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]

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])

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Rubi [A]  time = 0.0149745, antiderivative size = 43, normalized size of antiderivative = 1., 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 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
]

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 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]

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*}

Mathematica [A]  time = 0.0249409, 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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Maple [A]  time = 0.072, size = 29, normalized size = 0.7 \begin{align*}{\frac{\sin \left ( x \right ) \left ( 3\, \left ( \cos \left ( x \right ) \right ) ^{4}+4\, \left ( \cos \left ( x \right ) \right ) ^{2}+8 \right ) }{15\, \left ( \cos \left ( x \right ) \right ) ^{5}} \left ( \left ( \cos \left ( x \right ) \right ) ^{-2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

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 = 1.16518, size = 50, normalized size = 1.16 \begin{align*} \frac{8 \, \tan \left (x\right )}{15 \, \sqrt{\tan \left (x\right )^{2} + 1}} + \frac{4 \, \tan \left (x\right )}{15 \,{\left (\tan \left (x\right )^{2} + 1\right )}^{\frac{3}{2}}} + \frac{\tan \left (x\right )}{5 \,{\left (\tan \left (x\right )^{2} + 1\right )}^{\frac{5}{2}}} \end{align*}

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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Fricas [A]  time = 1.44419, size = 59, normalized size = 1.37 \begin{align*} -\frac{1}{15} \,{\left (3 \, \cos \left (x\right )^{4} + 4 \, \cos \left (x\right )^{2} + 8\right )} \sin \left (x\right ) \end{align*}

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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Sympy [A]  time = 17.704, size = 44, normalized size = 1.02 \begin{align*} \frac{8 \tan ^{5}{\left (x \right )}}{15 \left (\sec ^{2}{\left (x \right )}\right )^{\frac{5}{2}}} + \frac{4 \tan ^{3}{\left (x \right )}}{3 \left (\sec ^{2}{\left (x \right )}\right )^{\frac{5}{2}}} + \frac{\tan{\left (x \right )}}{\left (\sec ^{2}{\left (x \right )}\right )^{\frac{5}{2}}} \end{align*}

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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Giac [A]  time = 1.25735, size = 34, normalized size = 0.79 \begin{align*} \frac{1}{5} \, \mathrm{sgn}\left (\cos \left (x\right )\right ) \sin \left (x\right )^{5} - \frac{2}{3} \, \mathrm{sgn}\left (\cos \left (x\right )\right ) \sin \left (x\right )^{3} + \mathrm{sgn}\left (\cos \left (x\right )\right ) \sin \left (x\right ) \end{align*}

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)