### 3.117 $$\int \sec ^3(x) \tan ^3(x) \, dx$$

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

[Out]

-Sec[x]^3/3 + Sec[x]^5/5

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Rubi [A]  time = 0.0255822, antiderivative size = 17, normalized size of antiderivative = 1., 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 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])

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]]

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*}

Mathematica [A]  time = 0.0121843, size = 17, normalized size = 1. $\frac{\sec ^5(x)}{5}-\frac{\sec ^3(x)}{3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-Sec[x]^3/3 + Sec[x]^5/5

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Maple [B]  time = 0., size = 42, normalized size = 2.5 \begin{align*}{\frac{ \left ( \sin \left ( x \right ) \right ) ^{4}}{5\, \left ( \cos \left ( x \right ) \right ) ^{5}}}+{\frac{ \left ( \sin \left ( x \right ) \right ) ^{4}}{15\, \left ( \cos \left ( x \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( x \right ) \right ) ^{4}}{15\,\cos \left ( x \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( x \right ) \right ) ^{2} \right ) \cos \left ( x \right ) }{15}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.935191, size = 19, normalized size = 1.12 \begin{align*} -\frac{5 \, \cos \left (x\right )^{2} - 3}{15 \, \cos \left (x\right )^{5}} \end{align*}

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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Fricas [A]  time = 1.91428, size = 45, normalized size = 2.65 \begin{align*} -\frac{5 \, \cos \left (x\right )^{2} - 3}{15 \, \cos \left (x\right )^{5}} \end{align*}

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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Sympy [A]  time = 0.097077, size = 15, normalized size = 0.88 \begin{align*} - \frac{5 \cos ^{2}{\left (x \right )} - 3}{15 \cos ^{5}{\left (x \right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(5*cos(x)**2 - 3)/(15*cos(x)**5)

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Giac [A]  time = 1.05615, size = 19, normalized size = 1.12 \begin{align*} -\frac{5 \, \cos \left (x\right )^{2} - 3}{15 \, \cos \left (x\right )^{5}} \end{align*}

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