3.615 \(\int \frac{\sec ^{\frac{7}{2}}(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=279 \[ -\frac{a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{b d \left (a^2-b^2\right )}-\frac{a^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{\left (3 a^2-2 b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{b^2 d \left (a^2-b^2\right )}-\frac{\left (3 a^2-2 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b^2\right )}-\frac{a \left (3 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d (a-b) (a+b)^2} \]

[Out]

-(((3*a^2 - 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^2*(a^2 - b^2)*d)) - (a*
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b*(a^2 - b^2)*d) - (a*(3*a^2 - 5*b^2)*Sqrt[C
os[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a - b)*b^2*(a + b)^2*d) + ((3*a^2
- 2*b^2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(b^2*(a^2 - b^2)*d) - (a^2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(b*(a^2
- b^2)*d*(a + b*Sec[c + d*x]))

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Rubi [A]  time = 0.643241, antiderivative size = 279, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {3845, 4102, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ -\frac{a^2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{\left (3 a^2-2 b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{b^2 d \left (a^2-b^2\right )}-\frac{a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b d \left (a^2-b^2\right )}-\frac{\left (3 a^2-2 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b^2\right )}-\frac{a \left (3 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(7/2)/(a + b*Sec[c + d*x])^2,x]

[Out]

-(((3*a^2 - 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^2*(a^2 - b^2)*d)) - (a*
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b*(a^2 - b^2)*d) - (a*(3*a^2 - 5*b^2)*Sqrt[C
os[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/((a - b)*b^2*(a + b)^2*d) + ((3*a^2
- 2*b^2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(b^2*(a^2 - b^2)*d) - (a^2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(b*(a^2
- b^2)*d*(a + b*Sec[c + d*x]))

Rule 3845

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(a^2*
d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3))/(b*f*(m + 1)*(a^2 - b^2)), x] + Dist[d
^3/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3)*Simp[a^2*(n - 3) + a*b*(
m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*(m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && N
eQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n, 2]))

Rule 4102

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(d*Csc[e + f*x])^(n - 1))/(b*f*(m + n + 1)), x] + Dist[d/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + (A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C
*n)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 0]

Rule 4106

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2), Int[(d*Csc[
e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a*A - (A*b - a*B)*Csc[e + f*x])/Sqrt[d*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rule 3849

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sec ^{\frac{7}{2}}(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=-\frac{a^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\sqrt{\sec (c+d x)} \left (\frac{a^2}{2}-a b \sec (c+d x)-\frac{1}{2} \left (3 a^2-2 b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{\left (3 a^2-2 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{2 \int \frac{\frac{1}{4} a \left (3 a^2-2 b^2\right )+\frac{1}{2} b \left (2 a^2-b^2\right ) \sec (c+d x)+\frac{1}{4} a \left (3 a^2-4 b^2\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=\frac{\left (3 a^2-2 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{2 \int \frac{\frac{1}{4} a^2 \left (3 a^2-2 b^2\right )-\left (\frac{1}{4} a b \left (3 a^2-2 b^2\right )-\frac{1}{2} a b \left (2 a^2-b^2\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{a^2 b^2 \left (a^2-b^2\right )}-\frac{\left (a \left (3 a^2-5 b^2\right )\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac{\left (3 a^2-2 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{a \int \sqrt{\sec (c+d x)} \, dx}{2 b \left (a^2-b^2\right )}-\frac{\left (3 a^2-2 b^2\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 b^2 \left (a^2-b^2\right )}-\frac{\left (a \left (3 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=-\frac{a \left (3 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{(a-b) b^2 (a+b)^2 d}+\frac{\left (3 a^2-2 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 b \left (a^2-b^2\right )}-\frac{\left (\left (3 a^2-2 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=-\frac{\left (3 a^2-2 b^2\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{b^2 \left (a^2-b^2\right ) d}-\frac{a \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{b \left (a^2-b^2\right ) d}-\frac{a \left (3 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{(a-b) b^2 (a+b)^2 d}+\frac{\left (3 a^2-2 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 5.81816, size = 355, normalized size = 1.27 \[ \frac{\frac{\cot (c+d x) \left (-2 \left (3 a^2 b+3 a^3-4 a b^2-2 b^3\right ) \sqrt{-\tan ^2(c+d x)} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right ),-1\right )+2 b \left (3 a^2-2 b^2\right ) \sqrt{-\tan ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-3 a^2 b \sec ^{\frac{3}{2}}(c+d x)+3 a^2 b \cos (2 (c+d x)) \sec ^{\frac{3}{2}}(c+d x)-6 a^3 \sqrt{-\tan ^2(c+d x)} \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+10 a b^2 \sqrt{-\tan ^2(c+d x)} \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+2 b^3 \sec ^{\frac{3}{2}}(c+d x)-2 b^3 \cos (2 (c+d x)) \sec ^{\frac{3}{2}}(c+d x)\right )}{(a-b) (a+b)}+\frac{2 b \sin (c+d x) \left (2 b \left (a^2-b^2\right ) \sec (c+d x)+3 a^3-2 a b^2\right )}{\left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a \cos (c+d x)+b)}}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^(7/2)/(a + b*Sec[c + d*x])^2,x]

[Out]

((2*b*(3*a^3 - 2*a*b^2 + 2*b*(a^2 - b^2)*Sec[c + d*x])*Sin[c + d*x])/((a^2 - b^2)*(b + a*Cos[c + d*x])*Sqrt[Se
c[c + d*x]]) + (Cot[c + d*x]*(-3*a^2*b*Sec[c + d*x]^(3/2) + 2*b^3*Sec[c + d*x]^(3/2) + 3*a^2*b*Cos[2*(c + d*x)
]*Sec[c + d*x]^(3/2) - 2*b^3*Cos[2*(c + d*x)]*Sec[c + d*x]^(3/2) + 2*b*(3*a^2 - 2*b^2)*EllipticE[ArcSin[Sqrt[S
ec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] - 2*(3*a^3 + 3*a^2*b - 4*a*b^2 - 2*b^3)*EllipticF[ArcSin[Sqrt[Sec[c +
d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] - 6*a^3*EllipticPi[-(b/a), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d
*x]^2] + 10*a*b^2*EllipticPi[-(b/a), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2]))/((a - b)*(a + b)
))/(2*b^3*d)

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Maple [B]  time = 4.619, size = 868, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(7/2)/(a+b*sec(d*x+c))^2,x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*a^2/b^2/(a^2-a*b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(
-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1
/2*c),2*a/(a-b),2^(1/2))-2*a/b*(a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)
^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*a-a+b)-1/2/(a+b)/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^
(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*a/b/(a^2-
b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c
)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*
x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
-1/2/b/(a^2-b^2)/(a^2-a*b)*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+
1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b
)*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)
^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))+2/b^2*(-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c
)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1
/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [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(sec(d*x+c)^(7/2)/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [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(sec(d*x+c)^(7/2)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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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(sec(d*x+c)**(7/2)/(a+b*sec(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{\frac{7}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(7/2)/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^(7/2)/(b*sec(d*x + c) + a)^2, x)