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

Optimal. Leaf size=160 \[ -\frac{2 (2 A-5 B+8 C) \sin (c+d x)}{3 a^2 d}-\frac{(2 A-5 B+8 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac{(2 A-4 B+7 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{x (2 A-4 B+7 C)}{2 a^2}-\frac{(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

((2*A - 4*B + 7*C)*x)/(2*a^2) - (2*(2*A - 5*B + 8*C)*Sin[c + d*x])/(3*a^2*d) + ((2*A - 4*B + 7*C)*Cos[c + d*x]
*Sin[c + d*x])/(2*a^2*d) - ((2*A - 5*B + 8*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((A
- B + C)*Cos[c + d*x]^3*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

________________________________________________________________________________________

Rubi [A]  time = 0.316792, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.073, Rules used = {3041, 2977, 2734} \[ -\frac{2 (2 A-5 B+8 C) \sin (c+d x)}{3 a^2 d}-\frac{(2 A-5 B+8 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac{(2 A-4 B+7 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{x (2 A-4 B+7 C)}{2 a^2}-\frac{(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[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])^2,x]

[Out]

((2*A - 4*B + 7*C)*x)/(2*a^2) - (2*(2*A - 5*B + 8*C)*Sin[c + d*x])/(3*a^2*d) + ((2*A - 4*B + 7*C)*Cos[c + d*x]
*Sin[c + d*x])/(2*a^2*d) - ((2*A - 5*B + 8*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((A
- B + C)*Cos[c + d*x]^3*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

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)]

Rule 2977

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

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]

Rubi steps

\begin{align*} \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))^2} \, dx &=-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{\cos ^2(c+d x) (3 a (B-C)+a (2 A-2 B+5 C) \cos (c+d x))}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(2 A-5 B+8 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \cos (c+d x) \left (-2 a^2 (2 A-5 B+8 C)+3 a^2 (2 A-4 B+7 C) \cos (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac{(2 A-4 B+7 C) x}{2 a^2}-\frac{2 (2 A-5 B+8 C) \sin (c+d x)}{3 a^2 d}+\frac{(2 A-4 B+7 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac{(2 A-5 B+8 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end{align*}

Mathematica [B]  time = 0.844568, size = 385, normalized size = 2.41 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (36 d x (2 A-4 B+7 C) \cos \left (c+\frac{d x}{2}\right )+36 d x (2 A-4 B+7 C) \cos \left (\frac{d x}{2}\right )+96 A \sin \left (c+\frac{d x}{2}\right )-80 A \sin \left (c+\frac{3 d x}{2}\right )+24 A d x \cos \left (c+\frac{3 d x}{2}\right )+24 A d x \cos \left (2 c+\frac{3 d x}{2}\right )-144 A \sin \left (\frac{d x}{2}\right )-120 B \sin \left (c+\frac{d x}{2}\right )+164 B \sin \left (c+\frac{3 d x}{2}\right )+36 B \sin \left (2 c+\frac{3 d x}{2}\right )+12 B \sin \left (2 c+\frac{5 d x}{2}\right )+12 B \sin \left (3 c+\frac{5 d x}{2}\right )-48 B d x \cos \left (c+\frac{3 d x}{2}\right )-48 B d x \cos \left (2 c+\frac{3 d x}{2}\right )+264 B \sin \left (\frac{d x}{2}\right )+147 C \sin \left (c+\frac{d x}{2}\right )-239 C \sin \left (c+\frac{3 d x}{2}\right )-63 C \sin \left (2 c+\frac{3 d x}{2}\right )-15 C \sin \left (2 c+\frac{5 d x}{2}\right )-15 C \sin \left (3 c+\frac{5 d x}{2}\right )+3 C \sin \left (3 c+\frac{7 d x}{2}\right )+3 C \sin \left (4 c+\frac{7 d x}{2}\right )+84 C d x \cos \left (c+\frac{3 d x}{2}\right )+84 C d x \cos \left (2 c+\frac{3 d x}{2}\right )-381 C \sin \left (\frac{d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[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])^2,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(36*(2*A - 4*B + 7*C)*d*x*Cos[(d*x)/2] + 36*(2*A - 4*B + 7*C)*d*x*Cos[c + (d*x)/2]
+ 24*A*d*x*Cos[c + (3*d*x)/2] - 48*B*d*x*Cos[c + (3*d*x)/2] + 84*C*d*x*Cos[c + (3*d*x)/2] + 24*A*d*x*Cos[2*c +
(3*d*x)/2] - 48*B*d*x*Cos[2*c + (3*d*x)/2] + 84*C*d*x*Cos[2*c + (3*d*x)/2] - 144*A*Sin[(d*x)/2] + 264*B*Sin[(
d*x)/2] - 381*C*Sin[(d*x)/2] + 96*A*Sin[c + (d*x)/2] - 120*B*Sin[c + (d*x)/2] + 147*C*Sin[c + (d*x)/2] - 80*A*
Sin[c + (3*d*x)/2] + 164*B*Sin[c + (3*d*x)/2] - 239*C*Sin[c + (3*d*x)/2] + 36*B*Sin[2*c + (3*d*x)/2] - 63*C*Si
n[2*c + (3*d*x)/2] + 12*B*Sin[2*c + (5*d*x)/2] - 15*C*Sin[2*c + (5*d*x)/2] + 12*B*Sin[3*c + (5*d*x)/2] - 15*C*
Sin[3*c + (5*d*x)/2] + 3*C*Sin[3*c + (7*d*x)/2] + 3*C*Sin[4*c + (7*d*x)/2]))/(48*a^2*d*(1 + Cos[c + d*x])^2)

________________________________________________________________________________________

Maple [B]  time = 0.033, size = 309, normalized size = 1.9 \begin{align*}{\frac{A}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{3\,A}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{5\,B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{7\,C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-5\,{\frac{C \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}B}{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-3\,{\frac{C\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A}{d{a}^{2}}}-4\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{d{a}^{2}}}+7\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A-1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*B+1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3-3/2/d/a^2*A
*tan(1/2*d*x+1/2*c)+5/2/d/a^2*B*tan(1/2*d*x+1/2*c)-7/2/d/a^2*C*tan(1/2*d*x+1/2*c)-5/d/a^2/(tan(1/2*d*x+1/2*c)^
2+1)^2*C*tan(1/2*d*x+1/2*c)^3+2/d/a^2/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*c)^3*B-3/d/a^2/(tan(1/2*d*x+1
/2*c)^2+1)^2*C*tan(1/2*d*x+1/2*c)+2/d/a^2/(tan(1/2*d*x+1/2*c)^2+1)^2*B*tan(1/2*d*x+1/2*c)+2/d/a^2*arctan(tan(1
/2*d*x+1/2*c))*A-4/d/a^2*arctan(tan(1/2*d*x+1/2*c))*B+7/d/a^2*arctan(tan(1/2*d*x+1/2*c))*C

________________________________________________________________________________________

Maxima [B]  time = 1.5328, size = 475, normalized size = 2.97 \begin{align*} -\frac{C{\left (\frac{6 \,{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{42 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - B{\left (\frac{\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{24 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac{12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac{a^{2} \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 )}}\right )} + A{\left (\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{12 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))^2,x, algorithm="maxima")

[Out]

-1/6*(C*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - s
in(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) - B*((15*sin(d*x + c
)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a
^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))) + A*((9*sin(d*x + c
)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a
^2))/d

________________________________________________________________________________________

Fricas [A]  time = 1.91712, size = 383, normalized size = 2.39 \begin{align*} \frac{3 \,{\left (2 \, A - 4 \, B + 7 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \,{\left (2 \, A - 4 \, B + 7 \, C\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (2 \, A - 4 \, B + 7 \, C\right )} d x +{\left (3 \, C \cos \left (d x + c\right )^{3} + 6 \,{\left (B - C\right )} \cos \left (d x + c\right )^{2} -{\left (10 \, A - 28 \, B + 43 \, C\right )} \cos \left (d x + c\right ) - 8 \, A + 20 \, B - 32 \, C\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))^2,x, algorithm="fricas")

[Out]

1/6*(3*(2*A - 4*B + 7*C)*d*x*cos(d*x + c)^2 + 6*(2*A - 4*B + 7*C)*d*x*cos(d*x + c) + 3*(2*A - 4*B + 7*C)*d*x +
(3*C*cos(d*x + c)^3 + 6*(B - C)*cos(d*x + c)^2 - (10*A - 28*B + 43*C)*cos(d*x + c) - 8*A + 20*B - 32*C)*sin(d
*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

________________________________________________________________________________________

Sympy [A]  time = 20.0107, size = 1261, normalized size = 7.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))**2,x)

[Out]

Piecewise((6*A*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*
d) + 12*A*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) +
6*A*d*x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + A*tan(c/2 + d*x/2)**7/(6*a
**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 7*A*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(
c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 17*A*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x
/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 9*A*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a
**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 12*B*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d
*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 24*B*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(
c/2 + d*x/2)**2 + 6*a**2*d) - 12*B*d*x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*
d) - B*tan(c/2 + d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 13*B*ta
n(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 41*B*tan(c/2 + d
*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 27*B*tan(c/2 + d*x/2)/(6*
a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 21*C*d*x*tan(c/2 + d*x/2)**4/(6*a**2*
d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 42*C*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan
(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 21*C*d*x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**
2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + C*tan(c/2 + d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2
+ d*x/2)**2 + 6*a**2*d) - 19*C*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)*
*2 + 6*a**2*d) - 71*C*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a*
*2*d) - 39*C*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d), Ne(d,
0)), (x*(A + B*cos(c) + C*cos(c)**2)*cos(c)**2/(a*cos(c) + a)**2, True))

________________________________________________________________________________________

Giac [A]  time = 1.15979, size = 267, normalized size = 1.67 \begin{align*} \frac{\frac{3 \,{\left (d x + c\right )}{\left (2 \, A - 4 \, B + 7 \, C\right )}}{a^{2}} + \frac{6 \,{\left (2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, 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 ) - 3 \, 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}} + \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 21 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))^2,x, algorithm="giac")

[Out]

1/6*(3*(d*x + c)*(2*A - 4*B + 7*C)/a^2 + 6*(2*B*tan(1/2*d*x + 1/2*c)^3 - 5*C*tan(1/2*d*x + 1/2*c)^3 + 2*B*tan(
1/2*d*x + 1/2*c) - 3*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2) + (A*a^4*tan(1/2*d*x + 1/2*c
)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 - 9*A*a^4*tan(1/2*d*x + 1/2*c) + 15*B*a^4*ta
n(1/2*d*x + 1/2*c) - 21*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d