3.718 \(\int \frac{A+C \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\)
Optimal. Leaf size=270 \[ \frac{\left (a^2 C+A b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a b d \left (a^2-b^2\right )}+\frac{\left (3 A b^2-a^2 (2 A-C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac{\left (-a^2 b^2 (5 A+C)+a^4 (-C)+3 A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 b d (a-b) (a+b)^2}-\frac{\left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)}}+\frac{\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \]
[Out]
((3*A*b^2 - a^2*(2*A - C))*EllipticE[(c + d*x)/2, 2])/(a^2*(a^2 - b^2)*d) + ((A*b^2 + a^2*C)*EllipticF[(c + d*
x)/2, 2])/(a*b*(a^2 - b^2)*d) + ((3*A*b^4 - a^4*C - a^2*b^2*(5*A + C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2,
2])/(a^2*(a - b)*b*(a + b)^2*d) - ((3*A*b^2 - a^2*(2*A - C))*Sin[c + d*x])/(a^2*(a^2 - b^2)*d*Sqrt[Cos[c + d*x
]]) + ((A*b^2 + a^2*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x]))
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Rubi [A] time = 1.06579, antiderivative size = 270, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{3056, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac{\left (a^2 C+A b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a b d \left (a^2-b^2\right )}+\frac{\left (3 A b^2-a^2 (2 A-C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac{\left (-a^2 b^2 (5 A+C)+a^4 (-C)+3 A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 b d (a-b) (a+b)^2}-\frac{\left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)}}+\frac{\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
[In]
Int[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])^2),x]
[Out]
((3*A*b^2 - a^2*(2*A - C))*EllipticE[(c + d*x)/2, 2])/(a^2*(a^2 - b^2)*d) + ((A*b^2 + a^2*C)*EllipticF[(c + d*
x)/2, 2])/(a*b*(a^2 - b^2)*d) + ((3*A*b^4 - a^4*C - a^2*b^2*(5*A + C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2,
2])/(a^2*(a - b)*b*(a + b)^2*d) - ((3*A*b^2 - a^2*(2*A - C))*Sin[c + d*x])/(a^2*(a^2 - b^2)*d*Sqrt[Cos[c + d*x
]]) + ((A*b^2 + a^2*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x]))
Rule 3056
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Rule 3055
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))
Rule 3059
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]
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 3002
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
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]
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]
Rubi steps
\begin{align*} \int \frac{A+C \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx &=\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \cos (c+d x))}+\frac{\int \frac{\frac{1}{2} \left (-3 A b^2+2 a^2 \left (A-\frac{C}{2}\right )\right )-a b (A+C) \cos (c+d x)+\frac{1}{2} \left (A b^2+a^2 C\right ) \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{\left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \cos (c+d x))}+\frac{2 \int \frac{\frac{1}{4} b \left (3 A b^2-a^2 (4 A+C)\right )+\frac{1}{2} a \left (2 A b^2-a^2 (A-C)\right ) \cos (c+d x)+\frac{1}{4} b \left (3 A b^2-a^2 (2 A-C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=-\frac{\left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \cos (c+d x))}-\frac{2 \int \frac{-\frac{1}{4} b^2 \left (3 A b^2-a^2 (4 A+C)\right )-\frac{1}{4} a b \left (A b^2+a^2 C\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^2 b \left (a^2-b^2\right )}+\frac{\left (3 A b^2-a^2 (2 A-C)\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac{\left (3 A b^2-a^2 (2 A-C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}-\frac{\left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \cos (c+d x))}+\frac{\left (A b^2+a^2 C\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 a b \left (a^2-b^2\right )}+\frac{\left (3 A b^4-a^4 C-a^2 b^2 (5 A+C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a^2 b \left (a^2-b^2\right )}\\ &=\frac{\left (3 A b^2-a^2 (2 A-C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2+a^2 C\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a b \left (a^2-b^2\right ) d}+\frac{\left (3 A b^4-a^4 C-a^2 b^2 (5 A+C)\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 (a-b) b (a+b)^2 d}-\frac{\left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 4.04672, size = 310, normalized size = 1.15 \[ \frac{4 \sqrt{\cos (c+d x)} \left (\frac{\left (a^2 b C+A b^3\right ) \sin (c+d x)}{\left (b^2-a^2\right ) (a+b \cos (c+d x))}+2 A \tan (c+d x)\right )-\frac{-\frac{2 \left (a^2 b (10 A+C)-9 A b^3\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{\left (8 a A b^2-4 a^3 (A-C)\right ) \left (2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-\frac{2 a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{b}-\frac{2 \left (a^2 (2 A-C)-3 A b^2\right ) \sin (c+d x) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt{\sin ^2(c+d x)}}}{(b-a) (a+b)}}{4 a^2 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])^2),x]
[Out]
(-(((-2*(-9*A*b^3 + a^2*b*(10*A + C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + ((8*a*A*b^2 - 4*a^3
*(A - C))*(2*EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)))/b - (2*(-3*
A*b^2 + a^2*(2*A - C))*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[C
os[c + d*x]]], -1] + (2*a^2 - b^2)*EllipticPi[-(b/a), -ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b*Sqr
t[Sin[c + d*x]^2]))/((-a + b)*(a + b))) + 4*Sqrt[Cos[c + d*x]]*(((A*b^3 + a^2*b*C)*Sin[c + d*x])/((-a^2 + b^2)
*(a + b*Cos[c + d*x])) + 2*A*Tan[c + d*x]))/(4*a^2*d)
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Maple [B] time = 1.803, size = 899, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(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)*(-4*(-A*b^2+C*a^2)/a^2/(-2*a*b+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)*Ellipti
cPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+2*A/a^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)*(-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))+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)+2*(-A*b^2-C*a^2)/a/b*(-1/a*b^2/(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*b+a-b)-1/2/(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)*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))-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*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*b/(a-b),2^(1/2))+1/a/(a^2-
b^2)/(-2*a*b+2*b^2)*b^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*b/(a-b),2^(1/2))))/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((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(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((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(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((A+C*cos(d*x+c)**2)/cos(d*x+c)**(3/2)/(a+b*cos(d*x+c))**2,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^2,x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)/((b*cos(d*x + c) + a)^2*cos(d*x + c)^(3/2)), x)