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

Optimal. Leaf size=98 \[ -\frac{a^2 c (4 A+B) \cos ^3(e+f x)}{12 f}+\frac{a^2 c (4 A+B) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} a^2 c x (4 A+B)-\frac{B c \cos ^3(e+f x) \left (a^2 \sin (e+f x)+a^2\right )}{4 f} \]

[Out]

(a^2*(4*A + B)*c*x)/8 - (a^2*(4*A + B)*c*Cos[e + f*x]^3)/(12*f) + (a^2*(4*A + B)*c*Cos[e + f*x]*Sin[e + f*x])/
(8*f) - (B*c*Cos[e + f*x]^3*(a^2 + a^2*Sin[e + f*x]))/(4*f)

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Rubi [A]  time = 0.148919, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {2967, 2860, 2669, 2635, 8} \[ -\frac{a^2 c (4 A+B) \cos ^3(e+f x)}{12 f}+\frac{a^2 c (4 A+B) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} a^2 c x (4 A+B)-\frac{B c \cos ^3(e+f x) \left (a^2 \sin (e+f x)+a^2\right )}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.786035, size = 67, normalized size = 0.68 \[ -\frac{a^2 c (24 (A+B) \cos (e+f x)+8 (A+B) \cos (3 (e+f x))-12 f x (4 A+B)-24 A \sin (2 (e+f x))+3 B \sin (4 (e+f x)))}{96 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*c*(-12*(4*A + B)*f*x + 24*(A + B)*Cos[e + f*x] + 8*(A + B)*Cos[3*(e + f*x)] - 24*A*Sin[2*(e + f*x)] + 3*
B*Sin[4*(e + f*x)]))/(96*f)

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Maple [B]  time = 0.027, size = 186, normalized size = 1.9 \begin{align*}{\frac{1}{f} \left ({\frac{A{a}^{2}c \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}-A{a}^{2}c \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) -B{a}^{2}c \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{B{a}^{2}c \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}-A{a}^{2}c\cos \left ( fx+e \right ) +B{a}^{2}c \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) +A{a}^{2}c \left ( fx+e \right ) -B{a}^{2}c\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))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x)

[Out]

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

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Maxima [A]  time = 0.971595, size = 242, normalized size = 2.47 \begin{align*} -\frac{32 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a^{2} c + 24 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} c - 96 \,{\left (f x + e\right )} A a^{2} c + 32 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{2} c + 3 \,{\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^{2} c - 24 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} c + 96 \, A a^{2} c \cos \left (f x + e\right ) + 96 \, B a^{2} c \cos \left (f x + e\right )}{96 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(32*(cos(f*x + e)^3 - 3*cos(f*x + e))*A*a^2*c + 24*(2*f*x + 2*e - sin(2*f*x + 2*e))*A*a^2*c - 96*(f*x +
e)*A*a^2*c + 32*(cos(f*x + e)^3 - 3*cos(f*x + e))*B*a^2*c + 3*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x
+ 2*e))*B*a^2*c - 24*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a^2*c + 96*A*a^2*c*cos(f*x + e) + 96*B*a^2*c*cos(f*x +
e))/f

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Fricas [A]  time = 1.40412, size = 190, normalized size = 1.94 \begin{align*} -\frac{8 \,{\left (A + B\right )} a^{2} c \cos \left (f x + e\right )^{3} - 3 \,{\left (4 \, A + B\right )} a^{2} c f x + 3 \,{\left (2 \, B a^{2} c \cos \left (f x + e\right )^{3} -{\left (4 \, A + B\right )} a^{2} c \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(8*(A + B)*a^2*c*cos(f*x + e)^3 - 3*(4*A + B)*a^2*c*f*x + 3*(2*B*a^2*c*cos(f*x + e)^3 - (4*A + B)*a^2*c*
cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 1.79887, size = 396, normalized size = 4.04 \begin{align*} \begin{cases} - \frac{A a^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} - \frac{A a^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} + A a^{2} c x + \frac{A a^{2} c \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} + \frac{A a^{2} c \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} + \frac{2 A a^{2} c \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{A a^{2} c \cos{\left (e + f x \right )}}{f} - \frac{3 B a^{2} c x \sin ^{4}{\left (e + f x \right )}}{8} - \frac{3 B a^{2} c x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{B a^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} - \frac{3 B a^{2} c x \cos ^{4}{\left (e + f x \right )}}{8} + \frac{B a^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} + \frac{5 B a^{2} c \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} + \frac{B a^{2} c \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} + \frac{3 B a^{2} c \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac{B a^{2} c \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} + \frac{2 B a^{2} c \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{B a^{2} c \cos{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \left (A + B \sin{\left (e \right )}\right ) \left (a \sin{\left (e \right )} + a\right )^{2} \left (- c \sin{\left (e \right )} + c\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16806, size = 150, normalized size = 1.53 \begin{align*} -\frac{B a^{2} c \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac{A a^{2} c \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} + \frac{1}{8} \,{\left (4 \, A a^{2} c + B a^{2} c\right )} x - \frac{{\left (A a^{2} c + B a^{2} c\right )} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac{{\left (A a^{2} c + B a^{2} c\right )} \cos \left (f x + e\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*B*a^2*c*sin(4*f*x + 4*e)/f + 1/4*A*a^2*c*sin(2*f*x + 2*e)/f + 1/8*(4*A*a^2*c + B*a^2*c)*x - 1/12*(A*a^2*
c + B*a^2*c)*cos(3*f*x + 3*e)/f - 1/4*(A*a^2*c + B*a^2*c)*cos(f*x + e)/f