3.64 \(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=72 \[ \frac {c (A-7 B) \cos (e+f x)}{3 a^2 f (\sin (e+f x)+1)}-\frac {B c x}{a^2}-\frac {2 c (A-B) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2} \]

[Out]

-B*c*x/a^2+1/3*(A-7*B)*c*cos(f*x+e)/a^2/f/(1+sin(f*x+e))-2/3*(A-B)*c*cos(f*x+e)/f/(a+a*sin(f*x+e))^2

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Rubi [A]  time = 0.21, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2967, 2857, 2735, 2648} \[ \frac {c (A-7 B) \cos (e+f x)}{3 a^2 f (\sin (e+f x)+1)}-\frac {B c x}{a^2}-\frac {2 c (A-B) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((B*c*x)/a^2) + ((A - 7*B)*c*Cos[e + f*x])/(3*a^2*f*(1 + Sin[e + f*x])) - (2*(A - B)*c*Cos[e + f*x])/(3*f*(a
+ a*Sin[e + f*x])^2)

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2735

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

Rule 2857

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

Rule 2967

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 \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx &=(a c) \int \frac {\cos ^2(e+f x) (A+B \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx\\ &=-\frac {2 (A-B) c \cos (e+f x)}{3 f (a+a \sin (e+f x))^2}-\frac {c \int \frac {a A-4 a B+3 a B \sin (e+f x)}{a+a \sin (e+f x)} \, dx}{3 a^2}\\ &=-\frac {B c x}{a^2}-\frac {2 (A-B) c \cos (e+f x)}{3 f (a+a \sin (e+f x))^2}-\frac {((A-7 B) c) \int \frac {1}{a+a \sin (e+f x)} \, dx}{3 a}\\ &=-\frac {B c x}{a^2}-\frac {2 (A-B) c \cos (e+f x)}{3 f (a+a \sin (e+f x))^2}+\frac {(A-7 B) c \cos (e+f x)}{3 f \left (a^2+a^2 \sin (e+f x)\right )}\\ \end {align*}

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Mathematica [B]  time = 0.58, size = 156, normalized size = 2.17 \[ \frac {c \left (-6 (A-3 B) \cos \left (e+\frac {f x}{2}\right )+2 A \cos \left (e+\frac {3 f x}{2}\right )-9 B f x \sin \left (e+\frac {f x}{2}\right )-3 B f x \sin \left (e+\frac {3 f x}{2}\right )-14 B \cos \left (e+\frac {3 f x}{2}\right )+3 B f x \cos \left (2 e+\frac {3 f x}{2}\right )+24 B \sin \left (\frac {f x}{2}\right )-9 B f x \cos \left (\frac {f x}{2}\right )\right )}{6 a^2 f \left (\sin \left (\frac {e}{2}\right )+\cos \left (\frac {e}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(-9*B*f*x*Cos[(f*x)/2] - 6*(A - 3*B)*Cos[e + (f*x)/2] + 2*A*Cos[e + (3*f*x)/2] - 14*B*Cos[e + (3*f*x)/2] +
3*B*f*x*Cos[2*e + (3*f*x)/2] + 24*B*Sin[(f*x)/2] - 9*B*f*x*Sin[e + (f*x)/2] - 3*B*f*x*Sin[e + (3*f*x)/2]))/(6*
a^2*f*(Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

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fricas [B]  time = 0.44, size = 166, normalized size = 2.31 \[ \frac {6 \, B c f x - {\left (3 \, B c f x + {\left (A - 7 \, B\right )} c\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (A - B\right )} c + {\left (3 \, B c f x + {\left (A + 5 \, B\right )} c\right )} \cos \left (f x + e\right ) + {\left (6 \, B c f x - 2 \, {\left (A - B\right )} c + {\left (3 \, B c f x - {\left (A - 7 \, B\right )} c\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(6*B*c*f*x - (3*B*c*f*x + (A - 7*B)*c)*cos(f*x + e)^2 + 2*(A - B)*c + (3*B*c*f*x + (A + 5*B)*c)*cos(f*x +
e) + (6*B*c*f*x - 2*(A - B)*c + (3*B*c*f*x - (A - 7*B)*c)*cos(f*x + e))*sin(f*x + e))/(a^2*f*cos(f*x + e)^2 -
a^2*f*cos(f*x + e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e))

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giac [A]  time = 0.16, size = 92, normalized size = 1.28 \[ -\frac {\frac {3 \, {\left (f x + e\right )} B c}{a^{2}} + \frac {2 \, {\left (3 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 12 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + A c + 5 \, B c\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*(f*x + e)*B*c/a^2 + 2*(3*A*c*tan(1/2*f*x + 1/2*e)^2 + 3*B*c*tan(1/2*f*x + 1/2*e)^2 + 12*B*c*tan(1/2*f*
x + 1/2*e) + A*c + 5*B*c)/(a^2*(tan(1/2*f*x + 1/2*e) + 1)^3))/f

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maple [B]  time = 0.42, size = 160, normalized size = 2.22 \[ -\frac {2 c B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a^{2} f}+\frac {4 c A}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {4 c B}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 c A}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {2 c B}{a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {8 c A}{3 a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {8 c B}{3 a^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.46, size = 452, normalized size = 6.28 \[ -\frac {2 \, {\left (B c {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 4}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{2}}\right )} + \frac {A c {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} - \frac {A c {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {B c {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(B*c*((9*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4)/(a^2 + 3*a^2*sin(f*
x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^
3) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) + A*c*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)^
2/(cos(f*x + e) + 1)^2 + 2)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) - A*c*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 1)/(a^2 + 3*a^2*s
in(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e)
+ 1)^3) + B*c*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 1)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin
(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3))/f

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mupad [B]  time = 12.83, size = 133, normalized size = 1.85 \[ -\frac {B\,c\,x}{a^2}-\frac {\left (\frac {c\,\left (6\,A+6\,B+9\,B\,\left (e+f\,x\right )\right )}{3}-3\,B\,c\,\left (e+f\,x\right )\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+\left (\frac {c\,\left (24\,B+9\,B\,\left (e+f\,x\right )\right )}{3}-3\,B\,c\,\left (e+f\,x\right )\right )\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+\frac {c\,\left (2\,A+10\,B+3\,B\,\left (e+f\,x\right )\right )}{3}-B\,c\,\left (e+f\,x\right )}{a^2\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (B*c*x)/a^2 - (tan(e/2 + (f*x)/2)^2*((c*(6*A + 6*B + 9*B*(e + f*x)))/3 - 3*B*c*(e + f*x)) + tan(e/2 + (f*x)/
2)*((c*(24*B + 9*B*(e + f*x)))/3 - 3*B*c*(e + f*x)) + (c*(2*A + 10*B + 3*B*(e + f*x)))/3 - B*c*(e + f*x))/(a^2
*f*(tan(e/2 + (f*x)/2) + 1)^3)

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sympy [A]  time = 7.39, size = 702, normalized size = 9.75 \[ \begin {cases} - \frac {6 A c \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {2 A c}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {3 B c f x \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {9 B c f x \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {9 B c f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {3 B c f x}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {6 B c \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {24 B c \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {10 B c}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} & \text {for}\: f \neq 0 \\\frac {x \left (A + B \sin {\relax (e )}\right ) \left (- c \sin {\relax (e )} + c\right )}{\left (a \sin {\relax (e )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-6*A*c*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*
tan(e/2 + f*x/2) + 3*a**2*f) - 2*A*c/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*t
an(e/2 + f*x/2) + 3*a**2*f) - 3*B*c*f*x*tan(e/2 + f*x/2)**3/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 +
f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 9*B*c*f*x*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)*
*3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 9*B*c*f*x*tan(e/2 + f*x/2)/(3*a**2
*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 3*B*c*f*x/(3*a
**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 6*B*c*tan(e
/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a*
*2*f) - 24*B*c*tan(e/2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/
2 + f*x/2) + 3*a**2*f) - 10*B*c/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/
2 + f*x/2) + 3*a**2*f), Ne(f, 0)), (x*(A + B*sin(e))*(-c*sin(e) + c)/(a*sin(e) + a)**2, True))

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