Integrand size = 21, antiderivative size = 160 \[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=\frac {1}{8} a \left (4 a^2+9 b^2\right ) x-\frac {b \left (15 a^2+4 b^2\right ) \cos (e+f x)}{5 f}+\frac {b \left (15 a^2+4 b^2\right ) \cos ^3(e+f x)}{15 f}-\frac {a \left (4 a^2+9 b^2\right ) \cos (e+f x) \sin (e+f x)}{8 f}-\frac {11 a b^2 \cos (e+f x) \sin ^3(e+f x)}{20 f}-\frac {b^2 \cos (e+f x) \sin ^3(e+f x) (a+b \sin (e+f x))}{5 f} \] Output:
1/8*a*(4*a^2+9*b^2)*x-1/5*b*(15*a^2+4*b^2)*cos(f*x+e)/f+1/15*b*(15*a^2+4*b ^2)*cos(f*x+e)^3/f-1/8*a*(4*a^2+9*b^2)*cos(f*x+e)*sin(f*x+e)/f-11/20*a*b^2 *cos(f*x+e)*sin(f*x+e)^3/f-1/5*b^2*cos(f*x+e)*sin(f*x+e)^3*(a+b*sin(f*x+e) )/f
Time = 1.99 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.73 \[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=\frac {-60 b \left (18 a^2+5 b^2\right ) \cos (e+f x)+10 \left (12 a^2 b+5 b^3\right ) \cos (3 (e+f x))-6 b^3 \cos (5 (e+f x))+15 a \left (4 \left (4 a^2+9 b^2\right ) (e+f x)-8 \left (a^2+3 b^2\right ) \sin (2 (e+f x))+3 b^2 \sin (4 (e+f x))\right )}{480 f} \] Input:
Integrate[Sin[e + f*x]^2*(a + b*Sin[e + f*x])^3,x]
Output:
(-60*b*(18*a^2 + 5*b^2)*Cos[e + f*x] + 10*(12*a^2*b + 5*b^3)*Cos[3*(e + f* x)] - 6*b^3*Cos[5*(e + f*x)] + 15*a*(4*(4*a^2 + 9*b^2)*(e + f*x) - 8*(a^2 + 3*b^2)*Sin[2*(e + f*x)] + 3*b^2*Sin[4*(e + f*x)]))/(480*f)
Time = 0.69 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3270, 3042, 3232, 3042, 3232, 3042, 3213}
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.
| \(\displaystyle \int \sin ^2(e+f x) (a+b \sin (e+f x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (e+f x)^2 (a+b \sin (e+f x))^3dx\) |
| \(\Big \downarrow \) 3270 |
| \(\displaystyle \frac {\int (4 b-a \sin (e+f x)) (a+b \sin (e+f x))^3dx}{5 b}-\frac {\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (4 b-a \sin (e+f x)) (a+b \sin (e+f x))^3dx}{5 b}-\frac {\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}\) |
| \(\Big \downarrow \) 3232 |
| \(\displaystyle \frac {\frac {1}{4} \int (a+b \sin (e+f x))^2 \left (13 a b-\left (3 a^2-16 b^2\right ) \sin (e+f x)\right )dx+\frac {a \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}}{5 b}-\frac {\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \int (a+b \sin (e+f x))^2 \left (13 a b-\left (3 a^2-16 b^2\right ) \sin (e+f x)\right )dx+\frac {a \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}}{5 b}-\frac {\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}\) |
| \(\Big \downarrow \) 3232 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int (a+b \sin (e+f x)) \left (b \left (33 a^2+32 b^2\right )-a \left (6 a^2-71 b^2\right ) \sin (e+f x)\right )dx+\frac {\left (3 a^2-16 b^2\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{3 f}\right )+\frac {a \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}}{5 b}-\frac {\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int (a+b \sin (e+f x)) \left (b \left (33 a^2+32 b^2\right )-a \left (6 a^2-71 b^2\right ) \sin (e+f x)\right )dx+\frac {\left (3 a^2-16 b^2\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{3 f}\right )+\frac {a \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}}{5 b}-\frac {\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}\) |
| \(\Big \downarrow \) 3213 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {\left (3 a^2-16 b^2\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{3 f}+\frac {1}{3} \left (\frac {a b \left (6 a^2-71 b^2\right ) \sin (e+f x) \cos (e+f x)}{2 f}+\frac {15}{2} a b x \left (4 a^2+9 b^2\right )+\frac {2 \left (3 a^4-52 a^2 b^2-16 b^4\right ) \cos (e+f x)}{f}\right )\right )+\frac {a \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}}{5 b}-\frac {\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}\) |
Input:
Int[Sin[e + f*x]^2*(a + b*Sin[e + f*x])^3,x]
Output:
-1/5*(Cos[e + f*x]*(a + b*Sin[e + f*x])^4)/(b*f) + ((a*Cos[e + f*x]*(a + b *Sin[e + f*x])^3)/(4*f) + (((3*a^2 - 16*b^2)*Cos[e + f*x]*(a + b*Sin[e + f *x])^2)/(3*f) + ((15*a*b*(4*a^2 + 9*b^2)*x)/2 + (2*(3*a^4 - 52*a^2*b^2 - 1 6*b^4)*Cos[e + f*x])/f + (a*b*(6*a^2 - 71*b^2)*Cos[e + f*x]*Sin[e + f*x])/ (2*f))/3)/4)/(5*b)
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[1/(m + 1) Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[b* d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ [{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(-d^2)*Cos[e + f*x]*((a + b*Sin[e + f*x]) ^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x]) ^m*Simp[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x ], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && Ne Q[a^2 - b^2, 0] && !LtQ[m, -1]
Time = 15.90 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.78
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-a^{2} b \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )+3 a \,b^{2} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {b^{3} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5}}{f}\) | \(124\) |
| default | \(\frac {a^{3} \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-a^{2} b \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )+3 a \,b^{2} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {b^{3} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5}}{f}\) | \(124\) |
| parallelrisch | \(\frac {\left (120 a^{2} b +50 b^{3}\right ) \cos \left (3 f x +3 e \right )+\left (-120 a^{3}-360 a \,b^{2}\right ) \sin \left (2 f x +2 e \right )-6 b^{3} \cos \left (5 f x +5 e \right )+45 a \,b^{2} \sin \left (4 f x +4 e \right )+\left (-1080 a^{2} b -300 b^{3}\right ) \cos \left (f x +e \right )+240 a^{3} f x +540 a \,b^{2} f x -960 a^{2} b -256 b^{3}}{480 f}\) | \(125\) |
| parts | \(\frac {a^{3} \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {b^{3} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5 f}+\frac {3 a \,b^{2} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}-\frac {a^{2} b \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{f}\) | \(132\) |
| risch | \(\frac {a^{3} x}{2}+\frac {9 a \,b^{2} x}{8}-\frac {9 b \cos \left (f x +e \right ) a^{2}}{4 f}-\frac {5 b^{3} \cos \left (f x +e \right )}{8 f}-\frac {b^{3} \cos \left (5 f x +5 e \right )}{80 f}+\frac {3 a \,b^{2} \sin \left (4 f x +4 e \right )}{32 f}+\frac {b \cos \left (3 f x +3 e \right ) a^{2}}{4 f}+\frac {5 b^{3} \cos \left (3 f x +3 e \right )}{48 f}-\frac {\sin \left (2 f x +2 e \right ) a^{3}}{4 f}-\frac {3 \sin \left (2 f x +2 e \right ) a \,b^{2}}{4 f}\) | \(149\) |
| norman | \(\frac {-\frac {60 a^{2} b +16 b^{3}}{15 f}+\frac {a \left (4 a^{2}+9 b^{2}\right ) x}{8}-\frac {12 a^{2} b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{f}-\frac {2 \left (42 a^{2} b +16 b^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{3 f}-\frac {\left (60 a^{2} b +16 b^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3 f}-\frac {a \left (4 a^{2}+9 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}+\frac {a \left (4 a^{2}+9 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{4 f}+\frac {5 a \left (4 a^{2}+9 b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{8}+\frac {5 a \left (4 a^{2}+9 b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{4}+\frac {5 a \left (4 a^{2}+9 b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{4}+\frac {5 a \left (4 a^{2}+9 b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{8}+\frac {a \left (4 a^{2}+9 b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{8}-\frac {a \left (4 a^{2}+21 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 f}+\frac {a \left (4 a^{2}+21 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{2 f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{5}}\) | \(366\) |
| orering | \(\text {Expression too large to display}\) | \(2110\) |
Input:
int(sin(f*x+e)^2*(a+b*sin(f*x+e))^3,x,method=_RETURNVERBOSE)
Output:
1/f*(a^3*(-1/2*cos(f*x+e)*sin(f*x+e)+1/2*f*x+1/2*e)-a^2*b*(2+sin(f*x+e)^2) *cos(f*x+e)+3*a*b^2*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x +3/8*e)-1/5*b^3*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e))
Time = 0.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.74 \[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=-\frac {24 \, b^{3} \cos \left (f x + e\right )^{5} - 40 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (4 \, a^{3} + 9 \, a b^{2}\right )} f x + 120 \, {\left (3 \, a^{2} b + b^{3}\right )} \cos \left (f x + e\right ) - 15 \, {\left (6 \, a b^{2} \cos \left (f x + e\right )^{3} - {\left (4 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{120 \, f} \] Input:
integrate(sin(f*x+e)^2*(a+b*sin(f*x+e))^3,x, algorithm="fricas")
Output:
-1/120*(24*b^3*cos(f*x + e)^5 - 40*(3*a^2*b + 2*b^3)*cos(f*x + e)^3 - 15*( 4*a^3 + 9*a*b^2)*f*x + 120*(3*a^2*b + b^3)*cos(f*x + e) - 15*(6*a*b^2*cos( f*x + e)^3 - (4*a^3 + 15*a*b^2)*cos(f*x + e))*sin(f*x + e))/f
Time = 0.27 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.78 \[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=\begin {cases} \frac {a^{3} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {a^{3} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {a^{3} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {3 a^{2} b \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 a^{2} b \cos ^{3}{\left (e + f x \right )}}{f} + \frac {9 a b^{2} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {9 a b^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {9 a b^{2} x \cos ^{4}{\left (e + f x \right )}}{8} - \frac {15 a b^{2} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {9 a b^{2} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {b^{3} \sin ^{4}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {4 b^{3} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {8 b^{3} \cos ^{5}{\left (e + f x \right )}}{15 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin {\left (e \right )}\right )^{3} \sin ^{2}{\left (e \right )} & \text {otherwise} \end {cases} \] Input:
integrate(sin(f*x+e)**2*(a+b*sin(f*x+e))**3,x)
Output:
Piecewise((a**3*x*sin(e + f*x)**2/2 + a**3*x*cos(e + f*x)**2/2 - a**3*sin( e + f*x)*cos(e + f*x)/(2*f) - 3*a**2*b*sin(e + f*x)**2*cos(e + f*x)/f - 2* a**2*b*cos(e + f*x)**3/f + 9*a*b**2*x*sin(e + f*x)**4/8 + 9*a*b**2*x*sin(e + f*x)**2*cos(e + f*x)**2/4 + 9*a*b**2*x*cos(e + f*x)**4/8 - 15*a*b**2*si n(e + f*x)**3*cos(e + f*x)/(8*f) - 9*a*b**2*sin(e + f*x)*cos(e + f*x)**3/( 8*f) - b**3*sin(e + f*x)**4*cos(e + f*x)/f - 4*b**3*sin(e + f*x)**2*cos(e + f*x)**3/(3*f) - 8*b**3*cos(e + f*x)**5/(15*f), Ne(f, 0)), (x*(a + b*sin( e))**3*sin(e)**2, True))
Time = 0.04 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.76 \[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=\frac {120 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} + 480 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} b + 45 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a b^{2} - 32 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} b^{3}}{480 \, f} \] Input:
integrate(sin(f*x+e)^2*(a+b*sin(f*x+e))^3,x, algorithm="maxima")
Output:
1/480*(120*(2*f*x + 2*e - sin(2*f*x + 2*e))*a^3 + 480*(cos(f*x + e)^3 - 3* cos(f*x + e))*a^2*b + 45*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a*b^2 - 32*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e)) *b^3)/f
Time = 0.13 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.78 \[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=-\frac {b^{3} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac {3 \, a b^{2} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {1}{8} \, {\left (4 \, a^{3} + 9 \, a b^{2}\right )} x + \frac {{\left (12 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac {{\left (18 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (f x + e\right )}{8 \, f} - \frac {{\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \] Input:
integrate(sin(f*x+e)^2*(a+b*sin(f*x+e))^3,x, algorithm="giac")
Output:
-1/80*b^3*cos(5*f*x + 5*e)/f + 3/32*a*b^2*sin(4*f*x + 4*e)/f + 1/8*(4*a^3 + 9*a*b^2)*x + 1/48*(12*a^2*b + 5*b^3)*cos(3*f*x + 3*e)/f - 1/8*(18*a^2*b + 5*b^3)*cos(f*x + e)/f - 1/4*(a^3 + 3*a*b^2)*sin(2*f*x + 2*e)/f
Time = 18.12 (sec) , antiderivative size = 328, normalized size of antiderivative = 2.05 \[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,a^2+9\,b^2\right )}{4\,\left (a^3+\frac {9\,a\,b^2}{4}\right )}\right )\,\left (4\,a^2+9\,b^2\right )}{4\,f}-\frac {4\,a^2\,b-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9\,\left (a^3+\frac {9\,a\,b^2}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,a^3+\frac {21\,a\,b^2}{2}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (2\,a^3+\frac {21\,a\,b^2}{2}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (20\,a^2\,b+\frac {16\,b^3}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (28\,a^2\,b+\frac {32\,b^3}{3}\right )+\frac {16\,b^3}{15}+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a^3+\frac {9\,a\,b^2}{4}\right )+12\,a^2\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}-\frac {a\,\left (4\,a^2+9\,b^2\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {f\,x}{2}\right )}{4\,f} \] Input:
int(sin(e + f*x)^2*(a + b*sin(e + f*x))^3,x)
Output:
(a*atan((a*tan(e/2 + (f*x)/2)*(4*a^2 + 9*b^2))/(4*((9*a*b^2)/4 + a^3)))*(4 *a^2 + 9*b^2))/(4*f) - (4*a^2*b - tan(e/2 + (f*x)/2)^9*((9*a*b^2)/4 + a^3) + tan(e/2 + (f*x)/2)^3*((21*a*b^2)/2 + 2*a^3) - tan(e/2 + (f*x)/2)^7*((21 *a*b^2)/2 + 2*a^3) + tan(e/2 + (f*x)/2)^2*(20*a^2*b + (16*b^3)/3) + tan(e/ 2 + (f*x)/2)^4*(28*a^2*b + (32*b^3)/3) + (16*b^3)/15 + tan(e/2 + (f*x)/2)* ((9*a*b^2)/4 + a^3) + 12*a^2*b*tan(e/2 + (f*x)/2)^6)/(f*(5*tan(e/2 + (f*x) /2)^2 + 10*tan(e/2 + (f*x)/2)^4 + 10*tan(e/2 + (f*x)/2)^6 + 5*tan(e/2 + (f *x)/2)^8 + tan(e/2 + (f*x)/2)^10 + 1)) - (a*(4*a^2 + 9*b^2)*(atan(tan(e/2 + (f*x)/2)) - (f*x)/2))/(4*f)
Time = 0.18 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.05 \[ \int \sin ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=\frac {-24 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} b^{3}-90 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a \,b^{2}-120 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a^{2} b -32 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b^{3}-60 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{3}-135 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2}-240 \cos \left (f x +e \right ) a^{2} b -64 \cos \left (f x +e \right ) b^{3}+60 a^{3} f x +240 a^{2} b +135 a \,b^{2} f x +64 b^{3}}{120 f} \] Input:
int(sin(f*x+e)^2*(a+b*sin(f*x+e))^3,x)
Output:
( - 24*cos(e + f*x)*sin(e + f*x)**4*b**3 - 90*cos(e + f*x)*sin(e + f*x)**3 *a*b**2 - 120*cos(e + f*x)*sin(e + f*x)**2*a**2*b - 32*cos(e + f*x)*sin(e + f*x)**2*b**3 - 60*cos(e + f*x)*sin(e + f*x)*a**3 - 135*cos(e + f*x)*sin( e + f*x)*a*b**2 - 240*cos(e + f*x)*a**2*b - 64*cos(e + f*x)*b**3 + 60*a**3 *f*x + 240*a**2*b + 135*a*b**2*f*x + 64*b**3)/(120*f)