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

Optimal. Leaf size=74 \[ \frac{16 \tan (x)}{35 a^3 \sqrt{a \sec ^2(x)}}+\frac{8 \tan (x)}{35 a^2 \left (a \sec ^2(x)\right )^{3/2}}+\frac{6 \tan (x)}{35 a \left (a \sec ^2(x)\right )^{5/2}}+\frac{\tan (x)}{7 \left (a \sec ^2(x)\right )^{7/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0349597, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {4122, 192, 191} \[ \frac{16 \tan (x)}{35 a^3 \sqrt{a \sec ^2(x)}}+\frac{8 \tan (x)}{35 a^2 \left (a \sec ^2(x)\right )^{3/2}}+\frac{6 \tan (x)}{35 a \left (a \sec ^2(x)\right )^{5/2}}+\frac{\tan (x)}{7 \left (a \sec ^2(x)\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0347235, size = 42, normalized size = 0.57 \[ \frac{(1225 \sin (x)+245 \sin (3 x)+49 \sin (5 x)+5 \sin (7 x)) \cos (x) \sqrt{a \sec ^2(x)}}{2240 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[x]*Sqrt[a*Sec[x]^2]*(1225*Sin[x] + 245*Sin[3*x] + 49*Sin[5*x] + 5*Sin[7*x]))/(2240*a^4)

________________________________________________________________________________________

Maple [A]  time = 0.081, size = 37, normalized size = 0.5 \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 ({\frac{a}{ \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/(a*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/(a/cos(x)^2)^(7/2)

________________________________________________________________________________________

Maxima [A]  time = 1.7842, size = 38, normalized size = 0.51 \begin{align*} \frac{5 \, \sin \left (7 \, x\right ) + 49 \, \sin \left (5 \, x\right ) + 245 \, \sin \left (3 \, x\right ) + 1225 \, \sin \left (x\right )}{2240 \, a^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2240*(5*sin(7*x) + 49*sin(5*x) + 245*sin(3*x) + 1225*sin(x))/a^(7/2)

________________________________________________________________________________________

Fricas [A]  time = 1.45123, size = 115, normalized size = 1.55 \begin{align*} \frac{{\left (5 \, \cos \left (x\right )^{7} + 6 \, \cos \left (x\right )^{5} + 8 \, \cos \left (x\right )^{3} + 16 \, \cos \left (x\right )\right )} \sqrt{\frac{a}{\cos \left (x\right )^{2}}} \sin \left (x\right )}{35 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(5*cos(x)^7 + 6*cos(x)^5 + 8*cos(x)^3 + 16*cos(x))*sqrt(a/cos(x)^2)*sin(x)/a^4

________________________________________________________________________________________

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/(a*sec(x)**2)**(7/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.6085, size = 149, normalized size = 2.01 \begin{align*} \frac{2 \,{\left (35 \,{\left (\frac{1}{\tan \left (\frac{1}{2} \, x\right )} + \tan \left (\frac{1}{2} \, x\right )\right )}^{6} \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right ) - 140 \,{\left (\frac{1}{\tan \left (\frac{1}{2} \, x\right )} + \tan \left (\frac{1}{2} \, x\right )\right )}^{4} \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right ) + 336 \,{\left (\frac{1}{\tan \left (\frac{1}{2} \, x\right )} + \tan \left (\frac{1}{2} \, x\right )\right )}^{2} \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right ) - 320 \, \mathrm{sgn}\left (-\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )\right )}}{35 \, a^{\frac{7}{2}}{\left (\frac{1}{\tan \left (\frac{1}{2} \, x\right )} + \tan \left (\frac{1}{2} \, x\right )\right )}^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(35*(1/tan(1/2*x) + tan(1/2*x))^6*sgn(-tan(1/2*x)^2 + 1) - 140*(1/tan(1/2*x) + tan(1/2*x))^4*sgn(-tan(1/2
*x)^2 + 1) + 336*(1/tan(1/2*x) + tan(1/2*x))^2*sgn(-tan(1/2*x)^2 + 1) - 320*sgn(-tan(1/2*x)^2 + 1))/(a^(7/2)*(
1/tan(1/2*x) + tan(1/2*x))^7)