\(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{a+a \cos (c+d x)} \, dx\) [340]

Optimal result

Integrand size = 41, antiderivative size = 139 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=-\frac {(2 A-3 B+3 C) x}{2 a}+\frac {(3 A-3 B+4 C) \sin (c+d x)}{a d}-\frac {(2 A-3 B+3 C) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {(3 A-3 B+4 C) \sin ^3(c+d x)}{3 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3120, 2827, 2715, 8, 2713} \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=-\frac {(3 A-3 B+4 C) \sin ^3(c+d x)}{3 a d}+\frac {(3 A-3 B+4 C) \sin (c+d x)}{a d}-\frac {(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{d (a \cos (c+d x)+a)}-\frac {(2 A-3 B+3 C) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {x (2 A-3 B+3 C)}{2 a} \]

[In]

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

[Out]

-1/2*((2*A - 3*B + 3*C)*x)/a + ((3*A - 3*B + 4*C)*Sin[c + d*x])/(a*d) - ((2*A - 3*B + 3*C)*Cos[c + d*x]*Sin[c
+ d*x])/(2*a*d) - ((A - B + C)*Cos[c + d*x]^3*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])) - ((3*A - 3*B + 4*C)*Sin[
c + d*x]^3)/(3*a*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3120

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*} \text {integral}& = -\frac {(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac {\int \cos ^2(c+d x) (-a (2 A-3 B+3 C)+a (3 A-3 B+4 C) \cos (c+d x)) \, dx}{a^2} \\ & = -\frac {(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {(2 A-3 B+3 C) \int \cos ^2(c+d x) \, dx}{a}+\frac {(3 A-3 B+4 C) \int \cos ^3(c+d x) \, dx}{a} \\ & = -\frac {(2 A-3 B+3 C) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {(2 A-3 B+3 C) \int 1 \, dx}{2 a}-\frac {(3 A-3 B+4 C) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a d} \\ & = -\frac {(2 A-3 B+3 C) x}{2 a}+\frac {(3 A-3 B+4 C) \sin (c+d x)}{a d}-\frac {(2 A-3 B+3 C) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {(3 A-3 B+4 C) \sin ^3(c+d x)}{3 a d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(307\) vs. \(2(139)=278\).

Time = 2.46 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.21 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-12 (2 A-3 B+3 C) d x \cos \left (\frac {d x}{2}\right )-12 (2 A-3 B+3 C) d x \cos \left (c+\frac {d x}{2}\right )+60 A \sin \left (\frac {d x}{2}\right )-60 B \sin \left (\frac {d x}{2}\right )+69 C \sin \left (\frac {d x}{2}\right )+12 A \sin \left (c+\frac {d x}{2}\right )-12 B \sin \left (c+\frac {d x}{2}\right )+21 C \sin \left (c+\frac {d x}{2}\right )+12 A \sin \left (c+\frac {3 d x}{2}\right )-9 B \sin \left (c+\frac {3 d x}{2}\right )+18 C \sin \left (c+\frac {3 d x}{2}\right )+12 A \sin \left (2 c+\frac {3 d x}{2}\right )-9 B \sin \left (2 c+\frac {3 d x}{2}\right )+18 C \sin \left (2 c+\frac {3 d x}{2}\right )+3 B \sin \left (2 c+\frac {5 d x}{2}\right )-2 C \sin \left (2 c+\frac {5 d x}{2}\right )+3 B \sin \left (3 c+\frac {5 d x}{2}\right )-2 C \sin \left (3 c+\frac {5 d x}{2}\right )+C \sin \left (3 c+\frac {7 d x}{2}\right )+C \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{24 a d (1+\cos (c+d x))} \]

[In]

Integrate[(Cos[c + d*x]^2*(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]*(-12*(2*A - 3*B + 3*C)*d*x*Cos[(d*x)/2] - 12*(2*A - 3*B + 3*C)*d*x*Cos[c + (d*x)/2]
+ 60*A*Sin[(d*x)/2] - 60*B*Sin[(d*x)/2] + 69*C*Sin[(d*x)/2] + 12*A*Sin[c + (d*x)/2] - 12*B*Sin[c + (d*x)/2] +
21*C*Sin[c + (d*x)/2] + 12*A*Sin[c + (3*d*x)/2] - 9*B*Sin[c + (3*d*x)/2] + 18*C*Sin[c + (3*d*x)/2] + 12*A*Sin
[2*c + (3*d*x)/2] - 9*B*Sin[2*c + (3*d*x)/2] + 18*C*Sin[2*c + (3*d*x)/2] + 3*B*Sin[2*c + (5*d*x)/2] - 2*C*Sin[
2*c + (5*d*x)/2] + 3*B*Sin[3*c + (5*d*x)/2] - 2*C*Sin[3*c + (5*d*x)/2] + C*Sin[3*c + (7*d*x)/2] + C*Sin[4*c +
(7*d*x)/2]))/(24*a*d*(1 + Cos[c + d*x]))

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.63

method	result	size
parallelrisch	\(\frac {\left (\left (3 B -C \right ) \cos \left (2 d x +2 c \right )+C \cos \left (3 d x +3 c \right )+\left (12 A -6 B +17 C \right ) \cos \left (d x +c \right )+24 A -21 B +31 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-12 x d \left (-\frac {3 B}{2}+\frac {3 C}{2}+A \right )}{12 a d}\)	\(87\)
derivativedivides	\(\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {2 \left (\left (\frac {3 B}{2}-\frac {5 C}{2}-A \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 B -\frac {8 C}{3}-2 A \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {B}{2}-\frac {3 C}{2}-A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\left (2 A -3 B +3 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(147\)
default	\(\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {2 \left (\left (\frac {3 B}{2}-\frac {5 C}{2}-A \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 B -\frac {8 C}{3}-2 A \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {B}{2}-\frac {3 C}{2}-A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\left (2 A -3 B +3 C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(147\)
risch	\(-\frac {x A}{a}+\frac {3 B x}{2 a}-\frac {3 C x}{2 a}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{2 a d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} B}{2 a d}-\frac {7 i {\mathrm e}^{i \left (d x +c \right )} C}{8 a d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{2 a d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} B}{2 a d}+\frac {7 i {\mathrm e}^{-i \left (d x +c \right )} C}{8 a d}+\frac {2 i A}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {2 i B}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {2 i C}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {\sin \left (3 d x +3 c \right ) C}{12 a d}+\frac {\sin \left (2 d x +2 c \right ) B}{4 a d}-\frac {\sin \left (2 d x +2 c \right ) C}{4 a d}\)	\(260\)
norman	\(\frac {\frac {\left (A -B +C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (3 A -2 B +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {\left (6 A -7 B +9 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (2 A -3 B +3 C \right ) x}{2 a}-\frac {2 \left (2 A -3 B +3 C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {3 \left (2 A -3 B +3 C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 \left (2 A -3 B +3 C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {\left (2 A -3 B +3 C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {\left (30 A -27 B +37 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (36 A -39 B +49 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(278\)

[In]

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

[Out]

1/12*(((3*B-C)*cos(2*d*x+2*c)+C*cos(3*d*x+3*c)+(12*A-6*B+17*C)*cos(d*x+c)+24*A-21*B+31*C)*tan(1/2*d*x+1/2*c)-1
2*x*d*(-3/2*B+3/2*C+A))/a/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=-\frac {3 \, {\left (2 \, A - 3 \, B + 3 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (2 \, A - 3 \, B + 3 \, C\right )} d x - {\left (2 \, C \cos \left (d x + c\right )^{3} + {\left (3 \, B - C\right )} \cos \left (d x + c\right )^{2} + {\left (6 \, A - 3 \, B + 7 \, C\right )} \cos \left (d x + c\right ) + 12 \, A - 12 \, B + 16 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

[In]

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

[Out]

-1/6*(3*(2*A - 3*B + 3*C)*d*x*cos(d*x + c) + 3*(2*A - 3*B + 3*C)*d*x - (2*C*cos(d*x + c)^3 + (3*B - C)*cos(d*x
+ c)^2 + (6*A - 3*B + 7*C)*cos(d*x + c) + 12*A - 12*B + 16*C)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1739 vs. \(2 (116) = 232\).

Time = 1.64 (sec) , antiderivative size = 1739, normalized size of antiderivative = 12.51 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((-6*A*d*x*tan(c/2 + d*x/2)**6/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c
/2 + d*x/2)**2 + 6*a*d) - 18*A*d*x*tan(c/2 + d*x/2)**4/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4
+ 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 18*A*d*x*tan(c/2 + d*x/2)**2/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(
c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 6*A*d*x/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 +
d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 6*A*tan(c/2 + d*x/2)**7/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*
tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 30*A*tan(c/2 + d*x/2)**5/(6*a*d*tan(c/2 + d*x/2)**
6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 42*A*tan(c/2 + d*x/2)**3/(6*a*d*tan(c/2
+ d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 18*A*tan(c/2 + d*x/2)/(6*a*d
*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 9*B*d*x*tan(c/2 + d*
x/2)**6/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 27*B*d
*x*tan(c/2 + d*x/2)**4/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 +
6*a*d) + 27*B*d*x*tan(c/2 + d*x/2)**2/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2
+ d*x/2)**2 + 6*a*d) + 9*B*d*x/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x
/2)**2 + 6*a*d) - 6*B*tan(c/2 + d*x/2)**7/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan
(c/2 + d*x/2)**2 + 6*a*d) - 36*B*tan(c/2 + d*x/2)**5/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 +
18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 42*B*tan(c/2 + d*x/2)**3/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 +
d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 12*B*tan(c/2 + d*x/2)/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*ta
n(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 9*C*d*x*tan(c/2 + d*x/2)**6/(6*a*d*tan(c/2 + d*x/2)*
*6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 27*C*d*x*tan(c/2 + d*x/2)**4/(6*a*d*ta
n(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 27*C*d*x*tan(c/2 + d*x/
2)**2/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) - 9*C*d*x/
(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 6*C*tan(c/2 +
d*x/2)**7/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*a*d) + 48*C
*tan(c/2 + d*x/2)**5/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x/2)**2 + 6*
a*d) + 50*C*tan(c/2 + d*x/2)**3/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c/2 + d*x
/2)**2 + 6*a*d) + 24*C*tan(c/2 + d*x/2)/(6*a*d*tan(c/2 + d*x/2)**6 + 18*a*d*tan(c/2 + d*x/2)**4 + 18*a*d*tan(c
/2 + d*x/2)**2 + 6*a*d), Ne(d, 0)), (x*(A + B*cos(c) + C*cos(c)**2)*cos(c)**2/(a*cos(c) + a), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 400 vs. \(2 (133) = 266\).

Time = 0.31 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.88 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\frac {C {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a + \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {3 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, B {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, A {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{3 \, d} \]

[In]

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

[Out]

1/3*(C*((9*sin(d*x + c)/(cos(d*x + c) + 1) + 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5)/(a + 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a*sin
(d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 3*sin(d*x + c)/(a*(cos(d*x +
c) + 1))) - 3*B*((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))) - 3*A*(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))))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.49 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )} {\left (2 \, A - 3 \, B + 3 \, C\right )}}{a} - \frac {6 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a}}{6 \, d} \]

[In]

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

[Out]

-1/6*(3*(d*x + c)*(2*A - 3*B + 3*C)/a - 6*(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*(6*A*tan(1/2*d*x + 1/2*c)^5 - 9*B*tan(1/2*d*x + 1/2*c)^5 + 15*C*tan(1/2*d*x + 1/2*c)^5 + 12*A*tan
(1/2*d*x + 1/2*c)^3 - 12*B*tan(1/2*d*x + 1/2*c)^3 + 16*C*tan(1/2*d*x + 1/2*c)^3 + 6*A*tan(1/2*d*x + 1/2*c) - 3
*B*tan(1/2*d*x + 1/2*c) + 9*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a))/d

Mupad [B] (verification not implemented)

Time = 2.87 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\frac {\left (2\,A-3\,B+5\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (4\,A-4\,B+\frac {16\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A-B+3\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {x\,\left (2\,A-3\,B+3\,C\right )}{2\,a}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B+C\right )}{a\,d} \]

[In]

int((cos(c + d*x)^2*(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)*(2*A - B + 3*C) + tan(c/2 + (d*x)/2)^5*(2*A - 3*B + 5*C) + tan(c/2 + (d*x)/2)^3*(4*A - 4*B
+ (16*C)/3))/(d*(a + 3*a*tan(c/2 + (d*x)/2)^2 + 3*a*tan(c/2 + (d*x)/2)^4 + a*tan(c/2 + (d*x)/2)^6)) - (x*(2*A
- 3*B + 3*C))/(2*a) + (tan(c/2 + (d*x)/2)*(A - B + C))/(a*d)