3.271 \(\int \sec ^4(x) \tan ^3(x) \, dx\)

Optimal. Leaf size=17 \[ \frac{\sec ^6(x)}{6}-\frac{\sec ^4(x)}{4} \]

[Out]

-Sec[x]^4/4 + Sec[x]^6/6

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Rubi [A]  time = 0.0245593, 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} \[ \frac{\sec ^6(x)}{6}-\frac{\sec ^4(x)}{4} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^4*Tan[x]^3,x]

[Out]

-Sec[x]^4/4 + Sec[x]^6/6

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 ^4(x) \tan ^3(x) \, dx &=\operatorname{Subst}\left (\int x^3 \left (-1+x^2\right ) \, dx,x,\sec (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-x^3+x^5\right ) \, dx,x,\sec (x)\right )\\ &=-\frac{1}{4} \sec ^4(x)+\frac{\sec ^6(x)}{6}\\ \end{align*}

Mathematica [A]  time = 0.0109855, size = 17, normalized size = 1. \[ \frac{\sec ^6(x)}{6}-\frac{\sec ^4(x)}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^4*Tan[x]^3,x]

[Out]

-Sec[x]^4/4 + Sec[x]^6/6

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Maple [A]  time = 0.012, size = 22, normalized size = 1.3 \begin{align*}{\frac{ \left ( \sin \left ( x \right ) \right ) ^{4}}{6\, \left ( \cos \left ( x \right ) \right ) ^{6}}}+{\frac{ \left ( \sin \left ( x \right ) \right ) ^{4}}{12\, \left ( \cos \left ( x \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^4*tan(x)^3,x)

[Out]

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

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Maxima [B]  time = 0.942643, size = 41, normalized size = 2.41 \begin{align*} -\frac{3 \, \sin \left (x\right )^{2} - 1}{12 \,{\left (\sin \left (x\right )^{6} - 3 \, \sin \left (x\right )^{4} + 3 \, \sin \left (x\right )^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*sin(x)^2 - 1)/(sin(x)^6 - 3*sin(x)^4 + 3*sin(x)^2 - 1)

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Fricas [A]  time = 2.04556, size = 45, normalized size = 2.65 \begin{align*} -\frac{3 \, \cos \left (x\right )^{2} - 2}{12 \, \cos \left (x\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*cos(x)^2 - 2)/cos(x)^6

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Sympy [A]  time = 0.1049, size = 15, normalized size = 0.88 \begin{align*} - \frac{3 \cos ^{2}{\left (x \right )} - 2}{12 \cos ^{6}{\left (x \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*cos(x)**2 - 2)/(12*cos(x)**6)

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Giac [A]  time = 1.07188, size = 19, normalized size = 1.12 \begin{align*} -\frac{3 \, \cos \left (x\right )^{2} - 2}{12 \, \cos \left (x\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*cos(x)^2 - 2)/cos(x)^6