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

Optimal. Leaf size=398 \[ -\frac{\sin (c+d x) \left (-a^2 b^2 (6 A-7 C)+9 a^3 b B-12 a^4 C-6 a b^3 B+b^4 (3 A+2 C)\right )}{3 b^4 d \left (a^2-b^2\right )}+\frac{2 a^2 \left (2 a^2 A b^2-5 a^2 b^2 C-3 a^3 b B+4 a^4 C+4 a b^3 B-3 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\sin (c+d x) \cos ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}+\frac{\sin (c+d x) \cos (c+d x) \left (3 a^2 b B-4 a^3 C-2 a b^2 (A-C)-b^3 B\right )}{2 b^3 d \left (a^2-b^2\right )}+\frac{x \left (6 a^2 b B-8 a^3 C-2 a b^2 (2 A+C)+b^3 B\right )}{2 b^5} \]

[Out]

((6*a^2*b*B + b^3*B - 8*a^3*C - 2*a*b^2*(2*A + C))*x)/(2*b^5) + (2*a^2*(2*a^2*A*b^2 - 3*A*b^4 - 3*a^3*b*B + 4*
a*b^3*B + 4*a^4*C - 5*a^2*b^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*b^5*(a + b
)^(3/2)*d) - ((9*a^3*b*B - 6*a*b^3*B - a^2*b^2*(6*A - 7*C) - 12*a^4*C + b^4*(3*A + 2*C))*Sin[c + d*x])/(3*b^4*
(a^2 - b^2)*d) + ((3*a^2*b*B - b^3*B - 2*a*b^2*(A - C) - 4*a^3*C)*Cos[c + d*x]*Sin[c + d*x])/(2*b^3*(a^2 - b^2
)*d) + ((3*A*b^2 - 3*a*b*B + 4*a^2*C - b^2*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d) - ((A*b^2 - a
*(b*B - a*C))*Cos[c + d*x]^3*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

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Rubi [A]  time = 1.59703, antiderivative size = 398, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {3047, 3049, 3023, 2735, 2659, 205} \[ -\frac{\sin (c+d x) \left (-a^2 b^2 (6 A-7 C)+9 a^3 b B-12 a^4 C-6 a b^3 B+b^4 (3 A+2 C)\right )}{3 b^4 d \left (a^2-b^2\right )}+\frac{2 a^2 \left (2 a^2 A b^2-5 a^2 b^2 C-3 a^3 b B+4 a^4 C+4 a b^3 B-3 A b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\sin (c+d x) \cos ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{\sin (c+d x) \cos ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}+\frac{\sin (c+d x) \cos (c+d x) \left (3 a^2 b B-4 a^3 C-2 a b^2 (A-C)-b^3 B\right )}{2 b^3 d \left (a^2-b^2\right )}+\frac{x \left (6 a^2 b B-8 a^3 C-2 a b^2 (2 A+C)+b^3 B\right )}{2 b^5} \]

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])^2,x]

[Out]

((6*a^2*b*B + b^3*B - 8*a^3*C - 2*a*b^2*(2*A + C))*x)/(2*b^5) + (2*a^2*(2*a^2*A*b^2 - 3*A*b^4 - 3*a^3*b*B + 4*
a*b^3*B + 4*a^4*C - 5*a^2*b^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*b^5*(a + b
)^(3/2)*d) - ((9*a^3*b*B - 6*a*b^3*B - a^2*b^2*(6*A - 7*C) - 12*a^4*C + b^4*(3*A + 2*C))*Sin[c + d*x])/(3*b^4*
(a^2 - b^2)*d) + ((3*a^2*b*B - b^3*B - 2*a*b^2*(A - C) - 4*a^3*C)*Cos[c + d*x]*Sin[c + d*x])/(2*b^3*(a^2 - b^2
)*d) + ((3*A*b^2 - 3*a*b*B + 4*a^2*C - b^2*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d) - ((A*b^2 - a
*(b*B - a*C))*Cos[c + d*x]^3*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(
c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c
*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)
))*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] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

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))^2} \, dx &=-\frac{\left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{\cos ^2(c+d x) \left (3 \left (A b^2-a (b B-a C)\right )+b (b B-a (A+C)) \cos (c+d x)-\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (-2 a \left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right )+b \left (3 A b^2-3 a b B+a^2 C+2 b^2 C\right ) \cos (c+d x)-3 \left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=\frac{\left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{-3 a \left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right )+b \left (3 a^2 b B+3 b^3 B-4 a^3 C-2 a b^2 (3 A+C)\right ) \cos (c+d x)+2 \left (9 a^3 b B-6 a b^3 B-a^2 b^2 (6 A-7 C)-12 a^4 C+b^4 (3 A+2 C)\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (9 a^3 b B-6 a b^3 B-a^2 b^2 (6 A-7 C)-12 a^4 C+b^4 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}+\frac{\left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{-3 a b \left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right )-3 \left (a^2-b^2\right ) \left (6 a^2 b B+b^3 B-8 a^3 C-2 a b^2 (2 A+C)\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^4 \left (a^2-b^2\right )}\\ &=\frac{\left (6 a^2 b B+b^3 B-8 a^3 C-2 a b^2 (2 A+C)\right ) x}{2 b^5}-\frac{\left (9 a^3 b B-6 a b^3 B-a^2 b^2 (6 A-7 C)-12 a^4 C+b^4 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}+\frac{\left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (a^2 \left (3 A b^4+3 a^3 b B-4 a b^3 B-a^2 b^2 (2 A-5 C)-4 a^4 C\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^5 \left (a^2-b^2\right )}\\ &=\frac{\left (6 a^2 b B+b^3 B-8 a^3 C-2 a b^2 (2 A+C)\right ) x}{2 b^5}-\frac{\left (9 a^3 b B-6 a b^3 B-a^2 b^2 (6 A-7 C)-12 a^4 C+b^4 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}+\frac{\left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\left (2 a^2 \left (3 A b^4+3 a^3 b B-4 a b^3 B-a^2 b^2 (2 A-5 C)-4 a^4 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 \left (a^2-b^2\right ) d}\\ &=\frac{\left (6 a^2 b B+b^3 B-8 a^3 C-2 a b^2 (2 A+C)\right ) x}{2 b^5}+\frac{2 a^2 \left (2 a^2 A b^2-3 A b^4-3 a^3 b B+4 a b^3 B+4 a^4 C-5 a^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)^{3/2} b^5 (a+b)^{3/2} d}-\frac{\left (9 a^3 b B-6 a b^3 B-a^2 b^2 (6 A-7 C)-12 a^4 C+b^4 (3 A+2 C)\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}+\frac{\left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \cos ^3(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.74533, size = 256, normalized size = 0.64 \[ \frac{6 (c+d x) \left (6 a^2 b B-8 a^3 C-2 a b^2 (2 A+C)+b^3 B\right )+3 b \sin (c+d x) \left (12 a^2 C-8 a b B+4 A b^2+3 b^2 C\right )+\frac{12 a^3 b \sin (c+d x) \left (a (a C-b B)+A b^2\right )}{(a-b) (a+b) (a+b \cos (c+d x))}+\frac{24 a^2 \left (a^2 b^2 (2 A-5 C)-3 a^3 b B+4 a^4 C+4 a b^3 B-3 A b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+3 b^2 (b B-2 a C) \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 b^5 d} \]

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])^2,x]

[Out]

(6*(6*a^2*b*B + b^3*B - 8*a^3*C - 2*a*b^2*(2*A + C))*(c + d*x) + (24*a^2*(-3*A*b^4 - 3*a^3*b*B + 4*a*b^3*B + a
^2*b^2*(2*A - 5*C) + 4*a^4*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) + 3*b*(
4*A*b^2 - 8*a*b*B + 12*a^2*C + 3*b^2*C)*Sin[c + d*x] + (12*a^3*b*(A*b^2 + a*(-(b*B) + a*C))*Sin[c + d*x])/((a
- b)*(a + b)*(a + b*Cos[c + d*x])) + 3*b^2*(b*B - 2*a*C)*Sin[2*(c + d*x)] + b^3*C*Sin[3*(c + d*x)])/(12*b^5*d)

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Maple [B]  time = 0.047, size = 1229, normalized size = 3.1 \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))^2,x)

[Out]

8/d*a^3/b^2/(a+b)/(a-b)/((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/d*a^4/b^
3/(a+b)/(a-b)/((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^3/(tan(1/2*d*x+
1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)*C*a+6/d/b^4/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^5*a^2*C+2/d/b^3/(ta
n(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^5*C*a+12/d/b^4/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^3*a^2*
C-8/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^3*a*B-4/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1
/2*c)^5*a*B+6/d/b^4/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)*a^2*C-4/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^3*tan
(1/2*d*x+1/2*c)*a*B+1/d/b^2*B*arctan(tan(1/2*d*x+1/2*c))+4/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)
^3*A+1/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)*B-4/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a*A+4/3/d/b^2/
(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^3*C+2/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)*A+2/d/
b^2/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)*C+2/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^5*A-
8/d/b^5*arctan(tan(1/2*d*x+1/2*c))*a^3*C+6/d/b^4*arctan(tan(1/2*d*x+1/2*c))*a^2*B+2/d/b^2/(tan(1/2*d*x+1/2*c)^
2+1)^3*tan(1/2*d*x+1/2*c)^5*C-1/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^5*B-2/d/b^3*C*arctan(tan(1
/2*d*x+1/2*c))*a+8/d*a^6/b^5/(a+b)/(a-b)/((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*a^2/b/(a+b)/(a-b)/((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*a^5
/b^4/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B+4/d*a^4/b^3/(a+b)/
(a-b)/((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^4/(a^2-b^2)*tan(1/2
*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)*C+2/d*a^3/b^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(a*
tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)*A-2/d*a^4/b^3/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2
*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)*B

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.88757, size = 2986, normalized size = 7.5 \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))^2,x, algorithm="fricas")

[Out]

[-1/6*(3*(8*C*a^7*b - 6*B*a^6*b^2 + 2*(2*A - 7*C)*a^5*b^3 + 11*B*a^4*b^4 - 4*(2*A - C)*a^3*b^5 - 4*B*a^2*b^6 +
2*(2*A + C)*a*b^7 - B*b^8)*d*x*cos(d*x + c) + 3*(8*C*a^8 - 6*B*a^7*b + 2*(2*A - 7*C)*a^6*b^2 + 11*B*a^5*b^3 -
4*(2*A - C)*a^4*b^4 - 4*B*a^3*b^5 + 2*(2*A + C)*a^2*b^6 - B*a*b^7)*d*x + 3*(4*C*a^7 - 3*B*a^6*b + (2*A - 5*C)
*a^5*b^2 + 4*B*a^4*b^3 - 3*A*a^3*b^4 + (4*C*a^6*b - 3*B*a^5*b^2 + (2*A - 5*C)*a^4*b^3 + 4*B*a^3*b^4 - 3*A*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*cos(d*x + c) + a^2)) - (24*C*a^
7*b - 18*B*a^6*b^2 + 2*(6*A - 19*C)*a^5*b^3 + 30*B*a^4*b^4 - 2*(9*A - 5*C)*a^3*b^5 - 12*B*a^2*b^6 + 2*(3*A + 2
*C)*a*b^7 + 2*(C*a^4*b^4 - 2*C*a^2*b^6 + C*b^8)*cos(d*x + c)^3 - (4*C*a^5*b^3 - 3*B*a^4*b^4 - 8*C*a^3*b^5 + 6*
B*a^2*b^6 + 4*C*a*b^7 - 3*B*b^8)*cos(d*x + c)^2 + (12*C*a^6*b^2 - 9*B*a^5*b^3 + 2*(3*A - 10*C)*a^4*b^4 + 18*B*
a^3*b^5 - 4*(3*A - C)*a^2*b^6 - 9*B*a*b^7 + 2*(3*A + 2*C)*b^8)*cos(d*x + c))*sin(d*x + c))/((a^4*b^6 - 2*a^2*b
^8 + b^10)*d*cos(d*x + c) + (a^5*b^5 - 2*a^3*b^7 + a*b^9)*d), -1/6*(3*(8*C*a^7*b - 6*B*a^6*b^2 + 2*(2*A - 7*C)
*a^5*b^3 + 11*B*a^4*b^4 - 4*(2*A - C)*a^3*b^5 - 4*B*a^2*b^6 + 2*(2*A + C)*a*b^7 - B*b^8)*d*x*cos(d*x + c) + 3*
(8*C*a^8 - 6*B*a^7*b + 2*(2*A - 7*C)*a^6*b^2 + 11*B*a^5*b^3 - 4*(2*A - C)*a^4*b^4 - 4*B*a^3*b^5 + 2*(2*A + C)*
a^2*b^6 - B*a*b^7)*d*x - 6*(4*C*a^7 - 3*B*a^6*b + (2*A - 5*C)*a^5*b^2 + 4*B*a^4*b^3 - 3*A*a^3*b^4 + (4*C*a^6*b
- 3*B*a^5*b^2 + (2*A - 5*C)*a^4*b^3 + 4*B*a^3*b^4 - 3*A*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))) - (24*C*a^7*b - 18*B*a^6*b^2 + 2*(6*A - 19*C)*a^5*b^3 + 30*B*a^
4*b^4 - 2*(9*A - 5*C)*a^3*b^5 - 12*B*a^2*b^6 + 2*(3*A + 2*C)*a*b^7 + 2*(C*a^4*b^4 - 2*C*a^2*b^6 + C*b^8)*cos(d
*x + c)^3 - (4*C*a^5*b^3 - 3*B*a^4*b^4 - 8*C*a^3*b^5 + 6*B*a^2*b^6 + 4*C*a*b^7 - 3*B*b^8)*cos(d*x + c)^2 + (12
*C*a^6*b^2 - 9*B*a^5*b^3 + 2*(3*A - 10*C)*a^4*b^4 + 18*B*a^3*b^5 - 4*(3*A - C)*a^2*b^6 - 9*B*a*b^7 + 2*(3*A +
2*C)*b^8)*cos(d*x + c))*sin(d*x + c))/((a^4*b^6 - 2*a^2*b^8 + b^10)*d*cos(d*x + c) + (a^5*b^5 - 2*a^3*b^7 + a*
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)**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.24514, size = 760, normalized size = 1.91 \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))^2,x, algorithm="giac")

[Out]

-1/6*(12*(4*C*a^6 - 3*B*a^5*b + 2*A*a^4*b^2 - 5*C*a^4*b^2 + 4*B*a^3*b^3 - 3*A*a^2*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^2
*b^5 - b^7)*sqrt(a^2 - b^2)) - 12*(C*a^5*tan(1/2*d*x + 1/2*c) - B*a^4*b*tan(1/2*d*x + 1/2*c) + A*a^3*b^2*tan(1
/2*d*x + 1/2*c))/((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*(8*C*a^3
- 6*B*a^2*b + 4*A*a*b^2 + 2*C*a*b^2 - B*b^3)*(d*x + c)/b^5 - 2*(18*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 12*B*a*b*tan
(1/2*d*x + 1/2*c)^5 + 6*C*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^2*tan(1/2*d*x +
1/2*c)^5 + 6*C*b^2*tan(1/2*d*x + 1/2*c)^5 + 36*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 24*B*a*b*tan(1/2*d*x + 1/2*c)^3
+ 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 4*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 18*C*a^2*tan(1/2*d*x + 1/2*c) - 12*B*a*b*
tan(1/2*d*x + 1/2*c) - 6*C*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/2*d*x + 1/2*c) + 3*B*b^2*tan(1/2*d*x + 1/2
*c) + 6*C*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^4))/d