3.92 \(\int \sec (x) \tan ^5(x) \, dx\)

Optimal. Leaf size=19 \[ \frac{\sec ^5(x)}{5}-\frac{2 \sec ^3(x)}{3}+\sec (x) \]

[Out]

Sec[x] - (2*Sec[x]^3)/3 + Sec[x]^5/5

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Rubi [A]  time = 0.0163057, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2606, 194} \[ \frac{\sec ^5(x)}{5}-\frac{2 \sec ^3(x)}{3}+\sec (x) \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]*Tan[x]^5,x]

[Out]

Sec[x] - (2*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 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.0097963, size = 19, normalized size = 1. \[ \frac{\sec ^5(x)}{5}-\frac{2 \sec ^3(x)}{3}+\sec (x) \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]*Tan[x]^5,x]

[Out]

Sec[x] - (2*Sec[x]^3)/3 + Sec[x]^5/5

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Maple [B]  time = 0.008, size = 48, normalized size = 2.5 \begin{align*}{\frac{ \left ( \sin \left ( x \right ) \right ) ^{6}}{5\, \left ( \cos \left ( x \right ) \right ) ^{5}}}-{\frac{ \left ( \sin \left ( x \right ) \right ) ^{6}}{15\, \left ( \cos \left ( x \right ) \right ) ^{3}}}+{\frac{ \left ( \sin \left ( x \right ) \right ) ^{6}}{5\,\cos \left ( x \right ) }}+{\frac{\cos \left ( x \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( x \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( x \right ) \right ) ^{2}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)*tan(x)^5,x)

[Out]

1/5*sin(x)^6/cos(x)^5-1/15*sin(x)^6/cos(x)^3+1/5*sin(x)^6/cos(x)+1/5*(8/3+sin(x)^4+4/3*sin(x)^2)*cos(x)

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Maxima [A]  time = 0.937615, size = 27, normalized size = 1.42 \begin{align*} \frac{15 \, \cos \left (x\right )^{4} - 10 \, \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)*tan(x)^5,x, algorithm="maxima")

[Out]

1/15*(15*cos(x)^4 - 10*cos(x)^2 + 3)/cos(x)^5

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Fricas [A]  time = 2.02845, size = 63, normalized size = 3.32 \begin{align*} \frac{15 \, \cos \left (x\right )^{4} - 10 \, \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)*tan(x)^5,x, algorithm="fricas")

[Out]

1/15*(15*cos(x)^4 - 10*cos(x)^2 + 3)/cos(x)^5

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Sympy [A]  time = 0.103827, size = 20, normalized size = 1.05 \begin{align*} \frac{15 \cos ^{4}{\left (x \right )} - 10 \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)*tan(x)**5,x)

[Out]

(15*cos(x)**4 - 10*cos(x)**2 + 3)/(15*cos(x)**5)

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Giac [A]  time = 1.05896, size = 27, normalized size = 1.42 \begin{align*} \frac{15 \, \cos \left (x\right )^{4} - 10 \, \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)*tan(x)^5,x, algorithm="giac")

[Out]

1/15*(15*cos(x)^4 - 10*cos(x)^2 + 3)/cos(x)^5