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

Optimal. Leaf size=140 \[ -\frac{a^3 c (5 A+2 B) \cos ^3(e+f x)}{12 f}-\frac{c (5 A+2 B) \cos ^3(e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{20 f}+\frac{a^3 c (5 A+2 B) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} a^3 c x (5 A+2 B)-\frac{a B c \cos ^3(e+f x) (a \sin (e+f x)+a)^2}{5 f} \]

[Out]

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

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Rubi [A]  time = 0.222197, antiderivative size = 140, normalized size of antiderivative = 1., 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} \[ -\frac{a^3 c (5 A+2 B) \cos ^3(e+f x)}{12 f}-\frac{c (5 A+2 B) \cos ^3(e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{20 f}+\frac{a^3 c (5 A+2 B) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} a^3 c x (5 A+2 B)-\frac{a B c \cos ^3(e+f x) (a \sin (e+f x)+a)^2}{5 f} \]

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]),x]

[Out]

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

Mathematica [A]  time = 0.824205, size = 95, normalized size = 0.68 \[ \frac{a^3 c (15 (-(A+2 B) \sin (4 (e+f x))+4 f x (5 A+2 B)+8 A \sin (2 (e+f x)))-60 (4 A+3 B) \cos (e+f x)-10 (8 A+5 B) \cos (3 (e+f x))+6 B \cos (5 (e+f x)))}{480 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*c*(-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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Maple [A]  time = 0.031, size = 208, normalized size = 1.5 \begin{align*}{\frac{1}{f} \left ( -A{a}^{3}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{2\,A{a}^{3}c \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+{\frac{B{a}^{3}c\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 \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\cos \left ( fx+e \right ) +2\,B{a}^{3}c \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) +A{a}^{3}c \left ( fx+e \right ) -B{a}^{3}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))^3*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x)

[Out]

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

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Maxima [A]  time = 0.968768, size = 270, normalized size = 1.93 \begin{align*} -\frac{320 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a^{3} c + 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^{3} c - 480 \,{\left (f x + e\right )} A a^{3} c - 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^{3} c + 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^{3} c - 240 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{3} c + 960 \, A a^{3} c \cos \left (f x + e\right ) + 480 \, B a^{3} c \cos \left (f x + e\right )}{480 \, 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)),x, algorithm="maxima")

[Out]

-1/480*(320*(cos(f*x + e)^3 - 3*cos(f*x + e))*A*a^3*c + 15*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2
*e))*A*a^3*c - 480*(f*x + e)*A*a^3*c - 32*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*B*a^3*c + 3
0*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*B*a^3*c - 240*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a^3
*c + 960*A*a^3*c*cos(f*x + e) + 480*B*a^3*c*cos(f*x + e))/f

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Fricas [A]  time = 1.34974, size = 248, normalized size = 1.77 \begin{align*} \frac{24 \, B a^{3} c \cos \left (f x + e\right )^{5} - 80 \,{\left (A + B\right )} a^{3} c \cos \left (f x + e\right )^{3} + 15 \,{\left (5 \, A + 2 \, B\right )} a^{3} c f x - 15 \,{\left (2 \,{\left (A + 2 \, B\right )} a^{3} c \cos \left (f x + e\right )^{3} -{\left (5 \, A + 2 \, B\right )} a^{3} c \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{120 \, 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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 6.0621, size = 486, normalized size = 3.47 \begin{align*} \begin{cases} - \frac{3 A a^{3} c x \sin ^{4}{\left (e + f x \right )}}{8} - \frac{3 A a^{3} c x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} - \frac{3 A a^{3} c x \cos ^{4}{\left (e + f x \right )}}{8} + A a^{3} c x + \frac{5 A a^{3} c \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} + \frac{2 A a^{3} c \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} + \frac{3 A a^{3} c \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} + \frac{4 A a^{3} c \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{2 A a^{3} c \cos{\left (e + f x \right )}}{f} - \frac{3 B a^{3} c x \sin ^{4}{\left (e + f x \right )}}{4} - \frac{3 B a^{3} c x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{2} + B a^{3} c x \sin ^{2}{\left (e + f x \right )} - \frac{3 B a^{3} c x \cos ^{4}{\left (e + f x \right )}}{4} + B a^{3} c x \cos ^{2}{\left (e + f x \right )} + \frac{B a^{3} c \sin ^{4}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} + \frac{5 B a^{3} c \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{4 f} + \frac{4 B a^{3} c \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac{3 B a^{3} c \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{4 f} - \frac{B a^{3} c \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} + \frac{8 B a^{3} c \cos ^{5}{\left (e + f x \right )}}{15 f} - \frac{B a^{3} 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 )^{3} \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))**3*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x)

[Out]

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

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Giac [A]  time = 1.134, size = 196, normalized size = 1.4 \begin{align*} \frac{B a^{3} c \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac{A a^{3} c \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} + \frac{1}{8} \,{\left (5 \, A a^{3} c + 2 \, B a^{3} c\right )} x - \frac{{\left (8 \, A a^{3} c + 5 \, B a^{3} c\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac{{\left (4 \, A a^{3} c + 3 \, B a^{3} c\right )} \cos \left (f x + e\right )}{8 \, f} - \frac{{\left (A a^{3} c + 2 \, B a^{3} c\right )} \sin \left (4 \, f x + 4 \, e\right )}{32 \, 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)),x, algorithm="giac")

[Out]

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