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

Optimal result

Integrand size = 32, antiderivative size = 96 \[ \int \sin ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=\frac {1}{8} a^2 c x-\frac {a^2 c \cos ^3(e+f x)}{3 f}+\frac {a^2 c \cos ^5(e+f x)}{5 f}-\frac {a^2 c \cos (e+f x) \sin (e+f x)}{8 f}+\frac {a^2 c \cos (e+f x) \sin ^3(e+f x)}{4 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3045, 2715, 8, 2713} \[ \int \sin ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=\frac {a^2 c \cos ^5(e+f x)}{5 f}-\frac {a^2 c \cos ^3(e+f x)}{3 f}+\frac {a^2 c \sin ^3(e+f x) \cos (e+f x)}{4 f}-\frac {a^2 c \sin (e+f x) \cos (e+f x)}{8 f}+\frac {1}{8} a^2 c x \]

[In]

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

[Out]

(a^2*c*x)/8 - (a^2*c*Cos[e + f*x]^3)/(3*f) + (a^2*c*Cos[e + f*x]^5)/(5*f) - (a^2*c*Cos[e + f*x]*Sin[e + f*x])/
(8*f) + (a^2*c*Cos[e + f*x]*Sin[e + f*x]^3)/(4*f)

Rule 8

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

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 3045

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^2 c \sin ^2(e+f x)+a^2 c \sin ^3(e+f x)-a^2 c \sin ^4(e+f x)-a^2 c \sin ^5(e+f x)\right ) \, dx \\ & = \left (a^2 c\right ) \int \sin ^2(e+f x) \, dx+\left (a^2 c\right ) \int \sin ^3(e+f x) \, dx-\left (a^2 c\right ) \int \sin ^4(e+f x) \, dx-\left (a^2 c\right ) \int \sin ^5(e+f x) \, dx \\ & = -\frac {a^2 c \cos (e+f x) \sin (e+f x)}{2 f}+\frac {a^2 c \cos (e+f x) \sin ^3(e+f x)}{4 f}+\frac {1}{2} \left (a^2 c\right ) \int 1 \, dx-\frac {1}{4} \left (3 a^2 c\right ) \int \sin ^2(e+f x) \, dx-\frac {\left (a^2 c\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (e+f x)\right )}{f}+\frac {\left (a^2 c\right ) \text {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (e+f x)\right )}{f} \\ & = \frac {1}{2} a^2 c x-\frac {a^2 c \cos ^3(e+f x)}{3 f}+\frac {a^2 c \cos ^5(e+f x)}{5 f}-\frac {a^2 c \cos (e+f x) \sin (e+f x)}{8 f}+\frac {a^2 c \cos (e+f x) \sin ^3(e+f x)}{4 f}-\frac {1}{8} \left (3 a^2 c\right ) \int 1 \, dx \\ & = \frac {1}{8} a^2 c x-\frac {a^2 c \cos ^3(e+f x)}{3 f}+\frac {a^2 c \cos ^5(e+f x)}{5 f}-\frac {a^2 c \cos (e+f x) \sin (e+f x)}{8 f}+\frac {a^2 c \cos (e+f x) \sin ^3(e+f x)}{4 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.59 \[ \int \sin ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=\frac {a^2 c (60 e+60 f x-60 \cos (e+f x)-10 \cos (3 (e+f x))+6 \cos (5 (e+f x))-15 \sin (4 (e+f x)))}{480 f} \]

[In]

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

[Out]

(a^2*c*(60*e + 60*f*x - 60*Cos[e + f*x] - 10*Cos[3*(e + f*x)] + 6*Cos[5*(e + f*x)] - 15*Sin[4*(e + f*x)]))/(48
0*f)

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.59

method	result	size
parallelrisch	\(-\frac {a^{2} c \left (-60 f x +60 \cos \left (f x +e \right )-6 \cos \left (5 f x +5 e \right )+15 \sin \left (4 f x +4 e \right )+10 \cos \left (3 f x +3 e \right )+64\right )}{480 f}\)	\(57\)
risch	\(\frac {a^{2} c x}{8}-\frac {a^{2} c \cos \left (f x +e \right )}{8 f}+\frac {a^{2} c \cos \left (5 f x +5 e \right )}{80 f}-\frac {a^{2} c \sin \left (4 f x +4 e \right )}{32 f}-\frac {a^{2} c \cos \left (3 f x +3 e \right )}{48 f}\)	\(78\)
derivativedivides	\(\frac {\frac {a^{2} c \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}-a^{2} c \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {a^{2} c \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+a^{2} c \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\)	\(126\)
default	\(\frac {\frac {a^{2} c \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}-a^{2} c \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {a^{2} c \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+a^{2} c \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\)	\(126\)
parts	\(\frac {a^{2} c \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {a^{2} c \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3 f}-\frac {a^{2} c \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}+\frac {a^{2} c \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5 f}\)	\(134\)
norman	\(\frac {-\frac {4 a^{2} c}{15 f}+\frac {a^{2} c x}{8}-\frac {4 a^{2} c \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {4 a^{2} c \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}-\frac {4 a^{2} c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}-\frac {a^{2} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}+\frac {3 a^{2} c \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}-\frac {3 a^{2} c \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}+\frac {a^{2} c \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}+\frac {5 a^{2} c x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8}+\frac {5 a^{2} c x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}+\frac {5 a^{2} c x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}+\frac {5 a^{2} c x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8}+\frac {a^{2} c x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{5}}\)	\(262\)

[In]

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

[Out]

-1/480*a^2*c*(-60*f*x+60*cos(f*x+e)-6*cos(5*f*x+5*e)+15*sin(4*f*x+4*e)+10*cos(3*f*x+3*e)+64)/f

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.80 \[ \int \sin ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=\frac {24 \, a^{2} c \cos \left (f x + e\right )^{5} - 40 \, a^{2} c \cos \left (f x + e\right )^{3} + 15 \, a^{2} c f x - 15 \, {\left (2 \, a^{2} c \cos \left (f x + e\right )^{3} - a^{2} c \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{120 \, f} \]

[In]

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

[Out]

1/120*(24*a^2*c*cos(f*x + e)^5 - 40*a^2*c*cos(f*x + e)^3 + 15*a^2*c*f*x - 15*(2*a^2*c*cos(f*x + e)^3 - a^2*c*c
os(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. 301 vs. \(2 (87) = 174\).

Time = 0.26 (sec) , antiderivative size = 301, normalized size of antiderivative = 3.14 \[ \int \sin ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=\begin {cases} - \frac {3 a^{2} c x \sin ^{4}{\left (e + f x \right )}}{8} - \frac {3 a^{2} c x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {a^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} - \frac {3 a^{2} c x \cos ^{4}{\left (e + f x \right )}}{8} + \frac {a^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} + \frac {a^{2} c \sin ^{4}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {5 a^{2} c \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} + \frac {4 a^{2} c \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {a^{2} c \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {3 a^{2} c \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {a^{2} c \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} + \frac {8 a^{2} c \cos ^{5}{\left (e + f x \right )}}{15 f} - \frac {2 a^{2} c \cos ^{3}{\left (e + f x \right )}}{3 f} & \text {for}\: f \neq 0 \\x \left (a \sin {\left (e \right )} + a\right )^{2} \left (- c \sin {\left (e \right )} + c\right ) \sin ^{2}{\left (e \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.28 \[ \int \sin ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=\frac {32 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} a^{2} c + 160 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} c - 15 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c + 120 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c}{480 \, f} \]

[In]

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

[Out]

1/480*(32*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*a^2*c + 160*(cos(f*x + e)^3 - 3*cos(f*x + e
))*a^2*c - 15*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a^2*c + 120*(2*f*x + 2*e - sin(2*f*x + 2
*e))*a^2*c)/f

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.80 \[ \int \sin ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=\frac {1}{8} \, a^{2} c x + \frac {a^{2} c \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} - \frac {a^{2} c \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac {a^{2} c \cos \left (f x + e\right )}{8 \, f} - \frac {a^{2} c \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.24 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.21 \[ \int \sin ^2(e+f x) (a+a \sin (e+f x))^2 (c-c \sin (e+f x)) \, dx=\frac {a^2\,c\,\left (15\,e-30\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-160\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+180\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+160\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-480\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-180\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+30\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9+15\,f\,x+75\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (e+f\,x\right )+150\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (e+f\,x\right )+150\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (e+f\,x\right )+75\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (e+f\,x\right )+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}\,\left (e+f\,x\right )-32\right )}{120\,f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^5} \]

[In]

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

[Out]

(a^2*c*(15*e - 30*tan(e/2 + (f*x)/2) - 160*tan(e/2 + (f*x)/2)^2 + 180*tan(e/2 + (f*x)/2)^3 + 160*tan(e/2 + (f*
x)/2)^4 - 480*tan(e/2 + (f*x)/2)^6 - 180*tan(e/2 + (f*x)/2)^7 + 30*tan(e/2 + (f*x)/2)^9 + 15*f*x + 75*tan(e/2
+ (f*x)/2)^2*(e + f*x) + 150*tan(e/2 + (f*x)/2)^4*(e + f*x) + 150*tan(e/2 + (f*x)/2)^6*(e + f*x) + 75*tan(e/2
+ (f*x)/2)^8*(e + f*x) + 15*tan(e/2 + (f*x)/2)^10*(e + f*x) - 32))/(120*f*(tan(e/2 + (f*x)/2)^2 + 1)^5)