3.580 \(\int \frac{\cos (c+d x) (A+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^3} \, dx\)
Optimal. Leaf size=203 \[ -\frac{a \left (C \left (-5 a^2 b^2+2 a^4+6 b^4\right )+3 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{\left (a^2 b^2 (A+6 C)-3 a^4 C+2 A b^4\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{a \left (a^2 C+A b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{C x}{b^3} \]
[Out]
(C*x)/b^3 - (a*(3*A*b^4 + (2*a^4 - 5*a^2*b^2 + 6*b^4)*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(
(a - b)^(5/2)*b^3*(a + b)^(5/2)*d) + (a*(A*b^2 + a^2*C)*Sin[c + d*x])/(2*b^2*(a^2 - b^2)*d*(a + b*Cos[c + d*x]
)^2) + ((2*A*b^4 - 3*a^4*C + a^2*b^2*(A + 6*C))*Sin[c + d*x])/(2*b^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))
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Rubi [A] time = 0.464477, antiderivative size = 203, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used =
{3032, 3021, 2735, 2659, 205} \[ -\frac{a \left (C \left (-5 a^2 b^2+2 a^4+6 b^4\right )+3 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{\left (a^2 b^2 (A+6 C)-3 a^4 C+2 A b^4\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{a \left (a^2 C+A b^2\right ) \sin (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{C x}{b^3} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^3,x]
[Out]
(C*x)/b^3 - (a*(3*A*b^4 + (2*a^4 - 5*a^2*b^2 + 6*b^4)*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(
(a - b)^(5/2)*b^3*(a + b)^(5/2)*d) + (a*(A*b^2 + a^2*C)*Sin[c + d*x])/(2*b^2*(a^2 - b^2)*d*(a + b*Cos[c + d*x]
)^2) + ((2*A*b^4 - 3*a^4*C + a^2*b^2*(A + 6*C))*Sin[c + d*x])/(2*b^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))
Rule 3032
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 + 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)*(a*C*(b*c - a*d) + A*b*(a*c - b*d)) - ((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, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Rule 3021
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
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 (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx &=\frac{a \left (A b^2+a^2 C\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\int \frac{2 b \left (A b^2+a^2 C\right )-a \left (A b^2-\left (a^2-2 b^2\right ) C\right ) \cos (c+d x)-2 b \left (a^2-b^2\right ) C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac{a \left (A b^2+a^2 C\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{-a b^2 \left (3 A b^2-\left (a^2-4 b^2\right ) C\right )+2 b \left (a^2-b^2\right )^2 C \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{C x}{b^3}+\frac{a \left (A b^2+a^2 C\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (a \left (3 A b^4+\left (2 a^4-5 a^2 b^2+6 b^4\right ) C\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{C x}{b^3}+\frac{a \left (A b^2+a^2 C\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (a \left (3 A b^4+\left (2 a^4-5 a^2 b^2+6 b^4\right ) C\right )\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^3 \left (a^2-b^2\right )^2 d}\\ &=\frac{C x}{b^3}-\frac{a \left (3 A b^4+2 a^4 C-5 a^2 b^2 C+6 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^3 (a+b)^{5/2} d}+\frac{a \left (A b^2+a^2 C\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 1.28578, size = 194, normalized size = 0.96 \[ \frac{\frac{b \left (a^2 b^2 (A+6 C)-3 a^4 C+2 A b^4\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}+\frac{a b \left (a^2 C+A b^2\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+\frac{2 a \left (C \left (-5 a^2 b^2+2 a^4+6 b^4\right )+3 A b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+2 C (c+d x)}{2 b^3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^3,x]
[Out]
(2*C*(c + d*x) + (2*a*(3*A*b^4 + (2*a^4 - 5*a^2*b^2 + 6*b^4)*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 +
b^2]])/(-a^2 + b^2)^(5/2) + (a*b*(A*b^2 + a^2*C)*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) + (b*
(2*A*b^4 - 3*a^4*C + a^2*b^2*(A + 6*C))*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])))/(2*b^3*d)
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Maple [B] time = 0.04, size = 1093, normalized size = 5.4 \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)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x)
[Out]
2/d/b^3*arctan(tan(1/2*d*x+1/2*c))*C+2/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+1/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+2/d*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*A-2/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+1/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+6/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-b)/(a^2+2*a*b+b^2)*ta
n(1/2*d*x+1/2*c)^3*a^2*C+2/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-b)^2*tan(1/2*d
*x+1/2*c)*A-1/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+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*a*A+2
/d*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-b)^2*tan(1/2*d*x+1/2*c)*A-2/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-b)^2*tan(1/2*d*x+1/2*c)*C-1/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+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*a^3*C+6/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+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*a^2*C-3/d*a*b/(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/d*a^5/b^3/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*ar
ctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C+5/d*a^3/b/(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-6/d*b*a/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctan((a-b)*t
an(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C
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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)*(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 = 2.27732, size = 2279, normalized size = 11.23 \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)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
[Out]
[1/4*(4*(C*a^6*b^2 - 3*C*a^4*b^4 + 3*C*a^2*b^6 - C*b^8)*d*x*cos(d*x + c)^2 + 8*(C*a^7*b - 3*C*a^5*b^3 + 3*C*a^
3*b^5 - C*a*b^7)*d*x*cos(d*x + c) + 4*(C*a^8 - 3*C*a^6*b^2 + 3*C*a^4*b^4 - C*a^2*b^6)*d*x - (2*C*a^7 - 5*C*a^5
*b^2 + 3*(A + 2*C)*a^3*b^4 + (2*C*a^5*b^2 - 5*C*a^3*b^4 + 3*(A + 2*C)*a*b^6)*cos(d*x + c)^2 + 2*(2*C*a^6*b - 5
*C*a^4*b^3 + 3*(A + 2*C)*a^2*b^5)*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*co
s(d*x + c) + a^2)) - 2*(2*C*a^7*b - (2*A + 7*C)*a^5*b^3 + (A + 5*C)*a^3*b^5 + A*a*b^7 + (3*C*a^6*b^2 - (A + 9*
C)*a^4*b^4 - (A - 6*C)*a^2*b^6 + 2*A*b^8)*cos(d*x + c))*sin(d*x + c))/((a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11
)*d*cos(d*x + c)^2 + 2*(a^7*b^4 - 3*a^5*b^6 + 3*a^3*b^8 - a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 3*a^6*b^5 + 3*a^
4*b^7 - a^2*b^9)*d), 1/2*(2*(C*a^6*b^2 - 3*C*a^4*b^4 + 3*C*a^2*b^6 - C*b^8)*d*x*cos(d*x + c)^2 + 4*(C*a^7*b -
3*C*a^5*b^3 + 3*C*a^3*b^5 - C*a*b^7)*d*x*cos(d*x + c) + 2*(C*a^8 - 3*C*a^6*b^2 + 3*C*a^4*b^4 - C*a^2*b^6)*d*x
- (2*C*a^7 - 5*C*a^5*b^2 + 3*(A + 2*C)*a^3*b^4 + (2*C*a^5*b^2 - 5*C*a^3*b^4 + 3*(A + 2*C)*a*b^6)*cos(d*x + c)^
2 + 2*(2*C*a^6*b - 5*C*a^4*b^3 + 3*(A + 2*C)*a^2*b^5)*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))) - (2*C*a^7*b - (2*A + 7*C)*a^5*b^3 + (A + 5*C)*a^3*b^5 + A*a*b^7 + (3*C*a^6
*b^2 - (A + 9*C)*a^4*b^4 - (A - 6*C)*a^2*b^6 + 2*A*b^8)*cos(d*x + c))*sin(d*x + c))/((a^6*b^5 - 3*a^4*b^7 + 3*
a^2*b^9 - b^11)*d*cos(d*x + c)^2 + 2*(a^7*b^4 - 3*a^5*b^6 + 3*a^3*b^8 - a*b^10)*d*cos(d*x + c) + (a^8*b^3 - 3*
a^6*b^5 + 3*a^4*b^7 - a^2*b^9)*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)*(A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**3,x)
[Out]
Timed out
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Giac [B] time = 1.33529, size = 647, normalized size = 3.19 \begin{align*} -\frac{\frac{{\left (2 \, C a^{5} - 5 \, C a^{3} b^{2} + 3 \, A a b^{4} + 6 \, C a b^{4}\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^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \sqrt{a^{2} - b^{2}}} - \frac{{\left (d x + c\right )} C}{b^{3}} + \frac{2 \, C a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, C a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, A a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, C a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, C a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, A b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, C a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, A a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, C a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, C a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, A b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\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}}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")
[Out]
-((2*C*a^5 - 5*C*a^3*b^2 + 3*A*a*b^4 + 6*C*a*b^4)*(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^3 - 2*a^2*b^5 + b^7)*sqrt(a^2 - b^2)
) - (d*x + c)*C/b^3 + (2*C*a^5*tan(1/2*d*x + 1/2*c)^3 - 3*C*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 2*A*a^3*b^2*tan(1/2
*d*x + 1/2*c)^3 - 5*C*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + A*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^2*b^3*tan(1/2*
d*x + 1/2*c)^3 - A*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 2*A*b^5*tan(1/2*d*x + 1/2*c)^3 + 2*C*a^5*tan(1/2*d*x + 1/2*c
) + 3*C*a^4*b*tan(1/2*d*x + 1/2*c) - 2*A*a^3*b^2*tan(1/2*d*x + 1/2*c) - 5*C*a^3*b^2*tan(1/2*d*x + 1/2*c) - A*a
^2*b^3*tan(1/2*d*x + 1/2*c) - 6*C*a^2*b^3*tan(1/2*d*x + 1/2*c) - A*a*b^4*tan(1/2*d*x + 1/2*c) - 2*A*b^5*tan(1/
2*d*x + 1/2*c))/((a^4*b^2 - 2*a^2*b^4 + b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2))
/d