3.42 \(\int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=138 \[ -\frac{a^3 c^2 (6 A+B) \cos ^5(e+f x)}{30 f}+\frac{a^3 c^2 (6 A+B) \sin (e+f x) \cos ^3(e+f x)}{24 f}+\frac{a^3 c^2 (6 A+B) \sin (e+f x) \cos (e+f x)}{16 f}+\frac{1}{16} a^3 c^2 x (6 A+B)-\frac{B c^2 \cos ^5(e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{6 f} \]

[Out]

(a^3*(6*A + B)*c^2*x)/16 - (a^3*(6*A + B)*c^2*Cos[e + f*x]^5)/(30*f) + (a^3*(6*A + B)*c^2*Cos[e + f*x]*Sin[e +
f*x])/(16*f) + (a^3*(6*A + B)*c^2*Cos[e + f*x]^3*Sin[e + f*x])/(24*f) - (B*c^2*Cos[e + f*x]^5*(a^3 + a^3*Sin[
e + f*x]))/(6*f)

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Rubi [A]  time = 0.199824, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {2967, 2860, 2669, 2635, 8} \[ -\frac{a^3 c^2 (6 A+B) \cos ^5(e+f x)}{30 f}+\frac{a^3 c^2 (6 A+B) \sin (e+f x) \cos ^3(e+f x)}{24 f}+\frac{a^3 c^2 (6 A+B) \sin (e+f x) \cos (e+f x)}{16 f}+\frac{1}{16} a^3 c^2 x (6 A+B)-\frac{B c^2 \cos ^5(e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{6 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(6*A + B)*c^2*x)/16 - (a^3*(6*A + B)*c^2*Cos[e + f*x]^5)/(30*f) + (a^3*(6*A + B)*c^2*Cos[e + f*x]*Sin[e +
f*x])/(16*f) + (a^3*(6*A + B)*c^2*Cos[e + f*x]^3*Sin[e + f*x])/(24*f) - (B*c^2*Cos[e + f*x]^5*(a^3 + a^3*Sin[
e + f*x]))/(6*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 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))^3 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx &=\left (a^2 c^2\right ) \int \cos ^4(e+f x) (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx\\ &=-\frac{B c^2 \cos ^5(e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{6 f}+\frac{1}{6} \left (a^2 (6 A+B) c^2\right ) \int \cos ^4(e+f x) (a+a \sin (e+f x)) \, dx\\ &=-\frac{a^3 (6 A+B) c^2 \cos ^5(e+f x)}{30 f}-\frac{B c^2 \cos ^5(e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{6 f}+\frac{1}{6} \left (a^3 (6 A+B) c^2\right ) \int \cos ^4(e+f x) \, dx\\ &=-\frac{a^3 (6 A+B) c^2 \cos ^5(e+f x)}{30 f}+\frac{a^3 (6 A+B) c^2 \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac{B c^2 \cos ^5(e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{6 f}+\frac{1}{8} \left (a^3 (6 A+B) c^2\right ) \int \cos ^2(e+f x) \, dx\\ &=-\frac{a^3 (6 A+B) c^2 \cos ^5(e+f x)}{30 f}+\frac{a^3 (6 A+B) c^2 \cos (e+f x) \sin (e+f x)}{16 f}+\frac{a^3 (6 A+B) c^2 \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac{B c^2 \cos ^5(e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{6 f}+\frac{1}{16} \left (a^3 (6 A+B) c^2\right ) \int 1 \, dx\\ &=\frac{1}{16} a^3 (6 A+B) c^2 x-\frac{a^3 (6 A+B) c^2 \cos ^5(e+f x)}{30 f}+\frac{a^3 (6 A+B) c^2 \cos (e+f x) \sin (e+f x)}{16 f}+\frac{a^3 (6 A+B) c^2 \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac{B c^2 \cos ^5(e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{6 f}\\ \end{align*}

Mathematica [A]  time = 1.03372, size = 133, normalized size = 0.96 \[ \frac{a^3 c^2 (-120 (A+B) \cos (e+f x)-60 (A+B) \cos (3 (e+f x))+240 A \sin (2 (e+f x))+30 A \sin (4 (e+f x))-12 A \cos (5 (e+f x))+360 A e+360 A f x+15 B \sin (2 (e+f x))-15 B \sin (4 (e+f x))-5 B \sin (6 (e+f x))-12 B \cos (5 (e+f x))+60 B e+60 B f x)}{960 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*c^2*(360*A*e + 60*B*e + 360*A*f*x + 60*B*f*x - 120*(A + B)*Cos[e + f*x] - 60*(A + B)*Cos[3*(e + f*x)] - 1
2*A*Cos[5*(e + f*x)] - 12*B*Cos[5*(e + f*x)] + 240*A*Sin[2*(e + f*x)] + 15*B*Sin[2*(e + f*x)] + 30*A*Sin[4*(e
+ f*x)] - 15*B*Sin[4*(e + f*x)] - 5*B*Sin[6*(e + f*x)]))/(960*f)

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Maple [B]  time = 0.03, size = 364, normalized size = 2.6 \begin{align*}{\frac{1}{f} \left ( -{\frac{A{a}^{3}{c}^{2}\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 ) }+A{a}^{3}{c}^{2} \left ( -{\frac{\cos \left ( fx+e \right ) }{4} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\sin \left ( fx+e \right ) }{2}} \right ) }+{\frac{3\,fx}{8}}+{\frac{3\,e}{8}} \right ) +{\frac{2\,A{a}^{3}{c}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+B{a}^{3}{c}^{2} \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{B{a}^{3}{c}^{2}\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{a}^{3}{c}^{2} \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 ) -2\,A{a}^{3}{c}^{2} \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) +{\frac{2\,B{a}^{3}{c}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}-A{a}^{3}{c}^{2}\cos \left ( fx+e \right ) +B{a}^{3}{c}^{2} \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) +A{a}^{3}{c}^{2} \left ( fx+e \right ) -B{a}^{3}{c}^{2}\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))^3*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2,x)

[Out]

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

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Maxima [B]  time = 0.992559, size = 486, normalized size = 3.52 \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^{3} c^{2} + 640 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a^{3} c^{2} - 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 )} A a^{3} c^{2} + 480 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{3} c^{2} - 960 \,{\left (f x + e\right )} A a^{3} c^{2} + 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 )} B a^{3} c^{2} + 640 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{3} c^{2} - 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^{3} c^{2} + 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^{3} c^{2} - 240 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{3} c^{2} + 960 \, A a^{3} c^{2} \cos \left (f x + e\right ) + 960 \, B a^{3} c^{2} \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))^3*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2,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^3*c^2 + 640*(cos(f*x + e)^3 - 3*cos(f*
x + e))*A*a^3*c^2 - 30*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*A*a^3*c^2 + 480*(2*f*x + 2*e -
sin(2*f*x + 2*e))*A*a^3*c^2 - 960*(f*x + e)*A*a^3*c^2 + 64*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x
+ e))*B*a^3*c^2 + 640*(cos(f*x + e)^3 - 3*cos(f*x + e))*B*a^3*c^2 - 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^3*c^2 + 60*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e
))*B*a^3*c^2 - 240*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a^3*c^2 + 960*A*a^3*c^2*cos(f*x + e) + 960*B*a^3*c^2*cos
(f*x + e))/f

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Fricas [A]  time = 1.50649, size = 258, normalized size = 1.87 \begin{align*} -\frac{48 \,{\left (A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{5} - 15 \,{\left (6 \, A + B\right )} a^{3} c^{2} f x + 5 \,{\left (8 \, B a^{3} c^{2} \cos \left (f x + e\right )^{5} - 2 \,{\left (6 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{3} - 3 \,{\left (6 \, A + B\right )} a^{3} c^{2} \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))^3*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-1/240*(48*(A + B)*a^3*c^2*cos(f*x + e)^5 - 15*(6*A + B)*a^3*c^2*f*x + 5*(8*B*a^3*c^2*cos(f*x + e)^5 - 2*(6*A
+ B)*a^3*c^2*cos(f*x + e)^3 - 3*(6*A + B)*a^3*c^2*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 11.6746, size = 910, normalized size = 6.59 \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))**3*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**2,x)

[Out]

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

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Giac [A]  time = 1.17512, size = 275, normalized size = 1.99 \begin{align*} -\frac{B a^{3} c^{2} \sin \left (6 \, f x + 6 \, e\right )}{192 \, f} + \frac{1}{16} \,{\left (6 \, A a^{3} c^{2} + B a^{3} c^{2}\right )} x - \frac{{\left (A a^{3} c^{2} + B a^{3} c^{2}\right )} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} - \frac{{\left (A a^{3} c^{2} + B a^{3} c^{2}\right )} \cos \left (3 \, f x + 3 \, e\right )}{16 \, f} - \frac{{\left (A a^{3} c^{2} + B a^{3} c^{2}\right )} \cos \left (f x + e\right )}{8 \, f} + \frac{{\left (2 \, A a^{3} c^{2} - B a^{3} c^{2}\right )} \sin \left (4 \, f x + 4 \, e\right )}{64 \, f} + \frac{{\left (16 \, A a^{3} c^{2} + B a^{3} c^{2}\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))^3*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2,x, algorithm="giac")

[Out]

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