3.45 \(\int \frac {1}{\sec ^2(x)^{5/2}} \, dx\)

Optimal. Leaf size=43 \[ \frac {8 \tan (x)}{15 \sqrt {\sec ^2(x)}}+\frac {4 \tan (x)}{15 \sec ^2(x)^{3/2}}+\frac {\tan (x)}{5 \sec ^2(x)^{5/2}} \]

[Out]

1/5*tan(x)/(sec(x)^2)^(5/2)+4/15*tan(x)/(sec(x)^2)^(3/2)+8/15*tan(x)/(sec(x)^2)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4122, 192, 191} \[ \frac {8 \tan (x)}{15 \sqrt {\sec ^2(x)}}+\frac {4 \tan (x)}{15 \sec ^2(x)^{3/2}}+\frac {\tan (x)}{5 \sec ^2(x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[x]^2)^(-5/2),x]

[Out]

Tan[x]/(5*(Sec[x]^2)^(5/2)) + (4*Tan[x])/(15*(Sec[x]^2)^(3/2)) + (8*Tan[x])/(15*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)^{5/2}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^{7/2}} \, dx,x,\tan (x)\right )\\ &=\frac {\tan (x)}{5 \sec ^2(x)^{5/2}}+\frac {4}{5} \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^{5/2}} \, dx,x,\tan (x)\right )\\ &=\frac {\tan (x)}{5 \sec ^2(x)^{5/2}}+\frac {4 \tan (x)}{15 \sec ^2(x)^{3/2}}+\frac {8}{15} \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^{3/2}} \, dx,x,\tan (x)\right )\\ &=\frac {\tan (x)}{5 \sec ^2(x)^{5/2}}+\frac {4 \tan (x)}{15 \sec ^2(x)^{3/2}}+\frac {8 \tan (x)}{15 \sqrt {\sec ^2(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 31, normalized size = 0.72 \[ \frac {(150 \sin (x)+25 \sin (3 x)+3 \sin (5 x)) \sec (x)}{240 \sqrt {\sec ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x]^2)^(-5/2),x]

[Out]

(Sec[x]*(150*Sin[x] + 25*Sin[3*x] + 3*Sin[5*x]))/(240*Sqrt[Sec[x]^2])

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fricas [A]  time = 0.63, size = 18, normalized size = 0.42 \[ -\frac {1}{15} \, {\left (3 \, \cos \relax (x)^{4} + 4 \, \cos \relax (x)^{2} + 8\right )} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)^2)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.43, size = 25, normalized size = 0.58 \[ \frac {1}{5} \, \mathrm {sgn}\left (\cos \relax (x)\right ) \sin \relax (x)^{5} - \frac {2}{3} \, \mathrm {sgn}\left (\cos \relax (x)\right ) \sin \relax (x)^{3} + \mathrm {sgn}\left (\cos \relax (x)\right ) \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)^2)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.32, size = 29, normalized size = 0.67 \[ \frac {\sin \relax (x ) \left (3 \left (\cos ^{4}\relax (x )\right )+4 \left (\cos ^{2}\relax (x )\right )+8\right ) \left (\cos \left (2 x \right )+1\right )^{2} \sqrt {2}}{120 \cos \relax (x )^{5} \sqrt {\frac {1}{\cos \left (2 x \right )+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sec(x)^2)^(5/2),x)

[Out]

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

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maxima [A]  time = 0.42, size = 37, normalized size = 0.86 \[ \frac {8 \, \tan \relax (x)}{15 \, \sqrt {\tan \relax (x)^{2} + 1}} + \frac {4 \, \tan \relax (x)}{15 \, {\left (\tan \relax (x)^{2} + 1\right )}^{\frac {3}{2}}} + \frac {\tan \relax (x)}{5 \, {\left (\tan \relax (x)^{2} + 1\right )}^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)^2)^(5/2),x, algorithm="maxima")

[Out]

8/15*tan(x)/sqrt(tan(x)^2 + 1) + 4/15*tan(x)/(tan(x)^2 + 1)^(3/2) + 1/5*tan(x)/(tan(x)^2 + 1)^(5/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (\frac {1}{{\cos \relax (x)}^2}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1/cos(x)^2)^(5/2),x)

[Out]

int(1/(1/cos(x)^2)^(5/2), x)

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sympy [A]  time = 10.68, size = 44, normalized size = 1.02 \[ \frac {8 \tan ^{5}{\relax (x )}}{15 \left (\sec ^{2}{\relax (x )}\right )^{\frac {5}{2}}} + \frac {4 \tan ^{3}{\relax (x )}}{3 \left (\sec ^{2}{\relax (x )}\right )^{\frac {5}{2}}} + \frac {\tan {\relax (x )}}{\left (\sec ^{2}{\relax (x )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)**2)**(5/2),x)

[Out]

8*tan(x)**5/(15*(sec(x)**2)**(5/2)) + 4*tan(x)**3/(3*(sec(x)**2)**(5/2)) + tan(x)/(sec(x)**2)**(5/2)

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