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]

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])

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Rubi [A]  time = 0.0188078, antiderivative size = 57, normalized size of antiderivative = 1., 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 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)^{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*}

Mathematica [A]  time = 0.0376147, 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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Maple [A]  time = 0.088, size = 35, normalized size = 0.6 \begin{align*}{\frac{\sin \left ( x \right ) \left ( 5\, \left ( \cos \left ( x \right ) \right ) ^{6}+6\, \left ( \cos \left ( x \right ) \right ) ^{4}+8\, \left ( \cos \left ( x \right ) \right ) ^{2}+16 \right ) }{35\, \left ( \cos \left ( x \right ) \right ) ^{7}} \left ( \left ( \cos \left ( x \right ) \right ) ^{-2} \right ) ^{-{\frac{7}{2}}}} \end{align*}

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 = 1.06939, size = 66, normalized size = 1.16 \begin{align*} \frac{16 \, \tan \left (x\right )}{35 \, \sqrt{\tan \left (x\right )^{2} + 1}} + \frac{8 \, \tan \left (x\right )}{35 \,{\left (\tan \left (x\right )^{2} + 1\right )}^{\frac{3}{2}}} + \frac{6 \, \tan \left (x\right )}{35 \,{\left (\tan \left (x\right )^{2} + 1\right )}^{\frac{5}{2}}} + \frac{\tan \left (x\right )}{7 \,{\left (\tan \left (x\right )^{2} + 1\right )}^{\frac{7}{2}}} \end{align*}

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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Fricas [A]  time = 1.4212, size = 78, normalized size = 1.37 \begin{align*} -\frac{1}{35} \,{\left (5 \, \cos \left (x\right )^{6} + 6 \, \cos \left (x\right )^{4} + 8 \, \cos \left (x\right )^{2} + 16\right )} \sin \left (x\right ) \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.4046, size = 46, normalized size = 0.81 \begin{align*} -\frac{1}{7} \, \mathrm{sgn}\left (\cos \left (x\right )\right ) \sin \left (x\right )^{7} + \frac{3}{5} \, \mathrm{sgn}\left (\cos \left (x\right )\right ) \sin \left (x\right )^{5} - \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)^(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)