3.1.30 \(\int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx\) [30]

Optimal. Leaf size=98 \[ \frac {1}{8} a^2 (4 A+B) c x-\frac {a^2 (4 A+B) c \cos ^3(e+f x)}{12 f}+\frac {a^2 (4 A+B) c \cos (e+f x) \sin (e+f x)}{8 f}-\frac {B c \cos ^3(e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{4 f} \]

[Out]

1/8*a^2*(4*A+B)*c*x-1/12*a^2*(4*A+B)*c*cos(f*x+e)^3/f+1/8*a^2*(4*A+B)*c*cos(f*x+e)*sin(f*x+e)/f-1/4*B*c*cos(f*
x+e)^3*(a^2+a^2*sin(f*x+e))/f

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Rubi [A]
time = 0.11, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3046, 2939, 2748, 2715, 8} \begin {gather*} -\frac {a^2 c (4 A+B) \cos ^3(e+f x)}{12 f}+\frac {a^2 c (4 A+B) \sin (e+f x) \cos (e+f x)}{8 f}+\frac {1}{8} a^2 c x (4 A+B)-\frac {B c \cos ^3(e+f x) \left (a^2 \sin (e+f x)+a^2\right )}{4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x]),x]

[Out]

(a^2*(4*A + B)*c*x)/8 - (a^2*(4*A + B)*c*Cos[e + f*x]^3)/(12*f) + (a^2*(4*A + B)*c*Cos[e + f*x]*Sin[e + f*x])/
(8*f) - (B*c*Cos[e + f*x]^3*(a^2 + a^2*Sin[e + f*x]))/(4*f)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2939

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x
] + Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; F
reeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx &=(a c) \int \cos ^2(e+f x) (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx\\ &=-\frac {B c \cos ^3(e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{4 f}+\frac {1}{4} (a (4 A+B) c) \int \cos ^2(e+f x) (a+a \sin (e+f x)) \, dx\\ &=-\frac {a^2 (4 A+B) c \cos ^3(e+f x)}{12 f}-\frac {B c \cos ^3(e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{4 f}+\frac {1}{4} \left (a^2 (4 A+B) c\right ) \int \cos ^2(e+f x) \, dx\\ &=-\frac {a^2 (4 A+B) c \cos ^3(e+f x)}{12 f}+\frac {a^2 (4 A+B) c \cos (e+f x) \sin (e+f x)}{8 f}-\frac {B c \cos ^3(e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{4 f}+\frac {1}{8} \left (a^2 (4 A+B) c\right ) \int 1 \, dx\\ &=\frac {1}{8} a^2 (4 A+B) c x-\frac {a^2 (4 A+B) c \cos ^3(e+f x)}{12 f}+\frac {a^2 (4 A+B) c \cos (e+f x) \sin (e+f x)}{8 f}-\frac {B c \cos ^3(e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{4 f}\\ \end {align*}

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Mathematica [A]
time = 0.58, size = 67, normalized size = 0.68 \begin {gather*} -\frac {a^2 c (-12 (4 A+B) f x+24 (A+B) \cos (e+f x)+8 (A+B) \cos (3 (e+f x))-24 A \sin (2 (e+f x))+3 B \sin (4 (e+f x)))}{96 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x]),x]

[Out]

-1/96*(a^2*c*(-12*(4*A + B)*f*x + 24*(A + B)*Cos[e + f*x] + 8*(A + B)*Cos[3*(e + f*x)] - 24*A*Sin[2*(e + f*x)]
+ 3*B*Sin[4*(e + f*x)]))/f

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(185\) vs. \(2(90)=180\).
time = 0.13, size = 186, normalized size = 1.90

method	result	size
risch	\(\frac {a^{2} c x A}{2}+\frac {a^{2} c x B}{8}-\frac {a^{2} c \cos \left (f x +e \right ) A}{4 f}-\frac {a^{2} c \cos \left (f x +e \right ) B}{4 f}-\frac {B \,a^{2} c \sin \left (4 f x +4 e \right )}{32 f}-\frac {a^{2} c \cos \left (3 f x +3 e \right ) A}{12 f}-\frac {a^{2} c \cos \left (3 f x +3 e \right ) B}{12 f}+\frac {a^{2} A c \sin \left (2 f x +2 e \right )}{4 f}\)	\(126\)
derivativedivides	\(\frac {-a^{2} A c \cos \left (f x +e \right )+a^{2} A c \left (f x +e \right )+B \,a^{2} c \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-B \,a^{2} c \cos \left (f x +e \right )-a^{2} A c \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+\frac {B \,a^{2} c \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+\frac {a^{2} A c \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}-B \,a^{2} c \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}\)	\(186\)
default	\(\frac {-a^{2} A c \cos \left (f x +e \right )+a^{2} A c \left (f x +e \right )+B \,a^{2} c \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-B \,a^{2} c \cos \left (f x +e \right )-a^{2} A c \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+\frac {B \,a^{2} c \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+\frac {a^{2} A c \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}-B \,a^{2} c \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}\)	\(186\)
norman	\(\frac {\left (\frac {1}{2} a^{2} A c +\frac {1}{8} B \,a^{2} c \right ) x +\left (2 a^{2} A c +\frac {1}{2} B \,a^{2} c \right ) x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (2 a^{2} A c +\frac {1}{2} B \,a^{2} c \right ) x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (3 a^{2} A c +\frac {3}{4} B \,a^{2} c \right ) x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {1}{2} a^{2} A c +\frac {1}{8} B \,a^{2} c \right ) x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {2 a^{2} A c +2 B \,a^{2} c}{3 f}-\frac {2 \left (a^{2} A c +B \,a^{2} c \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}-\frac {2 \left (a^{2} A c +B \,a^{2} c \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {2 \left (a^{2} A c +B \,a^{2} c \right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {a^{2} c \left (4 A -B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}-\frac {a^{2} c \left (4 A -B \right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}+\frac {a^{2} c \left (4 A +7 B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}-\frac {a^{2} c \left (4 A +7 B \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{4}}\)	\(360\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/f*(-a^2*A*c*cos(f*x+e)+a^2*A*c*(f*x+e)+B*a^2*c*(-1/2*cos(f*x+e)*sin(f*x+e)+1/2*f*x+1/2*e)-B*a^2*c*cos(f*x+e)
-a^2*A*c*(-1/2*cos(f*x+e)*sin(f*x+e)+1/2*f*x+1/2*e)+1/3*B*a^2*c*(2+sin(f*x+e)^2)*cos(f*x+e)+1/3*a^2*A*c*(2+sin
(f*x+e)^2)*cos(f*x+e)-B*a^2*c*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (95) = 190\).
time = 0.27, size = 193, normalized size = 1.97 \begin {gather*} -\frac {32 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a^{2} c + 24 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} c - 96 \, {\left (f x + e\right )} A a^{2} c + 32 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{2} c + 3 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} c - 24 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} c + 96 \, A a^{2} c \cos \left (f x + e\right ) + 96 \, B a^{2} c \cos \left (f x + e\right )}{96 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x, algorithm="maxima")

[Out]

-1/96*(32*(cos(f*x + e)^3 - 3*cos(f*x + e))*A*a^2*c + 24*(2*f*x + 2*e - sin(2*f*x + 2*e))*A*a^2*c - 96*(f*x +
e)*A*a^2*c + 32*(cos(f*x + e)^3 - 3*cos(f*x + e))*B*a^2*c + 3*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x
+ 2*e))*B*a^2*c - 24*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a^2*c + 96*A*a^2*c*cos(f*x + e) + 96*B*a^2*c*cos(f*x +
e))/f

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Fricas [A]
time = 0.37, size = 81, normalized size = 0.83 \begin {gather*} -\frac {8 \, {\left (A + B\right )} a^{2} c \cos \left (f x + e\right )^{3} - 3 \, {\left (4 \, A + B\right )} a^{2} c f x + 3 \, {\left (2 \, B a^{2} c \cos \left (f x + e\right )^{3} - {\left (4 \, A + B\right )} a^{2} c \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x, algorithm="fricas")

[Out]

-1/24*(8*(A + B)*a^2*c*cos(f*x + e)^3 - 3*(4*A + B)*a^2*c*f*x + 3*(2*B*a^2*c*cos(f*x + e)^3 - (4*A + B)*a^2*c*
cos(f*x + e))*sin(f*x + e))/f

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 396 vs. \(2 (90) = 180\).
time = 0.24, size = 396, normalized size = 4.04 \begin {gather*} \begin {cases} - \frac {A a^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} - \frac {A a^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} + A a^{2} c x + \frac {A a^{2} c \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {A a^{2} c \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} + \frac {2 A a^{2} c \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {A a^{2} c \cos {\left (e + f x \right )}}{f} - \frac {3 B a^{2} c x \sin ^{4}{\left (e + f x \right )}}{8} - \frac {3 B a^{2} c x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {B a^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} - \frac {3 B a^{2} c x \cos ^{4}{\left (e + f x \right )}}{8} + \frac {B a^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} + \frac {5 B a^{2} c \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} + \frac {B a^{2} c \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {3 B a^{2} c \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {B a^{2} c \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} + \frac {2 B a^{2} c \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {B a^{2} c \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (A + B \sin {\left (e \right )}\right ) \left (a \sin {\left (e \right )} + a\right )^{2} \left (- c \sin {\left (e \right )} + c\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x)

[Out]

Piecewise((-A*a**2*c*x*sin(e + f*x)**2/2 - A*a**2*c*x*cos(e + f*x)**2/2 + A*a**2*c*x + A*a**2*c*sin(e + f*x)**
2*cos(e + f*x)/f + A*a**2*c*sin(e + f*x)*cos(e + f*x)/(2*f) + 2*A*a**2*c*cos(e + f*x)**3/(3*f) - A*a**2*c*cos(
e + f*x)/f - 3*B*a**2*c*x*sin(e + f*x)**4/8 - 3*B*a**2*c*x*sin(e + f*x)**2*cos(e + f*x)**2/4 + B*a**2*c*x*sin(
e + f*x)**2/2 - 3*B*a**2*c*x*cos(e + f*x)**4/8 + B*a**2*c*x*cos(e + f*x)**2/2 + 5*B*a**2*c*sin(e + f*x)**3*cos
(e + f*x)/(8*f) + B*a**2*c*sin(e + f*x)**2*cos(e + f*x)/f + 3*B*a**2*c*sin(e + f*x)*cos(e + f*x)**3/(8*f) - B*
a**2*c*sin(e + f*x)*cos(e + f*x)/(2*f) + 2*B*a**2*c*cos(e + f*x)**3/(3*f) - B*a**2*c*cos(e + f*x)/f, Ne(f, 0))
, (x*(A + B*sin(e))*(a*sin(e) + a)**2*(-c*sin(e) + c), True))

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Giac [A]
time = 0.39, size = 111, normalized size = 1.13 \begin {gather*} -\frac {B a^{2} c \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {A a^{2} c \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} + \frac {1}{8} \, {\left (4 \, A a^{2} c + B a^{2} c\right )} x - \frac {{\left (A a^{2} c + B a^{2} c\right )} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac {{\left (A a^{2} c + B a^{2} c\right )} \cos \left (f x + e\right )}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x, algorithm="giac")

[Out]

-1/32*B*a^2*c*sin(4*f*x + 4*e)/f + 1/4*A*a^2*c*sin(2*f*x + 2*e)/f + 1/8*(4*A*a^2*c + B*a^2*c)*x - 1/12*(A*a^2*
c + B*a^2*c)*cos(3*f*x + 3*e)/f - 1/4*(A*a^2*c + B*a^2*c)*cos(f*x + e)/f

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Mupad [B]
time = 13.70, size = 339, normalized size = 3.46 \begin {gather*} \frac {a^2\,c\,\mathrm {atan}\left (\frac {a^2\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,A+B\right )}{4\,\left (A\,a^2\,c+\frac {B\,a^2\,c}{4}\right )}\right )\,\left (4\,A+B\right )}{4\,f}-\frac {a^2\,c\,\left (4\,A+B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {f\,x}{2}\right )}{4\,f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (2\,A\,a^2\,c+2\,B\,a^2\,c\right )-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A\,a^2\,c-\frac {B\,a^2\,c}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (2\,A\,a^2\,c+2\,B\,a^2\,c\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {2\,A\,a^2\,c}{3}+\frac {2\,B\,a^2\,c}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (A\,a^2\,c-\frac {B\,a^2\,c}{4}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (A\,a^2\,c+\frac {7\,B\,a^2\,c}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (A\,a^2\,c+\frac {7\,B\,a^2\,c}{4}\right )+\frac {2\,A\,a^2\,c}{3}+\frac {2\,B\,a^2\,c}{3}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2*(c - c*sin(e + f*x)),x)

[Out]

(a^2*c*atan((a^2*c*tan(e/2 + (f*x)/2)*(4*A + B))/(4*(A*a^2*c + (B*a^2*c)/4)))*(4*A + B))/(4*f) - (a^2*c*(4*A +
B)*(atan(tan(e/2 + (f*x)/2)) - (f*x)/2))/(4*f) - (tan(e/2 + (f*x)/2)^4*(2*A*a^2*c + 2*B*a^2*c) - tan(e/2 + (f
*x)/2)*(A*a^2*c - (B*a^2*c)/4) + tan(e/2 + (f*x)/2)^6*(2*A*a^2*c + 2*B*a^2*c) + tan(e/2 + (f*x)/2)^2*((2*A*a^2
*c)/3 + (2*B*a^2*c)/3) + tan(e/2 + (f*x)/2)^7*(A*a^2*c - (B*a^2*c)/4) - tan(e/2 + (f*x)/2)^3*(A*a^2*c + (7*B*a
^2*c)/4) + tan(e/2 + (f*x)/2)^5*(A*a^2*c + (7*B*a^2*c)/4) + (2*A*a^2*c)/3 + (2*B*a^2*c)/3)/(f*(4*tan(e/2 + (f*
x)/2)^2 + 6*tan(e/2 + (f*x)/2)^4 + 4*tan(e/2 + (f*x)/2)^6 + tan(e/2 + (f*x)/2)^8 + 1))

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