3.67 \(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=93 \[ \frac {(4 A-B) \tan ^3(e+f x)}{15 a^2 c^3 f}+\frac {(4 A-B) \tan (e+f x)}{5 a^2 c^3 f}+\frac {(A+B) \sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )} \]

[Out]

1/5*(A+B)*sec(f*x+e)^3/a^2/f/(c^3-c^3*sin(f*x+e))+1/5*(4*A-B)*tan(f*x+e)/a^2/c^3/f+1/15*(4*A-B)*tan(f*x+e)^3/a
^2/c^3/f

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Rubi [A]  time = 0.22, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2967, 2859, 3767} \[ \frac {(4 A-B) \tan ^3(e+f x)}{15 a^2 c^3 f}+\frac {(4 A-B) \tan (e+f x)}{5 a^2 c^3 f}+\frac {(A+B) \sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^3),x]

[Out]

((A + B)*Sec[e + f*x]^3)/(5*a^2*f*(c^3 - c^3*Sin[e + f*x])) + ((4*A - B)*Tan[e + f*x])/(5*a^2*c^3*f) + ((4*A -
B)*Tan[e + f*x]^3)/(15*a^2*c^3*f)

Rule 2859

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

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]))

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3} \, dx &=\frac {\int \frac {\sec ^4(e+f x) (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx}{a^2 c^2}\\ &=\frac {(A+B) \sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {(4 A-B) \int \sec ^4(e+f x) \, dx}{5 a^2 c^3}\\ &=\frac {(A+B) \sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )}-\frac {(4 A-B) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{5 a^2 c^3 f}\\ &=\frac {(A+B) \sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )}+\frac {(4 A-B) \tan (e+f x)}{5 a^2 c^3 f}+\frac {(4 A-B) \tan ^3(e+f x)}{15 a^2 c^3 f}\\ \end {align*}

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Mathematica [B]  time = 1.02, size = 237, normalized size = 2.55 \[ \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) (54 (A+B) \cos (e+f x)-32 (4 A-B) \cos (2 (e+f x))-384 A \sin (e+f x)-18 A \sin (2 (e+f x))-128 A \sin (3 (e+f x))-9 A \sin (4 (e+f x))+18 A \cos (3 (e+f x))-64 A \cos (4 (e+f x))+96 B \sin (e+f x)-18 B \sin (2 (e+f x))+32 B \sin (3 (e+f x))-9 B \sin (4 (e+f x))+18 B \cos (3 (e+f x))+16 B \cos (4 (e+f x))-240 B)}{960 a^2 c^3 f (\sin (e+f x)-1)^3 (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^3),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-240*B + 54*(A + B)*Cos[e + f*x]
- 32*(4*A - B)*Cos[2*(e + f*x)] + 18*A*Cos[3*(e + f*x)] + 18*B*Cos[3*(e + f*x)] - 64*A*Cos[4*(e + f*x)] + 16*
B*Cos[4*(e + f*x)] - 384*A*Sin[e + f*x] + 96*B*Sin[e + f*x] - 18*A*Sin[2*(e + f*x)] - 18*B*Sin[2*(e + f*x)] -
128*A*Sin[3*(e + f*x)] + 32*B*Sin[3*(e + f*x)] - 9*A*Sin[4*(e + f*x)] - 9*B*Sin[4*(e + f*x)]))/(960*a^2*c^3*f*
(-1 + Sin[e + f*x])^3*(1 + Sin[e + f*x])^2)

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fricas [A]  time = 0.42, size = 116, normalized size = 1.25 \[ -\frac {2 \, {\left (4 \, A - B\right )} \cos \left (f x + e\right )^{4} - {\left (4 \, A - B\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, {\left (4 \, A - B\right )} \cos \left (f x + e\right )^{2} + 4 \, A - B\right )} \sin \left (f x + e\right ) - A + 4 \, B}{15 \, {\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - a^{2} c^{3} f \cos \left (f x + e\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/15*(2*(4*A - B)*cos(f*x + e)^4 - (4*A - B)*cos(f*x + e)^2 + (2*(4*A - B)*cos(f*x + e)^2 + 4*A - B)*sin(f*x
+ e) - A + 4*B)/(a^2*c^3*f*cos(f*x + e)^3*sin(f*x + e) - a^2*c^3*f*cos(f*x + e)^3)

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giac [B]  time = 0.20, size = 235, normalized size = 2.53 \[ -\frac {\frac {5 \, {\left (15 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 9 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 24 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 12 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 13 \, A - 7 \, B\right )}}{a^{2} c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} + \frac {165 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 45 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 480 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 60 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 650 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 70 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 400 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 20 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 113 \, A + 13 \, B}{a^{2} c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}}}{120 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/120*(5*(15*A*tan(1/2*f*x + 1/2*e)^2 - 9*B*tan(1/2*f*x + 1/2*e)^2 + 24*A*tan(1/2*f*x + 1/2*e) - 12*B*tan(1/2
*f*x + 1/2*e) + 13*A - 7*B)/(a^2*c^3*(tan(1/2*f*x + 1/2*e) + 1)^3) + (165*A*tan(1/2*f*x + 1/2*e)^4 + 45*B*tan(
1/2*f*x + 1/2*e)^4 - 480*A*tan(1/2*f*x + 1/2*e)^3 - 60*B*tan(1/2*f*x + 1/2*e)^3 + 650*A*tan(1/2*f*x + 1/2*e)^2
+ 70*B*tan(1/2*f*x + 1/2*e)^2 - 400*A*tan(1/2*f*x + 1/2*e) - 20*B*tan(1/2*f*x + 1/2*e) + 113*A + 13*B)/(a^2*c
^3*(tan(1/2*f*x + 1/2*e) - 1)^5))/f

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maple [B]  time = 0.50, size = 183, normalized size = 1.97 \[ \frac {-\frac {2 \left (A +B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2 A +2 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {\frac {3 A}{2}+B}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {5 A}{2}+2 B \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {2 \left (\frac {11 A}{16}+\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {-\frac {A}{4}+\frac {B}{4}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {A}{4}-\frac {B}{4}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {5 A}{16}-\frac {3 B}{16}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{f \,a^{2} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x)

[Out]

2/f/a^2/c^3*(-1/5*(A+B)/(tan(1/2*f*x+1/2*e)-1)^5-1/4*(2*A+2*B)/(tan(1/2*f*x+1/2*e)-1)^4-1/2*(3/2*A+B)/(tan(1/2
*f*x+1/2*e)-1)^2-1/3*(5/2*A+2*B)/(tan(1/2*f*x+1/2*e)-1)^3-(11/16*A+3/16*B)/(tan(1/2*f*x+1/2*e)-1)-1/2*(-1/4*A+
1/4*B)/(tan(1/2*f*x+1/2*e)+1)^2-1/3*(1/4*A-1/4*B)/(tan(1/2*f*x+1/2*e)+1)^3-(5/16*A-3/16*B)/(tan(1/2*f*x+1/2*e)
+1))

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maxima [B]  time = 0.47, size = 651, normalized size = 7.00 \[ \frac {2 \, {\left (\frac {A {\left (\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {13 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {25 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {5 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {15 \, \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + 3\right )}}{a^{2} c^{3} - \frac {2 \, a^{2} c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a^{2} c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {6 \, a^{2} c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {6 \, a^{2} c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {2 \, a^{2} c^{3} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {2 \, a^{2} c^{3} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac {a^{2} c^{3} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}} - \frac {B {\left (\frac {6 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {9 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {8 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {5 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {10 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 3\right )}}{a^{2} c^{3} - \frac {2 \, a^{2} c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a^{2} c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {6 \, a^{2} c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {6 \, a^{2} c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {2 \, a^{2} c^{3} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {2 \, a^{2} c^{3} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac {a^{2} c^{3} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}}\right )}}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

2/15*(A*(9*sin(f*x + e)/(cos(f*x + e) + 1) - 21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 13*sin(f*x + e)^3/(cos(f
*x + e) + 1)^3 + 25*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 15*sin(f*x +
e)^6/(cos(f*x + e) + 1)^6 + 15*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 3)/(a^2*c^3 - 2*a^2*c^3*sin(f*x + e)/(co
s(f*x + e) + 1) - 2*a^2*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 6*a^2*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^
3 - 6*a^2*c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2*a^2*c^3*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 2*a^2*c^3*
sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - a^2*c^3*sin(f*x + e)^8/(cos(f*x + e) + 1)^8) - B*(6*sin(f*x + e)/(cos(f*
x + e) + 1) - 9*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 8*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 5*sin(f*x + e)^4
/(cos(f*x + e) + 1)^4 + 10*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 15*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 3)/(
a^2*c^3 - 2*a^2*c^3*sin(f*x + e)/(cos(f*x + e) + 1) - 2*a^2*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 6*a^2*c^
3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 6*a^2*c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2*a^2*c^3*sin(f*x + e)
^6/(cos(f*x + e) + 1)^6 + 2*a^2*c^3*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - a^2*c^3*sin(f*x + e)^8/(cos(f*x + e)
+ 1)^8))/f

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mupad [B]  time = 12.45, size = 183, normalized size = 1.97 \[ \frac {\left (\frac {8\,A}{15}-\frac {2\,B}{15}-\frac {16\,A\,\sin \left (e+f\,x\right )}{15}+\frac {4\,B\,\sin \left (e+f\,x\right )}{15}\right )\,{\cos \left (e+f\,x\right )}^2+\frac {2\,A}{15}-\frac {8\,B}{15}-\frac {8\,A\,\sin \left (e+f\,x\right )}{15}+\frac {2\,B\,\sin \left (e+f\,x\right )}{15}}{a^2\,c^3\,f\,\left (2\,{\cos \left (e+f\,x\right )}^3\,\sin \left (e+f\,x\right )-2\,{\cos \left (e+f\,x\right )}^3\right )}-\frac {\frac {2\,A}{5}+\frac {2\,B}{5}-\frac {2\,A\,\sin \left (e+f\,x\right )}{5}-\frac {2\,B\,\sin \left (e+f\,x\right )}{5}}{a^2\,c^3\,f\,\left (2\,\sin \left (e+f\,x\right )-2\right )}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {16\,A}{15}-\frac {4\,B}{15}\right )}{a^2\,c^3\,f\,\left (2\,\sin \left (e+f\,x\right )-2\right )} \]

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))^3),x)

[Out]

((2*A)/15 - (8*B)/15 - (8*A*sin(e + f*x))/15 + (2*B*sin(e + f*x))/15 + cos(e + f*x)^2*((8*A)/15 - (2*B)/15 - (
16*A*sin(e + f*x))/15 + (4*B*sin(e + f*x))/15))/(a^2*c^3*f*(2*cos(e + f*x)^3*sin(e + f*x) - 2*cos(e + f*x)^3))
- ((2*A)/5 + (2*B)/5 - (2*A*sin(e + f*x))/5 - (2*B*sin(e + f*x))/5)/(a^2*c^3*f*(2*sin(e + f*x) - 2)) - (cos(e
+ f*x)*((16*A)/15 - (4*B)/15))/(a^2*c^3*f*(2*sin(e + f*x) - 2))

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sympy [A]  time = 27.73, size = 2674, normalized size = 28.75 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**3,x)

[Out]

Piecewise((-30*A*tan(e/2 + f*x/2)**7/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7
- 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3
+ 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) + 30*A*tan(e/2 + f*x
/2)**6/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x
/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*
x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) + 10*A*tan(e/2 + f*x/2)**5/(15*a**2*c**3*f*tan(e/2
+ f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/
2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e
/2 + f*x/2) - 15*a**2*c**3*f) - 50*A*tan(e/2 + f*x/2)**4/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*
tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f
*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f)
- 26*A*tan(e/2 + f*x/2)**3/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*
c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2
*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) + 42*A*tan(e/2 + f*x/2)**2/(15
*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 9
0*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 +
30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) - 18*A*tan(e/2 + f*x/2)/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8
- 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5
- 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) -
15*a**2*c**3*f) - 6*A/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3
*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**
3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) - 30*B*tan(e/2 + f*x/2)**6/(15*a**
2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a*
*2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a
**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) + 20*B*tan(e/2 + f*x/2)**5/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 -
30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5
- 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) -
15*a**2*c**3*f) - 10*B*tan(e/2 + f*x/2)**4/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/
2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x
/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) - 16*B*tan(e/2
+ f*x/2)**3/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2
+ f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/
2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f) - 18*B*tan(e/2 + f*x/2)**2/(15*a**2*c**3*f*t
an(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*
tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f
*tan(e/2 + f*x/2) - 15*a**2*c**3*f) + 12*B*tan(e/2 + f*x/2)/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3
*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**
3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 + f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*
f) - 6*B/(15*a**2*c**3*f*tan(e/2 + f*x/2)**8 - 30*a**2*c**3*f*tan(e/2 + f*x/2)**7 - 30*a**2*c**3*f*tan(e/2 + f
*x/2)**6 + 90*a**2*c**3*f*tan(e/2 + f*x/2)**5 - 90*a**2*c**3*f*tan(e/2 + f*x/2)**3 + 30*a**2*c**3*f*tan(e/2 +
f*x/2)**2 + 30*a**2*c**3*f*tan(e/2 + f*x/2) - 15*a**2*c**3*f), Ne(f, 0)), (x*(A + B*sin(e))/((a*sin(e) + a)**2
*(-c*sin(e) + c)**3), True))

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