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

Optimal. Leaf size=233 \[ -\frac{a \left (3 a^2 C+3 A b^2+2 b^2 C\right ) \sin (c+d x)}{3 b^4 d}-\frac{2 a^3 \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d \sqrt{a-b} \sqrt{a+b}}+\frac{\left (4 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{8 b^3 d}+\frac{x \left (4 a^2 b^2 (2 A+C)+8 a^4 C+b^4 (4 A+3 C)\right )}{8 b^5}-\frac{a C \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d}+\frac{C \sin (c+d x) \cos ^3(c+d x)}{4 b d} \]

[Out]

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

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Rubi [A]  time = 0.786423, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3050, 3049, 3023, 2735, 2659, 205} \[ -\frac{a \left (3 a^2 C+3 A b^2+2 b^2 C\right ) \sin (c+d x)}{3 b^4 d}-\frac{2 a^3 \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d \sqrt{a-b} \sqrt{a+b}}+\frac{\left (4 a^2 C+b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{8 b^3 d}+\frac{x \left (4 a^2 b^2 (2 A+C)+8 a^4 C+b^4 (4 A+3 C)\right )}{8 b^5}-\frac{a C \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d}+\frac{C \sin (c+d x) \cos ^3(c+d x)}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((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^2*C)*ArcTan[(Sqrt[a - b]*Tan
[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^5*Sqrt[a + b]*d) - (a*(3*A*b^2 + 3*a^2*C + 2*b^2*C)*Sin[c + d*x])/
(3*b^4*d) + ((4*a^2*C + b^2*(4*A + 3*C))*Cos[c + d*x]*Sin[c + d*x])/(8*b^3*d) - (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 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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 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)} \, 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 a C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{4 b}\\ &=-\frac{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^2 C+a b C \cos (c+d x)+3 \left (4 a^2 C+b^2 (4 A+3 C)\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{12 b^2}\\ &=\frac{\left (4 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac{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^2 C+b^2 (4 A+3 C)\right )+b \left (12 A b^2-4 a^2 C+9 b^2 C\right ) \cos (c+d x)-8 a \left (3 A b^2+3 a^2 C+2 b^2 C\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{24 b^3}\\ &=-\frac{a \left (3 A b^2+3 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 b^4 d}+\frac{\left (4 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac{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^2 C+b^2 (4 A+3 C)\right )+3 \left (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^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) x}{8 b^5}-\frac{a \left (3 A b^2+3 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 b^4 d}+\frac{\left (4 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac{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^2 C\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^5}\\ &=\frac{\left (8 a^4 C+4 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) x}{8 b^5}-\frac{a \left (3 A b^2+3 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 b^4 d}+\frac{\left (4 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac{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^2 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^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^2 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{a \left (3 A b^2+3 a^2 C+2 b^2 C\right ) \sin (c+d x)}{3 b^4 d}+\frac{\left (4 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac{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 = 0.636615, size = 194, normalized size = 0.83 \[ \frac{12 (c+d x) \left (4 a^2 b^2 (2 A+C)+8 a^4 C+b^4 (4 A+3 C)\right )+24 b^2 \left (C \left (a^2+b^2\right )+A b^2\right ) \sin (2 (c+d x))-24 a b \left (4 a^2 C+4 A b^2+3 b^2 C\right ) \sin (c+d x)+\frac{192 a^3 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}-8 a b^3 C \sin (3 (c+d x))+3 b^4 C \sin (4 (c+d x))}{96 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.062, size = 1060, normalized size = 4.6 \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)),x)

[Out]

-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/(t
an(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/(ta
n(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

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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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.77419, size = 1312, normalized size = 5.63 \begin{align*} \left [\frac{3 \,{\left (8 \, C a^{6} + 4 \,{\left (2 \, A - C\right )} a^{4} b^{2} -{\left (4 \, A + C\right )} a^{2} b^{4} -{\left (4 \, A + 3 \, C\right )} b^{6}\right )} d x - 12 \,{\left (C a^{5} + A a^{3} b^{2}\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) -{\left (24 \, C a^{5} b + 8 \,{\left (3 \, A - C\right )} a^{3} b^{3} - 8 \,{\left (3 \, A + 2 \, C\right )} a b^{5} - 6 \,{\left (C a^{2} b^{4} - C b^{6}\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left (C a^{3} b^{3} - C a b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (4 \, C a^{4} b^{2} +{\left (4 \, A - C\right )} a^{2} b^{4} -{\left (4 \, A + 3 \, C\right )} b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \,{\left (a^{2} b^{5} - b^{7}\right )} d}, \frac{3 \,{\left (8 \, C a^{6} + 4 \,{\left (2 \, A - C\right )} a^{4} b^{2} -{\left (4 \, A + C\right )} a^{2} b^{4} -{\left (4 \, A + 3 \, C\right )} b^{6}\right )} d x - 24 \,{\left (C a^{5} + A a^{3} b^{2}\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) -{\left (24 \, C a^{5} b + 8 \,{\left (3 \, A - C\right )} a^{3} b^{3} - 8 \,{\left (3 \, A + 2 \, C\right )} a b^{5} - 6 \,{\left (C a^{2} b^{4} - C b^{6}\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left (C a^{3} b^{3} - C a b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (4 \, C a^{4} b^{2} +{\left (4 \, A - C\right )} a^{2} b^{4} -{\left (4 \, A + 3 \, C\right )} b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \,{\left (a^{2} b^{5} - b^{7}\right )} d}\right ] \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)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.33939, size = 775, normalized size = 3.33 \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)),x, algorithm="giac")

[Out]

1/24*(3*(8*C*a^4 + 8*A*a^2*b^2 + 4*C*a^2*b^2 + 4*A*b^4 + 3*C*b^4)*(d*x + c)/b^5 + 48*(C*a^5 + A*a^3*b^2)*(pi*f
loor(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 + 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 + 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 + 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 + 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 + 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 - 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 - 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 + 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 + 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) - 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) + 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) - 15*C*b^3*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x +
1/2*c)^2 + 1)^4*b^4))/d