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)-2/3*sec(x)^3+1/5*sec(x)^5

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Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 1.00, 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 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]

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 (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*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.00 \[ \frac {\sec ^5(x)}{5}-\frac {2 \sec ^3(x)}{3}+\sec (x) \]

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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fricas [A]  time = 0.44, size = 20, normalized size = 1.05 \[ \frac {15 \, \cos \relax (x)^{4} - 10 \, \cos \relax (x)^{2} + 3}{15 \, \cos \relax (x)^{5}} \]

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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giac [A]  time = 1.03, size = 20, normalized size = 1.05 \[ \frac {15 \, \cos \relax (x)^{4} - 10 \, \cos \relax (x)^{2} + 3}{15 \, \cos \relax (x)^{5}} \]

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

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maple [B]  time = 0.02, size = 48, normalized size = 2.53 \[ \frac {\sin ^{6}\relax (x )}{5 \cos \relax (x )}-\frac {\sin ^{6}\relax (x )}{15 \cos \relax (x )^{3}}+\frac {\left (\sin ^{4}\relax (x )+\frac {4 \left (\sin ^{2}\relax (x )\right )}{3}+\frac {8}{3}\right ) \cos \relax (x )}{5}+\frac {\sin ^{6}\relax (x )}{5 \cos \relax (x )^{5}} \]

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*(sin(x)^4+4/3*sin(x)^2+8/3)*cos(x)

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maxima [A]  time = 0.43, size = 20, normalized size = 1.05 \[ \frac {15 \, \cos \relax (x)^{4} - 10 \, \cos \relax (x)^{2} + 3}{15 \, \cos \relax (x)^{5}} \]

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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mupad [B]  time = 0.27, size = 17, normalized size = 0.89 \[ \frac {{\cos \relax (x)}^4-\frac {2\,{\cos \relax (x)}^2}{3}+\frac {1}{5}}{{\cos \relax (x)}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^5/cos(x),x)

[Out]

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

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sympy [A]  time = 0.11, size = 22, normalized size = 1.16 \[ - \frac {- 15 \cos ^{4}{\relax (x )} + 10 \cos ^{2}{\relax (x )} - 3}{15 \cos ^{5}{\relax (x )}} \]

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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