3.1004 \(\int \frac {\cos (c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^4} \, dx\)

Optimal. Leaf size=314 \[ \frac {a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac {\left (a^3 B-a^2 b (4 A+3 C)+4 a b^2 B-b^3 (A+2 C)\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {\sin (c+d x) \left (-4 a^4 C+a^3 b B+a^2 b^2 (2 A+9 C)-6 a b^3 B+3 A b^4\right )}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\sin (c+d x) \left (2 a^5 C+a^4 b B+a^3 b^2 (2 A-5 C)-10 a^2 b^3 B+a b^4 (13 A+18 C)-6 b^5 B\right )}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))} \]

[Out]

(a^3*B+4*a*b^2*B-b^3*(A+2*C)-a^2*b*(4*A+3*C))*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(7/2)/(
a+b)^(7/2)/d+1/3*a*(A*b^2-a*(B*b-C*a))*sin(d*x+c)/b^2/(a^2-b^2)/d/(a+b*cos(d*x+c))^3+1/6*(3*A*b^4+a^3*b*B-6*a*
b^3*B-4*a^4*C+a^2*b^2*(2*A+9*C))*sin(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^2+1/6*(a^4*b*B-10*a^2*b^3*B-6*b
^5*B+a^3*b^2*(2*A-5*C)+2*a^5*C+a*b^4*(13*A+18*C))*sin(d*x+c)/b^2/(a^2-b^2)^3/d/(a+b*cos(d*x+c))

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Rubi [A]  time = 0.92, antiderivative size = 314, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3031, 3021, 2754, 12, 2659, 205} \[ \frac {\left (-a^2 b (4 A+3 C)+a^3 B+4 a b^2 B-b^3 (A+2 C)\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {\sin (c+d x) \left (a^3 b^2 (2 A-5 C)-10 a^2 b^3 B+a^4 b B+2 a^5 C+a b^4 (13 A+18 C)-6 b^5 B\right )}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))}+\frac {\sin (c+d x) \left (a^2 b^2 (2 A+9 C)+a^3 b B-4 a^4 C-6 a b^3 B+3 A b^4\right )}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {a \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3} \]

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])^4,x]

[Out]

((a^3*B + 4*a*b^2*B - b^3*(A + 2*C) - a^2*b*(4*A + 3*C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(
(a - b)^(7/2)*(a + b)^(7/2)*d) + (a*(A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Cos[c +
d*x])^3) + ((3*A*b^4 + a^3*b*B - 6*a*b^3*B - 4*a^4*C + a^2*b^2*(2*A + 9*C))*Sin[c + d*x])/(6*b^2*(a^2 - b^2)^2
*d*(a + b*Cos[c + d*x])^2) + ((a^4*b*B - 10*a^2*b^3*B - 6*b^5*B + a^3*b^2*(2*A - 5*C) + 2*a^5*C + a*b^4*(13*A
+ 18*C))*Sin[c + d*x])/(6*b^2*(a^2 - b^2)^3*d*(a + b*Cos[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

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 2754

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

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 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]

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))^4} \, dx &=\frac {a \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\int \frac {-3 b \left (A b^2-a (b B-a C)\right )+\left (a^2 b B-3 b^3 B-a^3 C+a b^2 (2 A+3 C)\right ) \cos (c+d x)+3 b \left (a^2-b^2\right ) C \cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=\frac {a \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac {\int \frac {-2 b^2 \left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right )-b \left (a^3 b B-6 a b^3 B+a^2 b^2 (2 A-3 C)+2 a^4 C+3 b^4 (A+2 C)\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{6 b^3 \left (a^2-b^2\right )^2}\\ &=\frac {a \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac {\int \frac {3 b^3 \left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 C)\right )}{a+b \cos (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )^3}\\ &=\frac {a \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac {\left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 C)\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=\frac {a \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}+\frac {\left (a^3 B+4 a b^2 B-b^3 (A+2 C)-a^2 b (4 A+3 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 )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac {\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B+3 a^2 b C+2 b^3 C\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac {a \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\left (3 A b^4+a^3 b B-6 a b^3 B-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4 b B-10 a^2 b^3 B-6 b^5 B+a^3 b^2 (2 A-5 C)+2 a^5 C+a b^4 (13 A+18 C)\right ) \sin (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.73, size = 307, normalized size = 0.98 \[ \frac {\frac {2 \sin (c+d x) \left (12 a^5 A+10 a^5 C-25 a^4 b B+22 a^3 A b^2+17 a^3 b^2 C-14 a^2 b^3 B+6 \cos (c+d x) \left (a^5 B+a^4 b (2 A+C)-9 a^3 b^2 B+9 a^2 b^3 (A+C)-2 a b^4 B-A b^5\right )+\cos (2 (c+d x)) \left (2 a^5 C+a^4 b B+a^3 b^2 (2 A-5 C)-10 a^2 b^3 B+a b^4 (13 A+18 C)-6 b^5 B\right )+11 a A b^4+18 a b^4 C-6 b^5 B\right )}{(a+b \cos (c+d x))^3}-\frac {24 \left (a^3 B-a^2 b (4 A+3 C)+4 a b^2 B-b^3 (A+2 C)\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}}{24 d \left (a^2-b^2\right )^3} \]

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])^4,x]

[Out]

((-24*(a^3*B + 4*a*b^2*B - b^3*(A + 2*C) - a^2*b*(4*A + 3*C))*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b
^2]])/Sqrt[-a^2 + b^2] + (2*(12*a^5*A + 22*a^3*A*b^2 + 11*a*A*b^4 - 25*a^4*b*B - 14*a^2*b^3*B - 6*b^5*B + 10*a
^5*C + 17*a^3*b^2*C + 18*a*b^4*C + 6*(-(A*b^5) + a^5*B - 9*a^3*b^2*B - 2*a*b^4*B + 9*a^2*b^3*(A + C) + a^4*b*(
2*A + C))*Cos[c + d*x] + (a^4*b*B - 10*a^2*b^3*B - 6*b^5*B + a^3*b^2*(2*A - 5*C) + 2*a^5*C + a*b^4*(13*A + 18*
C))*Cos[2*(c + d*x)])*Sin[c + d*x])/(a + b*Cos[c + d*x])^3)/(24*(a^2 - b^2)^3*d)

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fricas [B]  time = 1.05, size = 1410, normalized size = 4.49 \[ \text {result too large to display} \]

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))^4,x, algorithm="fricas")

[Out]

[-1/12*(3*(B*a^6 - (4*A + 3*C)*a^5*b + 4*B*a^4*b^2 - (A + 2*C)*a^3*b^3 + (B*a^3*b^3 - (4*A + 3*C)*a^2*b^4 + 4*
B*a*b^5 - (A + 2*C)*b^6)*cos(d*x + c)^3 + 3*(B*a^4*b^2 - (4*A + 3*C)*a^3*b^3 + 4*B*a^2*b^4 - (A + 2*C)*a*b^5)*
cos(d*x + c)^2 + 3*(B*a^5*b - (4*A + 3*C)*a^4*b^2 + 4*B*a^3*b^3 - (A + 2*C)*a^2*b^4)*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*(3*A + 2*C)*a^7 - 13*B*a^6*b + (4
*A + 7*C)*a^5*b^2 + 11*B*a^4*b^3 - 11*(A + C)*a^3*b^4 + 2*B*a^2*b^5 + A*a*b^6 + (2*C*a^7 + B*a^6*b + (2*A - 7*
C)*a^5*b^2 - 11*B*a^4*b^3 + (11*A + 23*C)*a^3*b^4 + 4*B*a^2*b^5 - (13*A + 18*C)*a*b^6 + 6*B*b^7)*cos(d*x + c)^
2 + 3*(B*a^7 + (2*A + C)*a^6*b - 10*B*a^5*b^2 + (7*A + 8*C)*a^4*b^3 + 7*B*a^3*b^4 - (10*A + 9*C)*a^2*b^5 + 2*B
*a*b^6 + A*b^7)*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*d*cos(d*x +
c)^3 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c)^2 + 3*(a^10*b - 4*a^8*b^3 + 6*a
^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c) + (a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*d), 1/6*(3
*(B*a^6 - (4*A + 3*C)*a^5*b + 4*B*a^4*b^2 - (A + 2*C)*a^3*b^3 + (B*a^3*b^3 - (4*A + 3*C)*a^2*b^4 + 4*B*a*b^5 -
(A + 2*C)*b^6)*cos(d*x + c)^3 + 3*(B*a^4*b^2 - (4*A + 3*C)*a^3*b^3 + 4*B*a^2*b^4 - (A + 2*C)*a*b^5)*cos(d*x +
c)^2 + 3*(B*a^5*b - (4*A + 3*C)*a^4*b^2 + 4*B*a^3*b^3 - (A + 2*C)*a^2*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arct
an(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) + (2*(3*A + 2*C)*a^7 - 13*B*a^6*b + (4*A + 7*C)*a^5*b
^2 + 11*B*a^4*b^3 - 11*(A + C)*a^3*b^4 + 2*B*a^2*b^5 + A*a*b^6 + (2*C*a^7 + B*a^6*b + (2*A - 7*C)*a^5*b^2 - 11
*B*a^4*b^3 + (11*A + 23*C)*a^3*b^4 + 4*B*a^2*b^5 - (13*A + 18*C)*a*b^6 + 6*B*b^7)*cos(d*x + c)^2 + 3*(B*a^7 +
(2*A + C)*a^6*b - 10*B*a^5*b^2 + (7*A + 8*C)*a^4*b^3 + 7*B*a^3*b^4 - (10*A + 9*C)*a^2*b^5 + 2*B*a*b^6 + A*b^7)
*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*d*cos(d*x + c)^3 + 3*(a^9*b
^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c)^2 + 3*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*
b^7 + a^2*b^9)*d*cos(d*x + c) + (a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*d)]

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giac [B]  time = 0.33, size = 966, normalized size = 3.08 \[ \text {result too large to display} \]

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))^4,x, algorithm="giac")

[Out]

-1/3*(3*(B*a^3 - 4*A*a^2*b - 3*C*a^2*b + 4*B*a*b^2 - A*b^3 - 2*C*b^3)*(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^6 - 3*a^4*b^2 + 3*
a^2*b^4 - b^6)*sqrt(a^2 - b^2)) - (6*A*a^5*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^5*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^5*t
an(1/2*d*x + 1/2*c)^5 - 6*A*a^4*b*tan(1/2*d*x + 1/2*c)^5 - 12*B*a^4*b*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^4*b*tan(1
/2*d*x + 1/2*c)^5 + 12*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 27*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^3*b^2*ta
n(1/2*d*x + 1/2*c)^5 - 27*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 12*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 27*C*a^2*b^
3*tan(1/2*d*x + 1/2*c)^5 + 12*A*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*B*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 18*C*a*b^4*t
an(1/2*d*x + 1/2*c)^5 + 3*A*b^5*tan(1/2*d*x + 1/2*c)^5 - 6*B*b^5*tan(1/2*d*x + 1/2*c)^5 + 12*A*a^5*tan(1/2*d*x
+ 1/2*c)^3 + 4*C*a^5*tan(1/2*d*x + 1/2*c)^3 - 28*B*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 16*A*a^3*b^2*tan(1/2*d*x +
1/2*c)^3 + 32*C*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 16*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 28*A*a*b^4*tan(1/2*d*x
+ 1/2*c)^3 - 36*C*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 12*B*b^5*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^5*tan(1/2*d*x + 1/2*c
) + 3*B*a^5*tan(1/2*d*x + 1/2*c) + 6*C*a^5*tan(1/2*d*x + 1/2*c) + 6*A*a^4*b*tan(1/2*d*x + 1/2*c) - 12*B*a^4*b*
tan(1/2*d*x + 1/2*c) + 3*C*a^4*b*tan(1/2*d*x + 1/2*c) + 12*A*a^3*b^2*tan(1/2*d*x + 1/2*c) - 27*B*a^3*b^2*tan(1
/2*d*x + 1/2*c) + 6*C*a^3*b^2*tan(1/2*d*x + 1/2*c) + 27*A*a^2*b^3*tan(1/2*d*x + 1/2*c) - 12*B*a^2*b^3*tan(1/2*
d*x + 1/2*c) + 27*C*a^2*b^3*tan(1/2*d*x + 1/2*c) + 12*A*a*b^4*tan(1/2*d*x + 1/2*c) - 6*B*a*b^4*tan(1/2*d*x + 1
/2*c) + 18*C*a*b^4*tan(1/2*d*x + 1/2*c) - 3*A*b^5*tan(1/2*d*x + 1/2*c) - 6*B*b^5*tan(1/2*d*x + 1/2*c))/((a^6 -
3*a^4*b^2 + 3*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)^3))/d

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maple [B]  time = 0.14, size = 2667, normalized size = 8.49 \[ \text {Expression too large to display} \]

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))^4,x)

[Out]

6/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*C
*a*b^2+6/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1
/2*c)^5*C*a*b^2+12/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a^2-2*a*b+b^2)/(a^2+2*a*b+b^2)*tan
(1/2*d*x+1/2*c)^3*C*a*b^2-4/d*b/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a
-b)/((a-b)*(a+b))^(1/2))*a^2*A+1/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)^3/(a-b)/(a^3+3*a^2*
b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*A+2/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a^3/(a-b)/(a^3
+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*C-2/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(
a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*b^3*B+6/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
)^3*a/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*A-6/d*b/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^
2*b+a+b)^3*a^2/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*B+6/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)^3*a/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*A-6/d*b/(a*tan(1/2*d*x+1/2*c)^2-tan(1
/2*d*x+1/2*c)^2*b+a+b)^3*a^2/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*B+28/3/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)^3*a/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A-28/3/d*b/(a*ta
n(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a^2/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*B+2/
d*a^2*b/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*
c)^5*A-2/d*a^2*b/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*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)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*ta
n(1/2*d*x+1/2*c)*a*B-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)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+
b^3)*tan(1/2*d*x+1/2*c)^5*a*B-3/d*b/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a+b)/(a^3-3*a^2*b+3
*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*C*a^2+3/d*b/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(a^3+3*
a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*C*a^2+1/d*a^3/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a
+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*B-2/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(
a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*b^3*B-4/d/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b
)^3/(a^2-2*a*b+b^2)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*b^3*B-1/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)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*A-3/d*b/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)
*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*C*a^2+4/d*b^2/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)
/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*a*B+2/d/(a*tan(1/2*d*x+1/2*c)^2-tan(
1/2*d*x+1/2*c)^2*b+a+b)^3*a^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*C+4/d/(a*tan(1/2*d*x+1/2*c)^2
-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a^3/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A+4/3/d/(a*tan(1/2*d*x
+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3*a^3/(a^2+2*a*b+b^2)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C+2/d*a^3/(a*
tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*d*x+1/2*c)*A+2/d*a^
3/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5*A
-1/d*a^3/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)^3/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2
*c)^5*B-2/d/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1
/2))*b^3*C+1/d/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))
^(1/2))*a^3*B-1/d*b^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)
*(a+b))^(1/2))*A

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 4.84, size = 516, normalized size = 1.64 \[ \frac {\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,a^3-A\,b^3+B\,a^3-2\,B\,b^3+2\,C\,a^3+6\,A\,a\,b^2-2\,A\,a^2\,b+2\,B\,a\,b^2-6\,B\,a^2\,b+6\,C\,a\,b^2-3\,C\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A\,a^3-3\,B\,b^3+C\,a^3+7\,A\,a\,b^2-7\,B\,a^2\,b+9\,C\,a\,b^2\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,A\,a^3+A\,b^3-B\,a^3-2\,B\,b^3+2\,C\,a^3+6\,A\,a\,b^2+2\,A\,a^2\,b-2\,B\,a\,b^2-6\,B\,a^2\,b+6\,C\,a\,b^2+3\,C\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}-\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (A\,b^3-B\,a^3+2\,C\,b^3+4\,A\,a^2\,b-4\,B\,a\,b^2+3\,C\,a^2\,b\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^4,x)

[Out]

((tan(c/2 + (d*x)/2)*(2*A*a^3 - A*b^3 + B*a^3 - 2*B*b^3 + 2*C*a^3 + 6*A*a*b^2 - 2*A*a^2*b + 2*B*a*b^2 - 6*B*a^
2*b + 6*C*a*b^2 - 3*C*a^2*b))/((a + b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (4*tan(c/2 + (d*x)/2)^3*(3*A*a^3 - 3
*B*b^3 + C*a^3 + 7*A*a*b^2 - 7*B*a^2*b + 9*C*a*b^2))/(3*(a + b)^2*(a^2 - 2*a*b + b^2)) + (tan(c/2 + (d*x)/2)^5
*(2*A*a^3 + A*b^3 - B*a^3 - 2*B*b^3 + 2*C*a^3 + 6*A*a*b^2 + 2*A*a^2*b - 2*B*a*b^2 - 6*B*a^2*b + 6*C*a*b^2 + 3*
C*a^2*b))/((a + b)^3*(a - b)))/(d*(3*a*b^2 - tan(c/2 + (d*x)/2)^4*(3*a*b^2 + 3*a^2*b - 3*a^3 - 3*b^3) - tan(c/
2 + (d*x)/2)^2*(3*a*b^2 - 3*a^2*b - 3*a^3 + 3*b^3) + 3*a^2*b + a^3 + b^3 + tan(c/2 + (d*x)/2)^6*(3*a*b^2 - 3*a
^2*b + a^3 - b^3))) - (atan((tan(c/2 + (d*x)/2)*(2*a - 2*b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3))/(2*(a + b)^(1/2)*
(a - b)^(7/2)))*(A*b^3 - B*a^3 + 2*C*b^3 + 4*A*a^2*b - 4*B*a*b^2 + 3*C*a^2*b))/(d*(a + b)^(7/2)*(a - b)^(7/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**4,x)

[Out]

Timed out

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