3.256 \(\int \sec ^3(x) \tan ^3(x) \, dx\)

Optimal. Leaf size=17 \[ \frac {\sec ^5(x)}{5}-\frac {\sec ^3(x)}{3} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2606, 14} \[ \frac {\sec ^5(x)}{5}-\frac {\sec ^3(x)}{3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sec[x]^3/3 + Sec[x]^5/5

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 2606

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

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Mathematica [A]  time = 0.02, size = 17, normalized size = 1.00 \[ \frac {\sec ^5(x)}{5}-\frac {\sec ^3(x)}{3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*Sec[x]^3 + Sec[x]^5/5

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fricas [A]  time = 0.41, size = 14, normalized size = 0.82 \[ -\frac {5 \, \cos \relax (x)^{2} - 3}{15 \, \cos \relax (x)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(5*cos(x)^2 - 3)/cos(x)^5

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giac [A]  time = 0.96, size = 14, normalized size = 0.82 \[ -\frac {5 \, \cos \relax (x)^{2} - 3}{15 \, \cos \relax (x)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(5*cos(x)^2 - 3)/cos(x)^5

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maple [B]  time = 0.00, size = 42, normalized size = 2.47 \[ -\frac {\sin ^{4}\relax (x )}{15 \cos \relax (x )}-\frac {\left (\sin ^{2}\relax (x )+2\right ) \cos \relax (x )}{15}+\frac {\sin ^{4}\relax (x )}{15 \cos \relax (x )^{3}}+\frac {\sin ^{4}\relax (x )}{5 \cos \relax (x )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^3*tan(x)^3,x)

[Out]

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

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maxima [A]  time = 0.56, size = 14, normalized size = 0.82 \[ -\frac {5 \, \cos \relax (x)^{2} - 3}{15 \, \cos \relax (x)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(5*cos(x)^2 - 3)/cos(x)^5

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mupad [B]  time = 0.00, size = 13, normalized size = 0.76 \[ \frac {1}{5\,{\cos \relax (x)}^5}-\frac {1}{3\,{\cos \relax (x)}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(5*cos(x)^5) - 1/(3*cos(x)^3)

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sympy [A]  time = 0.11, size = 14, normalized size = 0.82 \[ \frac {3 - 5 \cos ^{2}{\relax (x )}}{15 \cos ^{5}{\relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3 - 5*cos(x)**2)/(15*cos(x)**5)

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