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

Optimal. Leaf size=263 \[ \frac{\left (2 a^2 A b-3 a^3 B+2 a b^2 B-A b^3\right ) \sin (c+d x)}{b^3 d \left (a^2-b^2\right )}+\frac{2 a^2 \left (2 a^2 A b-3 a^3 B+4 a b^2 B-3 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{a (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{\left (-3 a^2 B+2 a A b+b^2 B\right ) \sin (c+d x) \cos (c+d x)}{2 b^2 d \left (a^2-b^2\right )}-\frac{x \left (-6 a^2 B+4 a A b-b^2 B\right )}{2 b^4} \]

[Out]

-((4*a*A*b - 6*a^2*B - b^2*B)*x)/(2*b^4) + (2*a^2*(2*a^2*A*b - 3*A*b^3 - 3*a^3*B + 4*a*b^2*B)*ArcTan[(Sqrt[a -
 b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*b^4*(a + b)^(3/2)*d) + ((2*a^2*A*b - A*b^3 - 3*a^3*B + 2*a*
b^2*B)*Sin[c + d*x])/(b^3*(a^2 - b^2)*d) - ((2*a*A*b - 3*a^2*B + b^2*B)*Cos[c + d*x]*Sin[c + d*x])/(2*b^2*(a^2
 - b^2)*d) + (a*(A*b - a*B)*Cos[c + d*x]^2*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

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Rubi [A]  time = 0.658914, antiderivative size = 263, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2989, 3049, 3023, 2735, 2659, 205} \[ \frac{\left (2 a^2 A b-3 a^3 B+2 a b^2 B-A b^3\right ) \sin (c+d x)}{b^3 d \left (a^2-b^2\right )}+\frac{2 a^2 \left (2 a^2 A b-3 a^3 B+4 a b^2 B-3 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{a (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{\left (-3 a^2 B+2 a A b+b^2 B\right ) \sin (c+d x) \cos (c+d x)}{2 b^2 d \left (a^2-b^2\right )}-\frac{x \left (-6 a^2 B+4 a A b-b^2 B\right )}{2 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*a*A*b - 6*a^2*B - b^2*B)*x)/(2*b^4) + (2*a^2*(2*a^2*A*b - 3*A*b^3 - 3*a^3*B + 4*a*b^2*B)*ArcTan[(Sqrt[a -
 b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(3/2)*b^4*(a + b)^(3/2)*d) + ((2*a^2*A*b - A*b^3 - 3*a^3*B + 2*a*
b^2*B)*Sin[c + d*x])/(b^3*(a^2 - b^2)*d) - ((2*a*A*b - 3*a^2*B + b^2*B)*Cos[c + d*x]*Sin[c + d*x])/(2*b^2*(a^2
 - b^2)*d) + (a*(A*b - a*B)*Cos[c + d*x]^2*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))

Rule 2989

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((b*c - a*d)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)
*(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 - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (
A*b + a*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)
*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*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] && GtQ[m, 1] && 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) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx &=\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (-2 a (A b-a B)+b (A b-a B) \cos (c+d x)+\left (2 a A b-3 a^2 B+b^2 B\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{a \left (2 a A b-3 a^2 B+b^2 B\right )-b \left (2 a A b-a^2 B-b^2 B\right ) \cos (c+d x)-2 \left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{a b \left (2 a A b-3 a^2 B+b^2 B\right )+\left (a^2-b^2\right ) \left (4 a A b-6 a^2 B-b^2 B\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (4 a A b-6 a^2 B-b^2 B\right ) x}{2 b^4}+\frac{\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (a^2 \left (2 a^2 A b-3 A b^3-3 a^3 B+4 a b^2 B\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^4 \left (a^2-b^2\right )}\\ &=-\frac{\left (4 a A b-6 a^2 B-b^2 B\right ) x}{2 b^4}+\frac{\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (2 a^2 \left (2 a^2 A b-3 A b^3-3 a^3 B+4 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^4 \left (a^2-b^2\right ) d}\\ &=-\frac{\left (4 a A b-6 a^2 B-b^2 B\right ) x}{2 b^4}+\frac{2 a^2 \left (2 a^2 A b-3 A b^3-3 a^3 B+4 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^4 (a+b)^{3/2} d}+\frac{\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \sin (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) \cos (c+d x) \sin (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.982682, size = 184, normalized size = 0.7 \[ \frac{2 (c+d x) \left (6 a^2 B-4 a A b+b^2 B\right )-\frac{8 a^2 \left (-2 a^2 A b+3 a^3 B-4 a b^2 B+3 A b^3\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac{4 a^3 b (A b-a B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+4 b (A b-2 a B) \sin (c+d x)+b^2 B \sin (2 (c+d x))}{4 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-4*a*A*b + 6*a^2*B + b^2*B)*(c + d*x) - (8*a^2*(-2*a^2*A*b + 3*A*b^3 + 3*a^3*B - 4*a*b^2*B)*ArcTanh[((a -
b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) + 4*b*(A*b - 2*a*B)*Sin[c + d*x] + (4*a^3*b*(A*b -
a*B)*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])) + b^2*B*Sin[2*(c + d*x)])/(4*b^4*d)

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Maple [B]  time = 0.145, size = 643, normalized size = 2.4 \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+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x)

[Out]

2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*A-4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c
)^3*B*a-1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*B+2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d
*x+1/2*c)*A-4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)*B*a+1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1
/2*d*x+1/2*c)*B-4/d/b^3*arctan(tan(1/2*d*x+1/2*c))*A*a+6/d/b^4*arctan(tan(1/2*d*x+1/2*c))*B*a^2+1/d/b^2*arctan
(tan(1/2*d*x+1/2*c))*B+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)*A-2/d*a^4/b^3/(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+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))*A-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))*A-6/d*a^5/b^4/(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+8/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))*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)^3*(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 [A]  time = 1.59442, size = 2120, normalized size = 8.06 \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+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/2*((6*B*a^6*b - 4*A*a^5*b^2 - 11*B*a^4*b^3 + 8*A*a^3*b^4 + 4*B*a^2*b^5 - 4*A*a*b^6 + B*b^7)*d*x*cos(d*x + c
) + (6*B*a^7 - 4*A*a^6*b - 11*B*a^5*b^2 + 8*A*a^4*b^3 + 4*B*a^3*b^4 - 4*A*a^2*b^5 + B*a*b^6)*d*x - (3*B*a^6 -
2*A*a^5*b - 4*B*a^4*b^2 + 3*A*a^3*b^3 + (3*B*a^5*b - 2*A*a^4*b^2 - 4*B*a^3*b^3 + 3*A*a^2*b^4)*cos(d*x + c))*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)) - (6*B*a^6*b - 4*A*a^5*b^2 - 10
*B*a^4*b^3 + 6*A*a^3*b^4 + 4*B*a^2*b^5 - 2*A*a*b^6 - (B*a^4*b^3 - 2*B*a^2*b^5 + B*b^7)*cos(d*x + c)^2 + (3*B*a
^5*b^2 - 2*A*a^4*b^3 - 6*B*a^3*b^4 + 4*A*a^2*b^5 + 3*B*a*b^6 - 2*A*b^7)*cos(d*x + c))*sin(d*x + c))/((a^4*b^5
- 2*a^2*b^7 + b^9)*d*cos(d*x + c) + (a^5*b^4 - 2*a^3*b^6 + a*b^8)*d), 1/2*((6*B*a^6*b - 4*A*a^5*b^2 - 11*B*a^4
*b^3 + 8*A*a^3*b^4 + 4*B*a^2*b^5 - 4*A*a*b^6 + B*b^7)*d*x*cos(d*x + c) + (6*B*a^7 - 4*A*a^6*b - 11*B*a^5*b^2 +
 8*A*a^4*b^3 + 4*B*a^3*b^4 - 4*A*a^2*b^5 + B*a*b^6)*d*x - 2*(3*B*a^6 - 2*A*a^5*b - 4*B*a^4*b^2 + 3*A*a^3*b^3 +
 (3*B*a^5*b - 2*A*a^4*b^2 - 4*B*a^3*b^3 + 3*A*a^2*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))) - (6*B*a^6*b - 4*A*a^5*b^2 - 10*B*a^4*b^3 + 6*A*a^3*b^4 + 4*B*a^2*b^5 - 2*
A*a*b^6 - (B*a^4*b^3 - 2*B*a^2*b^5 + B*b^7)*cos(d*x + c)^2 + (3*B*a^5*b^2 - 2*A*a^4*b^3 - 6*B*a^3*b^4 + 4*A*a^
2*b^5 + 3*B*a*b^6 - 2*A*b^7)*cos(d*x + c))*sin(d*x + c))/((a^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c) + (a^5*b^
4 - 2*a^3*b^6 + a*b^8)*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+B*cos(d*x+c))/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.62894, size = 456, normalized size = 1.73 \begin{align*} \frac{\frac{4 \,{\left (3 \, B a^{5} - 2 \, A a^{4} b - 4 \, B a^{3} b^{2} + 3 \, A a^{2} 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^{4} - b^{6}\right )} \sqrt{a^{2} - b^{2}}} - \frac{4 \,{\left (B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b\right )}} + \frac{{\left (6 \, B a^{2} - 4 \, A a b + B b^{2}\right )}{\left (d x + c\right )}}{b^{4}} - \frac{2 \,{\left (4 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B b \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 )}^{2} b^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

1/2*(4*(3*B*a^5 - 2*A*a^4*b - 4*B*a^3*b^2 + 3*A*a^2*b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + a
rctan(-(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^4 - b^6)*sqrt(a^2 - b^2)) -
 4*(B*a^4*tan(1/2*d*x + 1/2*c) - A*a^3*b*tan(1/2*d*x + 1/2*c))/((a^2*b^3 - b^5)*(a*tan(1/2*d*x + 1/2*c)^2 - b*
tan(1/2*d*x + 1/2*c)^2 + a + b)) + (6*B*a^2 - 4*A*a*b + B*b^2)*(d*x + c)/b^4 - 2*(4*B*a*tan(1/2*d*x + 1/2*c)^3
 - 2*A*b*tan(1/2*d*x + 1/2*c)^3 + B*b*tan(1/2*d*x + 1/2*c)^3 + 4*B*a*tan(1/2*d*x + 1/2*c) - 2*A*b*tan(1/2*d*x
+ 1/2*c) - B*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*b^3))/d