x]^(13/2)*Sin[x]^5,x]

[Out]

(2*Sec[x]^(3/2))/3 - (4*Sec[x]^(7/2))/7 + (2*Sec[x]^(11/2))/11

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 2702

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

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \sqrt {x} \left (-1+x^2\right )^2 \, dx,x,\sec (x)\right ) \\ & = \text {Subst}\left (\int \left (\sqrt {x}-2 x^{5/2}+x^{9/2}\right ) \, dx,x,\sec (x)\right ) \\ & = \frac {2}{3} \sec ^{\frac {3}{2}}(x)-\frac {4}{7} \sec ^{\frac {7}{2}}(x)+\frac {2}{11} \sec ^{\frac {11}{2}}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \sec ^{\frac {13}{2}}(x) \sin ^5(x) \, dx=\frac {1}{924} (135+44 \cos (2 x)+77 \cos (4 x)) \sec ^{\frac {11}{2}}(x) \]

[In]

Integrate[Sec[x]^(13/2)*Sin[x]^5,x]

[Out]

((135 + 44*Cos[2*x] + 77*Cos[4*x])*Sec[x]^(11/2))/924

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(\frac {2 \left (\sec ^{\frac {3}{2}}\left (x \right )\right )}{3}-\frac {4 \left (\sec ^{\frac {7}{2}}\left (x \right )\right )}{7}+\frac {2 \left (\sec ^{\frac {11}{2}}\left (x \right )\right )}{11}\)	\(20\)
default	\(\frac {2 \left (\sec ^{\frac {3}{2}}\left (x \right )\right )}{3}-\frac {4 \left (\sec ^{\frac {7}{2}}\left (x \right )\right )}{7}+\frac {2 \left (\sec ^{\frac {11}{2}}\left (x \right )\right )}{11}\)	\(20\)

[In]

int(sec(x)^(3/2)*tan(x)^5,x,method=_RETURNVERBOSE)

[Out]

2/3*sec(x)^(3/2)-4/7*sec(x)^(7/2)+2/11*sec(x)^(11/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \sec ^{\frac {13}{2}}(x) \sin ^5(x) \, dx=\frac {2 \, {\left (77 \, \cos \left (x\right )^{4} - 66 \, \cos \left (x\right )^{2} + 21\right )}}{231 \, \cos \left (x\right )^{\frac {11}{2}}} \]

[In]

integrate(sec(x)^(3/2)*tan(x)^5,x, algorithm="fricas")

[Out]

2/231*(77*cos(x)^4 - 66*cos(x)^2 + 21)/cos(x)^(11/2)

Sympy [A] (verification not implemented)

Time = 13.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \sec ^{\frac {13}{2}}(x) \sin ^5(x) \, dx=\frac {2 \tan ^{4}{\left (x \right )} \sec ^{\frac {3}{2}}{\left (x \right )}}{11} - \frac {16 \tan ^{2}{\left (x \right )} \sec ^{\frac {3}{2}}{\left (x \right )}}{77} + \frac {64 \sec ^{\frac {3}{2}}{\left (x \right )}}{231} \]

[In]

integrate(sec(x)**(3/2)*tan(x)**5,x)

[Out]

2*tan(x)**4*sec(x)**(3/2)/11 - 16*tan(x)**2*sec(x)**(3/2)/77 + 64*sec(x)**(3/2)/231

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.61 \[ \int \sec ^{\frac {13}{2}}(x) \sin ^5(x) \, dx=\frac {2}{3 \, \cos \left (x\right )^{\frac {3}{2}}} - \frac {4}{7 \, \cos \left (x\right )^{\frac {7}{2}}} + \frac {2}{11 \, \cos \left (x\right )^{\frac {11}{2}}} \]

[In]

integrate(sec(x)^(3/2)*tan(x)^5,x, algorithm="maxima")

[Out]

2/3/cos(x)^(3/2) - 4/7/cos(x)^(7/2) + 2/11/cos(x)^(11/2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \sec ^{\frac {13}{2}}(x) \sin ^5(x) \, dx=\frac {2 \, {\left (77 \, \cos \left (x\right )^{4} - 66 \, \cos \left (x\right )^{2} + 21\right )}}{231 \, \cos \left (x\right )^{\frac {11}{2}}} \]

[In]

integrate(sec(x)^(3/2)*tan(x)^5,x, algorithm="giac")

[Out]

2/231*(77*cos(x)^4 - 66*cos(x)^2 + 21)/cos(x)^(11/2)

Mupad [B] (verification not implemented)

Time = 1.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \sec ^{\frac {13}{2}}(x) \sin ^5(x) \, dx=\frac {2\,{\left (\frac {1}{\cos \left (x\right )}\right )}^{11/2}\,\left (77\,{\cos \left (x\right )}^4-66\,{\cos \left (x\right )}^2+21\right )}{231} \]

[In]

int(tan(x)^5*(1/cos(x))^(3/2),x)

[Out]

(2*(1/cos(x))^(11/2)*(77*cos(x)^4 - 66*cos(x)^2 + 21))/231