3.281 \(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx\)
Optimal. Leaf size=127 \[ -\frac {(2 A c+3 A d+3 B c+7 B d) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}-\frac {(2 A c+3 A d+3 B c-8 B d) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2}-\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a \sin (e+f x)+a)^3} \]
[Out]
-1/5*(A-B)*(c-d)*cos(f*x+e)/f/(a+a*sin(f*x+e))^3-1/15*(2*A*c+3*A*d+3*B*c-8*B*d)*cos(f*x+e)/a/f/(a+a*sin(f*x+e)
)^2-1/15*(2*A*c+3*A*d+3*B*c+7*B*d)*cos(f*x+e)/f/(a^3+a^3*sin(f*x+e))
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Rubi [A] time = 0.23, antiderivative size = 127, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used =
{2968, 3019, 2750, 2648} \[ -\frac {(2 A c+3 A d+3 B c+7 B d) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}-\frac {(2 A c+3 A d+3 B c-8 B d) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2}-\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a \sin (e+f x)+a)^3} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x]))/(a + a*Sin[e + f*x])^3,x]
[Out]
-((A - B)*(c - d)*Cos[e + f*x])/(5*f*(a + a*Sin[e + f*x])^3) - ((2*A*c + 3*B*c + 3*A*d - 8*B*d)*Cos[e + f*x])/
(15*a*f*(a + a*Sin[e + f*x])^2) - ((2*A*c + 3*B*c + 3*A*d + 7*B*d)*Cos[e + f*x])/(15*f*(a^3 + a^3*Sin[e + f*x]
))
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 2750
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b
*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]
Rule 2968
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Rule 3019
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
Rubi steps
\begin {align*} \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx &=\int \frac {A c+(B c+A d) \sin (e+f x)+B d \sin ^2(e+f x)}{(a+a \sin (e+f x))^3} \, dx\\ &=-\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac {\int \frac {-a (2 A c+3 B c+3 A d-3 B d)-5 a B d \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac {(2 A c+3 B c+3 A d-8 B d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}+\frac {(2 A c+3 B c+3 A d+7 B d) \int \frac {1}{a+a \sin (e+f x)} \, dx}{15 a^2}\\ &=-\frac {(A-B) (c-d) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac {(2 A c+3 B c+3 A d-8 B d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {(2 A c+3 B c+3 A d+7 B d) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )}\\ \end {align*}
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Mathematica [A] time = 0.69, size = 176, normalized size = 1.39 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (15 (A d+B (c+2 d)) \cos \left (\frac {1}{2} (e+f x)\right )-5 (2 A c+3 A d+3 B c+4 B d) \cos \left (\frac {3}{2} (e+f x)\right )-2 \sin \left (\frac {1}{2} (e+f x)\right ) ((2 A c+3 A d+3 B c-8 B d) \cos (e+f x)+(2 A c+3 A d+3 B c+7 B d) \cos (2 (e+f x))-3 (3 A c+2 A d+2 B c+8 B d))\right )}{30 a^3 f (\sin (e+f x)+1)^3} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x]))/(a + a*Sin[e + f*x])^3,x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(15*(A*d + B*(c + 2*d))*Cos[(e + f*x)/2] - 5*(2*A*c + 3*B*c + 3*A*d + 4
*B*d)*Cos[(3*(e + f*x))/2] - 2*(-3*(3*A*c + 2*B*c + 2*A*d + 8*B*d) + (2*A*c + 3*B*c + 3*A*d - 8*B*d)*Cos[e + f
*x] + (2*A*c + 3*B*c + 3*A*d + 7*B*d)*Cos[2*(e + f*x)])*Sin[(e + f*x)/2]))/(30*a^3*f*(1 + Sin[e + f*x])^3)
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fricas [B] time = 0.43, size = 271, normalized size = 2.13 \[ -\frac {{\left ({\left (2 \, A + 3 \, B\right )} c + {\left (3 \, A + 7 \, B\right )} d\right )} \cos \left (f x + e\right )^{3} - {\left (2 \, {\left (2 \, A + 3 \, B\right )} c + {\left (6 \, A - B\right )} d\right )} \cos \left (f x + e\right )^{2} - 3 \, {\left (A - B\right )} c + 3 \, {\left (A - B\right )} d - 3 \, {\left ({\left (3 \, A + 2 \, B\right )} c + {\left (2 \, A + 3 \, B\right )} d\right )} \cos \left (f x + e\right ) - {\left ({\left ({\left (2 \, A + 3 \, B\right )} c + {\left (3 \, A + 7 \, B\right )} d\right )} \cos \left (f x + e\right )^{2} - 3 \, {\left (A - B\right )} c + 3 \, {\left (A - B\right )} d + 3 \, {\left ({\left (2 \, A + 3 \, B\right )} c + {\left (3 \, A + 2 \, B\right )} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
-1/15*(((2*A + 3*B)*c + (3*A + 7*B)*d)*cos(f*x + e)^3 - (2*(2*A + 3*B)*c + (6*A - B)*d)*cos(f*x + e)^2 - 3*(A
- B)*c + 3*(A - B)*d - 3*((3*A + 2*B)*c + (2*A + 3*B)*d)*cos(f*x + e) - (((2*A + 3*B)*c + (3*A + 7*B)*d)*cos(f
*x + e)^2 - 3*(A - B)*c + 3*(A - B)*d + 3*((2*A + 3*B)*c + (3*A + 2*B)*d)*cos(f*x + e))*sin(f*x + e))/(a^3*f*c
os(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(
f*x + e) - 4*a^3*f)*sin(f*x + e))
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giac [A] time = 0.18, size = 223, normalized size = 1.76 \[ -\frac {2 \, {\left (15 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 10 \, B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, A c + 3 \, B c + 3 \, A d + 2 \, B d\right )}}{15 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^3,x, algorithm="giac")
[Out]
-2/15*(15*A*c*tan(1/2*f*x + 1/2*e)^4 + 30*A*c*tan(1/2*f*x + 1/2*e)^3 + 15*B*c*tan(1/2*f*x + 1/2*e)^3 + 15*A*d*
tan(1/2*f*x + 1/2*e)^3 + 40*A*c*tan(1/2*f*x + 1/2*e)^2 + 15*B*c*tan(1/2*f*x + 1/2*e)^2 + 15*A*d*tan(1/2*f*x +
1/2*e)^2 + 20*B*d*tan(1/2*f*x + 1/2*e)^2 + 20*A*c*tan(1/2*f*x + 1/2*e) + 15*B*c*tan(1/2*f*x + 1/2*e) + 15*A*d*
tan(1/2*f*x + 1/2*e) + 10*B*d*tan(1/2*f*x + 1/2*e) + 7*A*c + 3*B*c + 3*A*d + 2*B*d)/(a^3*f*(tan(1/2*f*x + 1/2*
e) + 1)^5)
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maple [A] time = 0.44, size = 151, normalized size = 1.19 \[ \frac {-\frac {2 \left (8 A c -6 A d -6 B c +4 B d \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {-8 A c +8 A d +8 B c -8 B d}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {-4 A c +2 A d +2 B c}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (4 A c -4 A d -4 B c +4 B d \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {2 A c}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{f \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^3,x)
[Out]
2/f/a^3*(-1/3*(8*A*c-6*A*d-6*B*c+4*B*d)/(tan(1/2*f*x+1/2*e)+1)^3-1/4*(-8*A*c+8*A*d+8*B*c-8*B*d)/(tan(1/2*f*x+1
/2*e)+1)^4-1/2*(-4*A*c+2*A*d+2*B*c)/(tan(1/2*f*x+1/2*e)+1)^2-1/5*(4*A*c-4*A*d-4*B*c+4*B*d)/(tan(1/2*f*x+1/2*e)
+1)^5-A*c/(tan(1/2*f*x+1/2*e)+1))
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maxima [B] time = 0.46, size = 733, normalized size = 5.77 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^3,x, algorithm="maxima")
[Out]
-2/15*(A*c*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 7)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)
^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 2*B*d*(5*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) +
1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*B*c*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x + e)^2/
(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e
)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*A*d*(5*sin(f*x + e)/(cos(f*x + e) + 1)
+ 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e
)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
+ 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f
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mupad [B] time = 14.26, size = 245, normalized size = 1.93 \[ \frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {53\,A\,c}{4}+3\,A\,d+3\,B\,c+\frac {13\,B\,d}{4}-4\,A\,c\,\cos \left (e+f\,x\right )+\frac {3\,A\,d\,\cos \left (e+f\,x\right )}{2}+\frac {3\,B\,c\,\cos \left (e+f\,x\right )}{2}+B\,d\,\cos \left (e+f\,x\right )+\frac {25\,A\,c\,\sin \left (e+f\,x\right )}{2}+\frac {15\,A\,d\,\sin \left (e+f\,x\right )}{2}+\frac {15\,B\,c\,\sin \left (e+f\,x\right )}{2}+\frac {5\,B\,d\,\sin \left (e+f\,x\right )}{2}-\frac {9\,A\,c\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {3\,A\,d\,\cos \left (2\,e+2\,f\,x\right )}{2}-\frac {3\,B\,c\,\cos \left (2\,e+2\,f\,x\right )}{2}-\frac {9\,B\,d\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {5\,A\,c\,\sin \left (2\,e+2\,f\,x\right )}{4}+\frac {5\,B\,d\,\sin \left (2\,e+2\,f\,x\right )}{4}\right )}{15\,a^3\,f\,\left (\frac {5\,\sqrt {2}\,\cos \left (\frac {3\,e}{2}+\frac {\pi }{4}+\frac {3\,f\,x}{2}\right )}{4}-\frac {5\,\sqrt {2}\,\cos \left (\frac {e}{2}-\frac {\pi }{4}+\frac {f\,x}{2}\right )}{2}+\frac {\sqrt {2}\,\cos \left (\frac {5\,e}{2}-\frac {\pi }{4}+\frac {5\,f\,x}{2}\right )}{4}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x)))/(a + a*sin(e + f*x))^3,x)
[Out]
(2*cos(e/2 + (f*x)/2)*((53*A*c)/4 + 3*A*d + 3*B*c + (13*B*d)/4 - 4*A*c*cos(e + f*x) + (3*A*d*cos(e + f*x))/2 +
(3*B*c*cos(e + f*x))/2 + B*d*cos(e + f*x) + (25*A*c*sin(e + f*x))/2 + (15*A*d*sin(e + f*x))/2 + (15*B*c*sin(e
+ f*x))/2 + (5*B*d*sin(e + f*x))/2 - (9*A*c*cos(2*e + 2*f*x))/4 - (3*A*d*cos(2*e + 2*f*x))/2 - (3*B*c*cos(2*e
+ 2*f*x))/2 - (9*B*d*cos(2*e + 2*f*x))/4 - (5*A*c*sin(2*e + 2*f*x))/4 + (5*B*d*sin(2*e + 2*f*x))/4))/(15*a^3*
f*((5*2^(1/2)*cos((3*e)/2 + pi/4 + (3*f*x)/2))/4 - (5*2^(1/2)*cos(e/2 - pi/4 + (f*x)/2))/2 + (2^(1/2)*cos((5*e
)/2 - pi/4 + (5*f*x)/2))/4))
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sympy [A] time = 15.10, size = 1819, normalized size = 14.32 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))/(a+a*sin(f*x+e))**3,x)
[Out]
Piecewise((-30*A*c*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a*
*3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*A*c*t
an(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2
)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 80*A*c*tan(e/2 + f*x/2)**2/(
15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*ta
n(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 40*A*c*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x
/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*
a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 14*A*c/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 +
150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 3
0*A*d*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2
+ f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30*A*d*tan(e/2 + f*x/
2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a*
*3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30*A*d*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/
2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**
2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 6*A*d/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2
)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*
f) - 30*B*c*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*ta
n(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30*B*c*tan(e/2
+ f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 +
150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30*B*c*tan(e/2 + f*x/2)/(15*a**3*f*
tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*
x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 6*B*c/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 +
f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15
*a**3*f) - 40*B*d*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**
3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 20*B*d*ta
n(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3
+ 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 4*B*d/(15*a**3*f*tan(e/2 + f*x/2
)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a*
*3*f*tan(e/2 + f*x/2) + 15*a**3*f), Ne(f, 0)), (x*(A + B*sin(e))*(c + d*sin(e))/(a*sin(e) + a)**3, True))
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