3.93 \(\int \sec ^3(x) \tan ^5(x) \, dx\)

Optimal. Leaf size=25 \[ \frac {\sec ^7(x)}{7}-\frac {2 \sec ^5(x)}{5}+\frac {\sec ^3(x)}{3} \]

[Out]

1/3*sec(x)^3-2/5*sec(x)^5+1/7*sec(x)^7

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Rubi [A]  time = 0.03, antiderivative size = 25, normalized size of antiderivative = 1.00, 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, 270} \[ \frac {\sec ^7(x)}{7}-\frac {2 \sec ^5(x)}{5}+\frac {\sec ^3(x)}{3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^3*Tan[x]^5,x]

[Out]

Sec[x]^3/3 - (2*Sec[x]^5)/5 + Sec[x]^7/7

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && 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 ^3(x) \tan ^5(x) \, dx &=\operatorname {Subst}\left (\int x^2 \left (-1+x^2\right )^2 \, dx,x,\sec (x)\right )\\ &=\operatorname {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sec (x)\right )\\ &=\frac {\sec ^3(x)}{3}-\frac {2 \sec ^5(x)}{5}+\frac {\sec ^7(x)}{7}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.00 \[ \frac {\sec ^7(x)}{7}-\frac {2 \sec ^5(x)}{5}+\frac {\sec ^3(x)}{3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^3*Tan[x]^5,x]

[Out]

Sec[x]^3/3 - (2*Sec[x]^5)/5 + Sec[x]^7/7

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fricas [A]  time = 0.42, size = 20, normalized size = 0.80 \[ \frac {35 \, \cos \relax (x)^{4} - 42 \, \cos \relax (x)^{2} + 15}{105 \, \cos \relax (x)^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3*tan(x)^5,x, algorithm="fricas")

[Out]

1/105*(35*cos(x)^4 - 42*cos(x)^2 + 15)/cos(x)^7

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giac [A]  time = 1.00, size = 20, normalized size = 0.80 \[ \frac {35 \, \cos \relax (x)^{4} - 42 \, \cos \relax (x)^{2} + 15}{105 \, \cos \relax (x)^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3*tan(x)^5,x, algorithm="giac")

[Out]

1/105*(35*cos(x)^4 - 42*cos(x)^2 + 15)/cos(x)^7

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maple [B]  time = 0.03, size = 58, normalized size = 2.32 \[ \frac {\sin ^{6}\relax (x )}{35 \cos \relax (x )}-\frac {\sin ^{6}\relax (x )}{105 \cos \relax (x )^{3}}+\frac {\left (\sin ^{4}\relax (x )+\frac {4 \left (\sin ^{2}\relax (x )\right )}{3}+\frac {8}{3}\right ) \cos \relax (x )}{35}+\frac {\sin ^{6}\relax (x )}{35 \cos \relax (x )^{5}}+\frac {\sin ^{6}\relax (x )}{7 \cos \relax (x )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^3*tan(x)^5,x)

[Out]

1/7*sin(x)^6/cos(x)^7+1/35/cos(x)^5*sin(x)^6-1/105/cos(x)^3*sin(x)^6+1/35/cos(x)*sin(x)^6+1/35*(sin(x)^4+4/3*s
in(x)^2+8/3)*cos(x)

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maxima [A]  time = 0.44, size = 20, normalized size = 0.80 \[ \frac {35 \, \cos \relax (x)^{4} - 42 \, \cos \relax (x)^{2} + 15}{105 \, \cos \relax (x)^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3*tan(x)^5,x, algorithm="maxima")

[Out]

1/105*(35*cos(x)^4 - 42*cos(x)^2 + 15)/cos(x)^7

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mupad [B]  time = 0.53, size = 19, normalized size = 0.76 \[ \frac {\frac {{\cos \relax (x)}^4}{3}-\frac {2\,{\cos \relax (x)}^2}{5}+\frac {1}{7}}{{\cos \relax (x)}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^5/cos(x)^3,x)

[Out]

(cos(x)^4/3 - (2*cos(x)^2)/5 + 1/7)/cos(x)^7

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sympy [A]  time = 0.12, size = 22, normalized size = 0.88 \[ - \frac {- 35 \cos ^{4}{\relax (x )} + 42 \cos ^{2}{\relax (x )} - 15}{105 \cos ^{7}{\relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**3*tan(x)**5,x)

[Out]

-(-35*cos(x)**4 + 42*cos(x)**2 - 15)/(105*cos(x)**7)

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