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

Optimal. Leaf size=261 \[ -\frac {b \left (a^2 (4 A+3 C)+b^2 (A+2 C)\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}+\frac {a \left (2 a^4 C+a^2 b^2 (2 A-5 C)+b^4 (13 A+18 C)\right ) \sin (c+d x)}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))}+\frac {\left (-4 a^4 C+a^2 b^2 (2 A+9 C)+3 A b^4\right ) \sin (c+d x)}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2} \]

[Out]

-b*(b^2*(A+2*C)+a^2*(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+C*a^2)*sin(d*x+c)/b^2/(a^2-b^2)/d/(a+b*cos(d*x+c))^3+1/6*(3*A*b^4-4*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*(a^2*b^2*(2*A-5*C)+2*a^4*C+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 = 0.57, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3032, 3021, 2754, 12, 2659, 205} \[ -\frac {b \left (a^2 (4 A+3 C)+b^2 (A+2 C)\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \sin (c+d x)}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))}+\frac {\left (a^2 b^2 (2 A+9 C)-4 a^4 C+3 A b^4\right ) \sin (c+d x)}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^4,x]

[Out]

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

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))^4} \, dx &=\frac {a \left (A b^2+a^2 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^2 C\right )-a \left (2 A b^2-\left (a^2-3 b^2\right ) 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^2 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^4-4 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 a b^2 \left (a^2 C-b^2 (5 A+6 C)\right )+b \left (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^2 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^4-4 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 {a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+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^4 \left (b^2 (A+2 C)+a^2 (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^2 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^4-4 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 {a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+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 (b \left (b^2 (A+2 C)+a^2 (4 A+3 C)\right )\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^2 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^4-4 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 {a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+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 (b \left (b^2 (A+2 C)+a^2 (4 A+3 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 )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac {b \left (4 a^2 A+A b^2+3 a^2 C+2 b^2 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^2 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^4-4 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 {a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+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 = 1.22, size = 224, normalized size = 0.86 \[ \frac {\frac {24 b \left (a^2 (4 A+3 C)+b^2 (A+2 C)\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+\frac {2 \sin (c+d x) \left (6 b \left (a^4 (2 A+C)+9 a^2 b^2 (A+C)-A b^4\right ) \cos (c+d x)+a \left (2 a^4 (6 A+5 C)+a^2 b^2 (22 A+17 C)+\left (2 a^4 C+a^2 b^2 (2 A-5 C)+b^4 (13 A+18 C)\right ) \cos (2 (c+d x))+b^4 (11 A+18 C)\right )\right )}{(a+b \cos (c+d x))^3}}{24 d \left (a^2-b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^4,x]

[Out]

((24*b*(b^2*(A + 2*C) + a^2*(4*A + 3*C))*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2
] + (2*(6*b*(-(A*b^4) + 9*a^2*b^2*(A + C) + a^4*(2*A + C))*Cos[c + d*x] + a*(2*a^4*(6*A + 5*C) + a^2*b^2*(22*A
+ 17*C) + b^4*(11*A + 18*C) + (a^2*b^2*(2*A - 5*C) + 2*a^4*C + 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 = 0.63, size = 1103, normalized size = 4.23 \[ \text {result too large to display} \]

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))^4,x, algorithm="fricas")

[Out]

[1/12*(3*((4*A + 3*C)*a^5*b + (A + 2*C)*a^3*b^3 + ((4*A + 3*C)*a^2*b^4 + (A + 2*C)*b^6)*cos(d*x + c)^3 + 3*((4
*A + 3*C)*a^3*b^3 + (A + 2*C)*a*b^5)*cos(d*x + c)^2 + 3*((4*A + 3*C)*a^4*b^2 + (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 + (4
*A + 7*C)*a^5*b^2 - 11*(A + C)*a^3*b^4 + A*a*b^6 + (2*C*a^7 + (2*A - 7*C)*a^5*b^2 + (11*A + 23*C)*a^3*b^4 - (1
3*A + 18*C)*a*b^6)*cos(d*x + c)^2 + 3*((2*A + C)*a^6*b + (7*A + 8*C)*a^4*b^3 - (10*A + 9*C)*a^2*b^5 + A*b^7)*c
os(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*((4*A + 3*C)*a^
5*b + (A + 2*C)*a^3*b^3 + ((4*A + 3*C)*a^2*b^4 + (A + 2*C)*b^6)*cos(d*x + c)^3 + 3*((4*A + 3*C)*a^3*b^3 + (A +
2*C)*a*b^5)*cos(d*x + c)^2 + 3*((4*A + 3*C)*a^4*b^2 + (A + 2*C)*a^2*b^4)*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*(3*A + 2*C)*a^7 + (4*A + 7*C)*a^5*b^2 - 11*(A + C)
*a^3*b^4 + A*a*b^6 + (2*C*a^7 + (2*A - 7*C)*a^5*b^2 + (11*A + 23*C)*a^3*b^4 - (13*A + 18*C)*a*b^6)*cos(d*x + c
)^2 + 3*((2*A + C)*a^6*b + (7*A + 8*C)*a^4*b^3 - (10*A + 9*C)*a^2*b^5 + 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 = 0.99, size = 689, normalized size = 2.64 \[ \frac {\frac {3 \, {\left (4 \, A a^{2} b + 3 \, C a^{2} b + A b^{3} + 2 \, C b^{3}\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^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {6 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, C a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 32 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 28 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, C a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{4} b \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 ) + 12 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, C a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, 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 )}^{3}}}{3 \, d} \]

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))^4,x, algorithm="giac")

[Out]

1/3*(3*(4*A*a^2*b + 3*C*a^2*b + 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 + 6*C*a^5*tan(1/2*d*x + 1/2*c)^5 - 6*A*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 + 6*C*a^3*b^2*tan(1/2*d*x + 1/2*c)
^5 - 27*A*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 + 18*C*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*A*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 + 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 - 2
8*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 + 6*A*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) + 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) + 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) + 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) + 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))/
((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 = 0.12, size = 1727, normalized size = 6.62 \[ \text {result too large to display} \]

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))^4,x)

[Out]

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+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+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+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+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+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+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+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+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+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/(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*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^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+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-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+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-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+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^2-4/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^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/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^2-2/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))*C

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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+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 = 4.35, size = 491, normalized size = 1.88 \[ \frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A\,a^3+C\,a^3+7\,A\,a\,b^2+9\,C\,a\,b^2\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,A\,a^3+A\,b^3+2\,C\,a^3+6\,A\,a\,b^2+2\,A\,a^2\,b+6\,C\,a\,b^2+3\,C\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,a^3-A\,b^3+2\,C\,a^3+6\,A\,a\,b^2-2\,A\,a^2\,b+6\,C\,a\,b^2-3\,C\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}-\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\,\left (4\,A\,a^2+A\,b^2+3\,C\,a^2+2\,C\,b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}\,\left (A\,b^3+2\,C\,b^3+4\,A\,a^2\,b+3\,C\,a^2\,b\right )}\right )\,\left (4\,A\,a^2+A\,b^2+3\,C\,a^2+2\,C\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((4*tan(c/2 + (d*x)/2)^3*(3*A*a^3 + C*a^3 + 7*A*a*b^2 + 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 + 2*C*a^3 + 6*A*a*b^2 + 2*A*a^2*b + 6*C*a*b^2 + 3*C*a^2*b))/((a + b)^3*(a - b
)) + (tan(c/2 + (d*x)/2)*(2*A*a^3 - A*b^3 + 2*C*a^3 + 6*A*a*b^2 - 2*A*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)))/(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) - ta
n(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))) - (b*atan((b*tan(c/2 + (d*x)/2)*(2*a - 2*b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)*(4*A*a^2 +
A*b^2 + 3*C*a^2 + 2*C*b^2))/(2*(a + b)^(1/2)*(a - b)^(7/2)*(A*b^3 + 2*C*b^3 + 4*A*a^2*b + 3*C*a^2*b)))*(4*A*a
^2 + A*b^2 + 3*C*a^2 + 2*C*b^2))/(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+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**4,x)

[Out]

Timed out

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