3.579 \(\int \frac{\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=262 \[ \frac{\left (3 a^2 C+A b^2-2 b^2 C\right ) \sin (c+d x)}{2 b^3 d \left (a^2-b^2\right )}+\frac{\left (a^2 b^4 (A+12 C)-15 a^4 b^2 C+6 a^6 C+2 A b^6\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{a \left (a^2 b^2 (A+6 C)-3 a^4 C+2 A b^4\right ) \sin (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{3 a C x}{b^4} \]

[Out]

(-3*a*C*x)/b^4 + ((2*A*b^6 + 6*a^6*C - 15*a^4*b^2*C + a^2*b^4*(A + 12*C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2]
)/Sqrt[a + b]])/((a - b)^(5/2)*b^4*(a + b)^(5/2)*d) + ((A*b^2 + 3*a^2*C - 2*b^2*C)*Sin[c + d*x])/(2*b^3*(a^2 -
b^2)*d) - ((A*b^2 + a^2*C)*Cos[c + d*x]^2*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - (a*(2*A*
b^4 - 3*a^4*C + a^2*b^2*(A + 6*C))*Sin[c + d*x])/(2*b^3*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))

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Rubi [A]  time = 0.8414, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3048, 3031, 3023, 2735, 2659, 205} \[ \frac{\left (3 a^2 C+A b^2-2 b^2 C\right ) \sin (c+d x)}{2 b^3 d \left (a^2-b^2\right )}+\frac{\left (a^2 b^4 (A+12 C)-15 a^4 b^2 C+6 a^6 C+2 A b^6\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{a \left (a^2 b^2 (A+6 C)-3 a^4 C+2 A b^4\right ) \sin (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{3 a C x}{b^4} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^3,x]

[Out]

(-3*a*C*x)/b^4 + ((2*A*b^6 + 6*a^6*C - 15*a^4*b^2*C + a^2*b^4*(A + 12*C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2]
)/Sqrt[a + b]])/((a - b)^(5/2)*b^4*(a + b)^(5/2)*d) + ((A*b^2 + 3*a^2*C - 2*b^2*C)*Sin[c + d*x])/(2*b^3*(a^2 -
b^2)*d) - ((A*b^2 + a^2*C)*Cos[c + d*x]^2*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - (a*(2*A*
b^4 - 3*a^4*C + a^2*b^2*(A + 6*C))*Sin[c + d*x])/(2*b^3*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))

Rule 3048

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

Rule 3031

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

Rule 3023

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx &=-\frac{\left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\int \frac{\cos (c+d x) \left (2 \left (A b^2+a^2 C\right )-2 a b (A+C) \cos (c+d x)-\left (A b^2+3 a^2 C-2 b^2 C\right ) \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\int \frac{-b \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right )-a \left (a^2-b^2\right ) \left (A b^2-3 a^2 C+4 b^2 C\right ) \cos (c+d x)-b \left (a^2-b^2\right ) \left (A b^2+3 a^2 C-2 b^2 C\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^2+3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\int \frac{-b^2 \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right )+6 a b \left (a^2-b^2\right )^2 C \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{3 a C x}{b^4}+\frac{\left (A b^2+3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (2 A b^6+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{3 a C x}{b^4}+\frac{\left (A b^2+3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (2 A b^6+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^4 \left (a^2-b^2\right )^2 d}\\ &=-\frac{3 a C x}{b^4}+\frac{\left (a^2 A b^4+2 A b^6+6 a^6 C-15 a^4 b^2 C+12 a^2 b^4 C\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} b^4 (a+b)^{5/2} d}+\frac{\left (A b^2+3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.70445, size = 214, normalized size = 0.82 \[ \frac{-\frac{a^2 b \left (a^2 C+A b^2\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+\frac{a b \left (a^2 b^2 (A-8 C)+5 a^4 C-4 A b^4\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}-\frac{2 \left (a^2 b^4 (A+12 C)-15 a^4 b^2 C+6 a^6 C+2 A b^6\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}-6 a C (c+d x)+2 b C \sin (c+d x)}{2 b^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^3,x]

[Out]

(-6*a*C*(c + d*x) - (2*(2*A*b^6 + 6*a^6*C - 15*a^4*b^2*C + a^2*b^4*(A + 12*C))*ArcTanh[((a - b)*Tan[(c + d*x)/
2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + 2*b*C*Sin[c + d*x] - (a^2*b*(A*b^2 + a^2*C)*Sin[c + d*x])/((a - b)
*(a + b)*(a + b*Cos[c + d*x])^2) + (a*b*(-4*A*b^4 + a^2*b^2*(A - 8*C) + 5*a^4*C)*Sin[c + d*x])/((a - b)^2*(a +
b)^2*(a + b*Cos[c + d*x])))/(2*b^4*d)

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Maple [B]  time = 0.04, size = 1094, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x)

[Out]

2/d*C/b^3*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2+1)-6/d*C/b^4*a*arctan(tan(1/2*d*x+1/2*c))-1/d*a^2/(a*tan(1/
2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A-4/d*b/(a*tan(1/2*d*x
+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*a/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A+4/d*a^5/b^3/(a*tan(1/2*
d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C-1/d*a^4/b^2/(a*tan(1/2
*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C-8/d/b/(a*tan(1/2*d*x+
1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*a^3/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C+1/d/(a*tan(1/2*d*x+1/2
*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*a^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*A-4/d*b/(a*tan(1/2*d*x+1/2*c)
^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^2*a/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*A+4/d/b^3/(a*tan(1/2*d*x+1/2*c)^2-
tan(1/2*d*x+1/2*c)^2*b+a+b)^2*a^5/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*C+1/d/b^2/(a*tan(1/2*d*x+1/2*c)^2-t
an(1/2*d*x+1/2*c)^2*b+a+b)^2*a^4/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*C-8/d/b/(a*tan(1/2*d*x+1/2*c)^2-tan(
1/2*d*x+1/2*c)^2*b+a+b)^2*a^3/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*C+1/d/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))
^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*a^2*A+2/d*b^2/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1
/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+6/d/b^4/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arc
tan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*a^6*C-15/d/b^2/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arcta
n((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*a^4*C+12/d/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctan((a-b
)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C*a^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.53466, size = 2638, normalized size = 10.07 \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+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*(12*(C*a^7*b^2 - 3*C*a^5*b^4 + 3*C*a^3*b^6 - C*a*b^8)*d*x*cos(d*x + c)^2 + 24*(C*a^8*b - 3*C*a^6*b^3 + 3
*C*a^4*b^5 - C*a^2*b^7)*d*x*cos(d*x + c) + 12*(C*a^9 - 3*C*a^7*b^2 + 3*C*a^5*b^4 - C*a^3*b^6)*d*x + (6*C*a^8 -
15*C*a^6*b^2 + (A + 12*C)*a^4*b^4 + 2*A*a^2*b^6 + (6*C*a^6*b^2 - 15*C*a^4*b^4 + (A + 12*C)*a^2*b^6 + 2*A*b^8)
*cos(d*x + c)^2 + 2*(6*C*a^7*b - 15*C*a^5*b^3 + (A + 12*C)*a^3*b^5 + 2*A*a*b^7)*cos(d*x + c))*sqrt(-a^2 + b^2)
*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c)
- a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - 2*(6*C*a^8*b - 17*C*a^6*b^3 - (3*A - 13*C)*
a^4*b^5 + (3*A - 2*C)*a^2*b^7 + 2*(C*a^6*b^3 - 3*C*a^4*b^5 + 3*C*a^2*b^7 - C*b^9)*cos(d*x + c)^2 + (9*C*a^7*b^
2 + (A - 25*C)*a^5*b^4 - 5*(A - 4*C)*a^3*b^6 + 4*(A - C)*a*b^8)*cos(d*x + c))*sin(d*x + c))/((a^6*b^6 - 3*a^4*
b^8 + 3*a^2*b^10 - b^12)*d*cos(d*x + c)^2 + 2*(a^7*b^5 - 3*a^5*b^7 + 3*a^3*b^9 - a*b^11)*d*cos(d*x + c) + (a^8
*b^4 - 3*a^6*b^6 + 3*a^4*b^8 - a^2*b^10)*d), -1/2*(6*(C*a^7*b^2 - 3*C*a^5*b^4 + 3*C*a^3*b^6 - C*a*b^8)*d*x*cos
(d*x + c)^2 + 12*(C*a^8*b - 3*C*a^6*b^3 + 3*C*a^4*b^5 - C*a^2*b^7)*d*x*cos(d*x + c) + 6*(C*a^9 - 3*C*a^7*b^2 +
3*C*a^5*b^4 - C*a^3*b^6)*d*x - (6*C*a^8 - 15*C*a^6*b^2 + (A + 12*C)*a^4*b^4 + 2*A*a^2*b^6 + (6*C*a^6*b^2 - 15
*C*a^4*b^4 + (A + 12*C)*a^2*b^6 + 2*A*b^8)*cos(d*x + c)^2 + 2*(6*C*a^7*b - 15*C*a^5*b^3 + (A + 12*C)*a^3*b^5 +
2*A*a*b^7)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (6*C*
a^8*b - 17*C*a^6*b^3 - (3*A - 13*C)*a^4*b^5 + (3*A - 2*C)*a^2*b^7 + 2*(C*a^6*b^3 - 3*C*a^4*b^5 + 3*C*a^2*b^7 -
C*b^9)*cos(d*x + c)^2 + (9*C*a^7*b^2 + (A - 25*C)*a^5*b^4 - 5*(A - 4*C)*a^3*b^6 + 4*(A - C)*a*b^8)*cos(d*x +
c))*sin(d*x + c))/((a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - b^12)*d*cos(d*x + c)^2 + 2*(a^7*b^5 - 3*a^5*b^7 + 3*a^3
*b^9 - a*b^11)*d*cos(d*x + c) + (a^8*b^4 - 3*a^6*b^6 + 3*a^4*b^8 - a^2*b^10)*d)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.72454, size = 660, normalized size = 2.52 \begin{align*} -\frac{\frac{{\left (6 \, C a^{6} - 15 \, C a^{4} b^{2} + A a^{2} b^{4} + 12 \, C a^{2} b^{4} + 2 \, A b^{6}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{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 )}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} \sqrt{a^{2} - b^{2}}} + \frac{3 \,{\left (d x + c\right )} C a}{b^{4}} - \frac{4 \, C a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, C a^{5} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, C a^{4} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 8 \, C a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, A a^{2} b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, A a b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, C a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, C a^{5} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 7 \, C a^{4} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 \, C a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, A a^{2} b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, A a b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b\right )}^{2}} - \frac{2 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} b^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

-((6*C*a^6 - 15*C*a^4*b^2 + A*a^2*b^4 + 12*C*a^2*b^4 + 2*A*b^6)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2
*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^4*b^4 - 2*a^2*b^6 + b^8)
*sqrt(a^2 - b^2)) + 3*(d*x + c)*C*a/b^4 - (4*C*a^6*tan(1/2*d*x + 1/2*c)^3 - 5*C*a^5*b*tan(1/2*d*x + 1/2*c)^3 -
7*C*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 - A*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 8*C*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 -
3*A*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 + 4*A*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 4*C*a^6*tan(1/2*d*x + 1/2*c) + 5*C*a^5
*b*tan(1/2*d*x + 1/2*c) - 7*C*a^4*b^2*tan(1/2*d*x + 1/2*c) + A*a^3*b^3*tan(1/2*d*x + 1/2*c) - 8*C*a^3*b^3*tan(
1/2*d*x + 1/2*c) - 3*A*a^2*b^4*tan(1/2*d*x + 1/2*c) - 4*A*a*b^5*tan(1/2*d*x + 1/2*c))/((a^4*b^3 - 2*a^2*b^5 +
b^7)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2) - 2*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x
+ 1/2*c)^2 + 1)*b^3))/d