\(\int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx\) [1]

Optimal result

Integrand size = 21, antiderivative size = 102 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=\frac {3 a^2 x}{4}-\frac {2 a^2 \cos (e+f x)}{f}+\frac {a^2 \cos ^3(e+f x)}{f}-\frac {a^2 \cos ^5(e+f x)}{5 f}-\frac {3 a^2 \cos (e+f x) \sin (e+f x)}{4 f}-\frac {a^2 \cos (e+f x) \sin ^3(e+f x)}{2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2836, 2713, 2715, 8} \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=-\frac {a^2 \cos ^5(e+f x)}{5 f}+\frac {a^2 \cos ^3(e+f x)}{f}-\frac {2 a^2 \cos (e+f x)}{f}-\frac {a^2 \sin ^3(e+f x) \cos (e+f x)}{2 f}-\frac {3 a^2 \sin (e+f x) \cos (e+f x)}{4 f}+\frac {3 a^2 x}{4} \]

[In]

Int[Sin[e + f*x]^3*(a + a*Sin[e + f*x])^2,x]

[Out]

(3*a^2*x)/4 - (2*a^2*Cos[e + f*x])/f + (a^2*Cos[e + f*x]^3)/f - (a^2*Cos[e + f*x]^5)/(5*f) - (3*a^2*Cos[e + f*
x]*Sin[e + f*x])/(4*f) - (a^2*Cos[e + f*x]*Sin[e + f*x]^3)/(2*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 2836

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.03 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=-\frac {a^2 \cos (e+f x) \left (30 \arcsin \left (\frac {\sqrt {1-\sin (e+f x)}}{\sqrt {2}}\right )+\sqrt {\cos ^2(e+f x)} \left (24+15 \sin (e+f x)+12 \sin ^2(e+f x)+10 \sin ^3(e+f x)+4 \sin ^4(e+f x)\right )\right )}{20 f \sqrt {\cos ^2(e+f x)}} \]

[In]

Integrate[Sin[e + f*x]^3*(a + a*Sin[e + f*x])^2,x]

[Out]

-1/20*(a^2*Cos[e + f*x]*(30*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt[2]] + Sqrt[Cos[e + f*x]^2]*(24 + 15*Sin[e + f*x
] + 12*Sin[e + f*x]^2 + 10*Sin[e + f*x]^3 + 4*Sin[e + f*x]^4)))/(f*Sqrt[Cos[e + f*x]^2])

Maple [A] (verified)

Time = 1.74 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.64

method	result	size
parallelrisch	\(-\frac {a^{2} \left (-60 f x +\cos \left (5 f x +5 e \right )-5 \sin \left (4 f x +4 e \right )-15 \cos \left (3 f x +3 e \right )+40 \sin \left (2 f x +2 e \right )+110 \cos \left (f x +e \right )+96\right )}{80 f}\)	\(65\)
risch	\(\frac {3 a^{2} x}{4}-\frac {11 a^{2} \cos \left (f x +e \right )}{8 f}-\frac {a^{2} \cos \left (5 f x +5 e \right )}{80 f}+\frac {a^{2} \sin \left (4 f x +4 e \right )}{16 f}+\frac {3 a^{2} \cos \left (3 f x +3 e \right )}{16 f}-\frac {a^{2} \sin \left (2 f x +2 e \right )}{2 f}\)	\(90\)
derivativedivides	\(\frac {-\frac {a^{2} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}+2 a^{2} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {a^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}}{f}\)	\(96\)
default	\(\frac {-\frac {a^{2} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}+2 a^{2} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {a^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}}{f}\)	\(96\)
parts	\(-\frac {a^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3 f}-\frac {a^{2} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5 f}+\frac {2 a^{2} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}\)	\(101\)
norman	\(\frac {\frac {3 a^{2} x}{4}-\frac {12 a^{2}}{5 f}-\frac {3 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 f}-\frac {7 a^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {7 a^{2} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {3 a^{2} \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}+\frac {15 a^{2} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}+\frac {15 a^{2} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {15 a^{2} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {15 a^{2} x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}+\frac {3 a^{2} x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}-\frac {4 a^{2} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {20 a^{2} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {12 a^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{5}}\)	\(248\)

[In]

int(sin(f*x+e)^3*(a+a*sin(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

-1/80*a^2*(-60*f*x+cos(5*f*x+5*e)-5*sin(4*f*x+4*e)-15*cos(3*f*x+3*e)+40*sin(2*f*x+2*e)+110*cos(f*x+e)+96)/f

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=-\frac {4 \, a^{2} \cos \left (f x + e\right )^{5} - 20 \, a^{2} \cos \left (f x + e\right )^{3} - 15 \, a^{2} f x + 40 \, a^{2} \cos \left (f x + e\right ) - 5 \, {\left (2 \, a^{2} \cos \left (f x + e\right )^{3} - 5 \, a^{2} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{20 \, f} \]

[In]

integrate(sin(f*x+e)^3*(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-1/20*(4*a^2*cos(f*x + e)^5 - 20*a^2*cos(f*x + e)^3 - 15*a^2*f*x + 40*a^2*cos(f*x + e) - 5*(2*a^2*cos(f*x + e)
^3 - 5*a^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. 221 vs. \(2 (94) = 188\).

Time = 0.26 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.17 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=\begin {cases} \frac {3 a^{2} x \sin ^{4}{\left (e + f x \right )}}{4} + \frac {3 a^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{2} + \frac {3 a^{2} x \cos ^{4}{\left (e + f x \right )}}{4} - \frac {a^{2} \sin ^{4}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {5 a^{2} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{4 f} - \frac {4 a^{2} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {a^{2} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {3 a^{2} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{4 f} - \frac {8 a^{2} \cos ^{5}{\left (e + f x \right )}}{15 f} - \frac {2 a^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} & \text {for}\: f \neq 0 \\x \left (a \sin {\left (e \right )} + a\right )^{2} \sin ^{3}{\left (e \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(sin(f*x+e)**3*(a+a*sin(f*x+e))**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.93 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=-\frac {16 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} a^{2} - 80 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} - 15 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2}}{240 \, f} \]

[In]

integrate(sin(f*x+e)^3*(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

-1/240*(16*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*a^2 - 80*(cos(f*x + e)^3 - 3*cos(f*x + e))
*a^2 - 15*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a^2)/f

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.87 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=\frac {3}{4} \, a^{2} x - \frac {a^{2} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac {3 \, a^{2} \cos \left (3 \, f x + 3 \, e\right )}{16 \, f} - \frac {11 \, a^{2} \cos \left (f x + e\right )}{8 \, f} + \frac {a^{2} \sin \left (4 \, f x + 4 \, e\right )}{16 \, f} - \frac {a^{2} \sin \left (2 \, f x + 2 \, e\right )}{2 \, f} \]

[In]

integrate(sin(f*x+e)^3*(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

3/4*a^2*x - 1/80*a^2*cos(5*f*x + 5*e)/f + 3/16*a^2*cos(3*f*x + 3*e)/f - 11/8*a^2*cos(f*x + e)/f + 1/16*a^2*sin
(4*f*x + 4*e)/f - 1/2*a^2*sin(2*f*x + 2*e)/f

Mupad [B] (verification not implemented)

Time = 9.63 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.21 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=\frac {3\,a^2\,x}{4}-\frac {\frac {3\,a^2\,\left (e+f\,x\right )}{4}+7\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-7\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7-\frac {3\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{2}-\frac {a^2\,\left (15\,e+15\,f\,x-48\right )}{20}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (\frac {15\,a^2\,\left (e+f\,x\right )}{2}-\frac {a^2\,\left (150\,e+150\,f\,x-80\right )}{20}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {15\,a^2\,\left (e+f\,x\right )}{4}-\frac {a^2\,\left (75\,e+75\,f\,x-240\right )}{20}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {15\,a^2\,\left (e+f\,x\right )}{2}-\frac {a^2\,\left (150\,e+150\,f\,x-400\right )}{20}\right )+\frac {3\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{2}}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^5} \]

[In]

int(sin(e + f*x)^3*(a + a*sin(e + f*x))^2,x)

[Out]

(3*a^2*x)/4 - ((3*a^2*(e + f*x))/4 + 7*a^2*tan(e/2 + (f*x)/2)^3 - 7*a^2*tan(e/2 + (f*x)/2)^7 - (3*a^2*tan(e/2
+ (f*x)/2)^9)/2 - (a^2*(15*e + 15*f*x - 48))/20 + tan(e/2 + (f*x)/2)^6*((15*a^2*(e + f*x))/2 - (a^2*(150*e + 1
50*f*x - 80))/20) + tan(e/2 + (f*x)/2)^2*((15*a^2*(e + f*x))/4 - (a^2*(75*e + 75*f*x - 240))/20) + tan(e/2 + (
f*x)/2)^4*((15*a^2*(e + f*x))/2 - (a^2*(150*e + 150*f*x - 400))/20) + (3*a^2*tan(e/2 + (f*x)/2))/2)/(f*(tan(e/
2 + (f*x)/2)^2 + 1)^5)