\(\int (a+b \sin (e+f x))^3 (c+d \sin (e+f x)) \, dx\) [695]

Optimal result

Integrand size = 23, antiderivative size = 171 \[ \int (a+b \sin (e+f x))^3 (c+d \sin (e+f x)) \, dx=\frac {1}{8} \left (8 a^3 c+12 a b^2 c+12 a^2 b d+3 b^3 d\right ) x-\frac {\left (16 a^2 b c+4 b^3 c+3 a^3 d+12 a b^2 d\right ) \cos (e+f x)}{6 f}-\frac {b \left (20 a b c+6 a^2 d+9 b^2 d\right ) \cos (e+f x) \sin (e+f x)}{24 f}-\frac {(4 b c+3 a d) \cos (e+f x) (a+b \sin (e+f x))^2}{12 f}-\frac {d \cos (e+f x) (a+b \sin (e+f x))^3}{4 f} \] Output:

1/8*(8*a^3*c+12*a^2*b*d+12*a*b^2*c+3*b^3*d)*x-1/6*(3*a^3*d+16*a^2*b*c+12*a 
*b^2*d+4*b^3*c)*cos(f*x+e)/f-1/24*b*(6*a^2*d+20*a*b*c+9*b^2*d)*cos(f*x+e)* 
sin(f*x+e)/f-1/12*(3*a*d+4*b*c)*cos(f*x+e)*(a+b*sin(f*x+e))^2/f-1/4*d*cos( 
f*x+e)*(a+b*sin(f*x+e))^3/f
 

Mathematica [A] (verified)

Time = 1.68 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.83 \[ \int (a+b \sin (e+f x))^3 (c+d \sin (e+f x)) \, dx=\frac {-24 \left (12 a^2 b c+3 b^3 c+4 a^3 d+9 a b^2 d\right ) \cos (e+f x)+8 b^2 (b c+3 a d) \cos (3 (e+f x))+3 \left (4 \left (8 a^3 c+12 a b^2 c+12 a^2 b d+3 b^3 d\right ) (e+f x)-8 b \left (3 a b c+3 a^2 d+b^2 d\right ) \sin (2 (e+f x))+b^3 d \sin (4 (e+f x))\right )}{96 f} \] Input:

Integrate[(a + b*Sin[e + f*x])^3*(c + d*Sin[e + f*x]),x]
 

Output:

(-24*(12*a^2*b*c + 3*b^3*c + 4*a^3*d + 9*a*b^2*d)*Cos[e + f*x] + 8*b^2*(b* 
c + 3*a*d)*Cos[3*(e + f*x)] + 3*(4*(8*a^3*c + 12*a*b^2*c + 12*a^2*b*d + 3* 
b^3*d)*(e + f*x) - 8*b*(3*a*b*c + 3*a^2*d + b^2*d)*Sin[2*(e + f*x)] + b^3* 
d*Sin[4*(e + f*x)]))/(96*f)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {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.

\(\displaystyle \int (a+b \sin (e+f x))^3 (c+d \sin (e+f x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \sin (e+f x))^3 (c+d \sin (e+f x))dx\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {1}{4} \int (a+b \sin (e+f x))^2 (4 a c+3 b d+(4 b c+3 a d) \sin (e+f x))dx-\frac {d \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int (a+b \sin (e+f x))^2 (4 a c+3 b d+(4 b c+3 a d) \sin (e+f x))dx-\frac {d \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int (a+b \sin (e+f x)) \left (12 c a^2+15 b d a+8 b^2 c+\left (6 d a^2+20 b c a+9 b^2 d\right ) \sin (e+f x)\right )dx-\frac {(3 a d+4 b c) \cos (e+f x) (a+b \sin (e+f x))^2}{3 f}\right )-\frac {d \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int (a+b \sin (e+f x)) \left (12 c a^2+15 b d a+8 b^2 c+\left (6 d a^2+20 b c a+9 b^2 d\right ) \sin (e+f x)\right )dx-\frac {(3 a d+4 b c) \cos (e+f x) (a+b \sin (e+f x))^2}{3 f}\right )-\frac {d \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}\)

\(\Big \downarrow \) 3213

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (-\frac {b \left (6 a^2 d+20 a b c+9 b^2 d\right ) \sin (e+f x) \cos (e+f x)}{2 f}-\frac {2 \left (3 a^3 d+16 a^2 b c+12 a b^2 d+4 b^3 c\right ) \cos (e+f x)}{f}+\frac {3}{2} x \left (8 a^3 c+12 a^2 b d+12 a b^2 c+3 b^3 d\right )\right )-\frac {(3 a d+4 b c) \cos (e+f x) (a+b \sin (e+f x))^2}{3 f}\right )-\frac {d \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}\)

Input:

Int[(a + b*Sin[e + f*x])^3*(c + d*Sin[e + f*x]),x]
 

Output:

-1/4*(d*Cos[e + f*x]*(a + b*Sin[e + f*x])^3)/f + (-1/3*((4*b*c + 3*a*d)*Co 
s[e + f*x]*(a + b*Sin[e + f*x])^2)/f + ((3*(8*a^3*c + 12*a*b^2*c + 12*a^2* 
b*d + 3*b^3*d)*x)/2 - (2*(16*a^2*b*c + 4*b^3*c + 3*a^3*d + 12*a*b^2*d)*Cos 
[e + f*x])/f - (b*(20*a*b*c + 6*a^2*d + 9*b^2*d)*Cos[e + f*x]*Sin[e + f*x] 
)/(2*f))/3)/4
 

Defintions of rubi rules used

rule 3042

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

rule 3213

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]
 

rule 3232

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]
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.06

\[\frac {a^{3} c \left (f x +e \right )-a^{3} d \cos \left (f x +e \right )-3 a^{2} b c \cos \left (f x +e \right )+3 a^{2} b d \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+3 a \,b^{2} c \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-a \,b^{2} d \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )-\frac {b^{3} c \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}+b^{3} d \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}\]

Input:

int((a+b*sin(f*x+e))^3*(c+d*sin(f*x+e)),x)
                                                                                    
                                                                                    
 

Output:

1/f*(a^3*c*(f*x+e)-a^3*d*cos(f*x+e)-3*a^2*b*c*cos(f*x+e)+3*a^2*b*d*(-1/2*s 
in(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)+3*a*b^2*c*(-1/2*sin(f*x+e)*cos(f*x+e)+ 
1/2*f*x+1/2*e)-a*b^2*d*(2+sin(f*x+e)^2)*cos(f*x+e)-1/3*b^3*c*(2+sin(f*x+e) 
^2)*cos(f*x+e)+b^3*d*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f* 
x+3/8*e))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.87 \[ \int (a+b \sin (e+f x))^3 (c+d \sin (e+f x)) \, dx=\frac {8 \, {\left (b^{3} c + 3 \, a b^{2} d\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (4 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} c + 3 \, {\left (4 \, a^{2} b + b^{3}\right )} d\right )} f x - 24 \, {\left ({\left (3 \, a^{2} b + b^{3}\right )} c + {\left (a^{3} + 3 \, a b^{2}\right )} d\right )} \cos \left (f x + e\right ) + 3 \, {\left (2 \, b^{3} d \cos \left (f x + e\right )^{3} - {\left (12 \, a b^{2} c + {\left (12 \, a^{2} b + 5 \, b^{3}\right )} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \] Input:

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

Output:

1/24*(8*(b^3*c + 3*a*b^2*d)*cos(f*x + e)^3 + 3*(4*(2*a^3 + 3*a*b^2)*c + 3* 
(4*a^2*b + b^3)*d)*f*x - 24*((3*a^2*b + b^3)*c + (a^3 + 3*a*b^2)*d)*cos(f* 
x + e) + 3*(2*b^3*d*cos(f*x + e)^3 - (12*a*b^2*c + (12*a^2*b + 5*b^3)*d)*c 
os(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. 386 vs. \(2 (170) = 340\).

Time = 0.23 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.26 \[ \int (a+b \sin (e+f x))^3 (c+d \sin (e+f x)) \, dx=\begin {cases} a^{3} c x - \frac {a^{3} d \cos {\left (e + f x \right )}}{f} - \frac {3 a^{2} b c \cos {\left (e + f x \right )}}{f} + \frac {3 a^{2} b d x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {3 a^{2} b d x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {3 a^{2} b d \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} + \frac {3 a b^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {3 a b^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {3 a b^{2} c \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {3 a b^{2} d \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 a b^{2} d \cos ^{3}{\left (e + f x \right )}}{f} - \frac {b^{3} c \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 b^{3} c \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac {3 b^{3} d x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {3 b^{3} d x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {3 b^{3} d x \cos ^{4}{\left (e + f x \right )}}{8} - \frac {5 b^{3} d \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {3 b^{3} d \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin {\left (e \right )}\right )^{3} \left (c + d \sin {\left (e \right )}\right ) & \text {otherwise} \end {cases} \] Input:

integrate((a+b*sin(f*x+e))**3*(c+d*sin(f*x+e)),x)
 

Output:

Piecewise((a**3*c*x - a**3*d*cos(e + f*x)/f - 3*a**2*b*c*cos(e + f*x)/f + 
3*a**2*b*d*x*sin(e + f*x)**2/2 + 3*a**2*b*d*x*cos(e + f*x)**2/2 - 3*a**2*b 
*d*sin(e + f*x)*cos(e + f*x)/(2*f) + 3*a*b**2*c*x*sin(e + f*x)**2/2 + 3*a* 
b**2*c*x*cos(e + f*x)**2/2 - 3*a*b**2*c*sin(e + f*x)*cos(e + f*x)/(2*f) - 
3*a*b**2*d*sin(e + f*x)**2*cos(e + f*x)/f - 2*a*b**2*d*cos(e + f*x)**3/f - 
 b**3*c*sin(e + f*x)**2*cos(e + f*x)/f - 2*b**3*c*cos(e + f*x)**3/(3*f) + 
3*b**3*d*x*sin(e + f*x)**4/8 + 3*b**3*d*x*sin(e + f*x)**2*cos(e + f*x)**2/ 
4 + 3*b**3*d*x*cos(e + f*x)**4/8 - 5*b**3*d*sin(e + f*x)**3*cos(e + f*x)/( 
8*f) - 3*b**3*d*sin(e + f*x)*cos(e + f*x)**3/(8*f), Ne(f, 0)), (x*(a + b*s 
in(e))**3*(c + d*sin(e)), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.02 \[ \int (a+b \sin (e+f x))^3 (c+d \sin (e+f x)) \, dx=\frac {96 \, {\left (f x + e\right )} a^{3} c + 72 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a b^{2} c + 32 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} b^{3} c + 72 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} b d + 96 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a b^{2} d + 3 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} b^{3} d - 288 \, a^{2} b c \cos \left (f x + e\right ) - 96 \, a^{3} d \cos \left (f x + e\right )}{96 \, f} \] Input:

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

Output:

1/96*(96*(f*x + e)*a^3*c + 72*(2*f*x + 2*e - sin(2*f*x + 2*e))*a*b^2*c + 3 
2*(cos(f*x + e)^3 - 3*cos(f*x + e))*b^3*c + 72*(2*f*x + 2*e - sin(2*f*x + 
2*e))*a^2*b*d + 96*(cos(f*x + e)^3 - 3*cos(f*x + e))*a*b^2*d + 3*(12*f*x + 
 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*b^3*d - 288*a^2*b*c*cos(f*x 
 + e) - 96*a^3*d*cos(f*x + e))/f
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.87 \[ \int (a+b \sin (e+f x))^3 (c+d \sin (e+f x)) \, dx=\frac {b^{3} d \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {1}{8} \, {\left (8 \, a^{3} c + 12 \, a b^{2} c + 12 \, a^{2} b d + 3 \, b^{3} d\right )} x + \frac {{\left (b^{3} c + 3 \, a b^{2} d\right )} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac {{\left (12 \, a^{2} b c + 3 \, b^{3} c + 4 \, a^{3} d + 9 \, a b^{2} d\right )} \cos \left (f x + e\right )}{4 \, f} - \frac {{\left (3 \, a b^{2} c + 3 \, a^{2} b d + b^{3} d\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \] Input:

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

Output:

1/32*b^3*d*sin(4*f*x + 4*e)/f + 1/8*(8*a^3*c + 12*a*b^2*c + 12*a^2*b*d + 3 
*b^3*d)*x + 1/12*(b^3*c + 3*a*b^2*d)*cos(3*f*x + 3*e)/f - 1/4*(12*a^2*b*c 
+ 3*b^3*c + 4*a^3*d + 9*a*b^2*d)*cos(f*x + e)/f - 1/4*(3*a*b^2*c + 3*a^2*b 
*d + b^3*d)*sin(2*f*x + 2*e)/f
 

Mupad [B] (verification not implemented)

Time = 17.75 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.07 \[ \int (a+b \sin (e+f x))^3 (c+d \sin (e+f x)) \, dx=\frac {2\,b^3\,c\,\cos \left (3\,e+3\,f\,x\right )-6\,b^3\,d\,\sin \left (2\,e+2\,f\,x\right )+\frac {3\,b^3\,d\,\sin \left (4\,e+4\,f\,x\right )}{4}-24\,a^3\,d\,\cos \left (e+f\,x\right )-18\,b^3\,c\,\cos \left (e+f\,x\right )-72\,a^2\,b\,c\,\cos \left (e+f\,x\right )-54\,a\,b^2\,d\,\cos \left (e+f\,x\right )+24\,a^3\,c\,f\,x+9\,b^3\,d\,f\,x+6\,a\,b^2\,d\,\cos \left (3\,e+3\,f\,x\right )-18\,a\,b^2\,c\,\sin \left (2\,e+2\,f\,x\right )-18\,a^2\,b\,d\,\sin \left (2\,e+2\,f\,x\right )+36\,a\,b^2\,c\,f\,x+36\,a^2\,b\,d\,f\,x}{24\,f} \] Input:

int((a + b*sin(e + f*x))^3*(c + d*sin(e + f*x)),x)
 

Output:

(2*b^3*c*cos(3*e + 3*f*x) - 6*b^3*d*sin(2*e + 2*f*x) + (3*b^3*d*sin(4*e + 
4*f*x))/4 - 24*a^3*d*cos(e + f*x) - 18*b^3*c*cos(e + f*x) - 72*a^2*b*c*cos 
(e + f*x) - 54*a*b^2*d*cos(e + f*x) + 24*a^3*c*f*x + 9*b^3*d*f*x + 6*a*b^2 
*d*cos(3*e + 3*f*x) - 18*a*b^2*c*sin(2*e + 2*f*x) - 18*a^2*b*d*sin(2*e + 2 
*f*x) + 36*a*b^2*c*f*x + 36*a^2*b*d*f*x)/(24*f)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.36 \[ \int (a+b \sin (e+f x))^3 (c+d \sin (e+f x)) \, dx=\frac {-6 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b^{3} d -24 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a \,b^{2} d -8 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b^{3} c -36 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{2} b d -36 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2} c -9 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b^{3} d -24 \cos \left (f x +e \right ) a^{3} d -72 \cos \left (f x +e \right ) a^{2} b c -48 \cos \left (f x +e \right ) a \,b^{2} d -16 \cos \left (f x +e \right ) b^{3} c +24 a^{3} c f x +24 a^{3} d +72 a^{2} b c +36 a^{2} b d f x +36 a \,b^{2} c f x +48 a \,b^{2} d +16 b^{3} c +9 b^{3} d f x}{24 f} \] Input:

int((a+b*sin(f*x+e))^3*(c+d*sin(f*x+e)),x)
 

Output:

( - 6*cos(e + f*x)*sin(e + f*x)**3*b**3*d - 24*cos(e + f*x)*sin(e + f*x)** 
2*a*b**2*d - 8*cos(e + f*x)*sin(e + f*x)**2*b**3*c - 36*cos(e + f*x)*sin(e 
 + f*x)*a**2*b*d - 36*cos(e + f*x)*sin(e + f*x)*a*b**2*c - 9*cos(e + f*x)* 
sin(e + f*x)*b**3*d - 24*cos(e + f*x)*a**3*d - 72*cos(e + f*x)*a**2*b*c - 
48*cos(e + f*x)*a*b**2*d - 16*cos(e + f*x)*b**3*c + 24*a**3*c*f*x + 24*a** 
3*d + 72*a**2*b*c + 36*a**2*b*d*f*x + 36*a*b**2*c*f*x + 48*a*b**2*d + 16*b 
**3*c + 9*b**3*d*f*x)/(24*f)