3.341 \(\int \frac {\cos (c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{a+a \cos (c+d x)} \, dx\)

Optimal. Leaf size=110 \[ -\frac {(A-2 B+2 C) \sin (c+d x)}{a d}-\frac {(A-B+C) \sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}+\frac {(2 A-2 B+3 C) \sin (c+d x) \cos (c+d x)}{2 a d}+\frac {x (2 A-2 B+3 C)}{2 a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {3041, 2734} \[ -\frac {(A-2 B+2 C) \sin (c+d x)}{a d}-\frac {(A-B+C) \sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}+\frac {(2 A-2 B+3 C) \sin (c+d x) \cos (c+d x)}{2 a d}+\frac {x (2 A-2 B+3 C)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2734

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)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 3041

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((a*A - b*B + a*C)*Cos[e + f*x]*(
a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m
+ 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c*(2*m + 1) - a*d*(m - n -
1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^
2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx &=-\frac {(A-B+C) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac {\int \cos (c+d x) (-a (A-2 B+2 C)+a (2 A-2 B+3 C) \cos (c+d x)) \, dx}{a^2}\\ &=\frac {(2 A-2 B+3 C) x}{2 a}-\frac {(A-2 B+2 C) \sin (c+d x)}{a d}+\frac {(2 A-2 B+3 C) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(A-B+C) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.48, size = 213, normalized size = 1.94 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (4 d x (2 A-2 B+3 C) \cos \left (c+\frac {d x}{2}\right )+4 d x (2 A-2 B+3 C) \cos \left (\frac {d x}{2}\right )-16 A \sin \left (\frac {d x}{2}\right )+4 B \sin \left (c+\frac {d x}{2}\right )+4 B \sin \left (c+\frac {3 d x}{2}\right )+4 B \sin \left (2 c+\frac {3 d x}{2}\right )+20 B \sin \left (\frac {d x}{2}\right )-4 C \sin \left (c+\frac {d x}{2}\right )-3 C \sin \left (c+\frac {3 d x}{2}\right )-3 C \sin \left (2 c+\frac {3 d x}{2}\right )+C \sin \left (2 c+\frac {5 d x}{2}\right )+C \sin \left (3 c+\frac {5 d x}{2}\right )-20 C \sin \left (\frac {d x}{2}\right )\right )}{8 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(4*(2*A - 2*B + 3*C)*d*x*Cos[(d*x)/2] + 4*(2*A - 2*B + 3*C)*d*x*Cos[c + (d*x)/2] -
16*A*Sin[(d*x)/2] + 20*B*Sin[(d*x)/2] - 20*C*Sin[(d*x)/2] + 4*B*Sin[c + (d*x)/2] - 4*C*Sin[c + (d*x)/2] + 4*B*
Sin[c + (3*d*x)/2] - 3*C*Sin[c + (3*d*x)/2] + 4*B*Sin[2*c + (3*d*x)/2] - 3*C*Sin[2*c + (3*d*x)/2] + C*Sin[2*c
+ (5*d*x)/2] + C*Sin[3*c + (5*d*x)/2]))/(8*a*d*(1 + Cos[c + d*x]))

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 91, normalized size = 0.83 \[ \frac {{\left (2 \, A - 2 \, B + 3 \, C\right )} d x \cos \left (d x + c\right ) + {\left (2 \, A - 2 \, B + 3 \, C\right )} d x + {\left (C \cos \left (d x + c\right )^{2} + {\left (2 \, B - C\right )} \cos \left (d x + c\right ) - 2 \, A + 4 \, B - 4 \, C\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.53, size = 138, normalized size = 1.25 \[ \frac {\frac {{\left (d x + c\right )} {\left (2 \, A - 2 \, B + 3 \, C\right )}}{a} - \frac {2 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} + \frac {2 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((d*x + c)*(2*A - 2*B + 3*C)/a - 2*(A*tan(1/2*d*x + 1/2*c) - B*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*
c))/a + 2*(2*B*tan(1/2*d*x + 1/2*c)^3 - 3*C*tan(1/2*d*x + 1/2*c)^3 + 2*B*tan(1/2*d*x + 1/2*c) - C*tan(1/2*d*x
+ 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a))/d

________________________________________________________________________________________

maple [B]  time = 0.14, size = 248, normalized size = 2.25 \[ -\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{a d}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{a d}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c)),x)

[Out]

-1/a/d*A*tan(1/2*d*x+1/2*c)+1/a/d*B*tan(1/2*d*x+1/2*c)-1/a/d*C*tan(1/2*d*x+1/2*c)-3/a/d/(1+tan(1/2*d*x+1/2*c)^
2)^2*tan(1/2*d*x+1/2*c)^3*C+2/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*B-1/a/d/(1+tan(1/2*d*x+1/2*c
)^2)^2*C*tan(1/2*d*x+1/2*c)+2/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*x+1/2*c)+2/a/d*arctan(tan(1/2*d*x+1/2
*c))*A-2/a/d*arctan(tan(1/2*d*x+1/2*c))*B+3/a/d*arctan(tan(1/2*d*x+1/2*c))*C

________________________________________________________________________________________

maxima [B]  time = 0.44, size = 273, normalized size = 2.48 \[ -\frac {C {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + B {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - A {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(C*((sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a + 2*a*sin(d*x + c)^2/(cos(d*
x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + sin(d*x
+ c)/(a*(cos(d*x + c) + 1))) + B*(2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - 2*sin(d*x + c)/((a + a*sin(d*
x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d*x + c) + 1))) - A*(2*arctan(sin(d*
x + c)/(cos(d*x + c) + 1))/a - sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d

________________________________________________________________________________________

mupad [B]  time = 1.47, size = 112, normalized size = 1.02 \[ \frac {\left (2\,B-3\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B-C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}+\frac {x\,\left (2\,A-2\,B+3\,C\right )}{2\,a}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B+C\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + a*cos(c + d*x)),x)

[Out]

(tan(c/2 + (d*x)/2)^3*(2*B - 3*C) + tan(c/2 + (d*x)/2)*(2*B - C))/(d*(a + 2*a*tan(c/2 + (d*x)/2)^2 + a*tan(c/2
+ (d*x)/2)^4)) + (x*(2*A - 2*B + 3*C))/(2*a) - (tan(c/2 + (d*x)/2)*(A - B + C))/(a*d)

________________________________________________________________________________________

sympy [A]  time = 4.83, size = 993, normalized size = 9.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+a*cos(d*x+c)),x)

[Out]

Piecewise((2*A*d*x*tan(c/2 + d*x/2)**4/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 4*A*d
*x*tan(c/2 + d*x/2)**2/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 2*A*d*x/(2*a*d*tan(c/
2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 2*A*tan(c/2 + d*x/2)**5/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*
d*tan(c/2 + d*x/2)**2 + 2*a*d) - 4*A*tan(c/2 + d*x/2)**3/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**
2 + 2*a*d) - 2*A*tan(c/2 + d*x/2)/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 2*B*d*x*ta
n(c/2 + d*x/2)**4/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 4*B*d*x*tan(c/2 + d*x/2)**
2/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 2*B*d*x/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d
*tan(c/2 + d*x/2)**2 + 2*a*d) + 2*B*tan(c/2 + d*x/2)**5/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2
+ 2*a*d) + 8*B*tan(c/2 + d*x/2)**3/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 6*B*tan(
c/2 + d*x/2)/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 3*C*d*x*tan(c/2 + d*x/2)**4/(2*
a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 6*C*d*x*tan(c/2 + d*x/2)**2/(2*a*d*tan(c/2 + d*
x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 3*C*d*x/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2
+ 2*a*d) - 2*C*tan(c/2 + d*x/2)**5/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 10*C*tan(
c/2 + d*x/2)**3/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 4*C*tan(c/2 + d*x/2)/(2*a*d*
tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d), Ne(d, 0)), (x*(A + B*cos(c) + C*cos(c)**2)*cos(c)/(a
*cos(c) + a), True))

________________________________________________________________________________________