\(\int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx\) [261]

Optimal result

Integrand size = 23, antiderivative size = 127 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=\frac {5}{8} a^3 (4 A+3 B) x-\frac {5 a^3 (4 A+3 B) \cos (e+f x)}{6 f}-\frac {5 a^3 (4 A+3 B) \cos (e+f x) \sin (e+f x)}{24 f}-\frac {a (4 A+3 B) \cos (e+f x) (a+a \sin (e+f x))^2}{12 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^3}{4 f} \] Output:

5/8*a^3*(4*A+3*B)*x-5/6*a^3*(4*A+3*B)*cos(f*x+e)/f-5/24*a^3*(4*A+3*B)*cos( 
f*x+e)*sin(f*x+e)/f-1/12*a*(4*A+3*B)*cos(f*x+e)*(a+a*sin(f*x+e))^2/f-1/4*B 
*cos(f*x+e)*(a+a*sin(f*x+e))^3/f
 

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.94 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=-\frac {a^3 \cos (e+f x) \left (30 (4 A+3 B) \arcsin \left (\frac {\sqrt {1-\sin (e+f x)}}{\sqrt {2}}\right )+\sqrt {\cos ^2(e+f x)} \left (88 A+72 B+9 (4 A+5 B) \sin (e+f x)+8 (A+3 B) \sin ^2(e+f x)+6 B \sin ^3(e+f x)\right )\right )}{24 f \sqrt {\cos ^2(e+f x)}} \] Input:

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

Output:

-1/24*(a^3*Cos[e + f*x]*(30*(4*A + 3*B)*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt 
[2]] + Sqrt[Cos[e + f*x]^2]*(88*A + 72*B + 9*(4*A + 5*B)*Sin[e + f*x] + 8* 
(A + 3*B)*Sin[e + f*x]^2 + 6*B*Sin[e + f*x]^3)))/(f*Sqrt[Cos[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3230, 3042, 3124, 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.

\(\displaystyle \int (a \sin (e+f x)+a)^3 (A+B \sin (e+f x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^3 (A+B \sin (e+f x))dx\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {1}{4} (4 A+3 B) \int (\sin (e+f x) a+a)^3dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^3}{4 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} (4 A+3 B) \int (\sin (e+f x) a+a)^3dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^3}{4 f}\)

\(\Big \downarrow \) 3124

\(\displaystyle \frac {1}{4} (4 A+3 B) \int \left (\sin ^3(e+f x) a^3+3 \sin ^2(e+f x) a^3+3 \sin (e+f x) a^3+a^3\right )dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^3}{4 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} (4 A+3 B) \left (\frac {a^3 \cos ^3(e+f x)}{3 f}-\frac {4 a^3 \cos (e+f x)}{f}-\frac {3 a^3 \sin (e+f x) \cos (e+f x)}{2 f}+\frac {5 a^3 x}{2}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^3}{4 f}\)

Input:

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

Output:

-1/4*(B*Cos[e + f*x]*(a + a*Sin[e + f*x])^3)/f + ((4*A + 3*B)*((5*a^3*x)/2 
 - (4*a^3*Cos[e + f*x])/f + (a^3*Cos[e + f*x]^3)/(3*f) - (3*a^3*Cos[e + f* 
x]*Sin[e + f*x])/(2*f)))/4
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3124

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTri 
g[(a + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - 
b^2, 0] && IGtQ[n, 0]
 

rule 3230

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[(a*d*m + b*c*(m + 1))/(b*(m + 1))   Int[(a + b*Sin[e 
 + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] 
&& EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]
 

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.40

\[\frac {-\frac {a^{3} A \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}+a^{3} B \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+3 a^{3} A \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-a^{3} B \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )-3 a^{3} A \cos \left (f x +e \right )+3 a^{3} B \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+a^{3} A \left (f x +e \right )-a^{3} B \cos \left (f x +e \right )}{f}\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.73 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=\frac {8 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} + 15 \, {\left (4 \, A + 3 \, B\right )} a^{3} f x - 96 \, {\left (A + B\right )} a^{3} \cos \left (f x + e\right ) + 3 \, {\left (2 \, B a^{3} \cos \left (f x + e\right )^{3} - {\left (12 \, A + 17 \, B\right )} a^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \] Input:

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

Output:

1/24*(8*(A + 3*B)*a^3*cos(f*x + e)^3 + 15*(4*A + 3*B)*a^3*f*x - 96*(A + B) 
*a^3*cos(f*x + e) + 3*(2*B*a^3*cos(f*x + e)^3 - (12*A + 17*B)*a^3*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. 371 vs. \(2 (119) = 238\).

Time = 0.21 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.92 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=\begin {cases} \frac {3 A a^{3} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {3 A a^{3} x \cos ^{2}{\left (e + f x \right )}}{2} + A a^{3} x - \frac {A a^{3} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {3 A a^{3} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {2 A a^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {3 A a^{3} \cos {\left (e + f x \right )}}{f} + \frac {3 B a^{3} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {3 B a^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {3 B a^{3} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {3 B a^{3} x \cos ^{4}{\left (e + f x \right )}}{8} + \frac {3 B a^{3} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {5 B a^{3} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {3 B a^{3} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {3 B a^{3} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {3 B a^{3} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {2 B a^{3} \cos ^{3}{\left (e + f x \right )}}{f} - \frac {B a^{3} \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (A + B \sin {\left (e \right )}\right ) \left (a \sin {\left (e \right )} + a\right )^{3} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((3*A*a**3*x*sin(e + f*x)**2/2 + 3*A*a**3*x*cos(e + f*x)**2/2 + A 
*a**3*x - A*a**3*sin(e + f*x)**2*cos(e + f*x)/f - 3*A*a**3*sin(e + f*x)*co 
s(e + f*x)/(2*f) - 2*A*a**3*cos(e + f*x)**3/(3*f) - 3*A*a**3*cos(e + f*x)/ 
f + 3*B*a**3*x*sin(e + f*x)**4/8 + 3*B*a**3*x*sin(e + f*x)**2*cos(e + f*x) 
**2/4 + 3*B*a**3*x*sin(e + f*x)**2/2 + 3*B*a**3*x*cos(e + f*x)**4/8 + 3*B* 
a**3*x*cos(e + f*x)**2/2 - 5*B*a**3*sin(e + f*x)**3*cos(e + f*x)/(8*f) - 3 
*B*a**3*sin(e + f*x)**2*cos(e + f*x)/f - 3*B*a**3*sin(e + f*x)*cos(e + f*x 
)**3/(8*f) - 3*B*a**3*sin(e + f*x)*cos(e + f*x)/(2*f) - 2*B*a**3*cos(e + f 
*x)**3/f - B*a**3*cos(e + f*x)/f, Ne(f, 0)), (x*(A + B*sin(e))*(a*sin(e) + 
 a)**3, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.35 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=\frac {32 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a^{3} + 72 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{3} + 96 \, {\left (f x + e\right )} A a^{3} + 96 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{3} + 3 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{3} + 72 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{3} - 288 \, A a^{3} \cos \left (f x + e\right ) - 96 \, B a^{3} \cos \left (f x + e\right )}{96 \, f} \] Input:

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

Output:

1/96*(32*(cos(f*x + e)^3 - 3*cos(f*x + e))*A*a^3 + 72*(2*f*x + 2*e - sin(2 
*f*x + 2*e))*A*a^3 + 96*(f*x + e)*A*a^3 + 96*(cos(f*x + e)^3 - 3*cos(f*x + 
 e))*B*a^3 + 3*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*B*a 
^3 + 72*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a^3 - 288*A*a^3*cos(f*x + e) - 
96*B*a^3*cos(f*x + e))/f
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.88 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=\frac {B a^{3} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {5}{8} \, {\left (4 \, A a^{3} + 3 \, B a^{3}\right )} x + \frac {{\left (A a^{3} + 3 \, B a^{3}\right )} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac {{\left (15 \, A a^{3} + 13 \, B a^{3}\right )} \cos \left (f x + e\right )}{4 \, f} - \frac {{\left (3 \, A a^{3} + 4 \, B a^{3}\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \] Input:

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

Output:

1/32*B*a^3*sin(4*f*x + 4*e)/f + 5/8*(4*A*a^3 + 3*B*a^3)*x + 1/12*(A*a^3 + 
3*B*a^3)*cos(3*f*x + 3*e)/f - 1/4*(15*A*a^3 + 13*B*a^3)*cos(f*x + e)/f - 1 
/4*(3*A*a^3 + 4*B*a^3)*sin(2*f*x + 2*e)/f
 

Mupad [B] (verification not implemented)

Time = 35.80 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.60 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=\frac {5\,a^3\,\mathrm {atan}\left (\frac {5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,A+3\,B\right )}{4\,\left (5\,A\,a^3+\frac {15\,B\,a^3}{4}\right )}\right )\,\left (4\,A+3\,B\right )}{4\,f}-\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,A\,a^3+\frac {15\,B\,a^3}{4}\right )+\frac {22\,A\,a^3}{3}+6\,B\,a^3+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (6\,A\,a^3+2\,B\,a^3\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (3\,A\,a^3+\frac {15\,B\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,A\,a^3+\frac {23\,B\,a^3}{4}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (3\,A\,a^3+\frac {23\,B\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (22\,A\,a^3+18\,B\,a^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {70\,A\,a^3}{3}+22\,B\,a^3\right )}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}-\frac {5\,a^3\,\left (4\,A+3\,B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {f\,x}{2}\right )}{4\,f} \] Input:

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

Output:

(5*a^3*atan((5*a^3*tan(e/2 + (f*x)/2)*(4*A + 3*B))/(4*(5*A*a^3 + (15*B*a^3 
)/4)))*(4*A + 3*B))/(4*f) - (tan(e/2 + (f*x)/2)*(3*A*a^3 + (15*B*a^3)/4) + 
 (22*A*a^3)/3 + 6*B*a^3 + tan(e/2 + (f*x)/2)^6*(6*A*a^3 + 2*B*a^3) - tan(e 
/2 + (f*x)/2)^7*(3*A*a^3 + (15*B*a^3)/4) + tan(e/2 + (f*x)/2)^3*(3*A*a^3 + 
 (23*B*a^3)/4) - tan(e/2 + (f*x)/2)^5*(3*A*a^3 + (23*B*a^3)/4) + tan(e/2 + 
 (f*x)/2)^4*(22*A*a^3 + 18*B*a^3) + tan(e/2 + (f*x)/2)^2*((70*A*a^3)/3 + 2 
2*B*a^3))/(f*(4*tan(e/2 + (f*x)/2)^2 + 6*tan(e/2 + (f*x)/2)^4 + 4*tan(e/2 
+ (f*x)/2)^6 + tan(e/2 + (f*x)/2)^8 + 1)) - (5*a^3*(4*A + 3*B)*(atan(tan(e 
/2 + (f*x)/2)) - (f*x)/2))/(4*f)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.98 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx=\frac {a^{3} \left (-6 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b -8 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a -24 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b -36 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a -45 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b -88 \cos \left (f x +e \right ) a -72 \cos \left (f x +e \right ) b +60 a f x +88 a +45 b f x +72 b \right )}{24 f} \] Input:

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

Output:

(a**3*( - 6*cos(e + f*x)*sin(e + f*x)**3*b - 8*cos(e + f*x)*sin(e + f*x)** 
2*a - 24*cos(e + f*x)*sin(e + f*x)**2*b - 36*cos(e + f*x)*sin(e + f*x)*a - 
 45*cos(e + f*x)*sin(e + f*x)*b - 88*cos(e + f*x)*a - 72*cos(e + f*x)*b + 
60*a*f*x + 88*a + 45*b*f*x + 72*b))/(24*f)