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

Optimal. Leaf size=279 \[ \frac{\sin (c+d x) \left (3 a^2 b B-3 a^3 C-a b^2 (3 A+2 C)+2 b^3 B\right )}{3 b^4 d}-\frac{2 a^3 \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d \sqrt{a-b} \sqrt{a+b}}+\frac{\sin (c+d x) \cos (c+d x) \left (4 a^2 C-4 a b B+4 A b^2+3 b^2 C\right )}{8 b^3 d}-\frac{x \left (-4 a^2 b^2 (2 A+C)+8 a^3 b B-8 a^4 C+4 a b^3 B-b^4 (4 A+3 C)\right )}{8 b^5}+\frac{(b B-a C) \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d}+\frac{C \sin (c+d x) \cos ^3(c+d x)}{4 b d} \]

[Out]

-((8*a^3*b*B + 4*a*b^3*B - 8*a^4*C - 4*a^2*b^2*(2*A + C) - b^4*(4*A + 3*C))*x)/(8*b^5) - (2*a^3*(A*b^2 - a*(b*
B - a*C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^5*Sqrt[a + b]*d) + ((3*a^2*b*B +
2*b^3*B - 3*a^3*C - a*b^2*(3*A + 2*C))*Sin[c + d*x])/(3*b^4*d) + ((4*A*b^2 - 4*a*b*B + 4*a^2*C + 3*b^2*C)*Cos[
c + d*x]*Sin[c + d*x])/(8*b^3*d) + ((b*B - a*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*b^2*d) + (C*Cos[c + d*x]^3*Sin
[c + d*x])/(4*b*d)

________________________________________________________________________________________

Rubi [A]  time = 0.935385, antiderivative size = 279, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3049, 3023, 2735, 2659, 205} \[ \frac{\sin (c+d x) \left (3 a^2 b B-3 a^3 C-a b^2 (3 A+2 C)+2 b^3 B\right )}{3 b^4 d}-\frac{2 a^3 \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d \sqrt{a-b} \sqrt{a+b}}+\frac{\sin (c+d x) \cos (c+d x) \left (4 a^2 C-4 a b B+4 A b^2+3 b^2 C\right )}{8 b^3 d}-\frac{x \left (-4 a^2 b^2 (2 A+C)+8 a^3 b B-8 a^4 C+4 a b^3 B-b^4 (4 A+3 C)\right )}{8 b^5}+\frac{(b B-a C) \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d}+\frac{C \sin (c+d x) \cos ^3(c+d x)}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*a^3*b*B + 4*a*b^3*B - 8*a^4*C - 4*a^2*b^2*(2*A + C) - b^4*(4*A + 3*C))*x)/(8*b^5) - (2*a^3*(A*b^2 - a*(b*
B - a*C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^5*Sqrt[a + b]*d) + ((3*a^2*b*B +
2*b^3*B - 3*a^3*C - a*b^2*(3*A + 2*C))*Sin[c + d*x])/(3*b^4*d) + ((4*A*b^2 - 4*a*b*B + 4*a^2*C + 3*b^2*C)*Cos[
c + d*x]*Sin[c + d*x])/(8*b^3*d) + ((b*B - a*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*b^2*d) + (C*Cos[c + d*x]^3*Sin
[c + d*x])/(4*b*d)

Rule 3049

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

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 ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac{\int \frac{\cos ^2(c+d x) \left (3 a C+b (4 A+3 C) \cos (c+d x)+4 (b B-a C) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{4 b}\\ &=\frac{(b B-a C) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac{\int \frac{\cos (c+d x) \left (8 a (b B-a C)+b (8 b B+a C) \cos (c+d x)+3 \left (4 A b^2-4 a b B+4 a^2 C+3 b^2 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{12 b^2}\\ &=\frac{\left (4 A b^2-4 a b B+4 a^2 C+3 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac{(b B-a C) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac{\int \frac{3 a \left (4 A b^2-4 a b B+4 a^2 C+3 b^2 C\right )+b \left (12 A b^2+4 a b B-4 a^2 C+9 b^2 C\right ) \cos (c+d x)+8 \left (3 a^2 b B+2 b^3 B-3 a^3 C-a b^2 (3 A+2 C)\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{24 b^3}\\ &=\frac{\left (3 a^2 b B+2 b^3 B-3 a^3 C-a b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 d}+\frac{\left (4 A b^2-4 a b B+4 a^2 C+3 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac{(b B-a C) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 b d}+\frac{\int \frac{3 a b \left (4 A b^2-4 a b B+4 a^2 C+3 b^2 C\right )-3 \left (8 a^3 b B+4 a b^3 B-8 a^4 C-4 a^2 b^2 (2 A+C)-b^4 (4 A+3 C)\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{24 b^4}\\ &=-\frac{\left (8 a^3 b B+4 a b^3 B-8 a^4 C-4 a^2 b^2 (2 A+C)-b^4 (4 A+3 C)\right ) x}{8 b^5}+\frac{\left (3 a^2 b B+2 b^3 B-3 a^3 C-a b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 d}+\frac{\left (4 A b^2-4 a b B+4 a^2 C+3 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac{(b B-a C) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 b d}-\frac{\left (a^3 \left (A b^2-a (b B-a C)\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^5}\\ &=-\frac{\left (8 a^3 b B+4 a b^3 B-8 a^4 C-4 a^2 b^2 (2 A+C)-b^4 (4 A+3 C)\right ) x}{8 b^5}+\frac{\left (3 a^2 b B+2 b^3 B-3 a^3 C-a b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 d}+\frac{\left (4 A b^2-4 a b B+4 a^2 C+3 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac{(b B-a C) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 b d}-\frac{\left (2 a^3 \left (A b^2-a (b B-a 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^5 d}\\ &=-\frac{\left (8 a^3 b B+4 a b^3 B-8 a^4 C-4 a^2 b^2 (2 A+C)-b^4 (4 A+3 C)\right ) x}{8 b^5}-\frac{2 a^3 \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} b^5 \sqrt{a+b} d}+\frac{\left (3 a^2 b B+2 b^3 B-3 a^3 C-a b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 d}+\frac{\left (4 A b^2-4 a b B+4 a^2 C+3 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}+\frac{(b B-a C) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 b d}\\ \end{align*}

Mathematica [A]  time = 0.87444, size = 238, normalized size = 0.85 \[ \frac{12 (c+d x) \left (4 a^2 b^2 (2 A+C)-8 a^3 b B+8 a^4 C-4 a b^3 B+b^4 (4 A+3 C)\right )+24 b^2 \sin (2 (c+d x)) \left (a^2 C-a b B+A b^2+b^2 C\right )+24 b \sin (c+d x) \left (4 a^2 b B-4 a^3 C-a b^2 (4 A+3 C)+3 b^3 B\right )+\frac{192 a^3 \left (a (a C-b B)+A b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+8 b^3 (b B-a C) \sin (3 (c+d x))+3 b^4 C \sin (4 (c+d x))}{96 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(-8*a^3*b*B - 4*a*b^3*B + 8*a^4*C + 4*a^2*b^2*(2*A + C) + b^4*(4*A + 3*C))*(c + d*x) + (192*a^3*(A*b^2 + a
*(-(b*B) + a*C))*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + 24*b*(4*a^2*b*B + 3*
b^3*B - 4*a^3*C - a*b^2*(4*A + 3*C))*Sin[c + d*x] + 24*b^2*(A*b^2 - a*b*B + a^2*C + b^2*C)*Sin[2*(c + d*x)] +
8*b^3*(b*B - a*C)*Sin[3*(c + d*x)] + 3*b^4*C*Sin[4*(c + d*x)])/(96*b^5*d)

________________________________________________________________________________________

Maple [B]  time = 0.042, size = 1580, normalized size = 5.7 \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)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x)

[Out]

2/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^7*a^2*B+1/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1
/2*c)^7*a*B+1/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^5*a*B+6/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^4*tan
(1/2*d*x+1/2*c)^3*a^2*B-1/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^3*a*B+6/d/b^3/(tan(1/2*d*x+1/2*c
)^2+1)^4*tan(1/2*d*x+1/2*c)^5*a^2*B-1/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)*a*B+2/d/b^3/(tan(1/2
*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)*a^2*B+2/d*a^4/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-10/3/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^5*C*a-10/3/d/b^2/(tan(1/2*d*x+1/
2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^3*C*a-6/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^5*a*A-6/d/b^4/(tan(
1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^5*a^3*C-1/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^7*a^2*C
-2/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^7*C*a-6/d/b^4/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/
2*c)^3*a^3*C-2/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)*C*a-2/d*a^3/b^3/((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^2/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^3*a*A-2/
d/b^4/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^7*a^3*C+1/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2
*c)*a^2*C-2/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)*a*A-1/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2
*d*x+1/2*c)^5*a^2*C+1/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^3*a^2*C-2/d*a^5/b^5/((a+b)*(a-b))^(1
/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C-2/d/b^4/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*
c)*a^3*C-2/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^7*a*A+1/d/b*arctan(tan(1/2*d*x+1/2*c))*A+3/4/d/
b*arctan(tan(1/2*d*x+1/2*c))*C-1/d/b/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^5*A+1/d/b/(tan(1/2*d*x+1/2*
c)^2+1)^4*tan(1/2*d*x+1/2*c)^3*A-3/4/d/b/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^3*C+1/d/b/(tan(1/2*d*x+
1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)*A+5/4/d/b/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)*C-1/d/b/(tan(1/2*d*x+
1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^7*A-5/4/d/b/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^7*C+1/d/b^3*arctan(
tan(1/2*d*x+1/2*c))*a^2*C+2/d/b^5*arctan(tan(1/2*d*x+1/2*c))*a^4*C+3/4/d/b/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*
d*x+1/2*c)^5*C+2/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a^2*A-1/d/b^2*arctan(tan(1/2*d*x+1/2*c))*a*B+10/3/d/b/(tan(1
/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^5*B+2/d/b/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^7*B+10/3/d/b/(
tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)^3*B+2/d/b/(tan(1/2*d*x+1/2*c)^2+1)^4*tan(1/2*d*x+1/2*c)*B-2/d/b^4
*arctan(tan(1/2*d*x+1/2*c))*a^3*B

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.23369, size = 1674, normalized size = 6. \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)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

[1/24*(3*(8*C*a^6 - 8*B*a^5*b + 4*(2*A - C)*a^4*b^2 + 4*B*a^3*b^3 - (4*A + C)*a^2*b^4 + 4*B*a*b^5 - (4*A + 3*C
)*b^6)*d*x - 12*(C*a^5 - B*a^4*b + A*a^3*b^2)*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)) - (24*C*a^5*b - 24*B*a^4*b^2 + 8*(3*A - C)*a^3*b^3 + 8*B*a^2*b^4 - 8*(3*A + 2*C)*a*b^5 + 16*B
*b^6 - 6*(C*a^2*b^4 - C*b^6)*cos(d*x + c)^3 + 8*(C*a^3*b^3 - B*a^2*b^4 - C*a*b^5 + B*b^6)*cos(d*x + c)^2 - 3*(
4*C*a^4*b^2 - 4*B*a^3*b^3 + (4*A - C)*a^2*b^4 + 4*B*a*b^5 - (4*A + 3*C)*b^6)*cos(d*x + c))*sin(d*x + c))/((a^2
*b^5 - b^7)*d), 1/24*(3*(8*C*a^6 - 8*B*a^5*b + 4*(2*A - C)*a^4*b^2 + 4*B*a^3*b^3 - (4*A + C)*a^2*b^4 + 4*B*a*b
^5 - (4*A + 3*C)*b^6)*d*x - 24*(C*a^5 - B*a^4*b + A*a^3*b^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqr
t(a^2 - b^2)*sin(d*x + c))) - (24*C*a^5*b - 24*B*a^4*b^2 + 8*(3*A - C)*a^3*b^3 + 8*B*a^2*b^4 - 8*(3*A + 2*C)*a
*b^5 + 16*B*b^6 - 6*(C*a^2*b^4 - C*b^6)*cos(d*x + c)^3 + 8*(C*a^3*b^3 - B*a^2*b^4 - C*a*b^5 + B*b^6)*cos(d*x +
c)^2 - 3*(4*C*a^4*b^2 - 4*B*a^3*b^3 + (4*A - C)*a^2*b^4 + 4*B*a*b^5 - (4*A + 3*C)*b^6)*cos(d*x + c))*sin(d*x
+ c))/((a^2*b^5 - b^7)*d)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.22737, size = 1081, normalized size = 3.87 \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)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

1/24*(3*(8*C*a^4 - 8*B*a^3*b + 8*A*a^2*b^2 + 4*C*a^2*b^2 - 4*B*a*b^3 + 4*A*b^4 + 3*C*b^4)*(d*x + c)/b^5 + 48*(
C*a^5 - B*a^4*b + A*a^3*b^2)*(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)))/(sqrt(a^2 - b^2)*b^5) - 2*(24*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 24
*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 12*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 24*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 12*B
*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 24*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 12*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 24*B*b^3
*tan(1/2*d*x + 1/2*c)^7 + 15*C*b^3*tan(1/2*d*x + 1/2*c)^7 + 72*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 72*B*a^2*b*tan(1
/2*d*x + 1/2*c)^5 + 12*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 72*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 12*B*a*b^2*tan(1/2
*d*x + 1/2*c)^5 + 40*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 12*A*b^3*tan(1/2*d*x + 1/2*c)^5 - 40*B*b^3*tan(1/2*d*x +
1/2*c)^5 - 9*C*b^3*tan(1/2*d*x + 1/2*c)^5 + 72*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 72*B*a^2*b*tan(1/2*d*x + 1/2*c)
^3 - 12*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 72*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 12*B*a*b^2*tan(1/2*d*x + 1/2*c)^3
+ 40*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 40*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 9*C
*b^3*tan(1/2*d*x + 1/2*c)^3 + 24*C*a^3*tan(1/2*d*x + 1/2*c) - 24*B*a^2*b*tan(1/2*d*x + 1/2*c) - 12*C*a^2*b*tan
(1/2*d*x + 1/2*c) + 24*A*a*b^2*tan(1/2*d*x + 1/2*c) + 12*B*a*b^2*tan(1/2*d*x + 1/2*c) + 24*C*a*b^2*tan(1/2*d*x
+ 1/2*c) - 12*A*b^3*tan(1/2*d*x + 1/2*c) - 24*B*b^3*tan(1/2*d*x + 1/2*c) - 15*C*b^3*tan(1/2*d*x + 1/2*c))/((t
an(1/2*d*x + 1/2*c)^2 + 1)^4*b^4))/d