\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx\) [1005]

Optimal result

Integrand size = 33, antiderivative size = 299 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=-\frac {\left (4 a^2 b B+b^3 B-a^3 (2 A+C)-a b^2 (3 A+4 C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (2 a^3 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \]

[Out]

-(4*B*a^2*b+B*b^3-a^3*(2*A+C)-a*b^2*(3*A+4*C))*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*b^2-a*(B*b-C*a))*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^3+1/6*(2*B*a^2*b+3*B*b^3+a^3*C
-a*b^2*(5*A+6*C))*sin(d*x+c)/b/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^2+1/6*(2*B*a^3*b+13*B*a*b^3+a^4*C-2*b^4*(2*A+3*C
)-a^2*b^2*(11*A+10*C))*sin(d*x+c)/b/(a^2-b^2)^3/d/(a+b*cos(d*x+c))

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3100, 2833, 12, 2738, 211} \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=-\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}-\frac {\left (-\left (a^3 (2 A+C)\right )+4 a^2 b B-a b^2 (3 A+4 C)+b^3 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac {\sin (c+d x) \left (a^3 C+2 a^2 b B-a b^2 (5 A+6 C)+3 b^3 B\right )}{6 b d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\sin (c+d x) \left (a^4 C+2 a^3 b B-a^2 b^2 (11 A+10 C)+13 a b^3 B-2 b^4 (2 A+3 C)\right )}{6 b d \left (a^2-b^2\right )^3 (a+b \cos (c+d x))} \]

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^4,x]

[Out]

-(((4*a^2*b*B + b^3*B - a^3*(2*A + C) - a*b^2*(3*A + 4*C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])
/((a - b)^(7/2)*(a + b)^(7/2)*d)) - ((A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Cos[c + d
*x])^3) + ((2*a^2*b*B + 3*b^3*B + a^3*C - a*b^2*(5*A + 6*C))*Sin[c + d*x])/(6*b*(a^2 - b^2)^2*d*(a + b*Cos[c +
d*x])^2) + ((2*a^3*b*B + 13*a*b^3*B + a^4*C - 2*b^4*(2*A + 3*C) - a^2*b^2*(11*A + 10*C))*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 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}& = -\frac {\left (A b^2-a (b B-a C)\right ) \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 (b B-a (A+C))+\left (2 A b^2-2 a b B-a^2 C+3 b^2 C\right ) \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx}{3 b \left (a^2-b^2\right )} \\ & = -\frac {\left (A b^2-a (b B-a C)\right ) \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 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\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 b B-a^2 (3 A+2 C)-b^2 (2 A+3 C)\right )+\left (2 a^2 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{6 b \left (a^2-b^2\right )^2} \\ & = -\frac {\left (A b^2-a (b B-a C)\right ) \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 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\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 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\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 b B+b^3 B-a^3 (2 A+C)-a b^2 (3 A+4 C)\right )}{a+b \cos (c+d x)} \, dx}{6 b \left (a^2-b^2\right )^3} \\ & = -\frac {\left (A b^2-a (b B-a C)\right ) \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 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\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 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\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 b B+b^3 B-a^3 (2 A+C)-a b^2 (3 A+4 C)\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3} \\ & = -\frac {\left (A b^2-a (b B-a C)\right ) \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 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\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 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\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 b B+b^3 B-a^3 (2 A+C)-a b^2 (3 A+4 C)\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 (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B+a^3 C+4 a b^2 C\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 {\left (A b^2-a (b B-a C)\right ) \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 b B+3 b^3 B+a^3 C-a b^2 (5 A+6 C)\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 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\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 = 1.93 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.01 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {-\frac {24 \left (-4 a^2 b B-b^3 B+a^3 (2 A+C)+a b^2 (3 A+4 C)\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {2 \left (-36 a^4 A b-a^2 A b^3-8 A b^5+12 a^5 B+22 a^3 b^2 B+11 a b^4 B-25 a^4 b C-14 a^2 b^3 C-6 b^5 C+6 \left (2 a^4 b B+9 a^2 b^3 B-b^5 B+a^5 C-9 a^3 b^2 (A+C)-a b^4 (A+2 C)\right ) \cos (c+d x)+b \left (2 a^3 b B+13 a b^3 B+a^4 C-2 b^4 (2 A+3 C)-a^2 b^2 (11 A+10 C)\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{(a+b \cos (c+d x))^3}}{24 \left (a^2-b^2\right )^3 d} \]

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^4,x]

[Out]

((-24*(-4*a^2*b*B - b^3*B + a^3*(2*A + C) + a*b^2*(3*A + 4*C))*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 +
b^2]])/Sqrt[-a^2 + b^2] + (2*(-36*a^4*A*b - a^2*A*b^3 - 8*A*b^5 + 12*a^5*B + 22*a^3*b^2*B + 11*a*b^4*B - 25*a^
4*b*C - 14*a^2*b^3*C - 6*b^5*C + 6*(2*a^4*b*B + 9*a^2*b^3*B - b^5*B + a^5*C - 9*a^3*b^2*(A + C) - a*b^4*(A + 2
*C))*Cos[c + d*x] + b*(2*a^3*b*B + 13*a*b^3*B + a^4*C - 2*b^4*(2*A + 3*C) - a^2*b^2*(11*A + 10*C))*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 = 0.56 (sec) , antiderivative size = 448, normalized size of antiderivative = 1.50

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (6 A \,a^{2} b +3 a A \,b^{2}+2 A \,b^{3}-2 B \,a^{3}-2 B \,a^{2} b -6 B a \,b^{2}-B \,b^{3}+a^{3} C +6 a^{2} b C +2 C a \,b^{2}+2 C \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {4 \left (9 A \,a^{2} b +A \,b^{3}-3 B \,a^{3}-7 B a \,b^{2}+7 a^{2} b C +3 C \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (6 A \,a^{2} b -3 a A \,b^{2}+2 A \,b^{3}-2 B \,a^{3}+2 B \,a^{2} b -6 B a \,b^{2}+B \,b^{3}-a^{3} C +6 a^{2} b C -2 C a \,b^{2}+2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}^{3}}+\frac {\left (2 A \,a^{3}+3 a A \,b^{2}-4 B \,a^{2} b -B \,b^{3}+a^{3} C +4 C a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(448\)
default	\(\frac {\frac {-\frac {\left (6 A \,a^{2} b +3 a A \,b^{2}+2 A \,b^{3}-2 B \,a^{3}-2 B \,a^{2} b -6 B a \,b^{2}-B \,b^{3}+a^{3} C +6 a^{2} b C +2 C a \,b^{2}+2 C \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {4 \left (9 A \,a^{2} b +A \,b^{3}-3 B \,a^{3}-7 B a \,b^{2}+7 a^{2} b C +3 C \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (6 A \,a^{2} b -3 a A \,b^{2}+2 A \,b^{3}-2 B \,a^{3}+2 B \,a^{2} b -6 B a \,b^{2}+B \,b^{3}-a^{3} C +6 a^{2} b C -2 C a \,b^{2}+2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}^{3}}+\frac {\left (2 A \,a^{3}+3 a A \,b^{2}-4 B \,a^{2} b -B \,b^{3}+a^{3} C +4 C a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(448\)
risch	\(\text {Expression too large to display}\)	\(1845\)

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(2*(-1/2*(6*A*a^2*b+3*A*a*b^2+2*A*b^3-2*B*a^3-2*B*a^2*b-6*B*a*b^2-B*b^3+C*a^3+6*C*a^2*b+2*C*a*b^2+2*C*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*(9*A*a^2*b+A*b^3-3*B*a^3-7*B*a*b^2+7*C*a^2*b+3*C*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*(6*A*a^2*b-3*A*a*b^2+2*A*b^3-2*B*a^3+2*B*a^2*b-6*B*
a*b^2+B*b^3-C*a^3+6*C*a^2*b-2*C*a*b^2+2*C*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-tan(1/2*d*x+1/2*c)^2*b+a+b)^3+(2*A*a^3+3*A*a*b^2-4*B*a^2*b-B*b^3+C*a^3+4*C*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. 667 vs. \(2 (281) = 562\).

Time = 0.37 (sec) , antiderivative size = 1404, normalized size of antiderivative = 4.70 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Too large to display} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

[-1/12*(3*((2*A + C)*a^6 - 4*B*a^5*b + (3*A + 4*C)*a^4*b^2 - B*a^3*b^3 + ((2*A + C)*a^3*b^3 - 4*B*a^2*b^4 + (3
*A + 4*C)*a*b^5 - B*b^6)*cos(d*x + c)^3 + 3*((2*A + C)*a^4*b^2 - 4*B*a^3*b^3 + (3*A + 4*C)*a^2*b^4 - B*a*b^5)*
cos(d*x + c)^2 + 3*((2*A + C)*a^5*b - 4*B*a^4*b^2 + (3*A + 4*C)*a^3*b^3 - B*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*B*a^7 - (18*A + 13*C)*a^6*b + 4*B
*a^5*b^2 + (23*A + 11*C)*a^4*b^3 - 11*B*a^3*b^4 - (7*A - 2*C)*a^2*b^5 + B*a*b^6 + 2*A*b^7 + (C*a^6*b + 2*B*a^5
*b^2 - 11*(A + C)*a^4*b^3 + 11*B*a^3*b^4 + (7*A + 4*C)*a^2*b^5 - 13*B*a*b^6 + 2*(2*A + 3*C)*b^7)*cos(d*x + c)^
2 + 3*(C*a^7 + 2*B*a^6*b - (9*A + 10*C)*a^5*b^2 + 7*B*a^4*b^3 + (8*A + 7*C)*a^3*b^4 - 10*B*a^2*b^5 + (A + 2*C)
*a*b^6 + B*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
*((2*A + C)*a^6 - 4*B*a^5*b + (3*A + 4*C)*a^4*b^2 - B*a^3*b^3 + ((2*A + C)*a^3*b^3 - 4*B*a^2*b^4 + (3*A + 4*C)
*a*b^5 - B*b^6)*cos(d*x + c)^3 + 3*((2*A + C)*a^4*b^2 - 4*B*a^3*b^3 + (3*A + 4*C)*a^2*b^4 - B*a*b^5)*cos(d*x +
c)^2 + 3*((2*A + C)*a^5*b - 4*B*a^4*b^2 + (3*A + 4*C)*a^3*b^3 - B*a^2*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arct
an(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) + (6*B*a^7 - (18*A + 13*C)*a^6*b + 4*B*a^5*b^2 + (23*
A + 11*C)*a^4*b^3 - 11*B*a^3*b^4 - (7*A - 2*C)*a^2*b^5 + B*a*b^6 + 2*A*b^7 + (C*a^6*b + 2*B*a^5*b^2 - 11*(A +
C)*a^4*b^3 + 11*B*a^3*b^4 + (7*A + 4*C)*a^2*b^5 - 13*B*a*b^6 + 2*(2*A + 3*C)*b^7)*cos(d*x + c)^2 + 3*(C*a^7 +
2*B*a^6*b - (9*A + 10*C)*a^5*b^2 + 7*B*a^4*b^3 + (8*A + 7*C)*a^3*b^4 - 10*B*a^2*b^5 + (A + 2*C)*a*b^6 + B*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. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**4,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(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. 966 vs. \(2 (281) = 562\).

Time = 0.38 (sec) , antiderivative size = 966, normalized size of antiderivative = 3.23 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Too large to display} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^4,x, algorithm="giac")

[Out]

-1/3*(3*(2*A*a^3 + C*a^3 - 4*B*a^2*b + 3*A*a*b^2 + 4*C*a*b^2 - B*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*B*a^5*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^5*tan(1/2*d*x + 1/2*c)^5 - 18*A*a^4*
b*tan(1/2*d*x + 1/2*c)^5 - 6*B*a^4*b*tan(1/2*d*x + 1/2*c)^5 - 12*C*a^4*b*tan(1/2*d*x + 1/2*c)^5 + 27*A*a^3*b^2
*tan(1/2*d*x + 1/2*c)^5 + 12*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 27*C*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^2*
b^3*tan(1/2*d*x + 1/2*c)^5 - 27*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 12*C*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*A*a
*b^4*tan(1/2*d*x + 1/2*c)^5 + 12*B*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*C*a*b^4*tan(1/2*d*x + 1/2*c)^5 - 6*A*b^5*t
an(1/2*d*x + 1/2*c)^5 + 3*B*b^5*tan(1/2*d*x + 1/2*c)^5 - 6*C*b^5*tan(1/2*d*x + 1/2*c)^5 + 12*B*a^5*tan(1/2*d*x
+ 1/2*c)^3 - 36*A*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 28*C*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 16*B*a^3*b^2*tan(1/2*d*x
+ 1/2*c)^3 + 32*A*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 16*C*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 28*B*a*b^4*tan(1/2*d
*x + 1/2*c)^3 + 4*A*b^5*tan(1/2*d*x + 1/2*c)^3 + 12*C*b^5*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^5*tan(1/2*d*x + 1/2*c
) + 3*C*a^5*tan(1/2*d*x + 1/2*c) - 18*A*a^4*b*tan(1/2*d*x + 1/2*c) + 6*B*a^4*b*tan(1/2*d*x + 1/2*c) - 12*C*a^4
*b*tan(1/2*d*x + 1/2*c) - 27*A*a^3*b^2*tan(1/2*d*x + 1/2*c) + 12*B*a^3*b^2*tan(1/2*d*x + 1/2*c) - 27*C*a^3*b^2
*tan(1/2*d*x + 1/2*c) - 6*A*a^2*b^3*tan(1/2*d*x + 1/2*c) + 27*B*a^2*b^3*tan(1/2*d*x + 1/2*c) - 12*C*a^2*b^3*ta
n(1/2*d*x + 1/2*c) - 3*A*a*b^4*tan(1/2*d*x + 1/2*c) + 12*B*a*b^4*tan(1/2*d*x + 1/2*c) - 6*C*a*b^4*tan(1/2*d*x
+ 1/2*c) - 6*A*b^5*tan(1/2*d*x + 1/2*c) - 3*B*b^5*tan(1/2*d*x + 1/2*c) - 6*C*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 = 5.20 (sec) , antiderivative size = 516, normalized size of antiderivative = 1.73 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (2\,A\,a^3-B\,b^3+C\,a^3+3\,A\,a\,b^2-4\,B\,a^2\,b+4\,C\,a\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,b^3-2\,B\,a^3+B\,b^3-C\,a^3+2\,C\,b^3-3\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2+2\,B\,a^2\,b-2\,C\,a\,b^2+6\,C\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A\,b^3-3\,B\,a^3+3\,C\,b^3+9\,A\,a^2\,b-7\,B\,a\,b^2+7\,C\,a^2\,b\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,A\,b^3-2\,B\,a^3-B\,b^3+C\,a^3+2\,C\,b^3+3\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2-2\,B\,a^2\,b+2\,C\,a\,b^2+6\,C\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )} \]

[In]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(a + b*cos(c + d*x))^4,x)

[Out]

(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)))*(2*A*a
^3 - B*b^3 + C*a^3 + 3*A*a*b^2 - 4*B*a^2*b + 4*C*a*b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2)) - ((tan(c/2 + (d*x)/2
)*(2*A*b^3 - 2*B*a^3 + B*b^3 - C*a^3 + 2*C*b^3 - 3*A*a*b^2 + 6*A*a^2*b - 6*B*a*b^2 + 2*B*a^2*b - 2*C*a*b^2 + 6
*C*a^2*b))/((a + b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (4*tan(c/2 + (d*x)/2)^3*(A*b^3 - 3*B*a^3 + 3*C*b^3 + 9*
A*a^2*b - 7*B*a*b^2 + 7*C*a^2*b))/(3*(a + b)^2*(a^2 - 2*a*b + b^2)) + (tan(c/2 + (d*x)/2)^5*(2*A*b^3 - 2*B*a^3
- B*b^3 + C*a^3 + 2*C*b^3 + 3*A*a*b^2 + 6*A*a^2*b - 6*B*a*b^2 - 2*B*a^2*b + 2*C*a*b^2 + 6*C*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)))