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

Optimal. Leaf size=280 \[ -\frac{\left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x)}{2 b^3 d \left (a^2-b^2\right )}-\frac{a \left (-5 a^2 A b^3+2 a^4 A b+15 a^3 b^2 B-6 a^5 B-12 a b^4 B+6 A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{a (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{a^2 \left (a^2 A b-3 a^3 B+6 a b^2 B-4 A b^3\right ) \sin (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{x (A b-3 a B)}{b^4} \]

[Out]

((A*b - 3*a*B)*x)/b^4 - (a*(2*a^4*A*b - 5*a^2*A*b^3 + 6*A*b^5 - 6*a^5*B + 15*a^3*b^2*B - 12*a*b^4*B)*ArcTan[(S
qrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^4*(a + b)^(5/2)*d) - ((a*A*b - 3*a^2*B + 2*b^2*B)*
Sin[c + d*x])/(2*b^3*(a^2 - b^2)*d) + (a*(A*b - a*B)*Cos[c + d*x]^2*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Co
s[c + d*x])^2) - (a^2*(a^2*A*b - 4*A*b^3 - 3*a^3*B + 6*a*b^2*B)*Sin[c + d*x])/(2*b^3*(a^2 - b^2)^2*d*(a + b*Co
s[c + d*x]))

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Rubi [A]  time = 1.22187, antiderivative size = 280, 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, 3031, 3023, 2735, 2659, 205} \[ -\frac{\left (-3 a^2 B+a A b+2 b^2 B\right ) \sin (c+d x)}{2 b^3 d \left (a^2-b^2\right )}-\frac{a \left (-5 a^2 A b^3+2 a^4 A b+15 a^3 b^2 B-6 a^5 B-12 a b^4 B+6 A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{a (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{a^2 \left (a^2 A b-3 a^3 B+6 a b^2 B-4 A b^3\right ) \sin (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{x (A b-3 a B)}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - 3*a*B)*x)/b^4 - (a*(2*a^4*A*b - 5*a^2*A*b^3 + 6*A*b^5 - 6*a^5*B + 15*a^3*b^2*B - 12*a*b^4*B)*ArcTan[(S
qrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^4*(a + b)^(5/2)*d) - ((a*A*b - 3*a^2*B + 2*b^2*B)*
Sin[c + d*x])/(2*b^3*(a^2 - b^2)*d) + (a*(A*b - a*B)*Cos[c + d*x]^2*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Co
s[c + d*x])^2) - (a^2*(a^2*A*b - 4*A*b^3 - 3*a^3*B + 6*a*b^2*B)*Sin[c + d*x])/(2*b^3*(a^2 - b^2)^2*d*(a + b*Co
s[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 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -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 ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx &=\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\int \frac{\cos (c+d x) \left (-2 a (A b-a B)+2 b (A b-a B) \cos (c+d x)+\left (a A b-3 a^2 B+2 b^2 B\right ) \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (a^2 A b-4 A b^3-3 a^3 B+6 a b^2 B\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\int \frac{-a b \left (a^2 A b-4 A b^3-3 a^3 B+6 a b^2 B\right )-\left (a^2-b^2\right ) \left (a^2 A b-2 A b^3-3 a^3 B+4 a b^2 B\right ) \cos (c+d x)+b \left (a^2-b^2\right ) \left (a A b-3 a^2 B+2 b^2 B\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (a A b-3 a^2 B+2 b^2 B\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (a^2 A b-4 A b^3-3 a^3 B+6 a b^2 B\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\int \frac{-a b^2 \left (a^2 A b-4 A b^3-3 a^3 B+6 a b^2 B\right )-2 b \left (a^2-b^2\right )^2 (A b-3 a B) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=\frac{(A b-3 a B) x}{b^4}-\frac{\left (a A b-3 a^2 B+2 b^2 B\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (a^2 A b-4 A b^3-3 a^3 B+6 a b^2 B\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (a \left (2 a^4 A b-5 a^2 A b^3+6 A b^5-6 a^5 B+15 a^3 b^2 B-12 a b^4 B\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=\frac{(A b-3 a B) x}{b^4}-\frac{\left (a A b-3 a^2 B+2 b^2 B\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (a^2 A b-4 A b^3-3 a^3 B+6 a b^2 B\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (a \left (2 a^4 A b-5 a^2 A b^3+6 A b^5-6 a^5 B+15 a^3 b^2 B-12 a b^4 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 )^2 d}\\ &=\frac{(A b-3 a B) x}{b^4}-\frac{a \left (2 a^4 A b-5 a^2 A b^3+6 A b^5-6 a^5 B+15 a^3 b^2 B-12 a b^4 B\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} b^4 (a+b)^{5/2} d}-\frac{\left (a A b-3 a^2 B+2 b^2 B\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{a^2 \left (a^2 A b-4 A b^3-3 a^3 B+6 a b^2 B\right ) \sin (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.05949, size = 232, normalized size = 0.83 \[ \frac{\frac{a^2 b \left (-3 a^2 A b+5 a^3 B-8 a b^2 B+6 A b^3\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}-\frac{2 a \left (5 a^2 A b^3-2 a^4 A b-15 a^3 b^2 B+6 a^5 B+12 a b^4 B-6 A b^5\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+\frac{a^3 b (A b-a B) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+2 (c+d x) (A b-3 a B)+2 b B \sin (c+d x)}{2 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b - 3*a*B)*(c + d*x) - (2*a*(-2*a^4*A*b + 5*a^2*A*b^3 - 6*A*b^5 + 6*a^5*B - 15*a^3*b^2*B + 12*a*b^4*B)*A
rcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + 2*b*B*Sin[c + d*x] + (a^3*b*(A*b - a
*B)*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) + (a^2*b*(-3*a^2*A*b + 6*A*b^3 + 5*a^3*B - 8*a*b^2*
B)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])))/(2*b^4*d)

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Maple [B]  time = 0.169, size = 1301, normalized size = 4.7 \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))^3,x)

[Out]

2/d/b^3*B*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+2/d/b^3*A*arctan(tan(1/2*d*x+1/2*c))-6/d/b^4*B*arctan(ta
n(1/2*d*x+1/2*c))*a-2/d*a^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*ta
n(1/2*d*x+1/2*c)^3*A+1/d*a^3/b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan
(1/2*d*x+1/2*c)^3*A+6/d*a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/
2*d*x+1/2*c)^3*A+4/d*a^5/b^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1
/2*d*x+1/2*c)^3*B-1/d*a^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(
1/2*d*x+1/2*c)^3*B-8/d*a^3/b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1
/2*d*x+1/2*c)^3*B-2/d*a^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+
1/2*c)*A-1/d*a^3/b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*A+6/
d*a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*A+4/d*a^5/b^3/(ta
n(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B+1/d*a^4/b^2/(tan(1/2*d*x
+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B-8/d*a^3/b/(tan(1/2*d*x+1/2*c)^2*a
-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B-2/d*a^5/b^3/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b)
)^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A+5/d*a^3/b/(a^4-2*a^2*b^2+b^4)/((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*b/(a^4-2*a^2*b^2+b^4)/((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+6/d*a^6/b^4/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(
tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B-15/d*a^4/b^2/(a^4-2*a^2*b^2+b^4)/((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+12/d*a^2/(a^4-2*a^2*b^2+b^4)/((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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.96437, size = 3380, normalized size = 12.07 \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))^3,x, algorithm="fricas")

[Out]

[-1/4*(4*(3*B*a^7*b^2 - A*a^6*b^3 - 9*B*a^5*b^4 + 3*A*a^4*b^5 + 9*B*a^3*b^6 - 3*A*a^2*b^7 - 3*B*a*b^8 + A*b^9)
*d*x*cos(d*x + c)^2 + 8*(3*B*a^8*b - A*a^7*b^2 - 9*B*a^6*b^3 + 3*A*a^5*b^4 + 9*B*a^4*b^5 - 3*A*a^3*b^6 - 3*B*a
^2*b^7 + A*a*b^8)*d*x*cos(d*x + c) + 4*(3*B*a^9 - A*a^8*b - 9*B*a^7*b^2 + 3*A*a^6*b^3 + 9*B*a^5*b^4 - 3*A*a^4*
b^5 - 3*B*a^3*b^6 + A*a^2*b^7)*d*x - (6*B*a^8 - 2*A*a^7*b - 15*B*a^6*b^2 + 5*A*a^5*b^3 + 12*B*a^4*b^4 - 6*A*a^
3*b^5 + (6*B*a^6*b^2 - 2*A*a^5*b^3 - 15*B*a^4*b^4 + 5*A*a^3*b^5 + 12*B*a^2*b^6 - 6*A*a*b^7)*cos(d*x + c)^2 + 2
*(6*B*a^7*b - 2*A*a^6*b^2 - 15*B*a^5*b^3 + 5*A*a^4*b^4 + 12*B*a^3*b^5 - 6*A*a^2*b^6)*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*(6*B*a^8*b - 2*A*a^7*b^2 - 17*B*a^6*
b^3 + 7*A*a^5*b^4 + 13*B*a^4*b^5 - 5*A*a^3*b^6 - 2*B*a^2*b^7 + 2*(B*a^6*b^3 - 3*B*a^4*b^5 + 3*B*a^2*b^7 - B*b^
9)*cos(d*x + c)^2 + (9*B*a^7*b^2 - 3*A*a^6*b^3 - 25*B*a^5*b^4 + 9*A*a^4*b^5 + 20*B*a^3*b^6 - 6*A*a^2*b^7 - 4*B
*a*b^8)*cos(d*x + c))*sin(d*x + c))/((a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - b^12)*d*cos(d*x + c)^2 + 2*(a^7*b^5 -
3*a^5*b^7 + 3*a^3*b^9 - a*b^11)*d*cos(d*x + c) + (a^8*b^4 - 3*a^6*b^6 + 3*a^4*b^8 - a^2*b^10)*d), -1/2*(2*(3*
B*a^7*b^2 - A*a^6*b^3 - 9*B*a^5*b^4 + 3*A*a^4*b^5 + 9*B*a^3*b^6 - 3*A*a^2*b^7 - 3*B*a*b^8 + A*b^9)*d*x*cos(d*x
+ c)^2 + 4*(3*B*a^8*b - A*a^7*b^2 - 9*B*a^6*b^3 + 3*A*a^5*b^4 + 9*B*a^4*b^5 - 3*A*a^3*b^6 - 3*B*a^2*b^7 + A*a
*b^8)*d*x*cos(d*x + c) + 2*(3*B*a^9 - A*a^8*b - 9*B*a^7*b^2 + 3*A*a^6*b^3 + 9*B*a^5*b^4 - 3*A*a^4*b^5 - 3*B*a^
3*b^6 + A*a^2*b^7)*d*x - (6*B*a^8 - 2*A*a^7*b - 15*B*a^6*b^2 + 5*A*a^5*b^3 + 12*B*a^4*b^4 - 6*A*a^3*b^5 + (6*B
*a^6*b^2 - 2*A*a^5*b^3 - 15*B*a^4*b^4 + 5*A*a^3*b^5 + 12*B*a^2*b^6 - 6*A*a*b^7)*cos(d*x + c)^2 + 2*(6*B*a^7*b
- 2*A*a^6*b^2 - 15*B*a^5*b^3 + 5*A*a^4*b^4 + 12*B*a^3*b^5 - 6*A*a^2*b^6)*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^8*b - 2*A*a^7*b^2 - 17*B*a^6*b^3 + 7*A*a^5*b^4
+ 13*B*a^4*b^5 - 5*A*a^3*b^6 - 2*B*a^2*b^7 + 2*(B*a^6*b^3 - 3*B*a^4*b^5 + 3*B*a^2*b^7 - B*b^9)*cos(d*x + c)^2
+ (9*B*a^7*b^2 - 3*A*a^6*b^3 - 25*B*a^5*b^4 + 9*A*a^4*b^5 + 20*B*a^3*b^6 - 6*A*a^2*b^7 - 4*B*a*b^8)*cos(d*x +
c))*sin(d*x + c))/((a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - b^12)*d*cos(d*x + c)^2 + 2*(a^7*b^5 - 3*a^5*b^7 + 3*a^3
*b^9 - a*b^11)*d*cos(d*x + c) + (a^8*b^4 - 3*a^6*b^6 + 3*a^4*b^8 - a^2*b^10)*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))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.70252, size = 733, normalized size = 2.62 \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))^3,x, algorithm="giac")

[Out]

-((6*B*a^6 - 2*A*a^5*b - 15*B*a^4*b^2 + 5*A*a^3*b^3 + 12*B*a^2*b^4 - 6*A*a*b^5)*(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^4*b^4 -
2*a^2*b^6 + b^8)*sqrt(a^2 - b^2)) - (4*B*a^6*tan(1/2*d*x + 1/2*c)^3 - 2*A*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 5*B*a
^5*b*tan(1/2*d*x + 1/2*c)^3 + 3*A*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 - 7*B*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 + 5*A*a^
3*b^3*tan(1/2*d*x + 1/2*c)^3 + 8*B*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 6*A*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 + 4*B*a
^6*tan(1/2*d*x + 1/2*c) - 2*A*a^5*b*tan(1/2*d*x + 1/2*c) + 5*B*a^5*b*tan(1/2*d*x + 1/2*c) - 3*A*a^4*b^2*tan(1/
2*d*x + 1/2*c) - 7*B*a^4*b^2*tan(1/2*d*x + 1/2*c) + 5*A*a^3*b^3*tan(1/2*d*x + 1/2*c) - 8*B*a^3*b^3*tan(1/2*d*x
+ 1/2*c) + 6*A*a^2*b^4*tan(1/2*d*x + 1/2*c))/((a^4*b^3 - 2*a^2*b^5 + b^7)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1
/2*d*x + 1/2*c)^2 + a + b)^2) + (3*B*a - A*b)*(d*x + c)/b^4 - 2*B*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^
2 + 1)*b^3))/d