\(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx\) [20]

Optimal result

Integrand size = 32, antiderivative size = 49 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\frac {1}{2} a A c x-\frac {a B c \cos ^3(e+f x)}{3 f}+\frac {a A c \cos (e+f x) \sin (e+f x)}{2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3046, 2748, 2715, 8} \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\frac {a A c \sin (e+f x) \cos (e+f x)}{2 f}+\frac {1}{2} a A c x-\frac {a B c \cos ^3(e+f x)}{3 f} \]

[In]

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

[Out]

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

Rule 8

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

Rule 2715

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] && Integ
erQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[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 3046

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*} \text {integral}& = (a c) \int \cos ^2(e+f x) (A+B \sin (e+f x)) \, dx \\ & = -\frac {a B c \cos ^3(e+f x)}{3 f}+(a A c) \int \cos ^2(e+f x) \, dx \\ & = -\frac {a B c \cos ^3(e+f x)}{3 f}+\frac {a A c \cos (e+f x) \sin (e+f x)}{2 f}+\frac {1}{2} (a A c) \int 1 \, dx \\ & = \frac {1}{2} a A c x-\frac {a B c \cos ^3(e+f x)}{3 f}+\frac {a A c \cos (e+f x) \sin (e+f x)}{2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=-\frac {a c (3 B \cos (e+f x)+B \cos (3 (e+f x))-3 A (-2 e+2 f x+\sin (2 (e+f x))))}{12 f} \]

[In]

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

[Out]

-1/12*(a*c*(3*B*Cos[e + f*x] + B*Cos[3*(e + f*x)] - 3*A*(-2*e + 2*f*x + Sin[2*(e + f*x)])))/f

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.02

method	result	size
parallelrisch	\(\frac {a c \left (6 f x A +3 A \sin \left (2 f x +2 e \right )-3 \cos \left (f x +e \right ) B -\cos \left (3 f x +3 e \right ) B -4 B \right )}{12 f}\)	\(50\)
risch	\(\frac {a A c x}{2}-\frac {B a c \cos \left (f x +e \right )}{4 f}-\frac {B a c \cos \left (3 f x +3 e \right )}{12 f}+\frac {A a c \sin \left (2 f x +2 e \right )}{4 f}\)	\(56\)
derivativedivides	\(\frac {\frac {B a c \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}-A a c \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-B a c \cos \left (f x +e \right )+A a c \left (f x +e \right )}{f}\)	\(74\)
default	\(\frac {\frac {B a c \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}-A a c \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-B a c \cos \left (f x +e \right )+A a c \left (f x +e \right )}{f}\)	\(74\)
parts	\(a A c x -\frac {B a c \cos \left (f x +e \right )}{f}-\frac {A a c \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {B a c \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3 f}\)	\(75\)
norman	\(\frac {\frac {A a c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {2 B a c}{3 f}-\frac {2 B a c \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {a A c x}{2}-\frac {A a c \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {3 a A c x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {3 a A c x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {a A c x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}\)	\(137\)

[In]

int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/12*a*c*(6*f*x*A+3*A*sin(2*f*x+2*e)-3*cos(f*x+e)*B-cos(3*f*x+3*e)*B-4*B)/f

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=-\frac {2 \, B a c \cos \left (f x + e\right )^{3} - 3 \, A a c f x - 3 \, A a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}{6 \, f} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (46) = 92\).

Time = 0.13 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.82 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\begin {cases} - \frac {A a c x \sin ^{2}{\left (e + f x \right )}}{2} - \frac {A a c x \cos ^{2}{\left (e + f x \right )}}{2} + A a c x + \frac {A a c \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} + \frac {B a c \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {2 B a c \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {B a c \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (A + B \sin {\left (e \right )}\right ) \left (a \sin {\left (e \right )} + a\right ) \left (- c \sin {\left (e \right )} + c\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.49 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=-\frac {3 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a c - 12 \, {\left (f x + e\right )} A a c + 4 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a c + 12 \, B a c \cos \left (f x + e\right )}{12 \, f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.12 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\frac {1}{2} \, A a c x - \frac {B a c \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac {B a c \cos \left (f x + e\right )}{4 \, f} + \frac {A a c \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.57 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.49 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\frac {A\,a\,c\,x}{2}-\frac {A\,a\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+\left (\frac {a\,c\,\left (12\,B-9\,A\,\left (e+f\,x\right )\right )}{6}+\frac {3\,A\,a\,c\,\left (e+f\,x\right )}{2}\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-A\,a\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+\frac {a\,c\,\left (4\,B-3\,A\,\left (e+f\,x\right )\right )}{6}+\frac {A\,a\,c\,\left (e+f\,x\right )}{2}}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^3} \]

[In]

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

[Out]

(A*a*c*x)/2 - (tan(e/2 + (f*x)/2)^4*((a*c*(12*B - 9*A*(e + f*x)))/6 + (3*A*a*c*(e + f*x))/2) + (a*c*(4*B - 3*A
*(e + f*x)))/6 - A*a*c*tan(e/2 + (f*x)/2) + (A*a*c*(e + f*x))/2 + A*a*c*tan(e/2 + (f*x)/2)^5)/(f*(tan(e/2 + (f
*x)/2)^2 + 1)^3)