Integrand size = 12, antiderivative size = 90 \[ \int (a+b \sin (e+f x))^3 \, dx=\frac {1}{2} a \left (2 a^2+3 b^2\right ) x-\frac {2 b \left (4 a^2+b^2\right ) \cos (e+f x)}{3 f}-\frac {5 a b^2 \cos (e+f x) \sin (e+f x)}{6 f}-\frac {b \cos (e+f x) (a+b \sin (e+f x))^2}{3 f} \] Output:
1/2*a*(2*a^2+3*b^2)*x-2/3*b*(4*a^2+b^2)*cos(f*x+e)/f-5/6*a*b^2*cos(f*x+e)* sin(f*x+e)/f-1/3*b*cos(f*x+e)*(a+b*sin(f*x+e))^2/f
Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.79 \[ \int (a+b \sin (e+f x))^3 \, dx=\frac {6 a \left (2 a^2+3 b^2\right ) (e+f x)-9 b \left (4 a^2+b^2\right ) \cos (e+f x)+b^3 \cos (3 (e+f x))-9 a b^2 \sin (2 (e+f x))}{12 f} \] Input:
Integrate[(a + b*Sin[e + f*x])^3,x]
Output:
(6*a*(2*a^2 + 3*b^2)*(e + f*x) - 9*b*(4*a^2 + b^2)*Cos[e + f*x] + b^3*Cos[ 3*(e + f*x)] - 9*a*b^2*Sin[2*(e + f*x)])/(12*f)
Time = 0.33 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3135, 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.
\(\displaystyle \int (a+b \sin (e+f x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \sin (e+f x))^3dx\) |
\(\Big \downarrow \) 3135 |
\(\displaystyle \frac {1}{3} \int (a+b \sin (e+f x)) \left (3 a^2+5 b \sin (e+f x) a+2 b^2\right )dx-\frac {b \cos (e+f x) (a+b \sin (e+f x))^2}{3 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \int (a+b \sin (e+f x)) \left (3 a^2+5 b \sin (e+f x) a+2 b^2\right )dx-\frac {b \cos (e+f x) (a+b \sin (e+f x))^2}{3 f}\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {1}{3} \left (-\frac {2 b \left (4 a^2+b^2\right ) \cos (e+f x)}{f}+\frac {3}{2} a x \left (2 a^2+3 b^2\right )-\frac {5 a b^2 \sin (e+f x) \cos (e+f x)}{2 f}\right )-\frac {b \cos (e+f x) (a+b \sin (e+f x))^2}{3 f}\) |
Input:
Int[(a + b*Sin[e + f*x])^3,x]
Output:
-1/3*(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^2)/f + ((3*a*(2*a^2 + 3*b^2)*x)/ 2 - (2*b*(4*a^2 + b^2)*Cos[e + f*x])/f - (5*a*b^2*Cos[e + f*x]*Sin[e + f*x ])/(2*f))/3
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[1/n Int[(a + b* Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c + d*x] , x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
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]
Time = 274.34 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84
method | result | size |
derivativedivides | \(\frac {-\frac {b^{3} \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}+3 a \,b^{2} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-3 a^{2} b \cos \left (f x +e \right )+a^{3} \left (f x +e \right )}{f}\) | \(76\) |
default | \(\frac {-\frac {b^{3} \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}+3 a \,b^{2} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-3 a^{2} b \cos \left (f x +e \right )+a^{3} \left (f x +e \right )}{f}\) | \(76\) |
parts | \(a^{3} x -\frac {b^{3} \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3 f}-\frac {3 b \cos \left (f x +e \right ) a^{2}}{f}+\frac {3 a \,b^{2} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\) | \(77\) |
risch | \(a^{3} x +\frac {3 a \,b^{2} x}{2}-\frac {3 b \cos \left (f x +e \right ) a^{2}}{f}-\frac {3 b^{3} \cos \left (f x +e \right )}{4 f}+\frac {b^{3} \cos \left (3 f x +3 e \right )}{12 f}-\frac {3 a \,b^{2} \sin \left (2 f x +2 e \right )}{4 f}\) | \(78\) |
parallelrisch | \(\frac {b^{3} \cos \left (3 f x +3 e \right )-9 a \,b^{2} \sin \left (2 f x +2 e \right )+\left (-36 a^{2} b -9 b^{3}\right ) \cos \left (f x +e \right )+12 a^{3} f x +18 a \,b^{2} f x -36 a^{2} b -8 b^{3}}{12 f}\) | \(80\) |
norman | \(\frac {\left (a^{3}+\frac {3}{2} a \,b^{2}\right ) x +\left (a^{3}+\frac {3}{2} a \,b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+\left (3 a^{3}+\frac {9}{2} a \,b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (3 a^{3}+\frac {9}{2} a \,b^{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-\frac {18 a^{2} b +4 b^{3}}{3 f}-\frac {6 a^{2} b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{f}-\frac {\left (12 a^{2} b +4 b^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{f}-\frac {3 a \,b^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {3 a \,b^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3}}\) | \(206\) |
Input:
int((a+b*sin(f*x+e))^3,x,method=_RETURNVERBOSE)
Output:
1/f*(-1/3*b^3*(2+sin(f*x+e)^2)*cos(f*x+e)+3*a*b^2*(-1/2*sin(f*x+e)*cos(f*x +e)+1/2*f*x+1/2*e)-3*a^2*b*cos(f*x+e)+a^3*(f*x+e))
Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.79 \[ \int (a+b \sin (e+f x))^3 \, dx=\frac {2 \, b^{3} \cos \left (f x + e\right )^{3} - 9 \, a b^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} f x - 6 \, {\left (3 \, a^{2} b + b^{3}\right )} \cos \left (f x + e\right )}{6 \, f} \] Input:
integrate((a+b*sin(f*x+e))^3,x, algorithm="fricas")
Output:
1/6*(2*b^3*cos(f*x + e)^3 - 9*a*b^2*cos(f*x + e)*sin(f*x + e) + 3*(2*a^3 + 3*a*b^2)*f*x - 6*(3*a^2*b + b^3)*cos(f*x + e))/f
Time = 0.14 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.42 \[ \int (a+b \sin (e+f x))^3 \, dx=\begin {cases} a^{3} x - \frac {3 a^{2} b \cos {\left (e + f x \right )}}{f} + \frac {3 a b^{2} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {3 a b^{2} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {3 a b^{2} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {b^{3} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 b^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin {\left (e \right )}\right )^{3} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*sin(f*x+e))**3,x)
Output:
Piecewise((a**3*x - 3*a**2*b*cos(e + f*x)/f + 3*a*b**2*x*sin(e + f*x)**2/2 + 3*a*b**2*x*cos(e + f*x)**2/2 - 3*a*b**2*sin(e + f*x)*cos(e + f*x)/(2*f) - b**3*sin(e + f*x)**2*cos(e + f*x)/f - 2*b**3*cos(e + f*x)**3/(3*f), Ne( f, 0)), (x*(a + b*sin(e))**3, True))
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.82 \[ \int (a+b \sin (e+f x))^3 \, dx=a^{3} x + \frac {3 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a b^{2}}{4 \, f} + \frac {{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} b^{3}}{3 \, f} - \frac {3 \, a^{2} b \cos \left (f x + e\right )}{f} \] Input:
integrate((a+b*sin(f*x+e))^3,x, algorithm="maxima")
Output:
a^3*x + 3/4*(2*f*x + 2*e - sin(2*f*x + 2*e))*a*b^2/f + 1/3*(cos(f*x + e)^3 - 3*cos(f*x + e))*b^3/f - 3*a^2*b*cos(f*x + e)/f
Time = 0.13 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.80 \[ \int (a+b \sin (e+f x))^3 \, dx=\frac {b^{3} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac {3 \, a b^{2} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} + \frac {1}{2} \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} x - \frac {3 \, {\left (4 \, a^{2} b + b^{3}\right )} \cos \left (f x + e\right )}{4 \, f} \] Input:
integrate((a+b*sin(f*x+e))^3,x, algorithm="giac")
Output:
1/12*b^3*cos(3*f*x + 3*e)/f - 3/4*a*b^2*sin(2*f*x + 2*e)/f + 1/2*(2*a^3 + 3*a*b^2)*x - 3/4*(4*a^2*b + b^3)*cos(f*x + e)/f
Time = 17.57 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.41 \[ \int (a+b \sin (e+f x))^3 \, dx=a^3\,x-\frac {4\,b^3\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{f}+\frac {8\,b^3\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6}{3\,f}+\frac {3\,a\,b^2\,x}{2}-\frac {6\,a^2\,b\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{f}-\frac {6\,a\,b^2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f}+\frac {3\,a\,b^2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f} \] Input:
int((a + b*sin(e + f*x))^3,x)
Output:
a^3*x - (4*b^3*cos(e/2 + (f*x)/2)^4)/f + (8*b^3*cos(e/2 + (f*x)/2)^6)/(3*f ) + (3*a*b^2*x)/2 - (6*a^2*b*cos(e/2 + (f*x)/2)^2)/f - (6*a*b^2*cos(e/2 + (f*x)/2)^3*sin(e/2 + (f*x)/2))/f + (3*a*b^2*cos(e/2 + (f*x)/2)*sin(e/2 + ( f*x)/2))/f
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int (a+b \sin (e+f x))^3 \, dx=\frac {-2 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b^{3}-9 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2}-18 \cos \left (f x +e \right ) a^{2} b -4 \cos \left (f x +e \right ) b^{3}+6 a^{3} f x +18 a^{2} b +9 a \,b^{2} f x +4 b^{3}}{6 f} \] Input:
int((a+b*sin(f*x+e))^3,x)
Output:
( - 2*cos(e + f*x)*sin(e + f*x)**2*b**3 - 9*cos(e + f*x)*sin(e + f*x)*a*b* *2 - 18*cos(e + f*x)*a**2*b - 4*cos(e + f*x)*b**3 + 6*a**3*f*x + 18*a**2*b + 9*a*b**2*f*x + 4*b**3)/(6*f)