3.46 \(\int \frac {1}{\sec ^2(x)^{7/2}} \, dx\)

Optimal. Leaf size=57 \[ \frac {16 \tan (x)}{35 \sqrt {\sec ^2(x)}}+\frac {8 \tan (x)}{35 \sec ^2(x)^{3/2}}+\frac {6 \tan (x)}{35 \sec ^2(x)^{5/2}}+\frac {\tan (x)}{7 \sec ^2(x)^{7/2}} \]

[Out]

1/7*tan(x)/(sec(x)^2)^(7/2)+6/35*tan(x)/(sec(x)^2)^(5/2)+8/35*tan(x)/(sec(x)^2)^(3/2)+16/35*tan(x)/(sec(x)^2)^
(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4122, 192, 191} \[ \frac {16 \tan (x)}{35 \sqrt {\sec ^2(x)}}+\frac {8 \tan (x)}{35 \sec ^2(x)^{3/2}}+\frac {6 \tan (x)}{35 \sec ^2(x)^{5/2}}+\frac {\tan (x)}{7 \sec ^2(x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[x]^2)^(-7/2),x]

[Out]

Tan[x]/(7*(Sec[x]^2)^(7/2)) + (6*Tan[x])/(35*(Sec[x]^2)^(5/2)) + (8*Tan[x])/(35*(Sec[x]^2)^(3/2)) + (16*Tan[x]
)/(35*Sqrt[Sec[x]^2])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 4122

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(b*ff)
/f, Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rubi steps

\begin {align*} \int \frac {1}{\sec ^2(x)^{7/2}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^{9/2}} \, dx,x,\tan (x)\right )\\ &=\frac {\tan (x)}{7 \sec ^2(x)^{7/2}}+\frac {6}{7} \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^{7/2}} \, dx,x,\tan (x)\right )\\ &=\frac {\tan (x)}{7 \sec ^2(x)^{7/2}}+\frac {6 \tan (x)}{35 \sec ^2(x)^{5/2}}+\frac {24}{35} \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^{5/2}} \, dx,x,\tan (x)\right )\\ &=\frac {\tan (x)}{7 \sec ^2(x)^{7/2}}+\frac {6 \tan (x)}{35 \sec ^2(x)^{5/2}}+\frac {8 \tan (x)}{35 \sec ^2(x)^{3/2}}+\frac {16}{35} \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^{3/2}} \, dx,x,\tan (x)\right )\\ &=\frac {\tan (x)}{7 \sec ^2(x)^{7/2}}+\frac {6 \tan (x)}{35 \sec ^2(x)^{5/2}}+\frac {8 \tan (x)}{35 \sec ^2(x)^{3/2}}+\frac {16 \tan (x)}{35 \sqrt {\sec ^2(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 37, normalized size = 0.65 \[ \frac {(1225 \sin (x)+245 \sin (3 x)+49 \sin (5 x)+5 \sin (7 x)) \sec (x)}{2240 \sqrt {\sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x]^2)^(-7/2),x]

[Out]

(Sec[x]*(1225*Sin[x] + 245*Sin[3*x] + 49*Sin[5*x] + 5*Sin[7*x]))/(2240*Sqrt[Sec[x]^2])

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fricas [A]  time = 0.54, size = 24, normalized size = 0.42 \[ -\frac {1}{35} \, {\left (5 \, \cos \relax (x)^{6} + 6 \, \cos \relax (x)^{4} + 8 \, \cos \relax (x)^{2} + 16\right )} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)^2)^(7/2),x, algorithm="fricas")

[Out]

-1/35*(5*cos(x)^6 + 6*cos(x)^4 + 8*cos(x)^2 + 16)*sin(x)

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giac [A]  time = 0.43, size = 34, normalized size = 0.60 \[ -\frac {1}{7} \, \mathrm {sgn}\left (\cos \relax (x)\right ) \sin \relax (x)^{7} + \frac {3}{5} \, \mathrm {sgn}\left (\cos \relax (x)\right ) \sin \relax (x)^{5} - \mathrm {sgn}\left (\cos \relax (x)\right ) \sin \relax (x)^{3} + \mathrm {sgn}\left (\cos \relax (x)\right ) \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)^2)^(7/2),x, algorithm="giac")

[Out]

-1/7*sgn(cos(x))*sin(x)^7 + 3/5*sgn(cos(x))*sin(x)^5 - sgn(cos(x))*sin(x)^3 + sgn(cos(x))*sin(x)

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maple [A]  time = 0.36, size = 35, normalized size = 0.61 \[ \frac {\sin \relax (x ) \left (5 \left (\cos ^{6}\relax (x )\right )+6 \left (\cos ^{4}\relax (x )\right )+8 \left (\cos ^{2}\relax (x )\right )+16\right ) \left (\cos \left (2 x \right )+1\right )^{3} \sqrt {2}}{560 \cos \relax (x )^{7} \sqrt {\frac {1}{\cos \left (2 x \right )+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sec(x)^2)^(7/2),x)

[Out]

1/35*sin(x)*(5*cos(x)^6+6*cos(x)^4+8*cos(x)^2+16)/cos(x)^7/(1/cos(x)^2)^(7/2)

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maxima [A]  time = 0.34, size = 49, normalized size = 0.86 \[ \frac {16 \, \tan \relax (x)}{35 \, \sqrt {\tan \relax (x)^{2} + 1}} + \frac {8 \, \tan \relax (x)}{35 \, {\left (\tan \relax (x)^{2} + 1\right )}^{\frac {3}{2}}} + \frac {6 \, \tan \relax (x)}{35 \, {\left (\tan \relax (x)^{2} + 1\right )}^{\frac {5}{2}}} + \frac {\tan \relax (x)}{7 \, {\left (\tan \relax (x)^{2} + 1\right )}^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)^2)^(7/2),x, algorithm="maxima")

[Out]

16/35*tan(x)/sqrt(tan(x)^2 + 1) + 8/35*tan(x)/(tan(x)^2 + 1)^(3/2) + 6/35*tan(x)/(tan(x)^2 + 1)^(5/2) + 1/7*ta
n(x)/(tan(x)^2 + 1)^(7/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (\frac {1}{{\cos \relax (x)}^2}\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1/cos(x)^2)^(7/2),x)

[Out]

int(1/(1/cos(x)^2)^(7/2), x)

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sympy [A]  time = 147.49, size = 60, normalized size = 1.05 \[ \frac {16 \tan ^{7}{\relax (x )}}{35 \left (\sec ^{2}{\relax (x )}\right )^{\frac {7}{2}}} + \frac {8 \tan ^{5}{\relax (x )}}{5 \left (\sec ^{2}{\relax (x )}\right )^{\frac {7}{2}}} + \frac {2 \tan ^{3}{\relax (x )}}{\left (\sec ^{2}{\relax (x )}\right )^{\frac {7}{2}}} + \frac {\tan {\relax (x )}}{\left (\sec ^{2}{\relax (x )}\right )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)**2)**(7/2),x)

[Out]

16*tan(x)**7/(35*(sec(x)**2)**(7/2)) + 8*tan(x)**5/(5*(sec(x)**2)**(7/2)) + 2*tan(x)**3/(sec(x)**2)**(7/2) + t
an(x)/(sec(x)**2)**(7/2)

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