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

Optimal. Leaf size=233 \[ \frac{\left (a^2 b^3 B+5 a^3 b^2 C-2 a^5 C-3 a b^4 (A+2 C)+2 b^5 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{\sin (c+d x) \left (a^2 b^2 (A+6 C)+a^3 b B-3 a^4 C-4 a b^3 B+2 A b^4\right )}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{C x}{b^3} \]

[Out]

(C*x)/b^3 + ((a^2*b^3*B + 2*b^5*B - 2*a^5*C + 5*a^3*b^2*C - 3*a*b^4*(A + 2*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*(b*B - a*C))*Sin[c + d*x])/(2*b^2*(a^
2 - b^2)*d*(a + b*Cos[c + d*x])^2) + ((2*A*b^4 + a^3*b*B - 4*a*b^3*B - 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 = 0.770619, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {3031, 3021, 2735, 2659, 205} \[ \frac{\left (a^2 b^3 B+5 a^3 b^2 C-2 a^5 C-3 a b^4 (A+2 C)+2 b^5 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{\sin (c+d x) \left (a^2 b^2 (A+6 C)+a^3 b B-3 a^4 C-4 a b^3 B+2 A b^4\right )}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{C x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*x)/b^3 + ((a^2*b^3*B + 2*b^5*B - 2*a^5*C + 5*a^3*b^2*C - 3*a*b^4*(A + 2*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*(b*B - a*C))*Sin[c + d*x])/(2*b^2*(a^
2 - b^2)*d*(a + b*Cos[c + d*x])^2) + ((2*A*b^4 + a^3*b*B - 4*a*b^3*B - 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 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 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+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx &=\frac{a \left (A b^2-a (b B-a 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 (b B-a C)\right )+\left (a^2 b B-2 b^3 B-a^3 C+a b^2 (A+2 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 (b B-a 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+a^3 b B-4 a b^3 B-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{-b^2 \left (a^2 b B+2 b^3 B+a^3 C-a b^2 (3 A+4 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 (b B-a 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+a^3 b B-4 a b^3 B-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^2 b^3 B+2 b^5 B-2 a^5 C+5 a^3 b^2 C-3 a b^4 (A+2 C)\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 (b B-a 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+a^3 b B-4 a b^3 B-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^2 b^3 B+2 b^5 B-2 a^5 C+5 a^3 b^2 C-3 a b^4 (A+2 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^3 \left (a^2-b^2\right )^2 d}\\ &=\frac{C x}{b^3}-\frac{\left (3 a A b^4-a^2 b^3 B-2 b^5 B+2 a^5 C-5 a^3 b^2 C+6 a 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 (b B-a 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+a^3 b B-4 a b^3 B-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 = 1.6684, size = 225, normalized size = 0.97 \[ \frac{\frac{b \sin (c+d x) \left (a^2 b^2 (A+6 C)+a^3 b B-3 a^4 C-4 a b^3 B+2 A b^4\right )}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}+\frac{2 \left (-a^2 b^3 B-5 a^3 b^2 C+2 a^5 C+3 a b^4 (A+2 C)-2 b^5 B\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+\frac{a b \sin (c+d x) \left (a (a C-b B)+A b^2\right )}{(a-b) (a+b) (a+b \cos (c+d x))^2}+2 C (c+d x)}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*C*(c + d*x) + (2*(-(a^2*b^3*B) - 2*b^5*B + 2*a^5*C - 5*a^3*b^2*C + 3*a*b^4*(A + 2*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*(-(b*B) + a*C))*Sin[c + d*x])/((a - b)*(
a + b)*(a + b*Cos[c + d*x])^2) + (b*(2*A*b^4 + a^3*b*B - 4*a*b^3*B - 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 = 0.044, size = 1485, normalized size = 6.4 \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+B*cos(d*x+c)+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-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*B-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)*t
an(1/2*d*x+1/2*c)^3*B-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)*t
an(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)*tan(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/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*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+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-b)^2*tan(1/2*d*x+1/2*c)*B-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-b)^2*tan(1/2*d*x+1/2*c)*B-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+1/d*a^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))*B+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))*B-2/d*a^5/b^3/(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+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)*tan(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+B*cos(d*x+c)+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 = 2.5683, size = 2651, normalized size = 11.38 \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+B*cos(d*x+c)+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 - B*a^4*b^3 + 3*(A + 2*C)*a^3*b^4 - 2*B*a^2*b^5 + (2*C*a^5*b^2 - 5*C*a^3*b^4 - B*a^2*b^5 + 3*(A + 2*C)*a*
b^6 - 2*B*b^7)*cos(d*x + c)^2 + 2*(2*C*a^6*b - 5*C*a^4*b^3 - B*a^3*b^4 + 3*(A + 2*C)*a^2*b^5 - 2*B*a*b^6)*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*(2*C*a^7*b - (2
*A + 7*C)*a^5*b^3 + 3*B*a^4*b^4 + (A + 5*C)*a^3*b^5 - 3*B*a^2*b^6 + A*a*b^7 + (3*C*a^6*b^2 - B*a^5*b^3 - (A +
9*C)*a^4*b^4 + 5*B*a^3*b^5 - (A - 6*C)*a^2*b^6 - 4*B*a*b^7 + 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*co
s(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 - B*a^4*b^3 + 3*(A + 2*C)*a^3*b^4 - 2*B*a^2*b^5 + (2*C*a^
5*b^2 - 5*C*a^3*b^4 - B*a^2*b^5 + 3*(A + 2*C)*a*b^6 - 2*B*b^7)*cos(d*x + c)^2 + 2*(2*C*a^6*b - 5*C*a^4*b^3 - B
*a^3*b^4 + 3*(A + 2*C)*a^2*b^5 - 2*B*a*b^6)*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 + 3*B*a^4*b^4 + (A + 5*C)*a^3*b^5 - 3*B*a^2*b^6 +
A*a*b^7 + (3*C*a^6*b^2 - B*a^5*b^3 - (A + 9*C)*a^4*b^4 + 5*B*a^3*b^5 - (A - 6*C)*a^2*b^6 - 4*B*a*b^7 + 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+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.26869, size = 814, normalized size = 3.49 \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+B*cos(d*x+c)+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 - B*a^2*b^3 + 3*A*a*b^4 + 6*C*a*b^4 - 2*B*b^5)*(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 + B*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 + 3*B*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 - 4*B*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) - B*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) + 3*B*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) + 4*B*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