3.95 \(\int \sec ^6(x) \tan ^3(x) \, dx\)

Optimal. Leaf size=17 \[ \frac{\sec ^8(x)}{8}-\frac{\sec ^6(x)}{6} \]

[Out]

-Sec[x]^6/6 + Sec[x]^8/8

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Rubi [A]  time = 0.0276463, 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} \[ \frac{\sec ^8(x)}{8}-\frac{\sec ^6(x)}{6} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^6*Tan[x]^3,x]

[Out]

-Sec[x]^6/6 + Sec[x]^8/8

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 ^6(x) \tan ^3(x) \, dx &=\operatorname{Subst}\left (\int x^5 \left (-1+x^2\right ) \, dx,x,\sec (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-x^5+x^7\right ) \, dx,x,\sec (x)\right )\\ &=-\frac{1}{6} \sec ^6(x)+\frac{\sec ^8(x)}{8}\\ \end{align*}

Mathematica [A]  time = 0.0125847, size = 17, normalized size = 1. \[ \frac{\sec ^8(x)}{8}-\frac{\sec ^6(x)}{6} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^6*Tan[x]^3,x]

[Out]

-Sec[x]^6/6 + Sec[x]^8/8

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Maple [B]  time = 0.013, size = 32, normalized size = 1.9 \begin{align*}{\frac{ \left ( \sin \left ( x \right ) \right ) ^{4}}{8\, \left ( \cos \left ( x \right ) \right ) ^{8}}}+{\frac{ \left ( \sin \left ( x \right ) \right ) ^{4}}{12\, \left ( \cos \left ( x \right ) \right ) ^{6}}}+{\frac{ \left ( \sin \left ( x \right ) \right ) ^{4}}{24\, \left ( \cos \left ( x \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^6*tan(x)^3,x)

[Out]

1/8*sin(x)^4/cos(x)^8+1/12*sin(x)^4/cos(x)^6+1/24*sin(x)^4/cos(x)^4

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Maxima [B]  time = 0.928485, size = 49, normalized size = 2.88 \begin{align*} \frac{4 \, \sin \left (x\right )^{2} - 1}{24 \,{\left (\sin \left (x\right )^{8} - 4 \, \sin \left (x\right )^{6} + 6 \, \sin \left (x\right )^{4} - 4 \, \sin \left (x\right )^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^6*tan(x)^3,x, algorithm="maxima")

[Out]

1/24*(4*sin(x)^2 - 1)/(sin(x)^8 - 4*sin(x)^6 + 6*sin(x)^4 - 4*sin(x)^2 + 1)

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Fricas [A]  time = 1.94501, size = 45, normalized size = 2.65 \begin{align*} -\frac{4 \, \cos \left (x\right )^{2} - 3}{24 \, \cos \left (x\right )^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^6*tan(x)^3,x, algorithm="fricas")

[Out]

-1/24*(4*cos(x)^2 - 3)/cos(x)^8

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Sympy [A]  time = 0.126972, size = 15, normalized size = 0.88 \begin{align*} - \frac{4 \cos ^{2}{\left (x \right )} - 3}{24 \cos ^{8}{\left (x \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**6*tan(x)**3,x)

[Out]

-(4*cos(x)**2 - 3)/(24*cos(x)**8)

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Giac [A]  time = 1.04927, size = 19, normalized size = 1.12 \begin{align*} -\frac{4 \, \cos \left (x\right )^{2} - 3}{24 \, \cos \left (x\right )^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^6*tan(x)^3,x, algorithm="giac")

[Out]

-1/24*(4*cos(x)^2 - 3)/cos(x)^8