\(\int (a+b \cos (c+d x)) (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)) \, dx\) [976]

Optimal result

Integrand size = 46, antiderivative size = 120 \[ \int (a+b \cos (c+d x)) \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=\frac {1}{2} \left (2 a^2 b B+b^3 B-2 a^3 C+a b^2 C\right ) x+\frac {2 b \left (3 a b B-2 a^2 C+b^2 C\right ) \sin (c+d x)}{3 d}+\frac {b^2 (3 b B-a C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {b C (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d} \] Output:

1/2*(2*B*a^2*b+B*b^3-2*C*a^3+C*a*b^2)*x+2/3*b*(3*B*a*b-2*C*a^2+C*b^2)*sin( 
d*x+c)/d+1/6*b^2*(3*B*b-C*a)*cos(d*x+c)*sin(d*x+c)/d+1/3*b*C*(a+b*cos(d*x+ 
c))^2*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.61 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.85 \[ \int (a+b \cos (c+d x)) \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=\frac {-6 \left (-2 a^2 b B-b^3 B+2 a^3 C-a b^2 C\right ) (c+d x)+3 b \left (8 a b B-4 a^2 C+3 b^2 C\right ) \sin (c+d x)+3 b^2 (b B+a C) \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 d} \] Input:

Integrate[(a + b*Cos[c + d*x])*(a*b*B - a^2*C + b^2*B*Cos[c + d*x] + b^2*C 
*Cos[c + d*x]^2),x]
 

Output:

(-6*(-2*a^2*b*B - b^3*B + 2*a^3*C - a*b^2*C)*(c + d*x) + 3*b*(8*a*b*B - 4* 
a^2*C + 3*b^2*C)*Sin[c + d*x] + 3*b^2*(b*B + a*C)*Sin[2*(c + d*x)] + b^3*C 
*Sin[3*(c + d*x)])/(12*d)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3494, 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.

\(\displaystyle \int (a+b \cos (c+d x)) \left (a^2 (-C)+a b B+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a^2 (-C)+a b B+b^2 B \sin \left (c+d x+\frac {\pi }{2}\right )+b^2 C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\)

\(\Big \downarrow \) 3494

\(\displaystyle \frac {\int (a+b \cos (c+d x))^2 \left (C \cos (c+d x) b^3+(b B-a C) b^2\right )dx}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (C \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+(b B-a C) b^2\right )dx}{b^2}\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {\frac {1}{3} \int (a+b \cos (c+d x)) \left ((3 b B-a C) \cos (c+d x) b^3+\left (2 C b^2+3 a (b B-a C)\right ) b^2\right )dx+\frac {b^3 C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left ((3 b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+\left (2 C b^2+3 a (b B-a C)\right ) b^2\right )dx+\frac {b^3 C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}}{b^2}\)

\(\Big \downarrow \) 3213

\(\displaystyle \frac {\frac {1}{3} \left (\frac {2 b^3 \left (-2 a^2 C+3 a b B+b^2 C\right ) \sin (c+d x)}{d}+\frac {3}{2} b^2 x \left (-2 a^3 C+2 a^2 b B+a b^2 C+b^3 B\right )+\frac {b^4 (3 b B-a C) \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {b^3 C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}}{b^2}\)

Input:

Int[(a + b*Cos[c + d*x])*(a*b*B - a^2*C + b^2*B*Cos[c + d*x] + b^2*C*Cos[c 
 + d*x]^2),x]
 

Output:

((b^3*C*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d) + ((3*b^2*(2*a^2*b*B + 
b^3*B - 2*a^3*C + a*b^2*C)*x)/2 + (2*b^3*(3*a*b*B - 2*a^2*C + b^2*C)*Sin[c 
 + d*x])/d + (b^4*(3*b*B - a*C)*Cos[c + d*x]*Sin[c + d*x])/(2*d))/3)/b^2
 

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]
 

rule 3494

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*B - a*C + b*C*Sin[e + f*x], x], x], 
 x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
 

Maple [A] (verified)

Time = 3.66 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.86

method	result	size
parallelrisch	\(\frac {3 \left (B \,b^{3}+C a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+C \sin \left (3 d x +3 c \right ) b^{3}+3 \left (8 B a \,b^{2}-4 a^{2} b C +3 C \,b^{3}\right ) \sin \left (d x +c \right )+12 d x \left (B \,a^{2} b +\frac {1}{2} B \,b^{3}-a^{3} C +\frac {1}{2} C a \,b^{2}\right )}{12 d}\)	\(103\)
parts	\(a^{2} \left (B b -C a \right ) x +\frac {\left (B \,b^{3}+C a \,b^{2}\right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (2 B a \,b^{2}-a^{2} b C \right ) \sin \left (d x +c \right )}{d}+\frac {C \,b^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3 d}\)	\(103\)
derivativedivides	\(\frac {\frac {C \,b^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 B a \,b^{2} \sin \left (d x +c \right )-C \sin \left (d x +c \right ) a^{2} b +B \,a^{2} b \left (d x +c \right )-a^{3} C \left (d x +c \right )}{d}\)	\(131\)
default	\(\frac {\frac {C \,b^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 B a \,b^{2} \sin \left (d x +c \right )-C \sin \left (d x +c \right ) a^{2} b +B \,a^{2} b \left (d x +c \right )-a^{3} C \left (d x +c \right )}{d}\)	\(131\)
risch	\(x B \,a^{2} b +\frac {x B \,b^{3}}{2}-a^{3} C x +\frac {a \,b^{2} C x}{2}+\frac {2 \sin \left (d x +c \right ) B a \,b^{2}}{d}-\frac {\sin \left (d x +c \right ) a^{2} b C}{d}+\frac {3 \sin \left (d x +c \right ) C \,b^{3}}{4 d}+\frac {\sin \left (3 d x +3 c \right ) C \,b^{3}}{12 d}+\frac {\sin \left (2 d x +2 c \right ) B \,b^{3}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C a \,b^{2}}{4 d}\)	\(133\)
norman	\(\frac {\left (B \,a^{2} b +\frac {1}{2} B \,b^{3}-a^{3} C +\frac {1}{2} C a \,b^{2}\right ) x +\left (B \,a^{2} b +\frac {1}{2} B \,b^{3}-a^{3} C +\frac {1}{2} C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (3 B \,a^{2} b +\frac {3}{2} B \,b^{3}-3 a^{3} C +\frac {3}{2} C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (3 B \,a^{2} b +\frac {3}{2} B \,b^{3}-3 a^{3} C +\frac {3}{2} C a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {b \left (4 B a b -B \,b^{2}-2 a^{2} C -a b C +2 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {b \left (4 B a b +B \,b^{2}-2 a^{2} C +a b C +2 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 b \left (6 B a b -3 a^{2} C +C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\)	\(285\)
orering	\(\text {Expression too large to display}\)	\(1465\)

Input:

int((a+b*cos(d*x+c))*(B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2),x,m 
ethod=_RETURNVERBOSE)
 

Output:

1/12*(3*(B*b^3+C*a*b^2)*sin(2*d*x+2*c)+C*sin(3*d*x+3*c)*b^3+3*(8*B*a*b^2-4 
*C*a^2*b+3*C*b^3)*sin(d*x+c)+12*d*x*(B*a^2*b+1/2*B*b^3-a^3*C+1/2*C*a*b^2)) 
/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.83 \[ \int (a+b \cos (c+d x)) \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=-\frac {3 \, {\left (2 \, C a^{3} - 2 \, B a^{2} b - C a b^{2} - B b^{3}\right )} d x - {\left (2 \, C b^{3} \cos \left (d x + c\right )^{2} - 6 \, C a^{2} b + 12 \, B a b^{2} + 4 \, C b^{3} + 3 \, {\left (C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:

integrate((a+b*cos(d*x+c))*(B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^ 
2),x, algorithm="fricas")
 

Output:

-1/6*(3*(2*C*a^3 - 2*B*a^2*b - C*a*b^2 - B*b^3)*d*x - (2*C*b^3*cos(d*x + c 
)^2 - 6*C*a^2*b + 12*B*a*b^2 + 4*C*b^3 + 3*(C*a*b^2 + B*b^3)*cos(d*x + c)) 
*sin(d*x + c))/d
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (114) = 228\).

Time = 0.14 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.01 \[ \int (a+b \cos (c+d x)) \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=\begin {cases} B a^{2} b x + \frac {2 B a b^{2} \sin {\left (c + d x \right )}}{d} + \frac {B b^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B b^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B b^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - C a^{3} x - \frac {C a^{2} b \sin {\left (c + d x \right )}}{d} + \frac {C a b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {C a b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {C a b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 C b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C b^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\left (c \right )}\right ) \left (B a b + B b^{2} \cos {\left (c \right )} - C a^{2} + C b^{2} \cos ^{2}{\left (c \right )}\right ) & \text {otherwise} \end {cases} \] Input:

integrate((a+b*cos(d*x+c))*(B*a*b-a**2*C+b**2*B*cos(d*x+c)+b**2*C*cos(d*x+ 
c)**2),x)
 

Output:

Piecewise((B*a**2*b*x + 2*B*a*b**2*sin(c + d*x)/d + B*b**3*x*sin(c + d*x)* 
*2/2 + B*b**3*x*cos(c + d*x)**2/2 + B*b**3*sin(c + d*x)*cos(c + d*x)/(2*d) 
 - C*a**3*x - C*a**2*b*sin(c + d*x)/d + C*a*b**2*x*sin(c + d*x)**2/2 + C*a 
*b**2*x*cos(c + d*x)**2/2 + C*a*b**2*sin(c + d*x)*cos(c + d*x)/(2*d) + 2*C 
*b**3*sin(c + d*x)**3/(3*d) + C*b**3*sin(c + d*x)*cos(c + d*x)**2/d, Ne(d, 
 0)), (x*(a + b*cos(c))*(B*a*b + B*b**2*cos(c) - C*a**2 + C*b**2*cos(c)**2 
), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.04 \[ \int (a+b \cos (c+d x)) \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=-\frac {12 \, {\left (d x + c\right )} C a^{3} - 12 \, {\left (d x + c\right )} B a^{2} b - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{2} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{3} + 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b^{3} + 12 \, C a^{2} b \sin \left (d x + c\right ) - 24 \, B a b^{2} \sin \left (d x + c\right )}{12 \, d} \] Input:

integrate((a+b*cos(d*x+c))*(B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^ 
2),x, algorithm="maxima")
 

Output:

-1/12*(12*(d*x + c)*C*a^3 - 12*(d*x + c)*B*a^2*b - 3*(2*d*x + 2*c + sin(2* 
d*x + 2*c))*C*a*b^2 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*b^3 + 4*(sin(d* 
x + c)^3 - 3*sin(d*x + c))*C*b^3 + 12*C*a^2*b*sin(d*x + c) - 24*B*a*b^2*si 
n(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.89 \[ \int (a+b \cos (c+d x)) \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=\frac {C b^{3} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {1}{2} \, {\left (2 \, C a^{3} - 2 \, B a^{2} b - C a b^{2} - B b^{3}\right )} x + \frac {{\left (C a b^{2} + B b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} - \frac {{\left (4 \, C a^{2} b - 8 \, B a b^{2} - 3 \, C b^{3}\right )} \sin \left (d x + c\right )}{4 \, d} \] Input:

integrate((a+b*cos(d*x+c))*(B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^ 
2),x, algorithm="giac")
 

Output:

1/12*C*b^3*sin(3*d*x + 3*c)/d - 1/2*(2*C*a^3 - 2*B*a^2*b - C*a*b^2 - B*b^3 
)*x + 1/4*(C*a*b^2 + B*b^3)*sin(2*d*x + 2*c)/d - 1/4*(4*C*a^2*b - 8*B*a*b^ 
2 - 3*C*b^3)*sin(d*x + c)/d
 

Mupad [B] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.10 \[ \int (a+b \cos (c+d x)) \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=\frac {B\,b^3\,x}{2}-C\,a^3\,x+B\,a^2\,b\,x+\frac {C\,a\,b^2\,x}{2}+\frac {3\,C\,b^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {2\,B\,a\,b^2\,\sin \left (c+d\,x\right )}{d}-\frac {C\,a^2\,b\,\sin \left (c+d\,x\right )}{d} \] Input:

int((a + b*cos(c + d*x))*(C*b^2*cos(c + d*x)^2 - C*a^2 + B*a*b + B*b^2*cos 
(c + d*x)),x)
 

Output:

(B*b^3*x)/2 - C*a^3*x + B*a^2*b*x + (C*a*b^2*x)/2 + (3*C*b^3*sin(c + d*x)) 
/(4*d) + (B*b^3*sin(2*c + 2*d*x))/(4*d) + (C*b^3*sin(3*c + 3*d*x))/(12*d) 
+ (C*a*b^2*sin(2*c + 2*d*x))/(4*d) + (2*B*a*b^2*sin(c + d*x))/d - (C*a^2*b 
*sin(c + d*x))/d
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.06 \[ \int (a+b \cos (c+d x)) \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=\frac {3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{2} c +3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4}-2 \sin \left (d x +c \right )^{3} b^{3} c -6 \sin \left (d x +c \right ) a^{2} b c +12 \sin \left (d x +c \right ) a \,b^{3}+6 \sin \left (d x +c \right ) b^{3} c -6 a^{3} c d x +6 a^{2} b^{2} d x +3 a \,b^{2} c d x +3 b^{4} d x}{6 d} \] Input:

int((a+b*cos(d*x+c))*(B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2),x)
 

Output:

(3*cos(c + d*x)*sin(c + d*x)*a*b**2*c + 3*cos(c + d*x)*sin(c + d*x)*b**4 - 
 2*sin(c + d*x)**3*b**3*c - 6*sin(c + d*x)*a**2*b*c + 12*sin(c + d*x)*a*b* 
*3 + 6*sin(c + d*x)*b**3*c - 6*a**3*c*d*x + 6*a**2*b**2*d*x + 3*a*b**2*c*d 
*x + 3*b**4*d*x)/(6*d)