\(\int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx\) [276]

Optimal result

Integrand size = 29, antiderivative size = 263 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=-\frac {\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac {a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \]

[Out]

-(4*A*a^2*b+A*b^3-B*a^3-4*B*a*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(7/2)/(a+b)^(7/2)/
d+1/3*a*(A*b-B*a)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^3+1/6*(2*A*a^2*b+3*A*b^3+B*a^3-6*B*a*b^2)*sin(d*x+
c)/b/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^2+1/6*(2*A*a^3*b+13*A*a*b^3+B*a^4-10*B*a^2*b^2-6*B*b^4)*sin(d*x+c)/b/(a^2-
b^2)^3/d/(a+b*cos(d*x+c))

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3047, 3100, 2833, 12, 2738, 211} \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\frac {a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\left (a^3 (-B)+4 a^2 A b-4 a b^2 B+A b^3\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {\left (a^3 B+2 a^2 A b-6 a b^2 B+3 A b^3\right ) \sin (c+d x)}{6 b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\left (a^4 B+2 a^3 A b-10 a^2 b^2 B+13 a A b^3-6 b^4 B\right ) \sin (c+d x)}{6 b d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))} \]

[In]

Int[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^4,x]

[Out]

-(((4*a^2*A*b + A*b^3 - a^3*B - 4*a*b^2*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(7/2)*
(a + b)^(7/2)*d)) + (a*(A*b - a*B)*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^3) + ((2*a^2*A*b + 3*
A*b^3 + a^3*B - 6*a*b^2*B)*Sin[c + d*x])/(6*b*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x])^2) + ((2*a^3*A*b + 13*a*A*b
^3 + a^4*B - 10*a^2*b^2*B - 6*b^4*B)*Sin[c + d*x])/(6*b*(a^2 - b^2)^3*d*(a + b*Cos[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2738

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 2833

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

Rule 3047

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

Rule 3100

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {A \cos (c+d x)+B \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx \\ & = \frac {a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {\int \frac {3 b (A b-a B)-\left (2 a A b+a^2 B-3 b^2 B\right ) \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx}{3 b \left (a^2-b^2\right )} \\ & = \frac {a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\int \frac {-2 b \left (5 a A b-2 a^2 B-3 b^2 B\right )+\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{6 b \left (a^2-b^2\right )^2} \\ & = \frac {a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}-\frac {\int \frac {3 b \left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\right )}{a+b \cos (c+d x)} \, dx}{6 b \left (a^2-b^2\right )^3} \\ & = \frac {a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}-\frac {\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3} \\ & = \frac {a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))}-\frac {\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d} \\ & = -\frac {\left (4 a^2 A b+A b^3-a^3 B-4 a b^2 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac {a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\left (2 a^2 A b+3 A b^3+a^3 B-6 a b^2 B\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.29 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.96 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\frac {-\frac {24 \left (-4 a^2 A b-A b^3+a^3 B+4 a b^2 B\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {2 \left (12 a^5 A+22 a^3 A b^2+11 a A b^4-25 a^4 b B-14 a^2 b^3 B-6 b^5 B+6 \left (2 a^4 A b+9 a^2 A b^3-A b^5+a^5 B-9 a^3 b^2 B-2 a b^4 B\right ) \cos (c+d x)+b \left (2 a^3 A b+13 a A b^3+a^4 B-10 a^2 b^2 B-6 b^4 B\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{(a+b \cos (c+d x))^3}}{24 \left (a^2-b^2\right )^3 d} \]

[In]

Integrate[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^4,x]

[Out]

((-24*(-4*a^2*A*b - A*b^3 + a^3*B + 4*a*b^2*B)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2
+ b^2] + (2*(12*a^5*A + 22*a^3*A*b^2 + 11*a*A*b^4 - 25*a^4*b*B - 14*a^2*b^3*B - 6*b^5*B + 6*(2*a^4*A*b + 9*a^
2*A*b^3 - A*b^5 + a^5*B - 9*a^3*b^2*B - 2*a*b^4*B)*Cos[c + d*x] + b*(2*a^3*A*b + 13*a*A*b^3 + a^4*B - 10*a^2*b
^2*B - 6*b^4*B)*Cos[2*(c + d*x)])*Sin[c + d*x])/(a + b*Cos[c + d*x])^3)/(24*(a^2 - b^2)^3*d)

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.46

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {\left (2 A \,a^{3}+2 A \,a^{2} b +6 A a \,b^{2}+A \,b^{3}-B \,a^{3}-6 B \,a^{2} b -2 B a \,b^{2}-2 B \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 \left (3 A \,a^{3}+7 A a \,b^{2}-7 B \,a^{2} b -3 B \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (2 A \,a^{3}-2 A \,a^{2} b +6 A a \,b^{2}-A \,b^{3}+B \,a^{3}-6 B \,a^{2} b +2 B a \,b^{2}-2 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\right )}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{3}}-\frac {\left (4 A \,a^{2} b +A \,b^{3}-B \,a^{3}-4 B a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(384\)
default	\(\frac {-\frac {2 \left (-\frac {\left (2 A \,a^{3}+2 A \,a^{2} b +6 A a \,b^{2}+A \,b^{3}-B \,a^{3}-6 B \,a^{2} b -2 B a \,b^{2}-2 B \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 \left (3 A \,a^{3}+7 A a \,b^{2}-7 B \,a^{2} b -3 B \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (2 A \,a^{3}-2 A \,a^{2} b +6 A a \,b^{2}-A \,b^{3}+B \,a^{3}-6 B \,a^{2} b +2 B a \,b^{2}-2 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}\right )}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}^{3}}-\frac {\left (4 A \,a^{2} b +A \,b^{3}-B \,a^{3}-4 B a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(384\)
risch	\(\text {Expression too large to display}\)	\(1282\)

[In]

int(cos(d*x+c)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 581 vs. \(2 (247) = 494\).

Time = 0.38 (sec) , antiderivative size = 1232, normalized size of antiderivative = 4.68 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))**4,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^4,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 de

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 722 vs. \(2 (247) = 494\).

Time = 0.33 (sec) , antiderivative size = 722, normalized size of antiderivative = 2.75 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=-\frac {\frac {3 \, {\left (B a^{3} - 4 \, A a^{2} b + 4 \, B a b^{2} - 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^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} - \frac {6 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 27 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 28 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 28 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, B a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 27 \, B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, A a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\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 )}^{3}}}{3 \, d} \]

[In]

integrate(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^4,x, algorithm="giac")

[Out]

-1/3*(3*(B*a^3 - 4*A*a^2*b + 4*B*a*b^2 - 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^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^
2 - b^2)) - (6*A*a^5*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^5*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^4*b*tan(1/2*d*x + 1/2*c)^
5 - 12*B*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 12*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 27*B*a^3*b^2*tan(1/2*d*x + 1/2*c
)^5 - 27*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 12*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 12*A*a*b^4*tan(1/2*d*x + 1/2
*c)^5 + 6*B*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*A*b^5*tan(1/2*d*x + 1/2*c)^5 - 6*B*b^5*tan(1/2*d*x + 1/2*c)^5 + 1
2*A*a^5*tan(1/2*d*x + 1/2*c)^3 - 28*B*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 16*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 16*
B*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 28*A*a*b^4*tan(1/2*d*x + 1/2*c)^3 + 12*B*b^5*tan(1/2*d*x + 1/2*c)^3 + 6*A*a
^5*tan(1/2*d*x + 1/2*c) + 3*B*a^5*tan(1/2*d*x + 1/2*c) + 6*A*a^4*b*tan(1/2*d*x + 1/2*c) - 12*B*a^4*b*tan(1/2*d
*x + 1/2*c) + 12*A*a^3*b^2*tan(1/2*d*x + 1/2*c) - 27*B*a^3*b^2*tan(1/2*d*x + 1/2*c) + 27*A*a^2*b^3*tan(1/2*d*x
+ 1/2*c) - 12*B*a^2*b^3*tan(1/2*d*x + 1/2*c) + 12*A*a*b^4*tan(1/2*d*x + 1/2*c) - 6*B*a*b^4*tan(1/2*d*x + 1/2*
c) - 3*A*b^5*tan(1/2*d*x + 1/2*c) - 6*B*b^5*tan(1/2*d*x + 1/2*c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(a*tan(
1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^3))/d

Mupad [B] (verification not implemented)

Time = 6.63 (sec) , antiderivative size = 451, normalized size of antiderivative = 1.71 \[ \int \frac {\cos (c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^4} \, dx=\frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A\,a^3-7\,B\,a^2\,b+7\,A\,a\,b^2-3\,B\,b^3\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,a^3-A\,b^3+B\,a^3-2\,B\,b^3+6\,A\,a\,b^2-2\,A\,a^2\,b+2\,B\,a\,b^2-6\,B\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,A\,a^3+A\,b^3-B\,a^3-2\,B\,b^3+6\,A\,a\,b^2+2\,A\,a^2\,b-2\,B\,a\,b^2-6\,B\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}-\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (-B\,a^3+4\,A\,a^2\,b-4\,B\,a\,b^2+A\,b^3\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \]

[In]

int((cos(c + d*x)*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x))^4,x)

[Out]

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