3.17 \(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx\)

Optimal. Leaf size=182 \[ \frac{7 a c^4 (2 A-B) \cos ^3(e+f x)}{24 f}+\frac{a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac{7 a (2 A-B) \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{40 f}+\frac{7 a c^4 (2 A-B) \sin (e+f x) \cos (e+f x)}{16 f}+\frac{7}{16} a c^4 x (2 A-B)-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f} \]

[Out]

(7*a*(2*A - B)*c^4*x)/16 + (7*a*(2*A - B)*c^4*Cos[e + f*x]^3)/(24*f) + (7*a*(2*A - B)*c^4*Cos[e + f*x]*Sin[e +
f*x])/(16*f) - (a*B*c*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^3)/(6*f) + (a*(2*A - B)*Cos[e + f*x]^3*(c^2 - c^2*S
in[e + f*x])^2)/(10*f) + (7*a*(2*A - B)*Cos[e + f*x]^3*(c^4 - c^4*Sin[e + f*x]))/(40*f)

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Rubi [A]  time = 0.295074, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{7 a c^4 (2 A-B) \cos ^3(e+f x)}{24 f}+\frac{a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac{7 a (2 A-B) \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{40 f}+\frac{7 a c^4 (2 A-B) \sin (e+f x) \cos (e+f x)}{16 f}+\frac{7}{16} a c^4 x (2 A-B)-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^4,x]

[Out]

(7*a*(2*A - B)*c^4*x)/16 + (7*a*(2*A - B)*c^4*Cos[e + f*x]^3)/(24*f) + (7*a*(2*A - B)*c^4*Cos[e + f*x]*Sin[e +
f*x])/(16*f) - (a*B*c*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^3)/(6*f) + (a*(2*A - B)*Cos[e + f*x]^3*(c^2 - c^2*S
in[e + f*x])^2)/(10*f) + (7*a*(2*A - B)*Cos[e + f*x]^3*(c^4 - c^4*Sin[e + f*x]))/(40*f)

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 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 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 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 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 8

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

Rubi steps

\begin{align*} \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx &=(a c) \int \cos ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^3 \, dx\\ &=-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f}+\frac{1}{2} (a (2 A-B) c) \int \cos ^2(e+f x) (c-c \sin (e+f x))^3 \, dx\\ &=-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f}+\frac{a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac{1}{10} \left (7 a (2 A-B) c^2\right ) \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))^3}{6 f}+\frac{a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac{7 a (2 A-B) \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{40 f}+\frac{1}{8} \left (7 a (2 A-B) c^3\right ) \int \cos ^2(e+f x) (c-c \sin (e+f x)) \, dx\\ &=\frac{7 a (2 A-B) c^4 \cos ^3(e+f x)}{24 f}-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f}+\frac{a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac{7 a (2 A-B) \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{40 f}+\frac{1}{8} \left (7 a (2 A-B) c^4\right ) \int \cos ^2(e+f x) \, dx\\ &=\frac{7 a (2 A-B) c^4 \cos ^3(e+f x)}{24 f}+\frac{7 a (2 A-B) c^4 \cos (e+f x) \sin (e+f x)}{16 f}-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f}+\frac{a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac{7 a (2 A-B) \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{40 f}+\frac{1}{16} \left (7 a (2 A-B) c^4\right ) \int 1 \, dx\\ &=\frac{7}{16} a (2 A-B) c^4 x+\frac{7 a (2 A-B) c^4 \cos ^3(e+f x)}{24 f}+\frac{7 a (2 A-B) c^4 \cos (e+f x) \sin (e+f x)}{16 f}-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f}+\frac{a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac{7 a (2 A-B) \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{40 f}\\ \end{align*}

Mathematica [A]  time = 0.934998, size = 131, normalized size = 0.72 \[ \frac{a c^4 (120 (7 A-5 B) \cos (e+f x)+20 (13 A-7 B) \cos (3 (e+f x))+240 A \sin (2 (e+f x))-90 A \sin (4 (e+f x))-12 A \cos (5 (e+f x))+840 A f x+15 B \sin (2 (e+f x))+105 B \sin (4 (e+f x))-5 B \sin (6 (e+f x))+36 B \cos (5 (e+f x))-420 B f x)}{960 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^4,x]

[Out]

(a*c^4*(840*A*f*x - 420*B*f*x + 120*(7*A - 5*B)*Cos[e + f*x] + 20*(13*A - 7*B)*Cos[3*(e + f*x)] - 12*A*Cos[5*(
e + f*x)] + 36*B*Cos[5*(e + f*x)] + 240*A*Sin[2*(e + f*x)] + 15*B*Sin[2*(e + f*x)] - 90*A*Sin[4*(e + f*x)] + 1
05*B*Sin[4*(e + f*x)] - 5*B*Sin[6*(e + f*x)]))/(960*f)

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Maple [B]  time = 0.036, size = 342, normalized size = 1.9 \begin{align*}{\frac{1}{f} \left ( -{\frac{A{c}^{4}a\cos \left ( fx+e \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) }-3\,A{c}^{4}a \left ( -1/4\, \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+3/2\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) +3/8\,fx+3/8\,e \right ) -{\frac{2\,A{c}^{4}a \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+2\,A{c}^{4}a \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) +3\,A{c}^{4}a\cos \left ( fx+e \right ) +B{c}^{4}a \left ( -{\frac{\cos \left ( fx+e \right ) }{6} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( fx+e \right ) }{8}} \right ) }+{\frac{5\,fx}{16}}+{\frac{5\,e}{16}} \right ) +{\frac{3\,B{c}^{4}a\cos \left ( fx+e \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) }+2\,B{c}^{4}a \left ( -1/4\, \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+3/2\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) +3/8\,fx+3/8\,e \right ) -{\frac{2\,B{c}^{4}a \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}-3\,B{c}^{4}a \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) +A{c}^{4}a \left ( fx+e \right ) -B{c}^{4}a\cos \left ( fx+e \right ) \right ) } \end{align*}

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))^4,x)

[Out]

1/f*(-1/5*A*c^4*a*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)-3*A*c^4*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^4*a*(2+sin(f*x+e)^2)*cos(f*x+e)+2*A*c^4*a*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f
*x+1/2*e)+3*A*c^4*a*cos(f*x+e)+B*c^4*a*(-1/6*(sin(f*x+e)^5+5/4*sin(f*x+e)^3+15/8*sin(f*x+e))*cos(f*x+e)+5/16*f
*x+5/16*e)+3/5*B*c^4*a*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)+2*B*c^4*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*B*c^4*a*(2+sin(f*x+e)^2)*cos(f*x+e)-3*B*c^4*a*(-1/2*sin(f*x+e)*cos(f*x+e)+
1/2*f*x+1/2*e)+A*c^4*a*(f*x+e)-B*c^4*a*cos(f*x+e))

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Maxima [A]  time = 0.977028, size = 454, normalized size = 2.49 \begin{align*} -\frac{64 \,{\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 )} A a c^{4} - 640 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a c^{4} + 90 \,{\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^{4} - 480 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a c^{4} - 960 \,{\left (f x + e\right )} A a c^{4} - 192 \,{\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^{4} - 640 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a c^{4} - 5 \,{\left (4 \, \sin \left (2 \, f x + 2 \, e\right )^{3} + 60 \, f x + 60 \, e + 9 \, \sin \left (4 \, f x + 4 \, e\right ) - 48 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a c^{4} - 60 \,{\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^{4} + 720 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a c^{4} - 2880 \, A a c^{4} \cos \left (f x + e\right ) + 960 \, B a c^{4} \cos \left (f x + e\right )}{960 \, f} \end{align*}

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))^4,x, algorithm="maxima")

[Out]

-1/960*(64*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*A*a*c^4 - 640*(cos(f*x + e)^3 - 3*cos(f*x
+ e))*A*a*c^4 + 90*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*A*a*c^4 - 480*(2*f*x + 2*e - sin(2*
f*x + 2*e))*A*a*c^4 - 960*(f*x + e)*A*a*c^4 - 192*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*B*a
*c^4 - 640*(cos(f*x + e)^3 - 3*cos(f*x + e))*B*a*c^4 - 5*(4*sin(2*f*x + 2*e)^3 + 60*f*x + 60*e + 9*sin(4*f*x +
4*e) - 48*sin(2*f*x + 2*e))*B*a*c^4 - 60*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*B*a*c^4 + 72
0*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a*c^4 - 2880*A*a*c^4*cos(f*x + e) + 960*B*a*c^4*cos(f*x + e))/f

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Fricas [A]  time = 1.48433, size = 302, normalized size = 1.66 \begin{align*} -\frac{48 \,{\left (A - 3 \, B\right )} a c^{4} \cos \left (f x + e\right )^{5} - 320 \,{\left (A - B\right )} a c^{4} \cos \left (f x + e\right )^{3} - 105 \,{\left (2 \, A - B\right )} a c^{4} f x + 5 \,{\left (8 \, B a c^{4} \cos \left (f x + e\right )^{5} + 2 \,{\left (18 \, A - 25 \, B\right )} a c^{4} \cos \left (f x + e\right )^{3} - 21 \,{\left (2 \, A - B\right )} a c^{4} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{240 \, f} \end{align*}

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))^4,x, algorithm="fricas")

[Out]

-1/240*(48*(A - 3*B)*a*c^4*cos(f*x + e)^5 - 320*(A - B)*a*c^4*cos(f*x + e)^3 - 105*(2*A - B)*a*c^4*f*x + 5*(8*
B*a*c^4*cos(f*x + e)^5 + 2*(18*A - 25*B)*a*c^4*cos(f*x + e)^3 - 21*(2*A - B)*a*c^4*cos(f*x + e))*sin(f*x + e))
/f

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Sympy [A]  time = 10.5282, size = 853, normalized size = 4.69 \begin{align*} \text{result too large to display} \end{align*}

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))**4,x)

[Out]

Piecewise((-9*A*a*c**4*x*sin(e + f*x)**4/8 - 9*A*a*c**4*x*sin(e + f*x)**2*cos(e + f*x)**2/4 + A*a*c**4*x*sin(e
+ f*x)**2 - 9*A*a*c**4*x*cos(e + f*x)**4/8 + A*a*c**4*x*cos(e + f*x)**2 + A*a*c**4*x - A*a*c**4*sin(e + f*x)*
*4*cos(e + f*x)/f + 15*A*a*c**4*sin(e + f*x)**3*cos(e + f*x)/(8*f) - 4*A*a*c**4*sin(e + f*x)**2*cos(e + f*x)**
3/(3*f) - 2*A*a*c**4*sin(e + f*x)**2*cos(e + f*x)/f + 9*A*a*c**4*sin(e + f*x)*cos(e + f*x)**3/(8*f) - A*a*c**4
*sin(e + f*x)*cos(e + f*x)/f - 8*A*a*c**4*cos(e + f*x)**5/(15*f) - 4*A*a*c**4*cos(e + f*x)**3/(3*f) + 3*A*a*c*
*4*cos(e + f*x)/f + 5*B*a*c**4*x*sin(e + f*x)**6/16 + 15*B*a*c**4*x*sin(e + f*x)**4*cos(e + f*x)**2/16 + 3*B*a
*c**4*x*sin(e + f*x)**4/4 + 15*B*a*c**4*x*sin(e + f*x)**2*cos(e + f*x)**4/16 + 3*B*a*c**4*x*sin(e + f*x)**2*co
s(e + f*x)**2/2 - 3*B*a*c**4*x*sin(e + f*x)**2/2 + 5*B*a*c**4*x*cos(e + f*x)**6/16 + 3*B*a*c**4*x*cos(e + f*x)
**4/4 - 3*B*a*c**4*x*cos(e + f*x)**2/2 - 11*B*a*c**4*sin(e + f*x)**5*cos(e + f*x)/(16*f) + 3*B*a*c**4*sin(e +
f*x)**4*cos(e + f*x)/f - 5*B*a*c**4*sin(e + f*x)**3*cos(e + f*x)**3/(6*f) - 5*B*a*c**4*sin(e + f*x)**3*cos(e +
f*x)/(4*f) + 4*B*a*c**4*sin(e + f*x)**2*cos(e + f*x)**3/f - 2*B*a*c**4*sin(e + f*x)**2*cos(e + f*x)/f - 5*B*a
*c**4*sin(e + f*x)*cos(e + f*x)**5/(16*f) - 3*B*a*c**4*sin(e + f*x)*cos(e + f*x)**3/(4*f) + 3*B*a*c**4*sin(e +
f*x)*cos(e + f*x)/(2*f) + 8*B*a*c**4*cos(e + f*x)**5/(5*f) - 4*B*a*c**4*cos(e + f*x)**3/(3*f) - B*a*c**4*cos(
e + f*x)/f, Ne(f, 0)), (x*(A + B*sin(e))*(a*sin(e) + a)*(-c*sin(e) + c)**4, True))

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Giac [A]  time = 1.20789, size = 248, normalized size = 1.36 \begin{align*} -\frac{B a c^{4} \sin \left (6 \, f x + 6 \, e\right )}{192 \, f} + \frac{7}{16} \,{\left (2 \, A a c^{4} - B a c^{4}\right )} x - \frac{{\left (A a c^{4} - 3 \, B a c^{4}\right )} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac{{\left (13 \, A a c^{4} - 7 \, B a c^{4}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} + \frac{{\left (7 \, A a c^{4} - 5 \, B a c^{4}\right )} \cos \left (f x + e\right )}{8 \, f} - \frac{{\left (6 \, A a c^{4} - 7 \, B a c^{4}\right )} \sin \left (4 \, f x + 4 \, e\right )}{64 \, f} + \frac{{\left (16 \, A a c^{4} + B a c^{4}\right )} \sin \left (2 \, f x + 2 \, e\right )}{64 \, f} \end{align*}

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))^4,x, algorithm="giac")

[Out]

-1/192*B*a*c^4*sin(6*f*x + 6*e)/f + 7/16*(2*A*a*c^4 - B*a*c^4)*x - 1/80*(A*a*c^4 - 3*B*a*c^4)*cos(5*f*x + 5*e)
/f + 1/48*(13*A*a*c^4 - 7*B*a*c^4)*cos(3*f*x + 3*e)/f + 1/8*(7*A*a*c^4 - 5*B*a*c^4)*cos(f*x + e)/f - 1/64*(6*A
*a*c^4 - 7*B*a*c^4)*sin(4*f*x + 4*e)/f + 1/64*(16*A*a*c^4 + B*a*c^4)*sin(2*f*x + 2*e)/f