\(\int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx\) [29]

Optimal result

Integrand size = 36, antiderivative size = 89 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\frac {3}{8} a^2 A c^2 x-\frac {a^2 B c^2 \cos ^5(e+f x)}{5 f}+\frac {3 a^2 A c^2 \cos (e+f x) \sin (e+f x)}{8 f}+\frac {a^2 A c^2 \cos ^3(e+f x) \sin (e+f x)}{4 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3046, 2748, 2715, 8} \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\frac {a^2 A c^2 \sin (e+f x) \cos ^3(e+f x)}{4 f}+\frac {3 a^2 A c^2 \sin (e+f x) \cos (e+f x)}{8 f}+\frac {3}{8} a^2 A c^2 x-\frac {a^2 B c^2 \cos ^5(e+f x)}{5 f} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.61 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\frac {a^2 c^2 \left (-32 B \cos ^5(e+f x)+5 A (12 (e+f x)+8 \sin (2 (e+f x))+\sin (4 (e+f x)))\right )}{160 f} \]

[In]

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

[Out]

(a^2*c^2*(-32*B*Cos[e + f*x]^5 + 5*A*(12*(e + f*x) + 8*Sin[2*(e + f*x)] + Sin[4*(e + f*x)])))/(160*f)

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.88

method	result	size
parallelrisch	\(\frac {a^{2} c^{2} \left (60 f x A +5 \sin \left (4 f x +4 e \right ) A +40 A \sin \left (2 f x +2 e \right )-20 \cos \left (f x +e \right ) B -2 \cos \left (5 f x +5 e \right ) B -10 \cos \left (3 f x +3 e \right ) B -32 B \right )}{160 f}\)	\(78\)
risch	\(\frac {3 a^{2} A \,c^{2} x}{8}-\frac {B \,a^{2} c^{2} \cos \left (f x +e \right )}{8 f}-\frac {B \,a^{2} c^{2} \cos \left (5 f x +5 e \right )}{80 f}+\frac {A \,a^{2} c^{2} \sin \left (4 f x +4 e \right )}{32 f}-\frac {B \,a^{2} c^{2} \cos \left (3 f x +3 e \right )}{16 f}+\frac {A \,a^{2} c^{2} \sin \left (2 f x +2 e \right )}{4 f}\)	\(114\)
derivativedivides	\(\frac {A \,a^{2} c^{2} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-2 A \,a^{2} c^{2} \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {B \,a^{2} c^{2} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}+\frac {2 B \,a^{2} c^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+A \,a^{2} c^{2} \left (f x +e \right )-B \,a^{2} c^{2} \cos \left (f x +e \right )}{f}\)	\(166\)
default	\(\frac {A \,a^{2} c^{2} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-2 A \,a^{2} c^{2} \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {B \,a^{2} c^{2} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}+\frac {2 B \,a^{2} c^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+A \,a^{2} c^{2} \left (f x +e \right )-B \,a^{2} c^{2} \cos \left (f x +e \right )}{f}\)	\(166\)
parts	\(a^{2} A \,c^{2} x +\frac {A \,a^{2} c^{2} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}-\frac {B \,a^{2} c^{2} \cos \left (f x +e \right )}{f}-\frac {B \,a^{2} c^{2} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5 f}-\frac {2 A \,a^{2} c^{2} \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {2 B \,a^{2} c^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3 f}\)	\(173\)
norman	\(\frac {-\frac {2 B \,a^{2} c^{2}}{5 f}-\frac {4 B \,a^{2} c^{2} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {2 B \,a^{2} c^{2} \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {3 a^{2} A \,c^{2} x}{8}+\frac {5 A \,a^{2} c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}+\frac {A \,a^{2} c^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}-\frac {A \,a^{2} c^{2} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}-\frac {5 A \,a^{2} c^{2} \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}+\frac {15 a^{2} A \,c^{2} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8}+\frac {15 a^{2} A \,c^{2} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}+\frac {15 a^{2} A \,c^{2} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}+\frac {15 a^{2} A \,c^{2} x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8}+\frac {3 a^{2} A \,c^{2} x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{5}}\)	\(281\)

[In]

int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

1/160*a^2*c^2*(60*f*x*A+5*sin(4*f*x+4*e)*A+40*A*sin(2*f*x+2*e)-20*cos(f*x+e)*B-2*cos(5*f*x+5*e)*B-10*cos(3*f*x
+3*e)*B-32*B)/f

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.84 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=-\frac {8 \, B a^{2} c^{2} \cos \left (f x + e\right )^{5} - 15 \, A a^{2} c^{2} f x - 5 \, {\left (2 \, A a^{2} c^{2} \cos \left (f x + e\right )^{3} + 3 \, A a^{2} c^{2} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{40 \, f} \]

[In]

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

[Out]

-1/40*(8*B*a^2*c^2*cos(f*x + e)^5 - 15*A*a^2*c^2*f*x - 5*(2*A*a^2*c^2*cos(f*x + e)^3 + 3*A*a^2*c^2*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. 372 vs. \(2 (87) = 174\).

Time = 0.28 (sec) , antiderivative size = 372, normalized size of antiderivative = 4.18 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\begin {cases} \frac {3 A a^{2} c^{2} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {3 A a^{2} c^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} - A a^{2} c^{2} x \sin ^{2}{\left (e + f x \right )} + \frac {3 A a^{2} c^{2} x \cos ^{4}{\left (e + f x \right )}}{8} - A a^{2} c^{2} x \cos ^{2}{\left (e + f x \right )} + A a^{2} c^{2} x - \frac {5 A a^{2} c^{2} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {3 A a^{2} c^{2} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} + \frac {A a^{2} c^{2} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {B a^{2} c^{2} \sin ^{4}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {4 B a^{2} c^{2} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac {2 B a^{2} c^{2} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {8 B a^{2} c^{2} \cos ^{5}{\left (e + f x \right )}}{15 f} + \frac {4 B a^{2} c^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {B a^{2} c^{2} \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (A + B \sin {\left (e \right )}\right ) \left (a \sin {\left (e \right )} + a\right )^{2} \left (- c \sin {\left (e \right )} + c\right )^{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (81) = 162\).

Time = 0.22 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.84 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\frac {15 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} c^{2} - 240 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} c^{2} + 480 \, {\left (f x + e\right )} A a^{2} c^{2} - 32 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} B a^{2} c^{2} - 320 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{2} c^{2} - 480 \, B a^{2} c^{2} \cos \left (f x + e\right )}{480 \, f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.27 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\frac {3}{8} \, A a^{2} c^{2} x - \frac {B a^{2} c^{2} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} - \frac {B a^{2} c^{2} \cos \left (3 \, f x + 3 \, e\right )}{16 \, f} - \frac {B a^{2} c^{2} \cos \left (f x + e\right )}{8 \, f} + \frac {A a^{2} c^{2} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {A a^{2} c^{2} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \]

[In]

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

[Out]

3/8*A*a^2*c^2*x - 1/80*B*a^2*c^2*cos(5*f*x + 5*e)/f - 1/16*B*a^2*c^2*cos(3*f*x + 3*e)/f - 1/8*B*a^2*c^2*cos(f*
x + e)/f + 1/32*A*a^2*c^2*sin(4*f*x + 4*e)/f + 1/4*A*a^2*c^2*sin(2*f*x + 2*e)/f

Mupad [B] (verification not implemented)

Time = 14.88 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.67 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\frac {3\,A\,a^2\,c^2\,x}{8}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (\frac {a^2\,c^2\,\left (80\,B-75\,A\,\left (e+f\,x\right )\right )}{40}+\frac {15\,A\,a^2\,c^2\,\left (e+f\,x\right )}{8}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {a^2\,c^2\,\left (160\,B-150\,A\,\left (e+f\,x\right )\right )}{40}+\frac {15\,A\,a^2\,c^2\,\left (e+f\,x\right )}{4}\right )+\frac {a^2\,c^2\,\left (16\,B-15\,A\,\left (e+f\,x\right )\right )}{40}+\frac {3\,A\,a^2\,c^2\,\left (e+f\,x\right )}{8}-\frac {A\,a^2\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{2}+\frac {A\,a^2\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{2}+\frac {5\,A\,a^2\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{4}-\frac {5\,A\,a^2\,c^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{4}}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^5} \]

[In]

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

[Out]

(3*A*a^2*c^2*x)/8 - (tan(e/2 + (f*x)/2)^8*((a^2*c^2*(80*B - 75*A*(e + f*x)))/40 + (15*A*a^2*c^2*(e + f*x))/8)
+ tan(e/2 + (f*x)/2)^4*((a^2*c^2*(160*B - 150*A*(e + f*x)))/40 + (15*A*a^2*c^2*(e + f*x))/4) + (a^2*c^2*(16*B
- 15*A*(e + f*x)))/40 + (3*A*a^2*c^2*(e + f*x))/8 - (A*a^2*c^2*tan(e/2 + (f*x)/2)^3)/2 + (A*a^2*c^2*tan(e/2 +
(f*x)/2)^7)/2 + (5*A*a^2*c^2*tan(e/2 + (f*x)/2)^9)/4 - (5*A*a^2*c^2*tan(e/2 + (f*x)/2))/4)/(f*(tan(e/2 + (f*x)
/2)^2 + 1)^5)