\(\int \sec ^4(x) \tan ^{\frac {3}{2}}(x) \, dx\) [360]

Optimal result

Integrand size = 11, antiderivative size = 21 \[ \int \sec ^4(x) \tan ^{\frac {3}{2}}(x) \, dx=\frac {2}{5} \tan ^{\frac {5}{2}}(x)+\frac {2}{9} \tan ^{\frac {9}{2}}(x) \]

[Out]

2/5*tan(x)^(5/2)+2/9*tan(x)^(9/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, 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 = {2687, 14} \[ \int \sec ^4(x) \tan ^{\frac {3}{2}}(x) \, dx=\frac {2}{9} \tan ^{\frac {9}{2}}(x)+\frac {2}{5} \tan ^{\frac {5}{2}}(x) \]

[In]

Int[Sec[x]^4*Tan[x]^(3/2),x]

[Out]

(2*Tan[x]^(5/2))/5 + (2*Tan[x]^(9/2))/9

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 2687

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \sec ^4(x) \tan ^{\frac {3}{2}}(x) \, dx=\frac {2}{45} (7+2 \cos (2 x)) \sec ^2(x) \tan ^{\frac {5}{2}}(x) \]

[In]

Integrate[Sec[x]^4*Tan[x]^(3/2),x]

[Out]

(2*(7 + 2*Cos[2*x])*Sec[x]^2*Tan[x]^(5/2))/45

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\frac {2 \left (\tan ^{\frac {5}{2}}\left (x \right )\right )}{5}+\frac {2 \left (\tan ^{\frac {9}{2}}\left (x \right )\right )}{9}\)	\(14\)
default	\(\frac {2 \left (\tan ^{\frac {5}{2}}\left (x \right )\right )}{5}+\frac {2 \left (\tan ^{\frac {9}{2}}\left (x \right )\right )}{9}\)	\(14\)

[In]

int(sec(x)^4*tan(x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/5*tan(x)^(5/2)+2/9*tan(x)^(9/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (13) = 26\).

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \sec ^4(x) \tan ^{\frac {3}{2}}(x) \, dx=-\frac {2 \, {\left (4 \, \cos \left (x\right )^{4} + \cos \left (x\right )^{2} - 5\right )} \sqrt {\frac {\sin \left (x\right )}{\cos \left (x\right )}}}{45 \, \cos \left (x\right )^{4}} \]

[In]

integrate(sec(x)^4*tan(x)^(3/2),x, algorithm="fricas")

[Out]

-2/45*(4*cos(x)^4 + cos(x)^2 - 5)*sqrt(sin(x)/cos(x))/cos(x)^4

Sympy [F]

\[ \int \sec ^4(x) \tan ^{\frac {3}{2}}(x) \, dx=\int \tan ^{\frac {3}{2}}{\left (x \right )} \sec ^{4}{\left (x \right )}\, dx \]

[In]

integrate(sec(x)**4*tan(x)**(3/2),x)

[Out]

Integral(tan(x)**(3/2)*sec(x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \sec ^4(x) \tan ^{\frac {3}{2}}(x) \, dx=\frac {2}{9} \, \tan \left (x\right )^{\frac {9}{2}} + \frac {2}{5} \, \tan \left (x\right )^{\frac {5}{2}} \]

[In]

integrate(sec(x)^4*tan(x)^(3/2),x, algorithm="maxima")

[Out]

2/9*tan(x)^(9/2) + 2/5*tan(x)^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \sec ^4(x) \tan ^{\frac {3}{2}}(x) \, dx=\frac {2}{9} \, \tan \left (x\right )^{\frac {9}{2}} + \frac {2}{5} \, \tan \left (x\right )^{\frac {5}{2}} \]

[In]

integrate(sec(x)^4*tan(x)^(3/2),x, algorithm="giac")

[Out]

2/9*tan(x)^(9/2) + 2/5*tan(x)^(5/2)

Mupad [B] (verification not implemented)

Time = 1.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \sec ^4(x) \tan ^{\frac {3}{2}}(x) \, dx=-\frac {4\,\sqrt {\sin \left (2\,x\right )}\,\left (2\,{\cos \left (2\,x\right )}^2+5\,\cos \left (2\,x\right )-7\right )}{45\,{\left (\cos \left (2\,x\right )+1\right )}^{5/2}} \]

[In]

int(tan(x)^(3/2)/cos(x)^4,x)

[Out]

-(4*sin(2*x)^(1/2)*(5*cos(2*x) + 2*cos(2*x)^2 - 7))/(45*(cos(2*x) + 1)^(5/2))