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

[Out]

(A*b-2*B*a)*x/b^3-2*a*(A*a^2*b-2*A*b^3-2*B*a^3+3*B*a*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(
a-b)^(3/2)/b^3/(a+b)^(3/2)/d+B*sin(d*x+c)/b^2/d-a^2*(A*b-B*a)*sin(d*x+c)/b^2/(a^2-b^2)/d/(a+b*cos(d*x+c))

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Rubi [A]  time = 0.44, antiderivative size = 155, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.85, size = 147, normalized size = 0.95 \[ \frac {\frac {a^2 b (a B-A b) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+\frac {2 a \left (2 a^3 B-a^2 A b-3 a b^2 B+2 A b^3\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+(c+d x) (A b-2 a B)+b B \sin (c+d x)}{b^3 d} \]

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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fricas [B]  time = 1.25, size = 788, normalized size = 5.08 \[ \left [-\frac {2 \, {\left (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}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (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}\right )} d x + {\left (2 \, B a^{5} - A a^{4} b - 3 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + {\left (2 \, B a^{4} b - A a^{3} b^{2} - 3 \, B a^{2} b^{3} + 2 \, A a b^{4}\right )} \cos \left (d x + c\right )\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 ) - 2 \, {\left (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} + {\left (B a^{4} b^{2} - 2 \, B a^{2} b^{4} + B b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d\right )}}, -\frac {{\left (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}\right )} d x \cos \left (d x + c\right ) + {\left (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}\right )} d x - {\left (2 \, B a^{5} - A a^{4} b - 3 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + {\left (2 \, B a^{4} b - A a^{3} b^{2} - 3 \, B a^{2} b^{3} + 2 \, A a b^{4}\right )} \cos \left (d x + c\right )\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 (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} + {\left (B a^{4} b^{2} - 2 \, B a^{2} b^{4} + B b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d}\right ] \]

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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giac [B]  time = 3.22, size = 1116, normalized size = 7.20 \[ \text {result too large to display} \]

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]

((4*B*a^6*b^2 - 2*A*a^5*b^3 - 2*B*a^5*b^3 + A*a^4*b^4 - 9*B*a^4*b^4 + 5*A*a^3*b^5 + 4*B*a^3*b^5 - 2*A*a^2*b^6
+ 5*B*a^2*b^6 - 3*A*a*b^7 - 2*B*a*b^7 + A*b^8 + 2*B*a^3*abs(-a^2*b^3 + b^5) - A*a^2*b*abs(-a^2*b^3 + b^5) - B*
a^2*b*abs(-a^2*b^3 + b^5) + A*a*b^2*abs(-a^2*b^3 + b^5) - 2*B*a*b^2*abs(-a^2*b^3 + b^5) + A*b^3*abs(-a^2*b^3 +
b^5))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2)*tan(1/2*d*x + 1/2*c)/sqrt((2*a^3*b^2 - 2*a*b^4 +
sqrt(-4*(a^3*b^2 + a^2*b^3 - a*b^4 - b^5)*(a^3*b^2 - a^2*b^3 - a*b^4 + b^5) + 4*(a^3*b^2 - a*b^4)^2))/(a^3*b^
2 - a^2*b^3 - a*b^4 + b^5))))/(a^3*b^2*abs(-a^2*b^3 + b^5) - a*b^4*abs(-a^2*b^3 + b^5) + (a^2*b^3 - b^5)^2) +
((a^2*b - a*b^2 - b^3)*sqrt(a^2 - b^2)*A*abs(-a^2*b^3 + b^5)*abs(-a + b) - (2*a^3 - a^2*b - 2*a*b^2)*sqrt(a^2
- b^2)*B*abs(-a^2*b^3 + b^5)*abs(-a + b) - (2*a^5*b^3 - a^4*b^4 - 5*a^3*b^5 + 2*a^2*b^6 + 3*a*b^7 - b^8)*sqrt(
a^2 - b^2)*A*abs(-a + b) + (4*a^6*b^2 - 2*a^5*b^3 - 9*a^4*b^4 + 4*a^3*b^5 + 5*a^2*b^6 - 2*a*b^7)*sqrt(a^2 - b^
2)*B*abs(-a + b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2)*tan(1/2*d*x + 1/2*c)/sqrt((2*a^3*b^2
- 2*a*b^4 - sqrt(-4*(a^3*b^2 + a^2*b^3 - a*b^4 - b^5)*(a^3*b^2 - a^2*b^3 - a*b^4 + b^5) + 4*(a^3*b^2 - a*b^4)^
2))/(a^3*b^2 - a^2*b^3 - a*b^4 + b^5))))/((a^2*b^3 - b^5)^2*(a^2 - 2*a*b + b^2) - (a^5*b^2 - 2*a^4*b^3 + 2*a^2
*b^5 - a*b^6)*abs(-a^2*b^3 + b^5)) + 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)))/d

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maple [B]  time = 0.10, size = 445, normalized size = 2.87 \[ -\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A}{d b \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{d \,b^{2} \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {2 a^{3} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d \,b^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {4 a \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {4 a^{4} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d \,b^{3} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {6 a^{2} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d b \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}}-\frac {4 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{3}} \]

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*a^2/b/(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^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)*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)*arct
an(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(tan(1/2*d*x+1/2*c)*(a
-b)/((a-b)*(a+b))^(1/2))*B+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(tan(1/2*d*x+1/2*c))*a

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 5.17, size = 3276, normalized size = 21.14 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(tan(c/2 + (d*x)/2) + 1i)*(A*b - 2*B*a)*1i)/(b^3*d) - ((2*tan(c/2 + (d*x)/2)^3*(A*a^2*b - B*b^3 - 2*B*a^3
+ B*a*b^2 + B*a^2*b))/(b^2*(a + b)*(a - b)) + (2*tan(c/2 + (d*x)/2)*(B*b^3 - 2*B*a^3 + A*a^2*b + B*a*b^2 - B*a
^2*b))/(b^2*(a + b)*(a - b)))/(d*(a + b + tan(c/2 + (d*x)/2)^4*(a - b) + 2*a*tan(c/2 + (d*x)/2)^2)) - (log(tan
(c/2 + (d*x)/2) - 1i)*(A*b*1i - B*a*2i))/(b^3*d) - (a*atan(((a*((32*tan(c/2 + (d*x)/2)*(A^2*b^8 + 8*B^2*a^8 -
2*A^2*a*b^7 - 8*B^2*a^7*b + 3*A^2*a^2*b^6 + 4*A^2*a^3*b^5 - 5*A^2*a^4*b^4 - 2*A^2*a^5*b^3 + 2*A^2*a^6*b^2 + 4*
B^2*a^2*b^6 - 8*B^2*a^3*b^5 + 5*B^2*a^4*b^4 + 16*B^2*a^5*b^3 - 16*B^2*a^6*b^2 - 4*A*B*a*b^7 - 8*A*B*a^7*b + 8*
A*B*a^2*b^6 - 8*A*B*a^3*b^5 - 16*A*B*a^4*b^4 + 18*A*B*a^5*b^3 + 8*A*B*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b
^4) + (a*((32*(A*a^2*b^10 - A*b^12 - 3*A*a^3*b^9 + A*a^5*b^7 - 3*B*a^2*b^10 - 3*B*a^3*b^9 + 5*B*a^4*b^8 + B*a^
5*b^7 - 2*B*a^6*b^6 + 2*A*a*b^11 + 2*B*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (32*a*tan(c/2 + (d*x)/2)*(
-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 - A*a^2*b - 3*B*a*b^2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a
^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/((a*b^6 + b^7 - a^2*b^5 - a^3*b^4)*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))*
(-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 - A*a^2*b - 3*B*a*b^2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3
))*(-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 - A*a^2*b - 3*B*a*b^2)*1i)/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a
^6*b^3) + (a*((32*tan(c/2 + (d*x)/2)*(A^2*b^8 + 8*B^2*a^8 - 2*A^2*a*b^7 - 8*B^2*a^7*b + 3*A^2*a^2*b^6 + 4*A^2*
a^3*b^5 - 5*A^2*a^4*b^4 - 2*A^2*a^5*b^3 + 2*A^2*a^6*b^2 + 4*B^2*a^2*b^6 - 8*B^2*a^3*b^5 + 5*B^2*a^4*b^4 + 16*B
^2*a^5*b^3 - 16*B^2*a^6*b^2 - 4*A*B*a*b^7 - 8*A*B*a^7*b + 8*A*B*a^2*b^6 - 8*A*B*a^3*b^5 - 16*A*B*a^4*b^4 + 18*
A*B*a^5*b^3 + 8*A*B*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) - (a*((32*(A*a^2*b^10 - A*b^12 - 3*A*a^3*b^9 +
A*a^5*b^7 - 3*B*a^2*b^10 - 3*B*a^3*b^9 + 5*B*a^4*b^8 + B*a^5*b^7 - 2*B*a^6*b^6 + 2*A*a*b^11 + 2*B*a*b^11))/(a
*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (32*a*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 - A
*a^2*b - 3*B*a*b^2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/((a*b^6 + b^7 - a
^2*b^5 - a^3*b^4)*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))*(-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 -
A*a^2*b - 3*B*a*b^2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))*(-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3
- A*a^2*b - 3*B*a*b^2)*1i)/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))/((64*(8*B^3*a^8 - 2*A^3*a*b^7 - 4*B^3*a^7
*b - 2*A^3*a^2*b^6 + 3*A^3*a^3*b^5 + A^3*a^4*b^4 - A^3*a^5*b^3 + 12*B^3*a^4*b^4 + 6*B^3*a^5*b^3 - 20*B^3*a^6*b
^2 - 12*A*B^2*a^7*b - 20*A*B^2*a^3*b^5 - 13*A*B^2*a^4*b^4 + 32*A*B^2*a^5*b^3 + 8*A*B^2*a^6*b^2 + 11*A^2*B*a^2*
b^6 + 9*A^2*B*a^3*b^5 - 17*A^2*B*a^4*b^4 - 5*A^2*B*a^5*b^3 + 6*A^2*B*a^6*b^2))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^
6) - (a*((32*tan(c/2 + (d*x)/2)*(A^2*b^8 + 8*B^2*a^8 - 2*A^2*a*b^7 - 8*B^2*a^7*b + 3*A^2*a^2*b^6 + 4*A^2*a^3*b
^5 - 5*A^2*a^4*b^4 - 2*A^2*a^5*b^3 + 2*A^2*a^6*b^2 + 4*B^2*a^2*b^6 - 8*B^2*a^3*b^5 + 5*B^2*a^4*b^4 + 16*B^2*a^
5*b^3 - 16*B^2*a^6*b^2 - 4*A*B*a*b^7 - 8*A*B*a^7*b + 8*A*B*a^2*b^6 - 8*A*B*a^3*b^5 - 16*A*B*a^4*b^4 + 18*A*B*a
^5*b^3 + 8*A*B*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) + (a*((32*(A*a^2*b^10 - A*b^12 - 3*A*a^3*b^9 + A*a^
5*b^7 - 3*B*a^2*b^10 - 3*B*a^3*b^9 + 5*B*a^4*b^8 + B*a^5*b^7 - 2*B*a^6*b^6 + 2*A*a*b^11 + 2*B*a*b^11))/(a*b^8
+ b^9 - a^2*b^7 - a^3*b^6) - (32*a*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 - A*a^2*
b - 3*B*a*b^2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/((a*b^6 + b^7 - a^2*b^
5 - a^3*b^4)*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3)))*(-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 - A*a^2
*b - 3*B*a*b^2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))*(-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 - A*
a^2*b - 3*B*a*b^2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3) + (a*((32*tan(c/2 + (d*x)/2)*(A^2*b^8 + 8*B^2*a^8
- 2*A^2*a*b^7 - 8*B^2*a^7*b + 3*A^2*a^2*b^6 + 4*A^2*a^3*b^5 - 5*A^2*a^4*b^4 - 2*A^2*a^5*b^3 + 2*A^2*a^6*b^2 +
4*B^2*a^2*b^6 - 8*B^2*a^3*b^5 + 5*B^2*a^4*b^4 + 16*B^2*a^5*b^3 - 16*B^2*a^6*b^2 - 4*A*B*a*b^7 - 8*A*B*a^7*b +
8*A*B*a^2*b^6 - 8*A*B*a^3*b^5 - 16*A*B*a^4*b^4 + 18*A*B*a^5*b^3 + 8*A*B*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3
*b^4) - (a*((32*(A*a^2*b^10 - A*b^12 - 3*A*a^3*b^9 + A*a^5*b^7 - 3*B*a^2*b^10 - 3*B*a^3*b^9 + 5*B*a^4*b^8 + B*
a^5*b^7 - 2*B*a^6*b^6 + 2*A*a*b^11 + 2*B*a*b^11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (32*a*tan(c/2 + (d*x)/2)
*(-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 - A*a^2*b - 3*B*a*b^2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4
*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6))/((a*b^6 + b^7 - a^2*b^5 - a^3*b^4)*(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b^3))
)*(-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 - A*a^2*b - 3*B*a*b^2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^6*b
^3))*(-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 - A*a^2*b - 3*B*a*b^2))/(b^9 - 3*a^2*b^7 + 3*a^4*b^5 - a^
6*b^3)))*(-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 - A*a^2*b - 3*B*a*b^2)*2i)/(d*(b^9 - 3*a^2*b^7 + 3*a^
4*b^5 - a^6*b^3))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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