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

Optimal. Leaf size=372 \[ -\frac{a \left (a^2 b^2 (2 A-21 C)+12 a^4 C-b^4 (5 A-6 C)\right ) \sin (c+d x)}{2 b^4 d \left (a^2-b^2\right )^2}-\frac{a \left (a^4 b^2 (2 A-29 C)-5 a^2 b^4 (A-4 C)+12 a^6 C+6 A b^6\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{\left (7 a^2 b^2 C-4 a^4 C+3 A b^4\right ) \sin (c+d x) \cos ^2(c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{\left (a^2 b^2 (A-10 C)+6 a^4 C-b^4 (4 A-C)\right ) \sin (c+d x) \cos (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2}+\frac{x \left (C \left (12 a^2+b^2\right )+2 A b^2\right )}{2 b^5} \]

[Out]

((2*A*b^2 + (12*a^2 + b^2)*C)*x)/(2*b^5) - (a*(6*A*b^6 + a^4*b^2*(2*A - 29*C) - 5*a^2*b^4*(A - 4*C) + 12*a^6*C
)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^5*(a + b)^(5/2)*d) - (a*(a^2*b^2*(2*A -
21*C) - b^4*(5*A - 6*C) + 12*a^4*C)*Sin[c + d*x])/(2*b^4*(a^2 - b^2)^2*d) + ((a^2*b^2*(A - 10*C) - b^4*(4*A -
C) + 6*a^4*C)*Cos[c + d*x]*Sin[c + d*x])/(2*b^3*(a^2 - b^2)^2*d) - ((A*b^2 + a^2*C)*Cos[c + d*x]^3*Sin[c + d*
x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + ((3*A*b^4 - 4*a^4*C + 7*a^2*b^2*C)*Cos[c + d*x]^2*Sin[c + d*x
])/(2*b^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.48927, antiderivative size = 372, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {3048, 3047, 3049, 3023, 2735, 2659, 205} \[ -\frac{a \left (a^2 b^2 (2 A-21 C)+12 a^4 C-b^4 (5 A-6 C)\right ) \sin (c+d x)}{2 b^4 d \left (a^2-b^2\right )^2}-\frac{a \left (a^4 b^2 (2 A-29 C)-5 a^2 b^4 (A-4 C)+12 a^6 C+6 A b^6\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{\left (7 a^2 b^2 C-4 a^4 C+3 A b^4\right ) \sin (c+d x) \cos ^2(c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{\left (a^2 b^2 (A-10 C)+6 a^4 C-b^4 (4 A-C)\right ) \sin (c+d x) \cos (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2}+\frac{x \left (C \left (12 a^2+b^2\right )+2 A b^2\right )}{2 b^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A*b^2 + (12*a^2 + b^2)*C)*x)/(2*b^5) - (a*(6*A*b^6 + a^4*b^2*(2*A - 29*C) - 5*a^2*b^4*(A - 4*C) + 12*a^6*C
)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^5*(a + b)^(5/2)*d) - (a*(a^2*b^2*(2*A -
21*C) - b^4*(5*A - 6*C) + 12*a^4*C)*Sin[c + d*x])/(2*b^4*(a^2 - b^2)^2*d) + ((a^2*b^2*(A - 10*C) - b^4*(4*A -
C) + 6*a^4*C)*Cos[c + d*x]*Sin[c + d*x])/(2*b^3*(a^2 - b^2)^2*d) - ((A*b^2 + a^2*C)*Cos[c + d*x]^3*Sin[c + d*
x])/(2*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + ((3*A*b^4 - 4*a^4*C + 7*a^2*b^2*C)*Cos[c + d*x]^2*Sin[c + d*x
])/(2*b^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))

Rule 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + 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*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(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, 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 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+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^3} \, dx &=-\frac{\left (A b^2+a^2 C\right ) \cos ^3(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\int \frac{\cos ^2(c+d x) \left (3 \left (A b^2+a^2 C\right )-2 a b (A+C) \cos (c+d x)-2 \left (A b^2+2 a^2 C-b^2 C\right ) \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2+a^2 C\right ) \cos ^3(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{\cos (c+d x) \left (2 \left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right )-a b \left (3 A b^2-\left (a^2-4 b^2\right ) C\right ) \cos (c+d x)+2 \left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^2+a^2 C\right ) \cos ^3(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{2 a \left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right )-2 b \left (2 a^4 C-b^4 (2 A+C)-a^2 b^2 (A+4 C)\right ) \cos (c+d x)-2 a \left (a^2 b^2 (2 A-21 C)-b^4 (5 A-6 C)+12 a^4 C\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{4 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{a \left (a^2 b^2 (2 A-21 C)-b^4 (5 A-6 C)+12 a^4 C\right ) \sin (c+d x)}{2 b^4 \left (a^2-b^2\right )^2 d}+\frac{\left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^2+a^2 C\right ) \cos ^3(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{2 a b \left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right )+2 \left (a^2-b^2\right )^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{4 b^4 \left (a^2-b^2\right )^2}\\ &=\frac{\left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) x}{2 b^5}-\frac{a \left (a^2 b^2 (2 A-21 C)-b^4 (5 A-6 C)+12 a^4 C\right ) \sin (c+d x)}{2 b^4 \left (a^2-b^2\right )^2 d}+\frac{\left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^2+a^2 C\right ) \cos ^3(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (a \left (6 A b^6+a^4 b^2 (2 A-29 C)-5 a^2 b^4 (A-4 C)+12 a^6 C\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 b^5 \left (a^2-b^2\right )^2}\\ &=\frac{\left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) x}{2 b^5}-\frac{a \left (a^2 b^2 (2 A-21 C)-b^4 (5 A-6 C)+12 a^4 C\right ) \sin (c+d x)}{2 b^4 \left (a^2-b^2\right )^2 d}+\frac{\left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^2+a^2 C\right ) \cos ^3(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (a \left (6 A b^6+a^4 b^2 (2 A-29 C)-5 a^2 b^4 (A-4 C)+12 a^6 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 )^2 d}\\ &=\frac{\left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) x}{2 b^5}-\frac{a \left (2 a^4 A b^2-5 a^2 A b^4+6 A b^6+12 a^6 C-29 a^4 b^2 C+20 a^2 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^5 (a+b)^{5/2} d}-\frac{a \left (a^2 b^2 (2 A-21 C)-b^4 (5 A-6 C)+12 a^4 C\right ) \sin (c+d x)}{2 b^4 \left (a^2-b^2\right )^2 d}+\frac{\left (a^2 b^2 (A-10 C)-b^4 (4 A-C)+6 a^4 C\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d}-\frac{\left (A b^2+a^2 C\right ) \cos ^3(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\left (3 A b^4-4 a^4 C+7 a^2 b^2 C\right ) \cos ^2(c+d x) \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 = 2.39855, size = 256, normalized size = 0.69 \[ \frac{2 (c+d x) \left (C \left (12 a^2+b^2\right )+2 A b^2\right )+\frac{2 a^3 b \left (a^2 C+A b^2\right ) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+\frac{2 a^2 b \left (a^2 b^2 (10 C-3 A)-7 a^4 C+6 A b^4\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}+\frac{4 a \left (a^4 b^2 (2 A-29 C)-5 a^2 b^4 (A-4 C)+12 a^6 C+6 A b^6\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}-12 a b C \sin (c+d x)+b^2 C \sin (2 (c+d x))}{4 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(2*A*b^2 + (12*a^2 + b^2)*C)*(c + d*x) + (4*a*(6*A*b^6 + a^4*b^2*(2*A - 29*C) - 5*a^2*b^4*(A - 4*C) + 12*a^
6*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) - 12*a*b*C*Sin[c + d*x] + (2*a^3
*b*(A*b^2 + a^2*C)*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) + (2*a^2*b*(6*A*b^4 - 7*a^4*C + a^2*
b^2*(-3*A + 10*C))*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])) + b^2*C*Sin[2*(c + d*x)])/(4*b^5*d
)

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Maple [B]  time = 0.046, size = 1428, normalized size = 3.8 \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+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x)

[Out]

-6/d/b^4/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*c)^3*a*C-6/d/b^4/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/
2*c)*a*C+2/d/b^3*arctan(tan(1/2*d*x+1/2*c))*A+1/d/b^3*arctan(tan(1/2*d*x+1/2*c))*C-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)*A-1/d*a^3/b/(a*tan(1/2*d*x+1/2*c)^2-ta
n(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^5/b^3/(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+10/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+10/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^3/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*c)*C+12/d/b
^5*arctan(tan(1/2*d*x+1/2*c))*a^2*C-1/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*c)^3*C-6/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-12/d*a^7/b^5/(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+29/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-20/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*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+6/d*a^2/(a*tan(1/2*d*x+1/2*c)^2-ta
n(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/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))*A+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))*A-6/d*a^6/b^4/(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*a^3/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^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A-6/d*a^6/b^4/(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-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*A+1/d*a^5/b^3/(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

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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+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 = 2.99712, size = 3391, normalized size = 9.12 \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+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(2*(12*C*a^8*b^2 + (2*A - 35*C)*a^6*b^4 - 3*(2*A - 11*C)*a^4*b^6 + 3*(2*A - 3*C)*a^2*b^8 - (2*A + C)*b^10
)*d*x*cos(d*x + c)^2 + 4*(12*C*a^9*b + (2*A - 35*C)*a^7*b^3 - 3*(2*A - 11*C)*a^5*b^5 + 3*(2*A - 3*C)*a^3*b^7 -
(2*A + C)*a*b^9)*d*x*cos(d*x + c) + 2*(12*C*a^10 + (2*A - 35*C)*a^8*b^2 - 3*(2*A - 11*C)*a^6*b^4 + 3*(2*A - 3
*C)*a^4*b^6 - (2*A + C)*a^2*b^8)*d*x - (12*C*a^9 + (2*A - 29*C)*a^7*b^2 - 5*(A - 4*C)*a^5*b^4 + 6*A*a^3*b^6 +
(12*C*a^7*b^2 + (2*A - 29*C)*a^5*b^4 - 5*(A - 4*C)*a^3*b^6 + 6*A*a*b^8)*cos(d*x + c)^2 + 2*(12*C*a^8*b + (2*A
- 29*C)*a^6*b^3 - 5*(A - 4*C)*a^4*b^5 + 6*A*a^2*b^7)*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*(12*C*a^9*b + (2*A - 33*C)*a^7*b^3 - (7*A - 27*C)*a^5*b^5 + (5*A - 6
*C)*a^3*b^7 - (C*a^6*b^4 - 3*C*a^4*b^6 + 3*C*a^2*b^8 - C*b^10)*cos(d*x + c)^3 + 4*(C*a^7*b^3 - 3*C*a^5*b^5 + 3
*C*a^3*b^7 - C*a*b^9)*cos(d*x + c)^2 + (18*C*a^8*b^2 + (3*A - 50*C)*a^6*b^4 - (9*A - 43*C)*a^4*b^6 + (6*A - 11
*C)*a^2*b^8)*cos(d*x + c))*sin(d*x + c))/((a^6*b^7 - 3*a^4*b^9 + 3*a^2*b^11 - b^13)*d*cos(d*x + c)^2 + 2*(a^7*
b^6 - 3*a^5*b^8 + 3*a^3*b^10 - a*b^12)*d*cos(d*x + c) + (a^8*b^5 - 3*a^6*b^7 + 3*a^4*b^9 - a^2*b^11)*d), 1/2*(
(12*C*a^8*b^2 + (2*A - 35*C)*a^6*b^4 - 3*(2*A - 11*C)*a^4*b^6 + 3*(2*A - 3*C)*a^2*b^8 - (2*A + C)*b^10)*d*x*co
s(d*x + c)^2 + 2*(12*C*a^9*b + (2*A - 35*C)*a^7*b^3 - 3*(2*A - 11*C)*a^5*b^5 + 3*(2*A - 3*C)*a^3*b^7 - (2*A +
C)*a*b^9)*d*x*cos(d*x + c) + (12*C*a^10 + (2*A - 35*C)*a^8*b^2 - 3*(2*A - 11*C)*a^6*b^4 + 3*(2*A - 3*C)*a^4*b^
6 - (2*A + C)*a^2*b^8)*d*x - (12*C*a^9 + (2*A - 29*C)*a^7*b^2 - 5*(A - 4*C)*a^5*b^4 + 6*A*a^3*b^6 + (12*C*a^7*
b^2 + (2*A - 29*C)*a^5*b^4 - 5*(A - 4*C)*a^3*b^6 + 6*A*a*b^8)*cos(d*x + c)^2 + 2*(12*C*a^8*b + (2*A - 29*C)*a^
6*b^3 - 5*(A - 4*C)*a^4*b^5 + 6*A*a^2*b^7)*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))) - (12*C*a^9*b + (2*A - 33*C)*a^7*b^3 - (7*A - 27*C)*a^5*b^5 + (5*A - 6*C)*a^3*b^7 - (C
*a^6*b^4 - 3*C*a^4*b^6 + 3*C*a^2*b^8 - C*b^10)*cos(d*x + c)^3 + 4*(C*a^7*b^3 - 3*C*a^5*b^5 + 3*C*a^3*b^7 - C*a
*b^9)*cos(d*x + c)^2 + (18*C*a^8*b^2 + (3*A - 50*C)*a^6*b^4 - (9*A - 43*C)*a^4*b^6 + (6*A - 11*C)*a^2*b^8)*cos
(d*x + c))*sin(d*x + c))/((a^6*b^7 - 3*a^4*b^9 + 3*a^2*b^11 - b^13)*d*cos(d*x + c)^2 + 2*(a^7*b^6 - 3*a^5*b^8
+ 3*a^3*b^10 - a*b^12)*d*cos(d*x + c) + (a^8*b^5 - 3*a^6*b^7 + 3*a^4*b^9 - a^2*b^11)*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+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.72114, size = 1544, normalized size = 4.15 \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+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(2*(12*C*a^7 + 2*A*a^5*b^2 - 29*C*a^5*b^2 - 5*A*a^3*b^4 + 20*C*a^3*b^4 + 6*A*a*b^6)*(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^5 - 2*a^2*b^7 + b^9)*sqrt(a^2 - b^2)) - 2*(12*C*a^7*tan(1/2*d*x + 1/2*c)^7 - 18*C*a^6*b*tan(1/2*d*x + 1/2*
c)^7 + 2*A*a^5*b^2*tan(1/2*d*x + 1/2*c)^7 - 17*C*a^5*b^2*tan(1/2*d*x + 1/2*c)^7 - 3*A*a^4*b^3*tan(1/2*d*x + 1/
2*c)^7 + 33*C*a^4*b^3*tan(1/2*d*x + 1/2*c)^7 - 5*A*a^3*b^4*tan(1/2*d*x + 1/2*c)^7 - 2*C*a^3*b^4*tan(1/2*d*x +
1/2*c)^7 + 6*A*a^2*b^5*tan(1/2*d*x + 1/2*c)^7 - 13*C*a^2*b^5*tan(1/2*d*x + 1/2*c)^7 + 4*C*a*b^6*tan(1/2*d*x +
1/2*c)^7 + C*b^7*tan(1/2*d*x + 1/2*c)^7 + 36*C*a^7*tan(1/2*d*x + 1/2*c)^5 - 18*C*a^6*b*tan(1/2*d*x + 1/2*c)^5
+ 6*A*a^5*b^2*tan(1/2*d*x + 1/2*c)^5 - 67*C*a^5*b^2*tan(1/2*d*x + 1/2*c)^5 - 3*A*a^4*b^3*tan(1/2*d*x + 1/2*c)^
5 + 29*C*a^4*b^3*tan(1/2*d*x + 1/2*c)^5 - 15*A*a^3*b^4*tan(1/2*d*x + 1/2*c)^5 + 26*C*a^3*b^4*tan(1/2*d*x + 1/2
*c)^5 + 6*A*a^2*b^5*tan(1/2*d*x + 1/2*c)^5 - 5*C*a^2*b^5*tan(1/2*d*x + 1/2*c)^5 - 4*C*a*b^6*tan(1/2*d*x + 1/2*
c)^5 - 3*C*b^7*tan(1/2*d*x + 1/2*c)^5 + 36*C*a^7*tan(1/2*d*x + 1/2*c)^3 + 18*C*a^6*b*tan(1/2*d*x + 1/2*c)^3 +
6*A*a^5*b^2*tan(1/2*d*x + 1/2*c)^3 - 67*C*a^5*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*A*a^4*b^3*tan(1/2*d*x + 1/2*c)^3
- 29*C*a^4*b^3*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^3*b^4*tan(1/2*d*x + 1/2*c)^3 + 26*C*a^3*b^4*tan(1/2*d*x + 1/2*c
)^3 - 6*A*a^2*b^5*tan(1/2*d*x + 1/2*c)^3 + 5*C*a^2*b^5*tan(1/2*d*x + 1/2*c)^3 - 4*C*a*b^6*tan(1/2*d*x + 1/2*c)
^3 + 3*C*b^7*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^7*tan(1/2*d*x + 1/2*c) + 18*C*a^6*b*tan(1/2*d*x + 1/2*c) + 2*A*a^
5*b^2*tan(1/2*d*x + 1/2*c) - 17*C*a^5*b^2*tan(1/2*d*x + 1/2*c) + 3*A*a^4*b^3*tan(1/2*d*x + 1/2*c) - 33*C*a^4*b
^3*tan(1/2*d*x + 1/2*c) - 5*A*a^3*b^4*tan(1/2*d*x + 1/2*c) - 2*C*a^3*b^4*tan(1/2*d*x + 1/2*c) - 6*A*a^2*b^5*ta
n(1/2*d*x + 1/2*c) + 13*C*a^2*b^5*tan(1/2*d*x + 1/2*c) + 4*C*a*b^6*tan(1/2*d*x + 1/2*c) - C*b^7*tan(1/2*d*x +
1/2*c))/((a^4*b^4 - 2*a^2*b^6 + b^8)*(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 + 2*a*tan(1/2*d*x +
1/2*c)^2 + a + b)^2) + (12*C*a^2 + 2*A*b^2 + C*b^2)*(d*x + c)/b^5)/d