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

Optimal. Leaf size=155 \[ -\frac{2 a \left (a^2 A b-2 a^3 B+3 a b^2 B-2 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{x (A b-2 a B)}{b^3}+\frac{B \sin (c+d x)}{b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.440118, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2988, 3023, 2735, 2659, 205} \[ -\frac{2 a \left (a^2 A b-2 a^3 B+3 a b^2 B-2 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{x (A b-2 a B)}{b^3}+\frac{B \sin (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2988

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

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

Mathematica [A]  time = 0.781552, size = 147, normalized size = 0.95 \[ \frac{\frac{2 a \left (-a^2 A b+2 a^3 B-3 a b^2 B+2 A b^3\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac{a^2 b (a B-A b) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+(c+d x) (A b-2 a B)+b B \sin (c+d x)}{b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.125, size = 445, normalized size = 2.9 \begin{align*} 2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{{b}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+2\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{{b}^{2}d}}-4\,{\frac{B\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) a}{d{b}^{3}}}-2\,{\frac{{a}^{2}\tan \left ( 1/2\,dx+c/2 \right ) A}{bd \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}+2\,{\frac{{a}^{3}\tan \left ( 1/2\,dx+c/2 \right ) B}{{b}^{2}d \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}-2\,{\frac{A{a}^{3}}{{b}^{2}d \left ( a-b \right ) \left ( a+b \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) }+4\,{\frac{aA}{d \left ( a-b \right ) \left ( a+b \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) }+4\,{\frac{{a}^{4}B}{d{b}^{3} \left ( a-b \right ) \left ( a+b \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) }-6\,{\frac{B{a}^{2}}{bd \left ( a-b \right ) \left ( a+b \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/b^2*B*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+2/d/b^2*A*arctan(tan(1/2*d*x+1/2*c))-4/d/b^3*B*arctan(ta
n(1/2*d*x+1/2*c))*a-2/d*a^2/b/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)
*A+2/d*a^3/b^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)*B-2/d*a^3/b^2/
(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A+4/d*a/(a-b)/(a+b)/((a-b
)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A+4/d*a^4/b^3/(a-b)/(a+b)/((a-b)*(a+b))^(1
/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B-6/d*a^2/b/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctan(ta
n(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*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)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.43307, size = 1701, normalized size = 10.97 \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)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/2*(2*(2*B*a^5*b - A*a^4*b^2 - 4*B*a^3*b^3 + 2*A*a^2*b^4 + 2*B*a*b^5 - A*b^6)*d*x*cos(d*x + c) + 2*(2*B*a^6
- A*a^5*b - 4*B*a^4*b^2 + 2*A*a^3*b^3 + 2*B*a^2*b^4 - A*a*b^5)*d*x + (2*B*a^5 - A*a^4*b - 3*B*a^3*b^2 + 2*A*a
^2*b^3 + (2*B*a^4*b - A*a^3*b^2 - 3*B*a^2*b^3 + 2*A*a*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*B*a^5*b - A*a^4*b^2 - 3*B*a^3*b^3 + A*a^2*b^4 + B*a*b^5 + (
B*a^4*b^2 - 2*B*a^2*b^4 + B*b^6)*cos(d*x + c))*sin(d*x + c))/((a^4*b^4 - 2*a^2*b^6 + b^8)*d*cos(d*x + c) + (a^
5*b^3 - 2*a^3*b^5 + a*b^7)*d), -((2*B*a^5*b - A*a^4*b^2 - 4*B*a^3*b^3 + 2*A*a^2*b^4 + 2*B*a*b^5 - A*b^6)*d*x*c
os(d*x + c) + (2*B*a^6 - A*a^5*b - 4*B*a^4*b^2 + 2*A*a^3*b^3 + 2*B*a^2*b^4 - A*a*b^5)*d*x - (2*B*a^5 - A*a^4*b
- 3*B*a^3*b^2 + 2*A*a^2*b^3 + (2*B*a^4*b - A*a^3*b^2 - 3*B*a^2*b^3 + 2*A*a*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*B*a^5*b - A*a^4*b^2 - 3*B*a^3*b^3 + A*a^2*b
^4 + B*a*b^5 + (B*a^4*b^2 - 2*B*a^2*b^4 + B*b^6)*cos(d*x + c))*sin(d*x + c))/((a^4*b^4 - 2*a^2*b^6 + b^8)*d*co
s(d*x + c) + (a^5*b^3 - 2*a^3*b^5 + a*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)**2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.30676, size = 502, normalized size = 3.24 \begin{align*} -\frac{\frac{2 \,{\left (2 \, B a^{4} - A a^{3} b - 3 \, B a^{2} b^{2} + 2 \, A a 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^{2} b^{3} - b^{5}\right )} \sqrt{a^{2} - b^{2}}} - \frac{2 \,{\left (2 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b\right )}{\left (a^{2} b^{2} - b^{4}\right )}} + \frac{{\left (2 \, B a - A b\right )}{\left (d x + c\right )}}{b^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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