Integrand size = 21, antiderivative size = 102 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=\frac {3 a^2 x}{4}-\frac {2 a^2 \cos (e+f x)}{f}+\frac {a^2 \cos ^3(e+f x)}{f}-\frac {a^2 \cos ^5(e+f x)}{5 f}-\frac {3 a^2 \cos (e+f x) \sin (e+f x)}{4 f}-\frac {a^2 \cos (e+f x) \sin ^3(e+f x)}{2 f} \] Output:
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
Time = 0.33 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.03 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=-\frac {a^2 \cos (e+f x) \left (30 \arcsin \left (\frac {\sqrt {1-\sin (e+f x)}}{\sqrt {2}}\right )+\sqrt {\cos ^2(e+f x)} \left (24+15 \sin (e+f x)+12 \sin ^2(e+f x)+10 \sin ^3(e+f x)+4 \sin ^4(e+f x)\right )\right )}{20 f \sqrt {\cos ^2(e+f x)}} \] Input:
Integrate[Sin[e + f*x]^3*(a + a*Sin[e + f*x])^2,x]
Output:
-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])
Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3236, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sin ^3(e+f x) (a \sin (e+f x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (e+f x)^3 (a \sin (e+f x)+a)^2dx\) |
\(\Big \downarrow \) 3236 |
\(\displaystyle \int \left (a^2 \sin ^5(e+f x)+2 a^2 \sin ^4(e+f x)+a^2 \sin ^3(e+f x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 \cos ^5(e+f x)}{5 f}+\frac {a^2 \cos ^3(e+f x)}{f}-\frac {2 a^2 \cos (e+f x)}{f}-\frac {a^2 \sin ^3(e+f x) \cos (e+f x)}{2 f}-\frac {3 a^2 \sin (e+f x) \cos (e+f x)}{4 f}+\frac {3 a^2 x}{4}\) |
Input:
Int[Sin[e + f*x]^3*(a + a*Sin[e + f*x])^2,x]
Output:
(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)
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(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] && IGt Q[m, 0] && RationalQ[n]
Time = 8.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.64
method | result | size |
parallelrisch | \(-\frac {a^{2} \left (-60 f x +110 \cos \left (f x +e \right )+\cos \left (5 f x +5 e \right )-5 \sin \left (4 f x +4 e \right )-15 \cos \left (3 f x +3 e \right )+40 \sin \left (2 f x +2 e \right )+96\right )}{80 f}\) | \(65\) |
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 | \(\frac {-\frac {a^{2} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5}+2 a^{2} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {a^{2} \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}}{f}\) | \(96\) |
default | \(\frac {-\frac {a^{2} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5}+2 a^{2} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {a^{2} \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}}{f}\) | \(96\) |
parts | \(-\frac {a^{2} \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3 f}-\frac {a^{2} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5 f}+\frac {2 a^{2} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}\) | \(101\) |
norman | \(\frac {\frac {3 a^{2} x}{4}-\frac {12 a^{2}}{5 f}-\frac {3 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 f}-\frac {7 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f}+\frac {7 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f}+\frac {3 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{2 f}+\frac {15 a^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{4}+\frac {15 a^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{2}+\frac {15 a^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{2}+\frac {15 a^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{4}+\frac {3 a^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{4}-\frac {4 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{f}-\frac {20 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{f}-\frac {12 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{5}}\) | \(248\) |
orering | \(\text {Expression too large to display}\) | \(1770\) |
Input:
int(sin(f*x+e)^3*(a+sin(f*x+e)*a)^2,x,method=_RETURNVERBOSE)
Output:
-1/80*a^2*(-60*f*x+110*cos(f*x+e)+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)+96)/f
Time = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=-\frac {4 \, a^{2} \cos \left (f x + e\right )^{5} - 20 \, a^{2} \cos \left (f x + e\right )^{3} - 15 \, a^{2} f x + 40 \, a^{2} \cos \left (f x + e\right ) - 5 \, {\left (2 \, a^{2} \cos \left (f x + e\right )^{3} - 5 \, a^{2} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{20 \, f} \] Input:
integrate(sin(f*x+e)^3*(a+a*sin(f*x+e))^2,x, algorithm="fricas")
Output:
-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
Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (94) = 188\).
Time = 0.25 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.17 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=\begin {cases} \frac {3 a^{2} x \sin ^{4}{\left (e + f x \right )}}{4} + \frac {3 a^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{2} + \frac {3 a^{2} x \cos ^{4}{\left (e + f x \right )}}{4} - \frac {a^{2} \sin ^{4}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {5 a^{2} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{4 f} - \frac {4 a^{2} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {a^{2} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {3 a^{2} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{4 f} - \frac {8 a^{2} \cos ^{5}{\left (e + f x \right )}}{15 f} - \frac {2 a^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} & \text {for}\: f \neq 0 \\x \left (a \sin {\left (e \right )} + a\right )^{2} \sin ^{3}{\left (e \right )} & \text {otherwise} \end {cases} \] Input:
integrate(sin(f*x+e)**3*(a+a*sin(f*x+e))**2,x)
Output:
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*cos (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))
Time = 0.04 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.93 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=-\frac {16 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} a^{2} - 80 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} - 15 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2}}{240 \, f} \] Input:
integrate(sin(f*x+e)^3*(a+a*sin(f*x+e))^2,x, algorithm="maxima")
Output:
-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
Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.87 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=\frac {3}{4} \, a^{2} x - \frac {a^{2} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac {3 \, a^{2} \cos \left (3 \, f x + 3 \, e\right )}{16 \, f} - \frac {11 \, a^{2} \cos \left (f x + e\right )}{8 \, f} + \frac {a^{2} \sin \left (4 \, f x + 4 \, e\right )}{16 \, f} - \frac {a^{2} \sin \left (2 \, f x + 2 \, e\right )}{2 \, f} \] Input:
integrate(sin(f*x+e)^3*(a+a*sin(f*x+e))^2,x, algorithm="giac")
Output:
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
Time = 20.19 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.21 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=\frac {3\,a^2\,x}{4}-\frac {\frac {3\,a^2\,\left (e+f\,x\right )}{4}+7\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-7\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7-\frac {3\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{2}-\frac {a^2\,\left (15\,e+15\,f\,x-48\right )}{20}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (\frac {15\,a^2\,\left (e+f\,x\right )}{2}-\frac {a^2\,\left (150\,e+150\,f\,x-80\right )}{20}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {15\,a^2\,\left (e+f\,x\right )}{4}-\frac {a^2\,\left (75\,e+75\,f\,x-240\right )}{20}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {15\,a^2\,\left (e+f\,x\right )}{2}-\frac {a^2\,\left (150\,e+150\,f\,x-400\right )}{20}\right )+\frac {3\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{2}}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^5} \] Input:
int(sin(e + f*x)^3*(a + a*sin(e + f*x))^2,x)
Output:
(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*ta n(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 + 150 *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)
Time = 0.18 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82 \[ \int \sin ^3(e+f x) (a+a \sin (e+f x))^2 \, dx=\frac {a^{2} \left (-4 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4}-10 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3}-12 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}-15 \cos \left (f x +e \right ) \sin \left (f x +e \right )-24 \cos \left (f x +e \right )+15 f x +24\right )}{20 f} \] Input:
int(sin(f*x+e)^3*(a+a*sin(f*x+e))^2,x)
Output:
(a**2*( - 4*cos(e + f*x)*sin(e + f*x)**4 - 10*cos(e + f*x)*sin(e + f*x)**3 - 12*cos(e + f*x)*sin(e + f*x)**2 - 15*cos(e + f*x)*sin(e + f*x) - 24*cos (e + f*x) + 15*f*x + 24))/(20*f)