\(\int \cos (c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [302]

Optimal result

Integrand size = 37, antiderivative size = 118 \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{8} a (4 A+4 B+3 C) x+\frac {a (3 A+2 (B+C)) \sin (c+d x)}{3 d}+\frac {a (4 A+4 B+3 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (B+C) \cos ^2(c+d x) \sin (c+d x)}{3 d}+\frac {a C \cos ^3(c+d x) \sin (c+d x)}{4 d} \]

[Out]

1/8*a*(4*A+4*B+3*C)*x+1/3*a*(3*A+2*B+2*C)*sin(d*x+c)/d+1/8*a*(4*A+4*B+3*C)*cos(d*x+c)*sin(d*x+c)/d+1/3*a*(B+C)
*cos(d*x+c)^2*sin(d*x+c)/d+1/4*a*C*cos(d*x+c)^3*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {3112, 3102, 2813} \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {a (3 A+2 (B+C)) \sin (c+d x)}{3 d}+\frac {a (4 A+4 B+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a x (4 A+4 B+3 C)+\frac {a (B+C) \sin (c+d x) \cos ^2(c+d x)}{3 d}+\frac {a C \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[In]

Int[Cos[c + d*x]*(a + a*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(a*(4*A + 4*B + 3*C)*x)/8 + (a*(3*A + 2*(B + C))*Sin[c + d*x])/(3*d) + (a*(4*A + 4*B + 3*C)*Cos[c + d*x]*Sin[c
+ d*x])/(8*d) + (a*(B + C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d) + (a*C*Cos[c + d*x]^3*Sin[c + d*x])/(4*d)

Rule 2813

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

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3112

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a +
b*Sin[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*
c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {a C \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{4} \int \cos (c+d x) \left (4 a A+a (4 A+4 B+3 C) \cos (c+d x)+4 a (B+C) \cos ^2(c+d x)\right ) \, dx \\ & = \frac {a (B+C) \cos ^2(c+d x) \sin (c+d x)}{3 d}+\frac {a C \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{12} \int \cos (c+d x) (4 a (3 A+2 (B+C))+3 a (4 A+4 B+3 C) \cos (c+d x)) \, dx \\ & = \frac {1}{8} a (4 A+4 B+3 C) x+\frac {a (3 A+2 (B+C)) \sin (c+d x)}{3 d}+\frac {a (4 A+4 B+3 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (B+C) \cos ^2(c+d x) \sin (c+d x)}{3 d}+\frac {a C \cos ^3(c+d x) \sin (c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.81 \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {a (48 B c+24 c C+48 A d x+48 B d x+36 C d x+24 (4 A+3 (B+C)) \sin (c+d x)+24 (A+B+C) \sin (2 (c+d x))+8 B \sin (3 (c+d x))+8 C \sin (3 (c+d x))+3 C \sin (4 (c+d x)))}{96 d} \]

[In]

Integrate[Cos[c + d*x]*(a + a*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(a*(48*B*c + 24*c*C + 48*A*d*x + 48*B*d*x + 36*C*d*x + 24*(4*A + 3*(B + C))*Sin[c + d*x] + 24*(A + B + C)*Sin[
2*(c + d*x)] + 8*B*Sin[3*(c + d*x)] + 8*C*Sin[3*(c + d*x)] + 3*C*Sin[4*(c + d*x)]))/(96*d)

Maple [A] (verified)

Time = 3.40 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.64

method	result	size
parallelrisch	\(\frac {\left (\frac {\left (A +B +C \right ) \sin \left (2 d x +2 c \right )}{2}+\frac {\left (B +C \right ) \sin \left (3 d x +3 c \right )}{6}+\frac {\sin \left (4 d x +4 c \right ) C}{16}+\left (2 A +\frac {3 B}{2}+\frac {3 C}{2}\right ) \sin \left (d x +c \right )+d x \left (A +B +\frac {3 C}{4}\right )\right ) a}{2 d}\)	\(75\)
parts	\(\frac {\left (a A +B a \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (B a +a C \right ) \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {\sin \left (d x +c \right ) a A}{d}+\frac {a C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(114\)
derivativedivides	\(\frac {a C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a A \sin \left (d x +c \right )}{d}\)	\(141\)
default	\(\frac {a C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a A \sin \left (d x +c \right )}{d}\)	\(141\)
risch	\(\frac {a x A}{2}+\frac {a B x}{2}+\frac {3 a C x}{8}+\frac {\sin \left (d x +c \right ) a A}{d}+\frac {3 a B \sin \left (d x +c \right )}{4 d}+\frac {3 a C \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (4 d x +4 c \right ) a C}{32 d}+\frac {\sin \left (3 d x +3 c \right ) B a}{12 d}+\frac {\sin \left (3 d x +3 c \right ) a C}{12 d}+\frac {\sin \left (2 d x +2 c \right ) a A}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B a}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a C}{4 d}\)	\(151\)
norman	\(\frac {\frac {a \left (4 A +4 B +3 C \right ) x}{8}+\frac {a \left (4 A +4 B +3 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a \left (4 A +4 B +3 C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {3 a \left (4 A +4 B +3 C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {a \left (4 A +4 B +3 C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {a \left (4 A +4 B +3 C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {a \left (12 A +12 B +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (60 A +28 B +49 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {a \left (84 A +52 B +31 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(238\)

[In]

int(cos(d*x+c)*(a+cos(d*x+c)*a)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/2*(1/2*(A+B+C)*sin(2*d*x+2*c)+1/6*(B+C)*sin(3*d*x+3*c)+1/16*sin(4*d*x+4*c)*C+(2*A+3/2*B+3/2*C)*sin(d*x+c)+d*
x*(A+B+3/4*C))*a/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.74 \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (4 \, A + 4 \, B + 3 \, C\right )} a d x + {\left (6 \, C a \cos \left (d x + c\right )^{3} + 8 \, {\left (B + C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 4 \, B + 3 \, C\right )} a \cos \left (d x + c\right ) + 8 \, {\left (3 \, A + 2 \, B + 2 \, C\right )} a\right )} \sin \left (d x + c\right )}{24 \, d} \]

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/24*(3*(4*A + 4*B + 3*C)*a*d*x + (6*C*a*cos(d*x + c)^3 + 8*(B + C)*a*cos(d*x + c)^2 + 3*(4*A + 4*B + 3*C)*a*c
os(d*x + c) + 8*(3*A + 2*B + 2*C)*a)*sin(d*x + c))/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (110) = 220\).

Time = 0.19 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.71 \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} \frac {A a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {A a \sin {\left (c + d x \right )}}{d} + \frac {B a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {2 B a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {3 C a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 C a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 C a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {C a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + a\right ) \left (A + B \cos {\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Piecewise((A*a*x*sin(c + d*x)**2/2 + A*a*x*cos(c + d*x)**2/2 + A*a*sin(c + d*x)*cos(c + d*x)/(2*d) + A*a*sin(c
+ d*x)/d + B*a*x*sin(c + d*x)**2/2 + B*a*x*cos(c + d*x)**2/2 + 2*B*a*sin(c + d*x)**3/(3*d) + B*a*sin(c + d*x)
*cos(c + d*x)**2/d + B*a*sin(c + d*x)*cos(c + d*x)/(2*d) + 3*C*a*x*sin(c + d*x)**4/8 + 3*C*a*x*sin(c + d*x)**2
*cos(c + d*x)**2/4 + 3*C*a*x*cos(c + d*x)**4/8 + 3*C*a*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 2*C*a*sin(c + d*x)
**3/(3*d) + 5*C*a*sin(c + d*x)*cos(c + d*x)**3/(8*d) + C*a*sin(c + d*x)*cos(c + d*x)**2/d, Ne(d, 0)), (x*(a*co
s(c) + a)*(A + B*cos(c) + C*cos(c)**2)*cos(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.12 \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a + 96 \, A a \sin \left (d x + c\right )}{96 \, d} \]

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

1/96*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a + 24*(2*d*x + 2*c + s
in(2*d*x + 2*c))*B*a - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a + 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(
2*d*x + 2*c))*C*a + 96*A*a*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.86 \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{8} \, {\left (4 \, A a + 4 \, B a + 3 \, C a\right )} x + \frac {C a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (B a + C a\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (A a + B a + C a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, A a + 3 \, B a + 3 \, C a\right )} \sin \left (d x + c\right )}{4 \, d} \]

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

1/8*(4*A*a + 4*B*a + 3*C*a)*x + 1/32*C*a*sin(4*d*x + 4*c)/d + 1/12*(B*a + C*a)*sin(3*d*x + 3*c)/d + 1/4*(A*a +
B*a + C*a)*sin(2*d*x + 2*c)/d + 1/4*(4*A*a + 3*B*a + 3*C*a)*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 2.67 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.03 \[ \int \cos (c+d x) (a+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\left (A\,a+B\,a+\frac {3\,C\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (5\,A\,a+\frac {7\,B\,a}{3}+\frac {49\,C\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (7\,A\,a+\frac {13\,B\,a}{3}+\frac {31\,C\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A\,a+3\,B\,a+\frac {13\,C\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )\,\left (4\,A+4\,B+3\,C\right )}{4\,d}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,A+4\,B+3\,C\right )}{4\,\left (A\,a+B\,a+\frac {3\,C\,a}{4}\right )}\right )\,\left (4\,A+4\,B+3\,C\right )}{4\,d} \]

[In]

int(cos(c + d*x)*(a + a*cos(c + d*x))*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

(tan(c/2 + (d*x)/2)*(3*A*a + 3*B*a + (13*C*a)/4) + tan(c/2 + (d*x)/2)^7*(A*a + B*a + (3*C*a)/4) + tan(c/2 + (d
*x)/2)^3*(7*A*a + (13*B*a)/3 + (31*C*a)/12) + tan(c/2 + (d*x)/2)^5*(5*A*a + (7*B*a)/3 + (49*C*a)/12))/(d*(4*ta
n(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (a*(atan(t
an(c/2 + (d*x)/2)) - (d*x)/2)*(4*A + 4*B + 3*C))/(4*d) + (a*atan((a*tan(c/2 + (d*x)/2)*(4*A + 4*B + 3*C))/(4*(
A*a + B*a + (3*C*a)/4)))*(4*A + 4*B + 3*C))/(4*d)