3.18 \(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=142 \[ \frac {a c^3 (5 A-2 B) \cos ^3(e+f x)}{12 f}+\frac {a (5 A-2 B) \cos ^3(e+f x) \left (c^3-c^3 \sin (e+f x)\right )}{20 f}+\frac {a c^3 (5 A-2 B) \sin (e+f x) \cos (e+f x)}{8 f}+\frac {1}{8} a c^3 x (5 A-2 B)-\frac {a B c \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f} \]

[Out]

1/8*a*(5*A-2*B)*c^3*x+1/12*a*(5*A-2*B)*c^3*cos(f*x+e)^3/f+1/8*a*(5*A-2*B)*c^3*cos(f*x+e)*sin(f*x+e)/f-1/5*a*B*
c*cos(f*x+e)^3*(c-c*sin(f*x+e))^2/f+1/20*a*(5*A-2*B)*cos(f*x+e)^3*(c^3-c^3*sin(f*x+e))/f

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Rubi [A]  time = 0.25, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2967, 2860, 2678, 2669, 2635, 8} \[ \frac {a c^3 (5 A-2 B) \cos ^3(e+f x)}{12 f}+\frac {a (5 A-2 B) \cos ^3(e+f x) \left (c^3-c^3 \sin (e+f x)\right )}{20 f}+\frac {a c^3 (5 A-2 B) \sin (e+f x) \cos (e+f x)}{8 f}+\frac {1}{8} a c^3 x (5 A-2 B)-\frac {a B c \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(5*A - 2*B)*c^3*x)/8 + (a*(5*A - 2*B)*c^3*Cos[e + f*x]^3)/(12*f) + (a*(5*A - 2*B)*c^3*Cos[e + f*x]*Sin[e +
f*x])/(8*f) - (a*B*c*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^2)/(5*f) + (a*(5*A - 2*B)*Cos[e + f*x]^3*(c^3 - c^3*S
in[e + f*x]))/(20*f)

Rule 8

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

Rule 2635

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] && Integer
Q[2*n]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
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 2678

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

Rule 2860

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] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[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]))

Rubi steps

\begin {align*} \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^3 \, dx &=(a c) \int \cos ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx\\ &=-\frac {a B c \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f}+\frac {1}{5} (a (5 A-2 B) c) \int \cos ^2(e+f x) (c-c \sin (e+f x))^2 \, dx\\ &=-\frac {a B c \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f}+\frac {a (5 A-2 B) \cos ^3(e+f x) \left (c^3-c^3 \sin (e+f x)\right )}{20 f}+\frac {1}{4} \left (a (5 A-2 B) c^2\right ) \int \cos ^2(e+f x) (c-c \sin (e+f x)) \, dx\\ &=\frac {a (5 A-2 B) c^3 \cos ^3(e+f x)}{12 f}-\frac {a B c \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f}+\frac {a (5 A-2 B) \cos ^3(e+f x) \left (c^3-c^3 \sin (e+f x)\right )}{20 f}+\frac {1}{4} \left (a (5 A-2 B) c^3\right ) \int \cos ^2(e+f x) \, dx\\ &=\frac {a (5 A-2 B) c^3 \cos ^3(e+f x)}{12 f}+\frac {a (5 A-2 B) c^3 \cos (e+f x) \sin (e+f x)}{8 f}-\frac {a B c \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f}+\frac {a (5 A-2 B) \cos ^3(e+f x) \left (c^3-c^3 \sin (e+f x)\right )}{20 f}+\frac {1}{8} \left (a (5 A-2 B) c^3\right ) \int 1 \, dx\\ &=\frac {1}{8} a (5 A-2 B) c^3 x+\frac {a (5 A-2 B) c^3 \cos ^3(e+f x)}{12 f}+\frac {a (5 A-2 B) c^3 \cos (e+f x) \sin (e+f x)}{8 f}-\frac {a B c \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f}+\frac {a (5 A-2 B) \cos ^3(e+f x) \left (c^3-c^3 \sin (e+f x)\right )}{20 f}\\ \end {align*}

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Mathematica [A]  time = 0.84, size = 95, normalized size = 0.67 \[ \frac {a c^3 (15 (-(A-2 B) \sin (4 (e+f x))+4 f x (5 A-2 B)+8 A \sin (2 (e+f x)))+60 (4 A-3 B) \cos (e+f x)+10 (8 A-5 B) \cos (3 (e+f x))+6 B \cos (5 (e+f x)))}{480 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c^3*(60*(4*A - 3*B)*Cos[e + f*x] + 10*(8*A - 5*B)*Cos[3*(e + f*x)] + 6*B*Cos[5*(e + f*x)] + 15*(4*(5*A - 2*
B)*f*x + 8*A*Sin[2*(e + f*x)] - (A - 2*B)*Sin[4*(e + f*x)])))/(480*f)

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fricas [A]  time = 0.44, size = 102, normalized size = 0.72 \[ \frac {24 \, B a c^{3} \cos \left (f x + e\right )^{5} + 80 \, {\left (A - B\right )} a c^{3} \cos \left (f x + e\right )^{3} + 15 \, {\left (5 \, A - 2 \, B\right )} a c^{3} f x - 15 \, {\left (2 \, {\left (A - 2 \, B\right )} a c^{3} \cos \left (f x + e\right )^{3} - {\left (5 \, A - 2 \, B\right )} a c^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{120 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(24*B*a*c^3*cos(f*x + e)^5 + 80*(A - B)*a*c^3*cos(f*x + e)^3 + 15*(5*A - 2*B)*a*c^3*f*x - 15*(2*(A - 2*B
)*a*c^3*cos(f*x + e)^3 - (5*A - 2*B)*a*c^3*cos(f*x + e))*sin(f*x + e))/f

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giac [A]  time = 0.18, size = 145, normalized size = 1.02 \[ \frac {B a c^{3} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac {A a c^{3} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} + \frac {1}{8} \, {\left (5 \, A a c^{3} - 2 \, B a c^{3}\right )} x + \frac {{\left (8 \, A a c^{3} - 5 \, B a c^{3}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} + \frac {{\left (4 \, A a c^{3} - 3 \, B a c^{3}\right )} \cos \left (f x + e\right )}{8 \, f} - \frac {{\left (A a c^{3} - 2 \, B a c^{3}\right )} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*B*a*c^3*cos(5*f*x + 5*e)/f + 1/4*A*a*c^3*sin(2*f*x + 2*e)/f + 1/8*(5*A*a*c^3 - 2*B*a*c^3)*x + 1/48*(8*A*a
*c^3 - 5*B*a*c^3)*cos(3*f*x + 3*e)/f + 1/8*(4*A*a*c^3 - 3*B*a*c^3)*cos(f*x + e)/f - 1/32*(A*a*c^3 - 2*B*a*c^3)
*sin(4*f*x + 4*e)/f

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maple [A]  time = 0.48, size = 208, normalized size = 1.46 \[ \frac {-A \,c^{3} a \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {2 A \,c^{3} a \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+2 A \,c^{3} a \cos \left (f x +e \right )+\frac {B \,c^{3} a \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}+2 B \,c^{3} a \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-2 B \,c^{3} a \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+A \,c^{3} a \left (f x +e \right )-B \,c^{3} a \cos \left (f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-A*c^3*a*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)-2/3*A*c^3*a*(2+sin(f*x+e)^2)*cos(f
*x+e)+2*A*c^3*a*cos(f*x+e)+1/5*B*c^3*a*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)+2*B*c^3*a*(-1/4*(sin(f*x
+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)-2*B*c^3*a*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)+A*c^3*a*(
f*x+e)-B*c^3*a*cos(f*x+e))

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maxima [A]  time = 0.33, size = 200, normalized size = 1.41 \[ \frac {320 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a c^{3} - 15 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} A a c^{3} + 480 \, {\left (f x + e\right )} A a c^{3} + 32 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} B a c^{3} + 30 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a c^{3} - 240 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a c^{3} + 960 \, A a c^{3} \cos \left (f x + e\right ) - 480 \, B a c^{3} \cos \left (f x + e\right )}{480 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(320*(cos(f*x + e)^3 - 3*cos(f*x + e))*A*a*c^3 - 15*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*
e))*A*a*c^3 + 480*(f*x + e)*A*a*c^3 + 32*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*B*a*c^3 + 30
*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*B*a*c^3 - 240*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a*c^
3 + 960*A*a*c^3*cos(f*x + e) - 480*B*a*c^3*cos(f*x + e))/f

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mupad [B]  time = 13.80, size = 389, normalized size = 2.74 \[ \frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {3\,A\,a\,c^3}{4}+\frac {B\,a\,c^3}{2}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (4\,A\,a\,c^3-2\,B\,a\,c^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {7\,A\,a\,c^3}{2}-3\,B\,a\,c^3\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (\frac {7\,A\,a\,c^3}{2}-3\,B\,a\,c^3\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9\,\left (\frac {3\,A\,a\,c^3}{4}+\frac {B\,a\,c^3}{2}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (8\,A\,a\,c^3-8\,B\,a\,c^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {8\,A\,a\,c^3}{3}-\frac {8\,B\,a\,c^3}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {16\,A\,a\,c^3}{3}-\frac {4\,B\,a\,c^3}{3}\right )+\frac {4\,A\,a\,c^3}{3}-\frac {14\,B\,a\,c^3}{15}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}+\frac {a\,c^3\,\mathrm {atan}\left (\frac {a\,c^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (5\,A-2\,B\right )}{4\,\left (\frac {5\,A\,a\,c^3}{4}-\frac {B\,a\,c^3}{2}\right )}\right )\,\left (5\,A-2\,B\right )}{4\,f}-\frac {a\,c^3\,\left (5\,A-2\,B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {f\,x}{2}\right )}{4\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(e/2 + (f*x)/2)*((3*A*a*c^3)/4 + (B*a*c^3)/2) + tan(e/2 + (f*x)/2)^8*(4*A*a*c^3 - 2*B*a*c^3) + tan(e/2 + (
f*x)/2)^3*((7*A*a*c^3)/2 - 3*B*a*c^3) - tan(e/2 + (f*x)/2)^7*((7*A*a*c^3)/2 - 3*B*a*c^3) - tan(e/2 + (f*x)/2)^
9*((3*A*a*c^3)/4 + (B*a*c^3)/2) + tan(e/2 + (f*x)/2)^6*(8*A*a*c^3 - 8*B*a*c^3) + tan(e/2 + (f*x)/2)^2*((8*A*a*
c^3)/3 - (8*B*a*c^3)/3) + tan(e/2 + (f*x)/2)^4*((16*A*a*c^3)/3 - (4*B*a*c^3)/3) + (4*A*a*c^3)/3 - (14*B*a*c^3)
/15)/(f*(5*tan(e/2 + (f*x)/2)^2 + 10*tan(e/2 + (f*x)/2)^4 + 10*tan(e/2 + (f*x)/2)^6 + 5*tan(e/2 + (f*x)/2)^8 +
tan(e/2 + (f*x)/2)^10 + 1)) + (a*c^3*atan((a*c^3*tan(e/2 + (f*x)/2)*(5*A - 2*B))/(4*((5*A*a*c^3)/4 - (B*a*c^3
)/2)))*(5*A - 2*B))/(4*f) - (a*c^3*(5*A - 2*B)*(atan(tan(e/2 + (f*x)/2)) - (f*x)/2))/(4*f)

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sympy [A]  time = 4.09, size = 486, normalized size = 3.42 \[ \begin {cases} - \frac {3 A a c^{3} x \sin ^{4}{\left (e + f x \right )}}{8} - \frac {3 A a c^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} - \frac {3 A a c^{3} x \cos ^{4}{\left (e + f x \right )}}{8} + A a c^{3} x + \frac {5 A a c^{3} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {2 A a c^{3} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {3 A a c^{3} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {4 A a c^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac {2 A a c^{3} \cos {\left (e + f x \right )}}{f} + \frac {3 B a c^{3} x \sin ^{4}{\left (e + f x \right )}}{4} + \frac {3 B a c^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{2} - B a c^{3} x \sin ^{2}{\left (e + f x \right )} + \frac {3 B a c^{3} x \cos ^{4}{\left (e + f x \right )}}{4} - B a c^{3} x \cos ^{2}{\left (e + f x \right )} + \frac {B a c^{3} \sin ^{4}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {5 B a c^{3} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{4 f} + \frac {4 B a c^{3} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {3 B a c^{3} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{4 f} + \frac {B a c^{3} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {8 B a c^{3} \cos ^{5}{\left (e + f x \right )}}{15 f} - \frac {B a c^{3} \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (A + B \sin {\relax (e )}\right ) \left (a \sin {\relax (e )} + a\right ) \left (- c \sin {\relax (e )} + c\right )^{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-3*A*a*c**3*x*sin(e + f*x)**4/8 - 3*A*a*c**3*x*sin(e + f*x)**2*cos(e + f*x)**2/4 - 3*A*a*c**3*x*cos
(e + f*x)**4/8 + A*a*c**3*x + 5*A*a*c**3*sin(e + f*x)**3*cos(e + f*x)/(8*f) - 2*A*a*c**3*sin(e + f*x)**2*cos(e
+ f*x)/f + 3*A*a*c**3*sin(e + f*x)*cos(e + f*x)**3/(8*f) - 4*A*a*c**3*cos(e + f*x)**3/(3*f) + 2*A*a*c**3*cos(
e + f*x)/f + 3*B*a*c**3*x*sin(e + f*x)**4/4 + 3*B*a*c**3*x*sin(e + f*x)**2*cos(e + f*x)**2/2 - B*a*c**3*x*sin(
e + f*x)**2 + 3*B*a*c**3*x*cos(e + f*x)**4/4 - B*a*c**3*x*cos(e + f*x)**2 + B*a*c**3*sin(e + f*x)**4*cos(e + f
*x)/f - 5*B*a*c**3*sin(e + f*x)**3*cos(e + f*x)/(4*f) + 4*B*a*c**3*sin(e + f*x)**2*cos(e + f*x)**3/(3*f) - 3*B
*a*c**3*sin(e + f*x)*cos(e + f*x)**3/(4*f) + B*a*c**3*sin(e + f*x)*cos(e + f*x)/f + 8*B*a*c**3*cos(e + f*x)**5
/(15*f) - B*a*c**3*cos(e + f*x)/f, Ne(f, 0)), (x*(A + B*sin(e))*(a*sin(e) + a)*(-c*sin(e) + c)**3, True))

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