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

Optimal. Leaf size=21 \[ \frac {2}{5} \tan ^{\frac {5}{2}}(x)+\frac {2}{9} \tan ^{\frac {9}{2}}(x) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2687, 14} \begin {gather*} \frac {2}{9} \tan ^{\frac {9}{2}}(x)+\frac {2}{5} \tan ^{\frac {5}{2}}(x) \end {gather*}

Antiderivative was successfully verified.

[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*} \int \sec ^4(x) \tan ^{\frac {3}{2}}(x) \, dx &=\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*}

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Mathematica [A]
time = 0.03, size = 22, normalized size = 1.05 \begin {gather*} \frac {2}{45} (7+2 \cos (2 x)) \sec ^2(x) \tan ^{\frac {5}{2}}(x) \end {gather*}

Antiderivative was successfully verified.

[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

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Maple [A]
time = 0.28, size = 26, normalized size = 1.24

method	result	size
default	\(\frac {2 \left (4 \left (\cos ^{2}\left (x \right )\right )+5\right ) \sin \left (x \right ) \left (\frac {\sin \left (x \right )}{\cos \left (x \right )}\right )^{\frac {3}{2}}}{45 \cos \left (x \right )^{3}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(4*cos(x)^2+5)*sin(x)*(sin(x)/cos(x))^(3/2)/cos(x)^3

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Maxima [A]
time = 2.13, size = 13, normalized size = 0.62 \begin {gather*} \frac {2}{9} \, \tan \left (x\right )^{\frac {9}{2}} + \frac {2}{5} \, \tan \left (x\right )^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (13) = 26\).
time = 1.07, size = 27, normalized size = 1.29 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \tan ^{\frac {3}{2}}{\left (x \right )} \sec ^{4}{\left (x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(x)**(3/2)*sec(x)**4, x)

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Giac [A]
time = 1.27, size = 13, normalized size = 0.62 \begin {gather*} \frac {2}{9} \, \tan \left (x\right )^{\frac {9}{2}} + \frac {2}{5} \, \tan \left (x\right )^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Mupad [B]
time = 1.15, size = 32, normalized size = 1.52 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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