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

Optimal. Leaf size=177 \[ \frac{\left (3 a^2 C+b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b^3 d}+\frac{2 a^2 \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{a x \left (C \left (2 a^2+b^2\right )+2 A b^2\right )}{2 b^4}-\frac{a C \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac{C \sin (c+d x) \cos ^2(c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.47393, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{\left (3 a^2 C+b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b^3 d}+\frac{2 a^2 \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{a x \left (C \left (2 a^2+b^2\right )+2 A b^2\right )}{2 b^4}-\frac{a C \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac{C \sin (c+d x) \cos ^2(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(2*A*b^2 + (2*a^2 + b^2)*C)*x)/(2*b^4) + (2*a^2*(A*b^2 + a^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt
[a + b]])/(Sqrt[a - b]*b^4*Sqrt[a + b]*d) + ((3*a^2*C + b^2*(3*A + 2*C))*Sin[c + d*x])/(3*b^3*d) - (a*C*Cos[c
+ d*x]*Sin[c + d*x])/(2*b^2*d) + (C*Cos[c + d*x]^2*Sin[c + d*x])/(3*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 ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=\frac{C \cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac{\int \frac{\cos (c+d x) \left (2 a C+b (3 A+2 C) \cos (c+d x)-3 a C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{3 b}\\ &=-\frac{a C \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{C \cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac{\int \frac{-3 a^2 C+a b C \cos (c+d x)+2 \left (3 a^2 C+b^2 (3 A+2 C)\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^2}\\ &=\frac{\left (3 a^2 C+b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b^3 d}-\frac{a C \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{C \cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac{\int \frac{-3 a^2 b C-3 a \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^3}\\ &=-\frac{a \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) x}{2 b^4}+\frac{\left (3 a^2 C+b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b^3 d}-\frac{a C \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{C \cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac{\left (a^2 \left (A b^2+a^2 C\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^4}\\ &=-\frac{a \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) x}{2 b^4}+\frac{\left (3 a^2 C+b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b^3 d}-\frac{a C \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{C \cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac{\left (2 a^2 \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^4 d}\\ &=-\frac{a \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) x}{2 b^4}+\frac{2 a^2 \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^4 \sqrt{a+b} d}+\frac{\left (3 a^2 C+b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b^3 d}-\frac{a C \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{C \cos ^2(c+d x) \sin (c+d x)}{3 b d}\\ \end{align*}

Mathematica [A]  time = 0.435716, size = 152, normalized size = 0.86 \[ \frac{-6 a (c+d x) \left (C \left (2 a^2+b^2\right )+2 A b^2\right )+3 b \left (4 a^2 C+4 A b^2+3 b^2 C\right ) \sin (c+d x)-\frac{24 a^2 \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}-3 a b^2 C \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.033, size = 551, 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)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x)

[Out]

2/d/b/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^5*A+2/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^
5*a^2*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*C*a+2/d/b/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d
*x+1/2*c)^5*C+4/d/b/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^3*A+4/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1
/2*d*x+1/2*c)^3*a^2*C+4/3/d/b/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^3*C+2/d/b/(tan(1/2*d*x+1/2*c)^2+1)
^3*tan(1/2*d*x+1/2*c)*A+2/d/b^3/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)*a^2*C+2/d/b/(tan(1/2*d*x+1/2*c)^
2+1)^3*tan(1/2*d*x+1/2*c)*C-1/d/b^2/(tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)*C*a-2/d/b^2*A*arctan(tan(1/2
*d*x+1/2*c))*a-2/d/b^4*C*arctan(tan(1/2*d*x+1/2*c))*a^3-1/d/b^2*C*arctan(tan(1/2*d*x+1/2*c))*a+2/d*a^2/b^2/((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^4/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

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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)^2*(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 = 1.68917, size = 1050, normalized size = 5.93 \begin{align*} \left [-\frac{3 \,{\left (2 \, C a^{5} +{\left (2 \, A - C\right )} a^{3} b^{2} -{\left (2 \, A + C\right )} a b^{4}\right )} d x + 3 \,{\left (C a^{4} + A a^{2} 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 (6 \, C a^{4} b + 2 \,{\left (3 \, A - C\right )} a^{2} b^{3} - 2 \,{\left (3 \, A + 2 \, C\right )} b^{5} + 2 \,{\left (C a^{2} b^{3} - C b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (C a^{3} b^{2} - C a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} b^{4} - b^{6}\right )} d}, -\frac{3 \,{\left (2 \, C a^{5} +{\left (2 \, A - C\right )} a^{3} b^{2} -{\left (2 \, A + C\right )} a b^{4}\right )} d x - 6 \,{\left (C a^{4} + A a^{2} 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 (6 \, C a^{4} b + 2 \,{\left (3 \, A - C\right )} a^{2} b^{3} - 2 \,{\left (3 \, A + 2 \, C\right )} b^{5} + 2 \,{\left (C a^{2} b^{3} - C b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (C a^{3} b^{2} - C a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} b^{4} - b^{6}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

[-1/6*(3*(2*C*a^5 + (2*A - C)*a^3*b^2 - (2*A + C)*a*b^4)*d*x + 3*(C*a^4 + A*a^2*b^2)*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)) - (6*C*a^4*b + 2*(3*A - C)*a^2*b^3 - 2*(3*A + 2*C)*b^5
+ 2*(C*a^2*b^3 - C*b^5)*cos(d*x + c)^2 - 3*(C*a^3*b^2 - C*a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)
*d), -1/6*(3*(2*C*a^5 + (2*A - C)*a^3*b^2 - (2*A + C)*a*b^4)*d*x - 6*(C*a^4 + A*a^2*b^2)*sqrt(a^2 - b^2)*arcta
n(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (6*C*a^4*b + 2*(3*A - C)*a^2*b^3 - 2*(3*A + 2*C)*b^5
+ 2*(C*a^2*b^3 - C*b^5)*cos(d*x + c)^2 - 3*(C*a^3*b^2 - C*a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)
*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)**2*(A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.279, size = 440, normalized size = 2.49 \begin{align*} -\frac{\frac{3 \,{\left (2 \, C a^{3} + 2 \, A a b^{2} + C a b^{2}\right )}{\left (d x + c\right )}}{b^{4}} + \frac{12 \,{\left (C a^{4} + A a^{2} b^{2}\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 )}}{\sqrt{a^{2} - b^{2}} b^{4}} - \frac{2 \,{\left (6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(3*(2*C*a^3 + 2*A*a*b^2 + C*a*b^2)*(d*x + c)/b^4 + 12*(C*a^4 + A*a^2*b^2)*(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)))/(sqrt(a^2 - b
^2)*b^4) - 2*(6*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*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
+ 6*C*b^2*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^2*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 + 6*C*a^2*tan(1/2*d*x + 1/2*c) - 3*C*a*b*tan(1/2*d*x + 1/2*c) + 6*A*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^3))/d