\(\int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^4} \, dx\) [21]

Optimal result

Integrand size = 31, antiderivative size = 260 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^4} \, dx=\frac {\left (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} e}+\frac {C}{3 b e (a+b \cos (d+e x))^3}-\frac {(A b-a B) \sin (d+e x)}{3 \left (a^2-b^2\right ) e (a+b \cos (d+e x))^3}-\frac {\left (5 a A b-2 a^2 B-3 b^2 B\right ) \sin (d+e x)}{6 \left (a^2-b^2\right )^2 e (a+b \cos (d+e x))^2}-\frac {\left (11 a^2 A b+4 A b^3-2 a^3 B-13 a b^2 B\right ) \sin (d+e x)}{6 \left (a^2-b^2\right )^3 e (a+b \cos (d+e x))} \] Output:

(2*A*a^3+3*A*a*b^2-4*B*a^2*b-B*b^3)*arctan((a-b)^(1/2)*tan(1/2*e*x+1/2*d)/ 
(a+b)^(1/2))/(a-b)^(7/2)/(a+b)^(7/2)/e+1/3*C/b/e/(a+b*cos(e*x+d))^3-1/3*(A 
*b-B*a)*sin(e*x+d)/(a^2-b^2)/e/(a+b*cos(e*x+d))^3-1/6*(5*A*a*b-2*B*a^2-3*B 
*b^2)*sin(e*x+d)/(a^2-b^2)^2/e/(a+b*cos(e*x+d))^2-1/6*(11*A*a^2*b+4*A*b^3- 
2*B*a^3-13*B*a*b^2)*sin(e*x+d)/(a^2-b^2)^3/e/(a+b*cos(e*x+d))
 

Mathematica [A] (verified)

Time = 1.27 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.16 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^4} \, dx=\frac {\frac {24 \left (2 a^3 A+3 a A b^2-4 a^2 b B-b^3 B\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{7/2}}+\frac {-8 a^6 C+24 a^4 b^2 C-24 a^2 b^4 C+8 b^6 C-3 b \left (-24 a^4 A b+3 a^2 A b^3-4 A b^5+8 a^5 B+14 a^3 b^2 B+3 a b^4 B\right ) \sin (d+e x)+6 b^2 \left (9 a^3 A b+a A b^3-2 a^4 B-9 a^2 b^2 B+b^4 B\right ) \sin (2 (d+e x))+11 a^2 A b^4 \sin (3 (d+e x))+4 A b^6 \sin (3 (d+e x))-2 a^3 b^3 B \sin (3 (d+e x))-13 a b^5 B \sin (3 (d+e x))}{b \left (-a^2+b^2\right )^3 (a+b \cos (d+e x))^3}}{24 e} \] Input:

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

Output:

((24*(2*a^3*A + 3*a*A*b^2 - 4*a^2*b*B - b^3*B)*ArcTanh[((a - b)*Tan[(d + e 
*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(7/2) + (-8*a^6*C + 24*a^4*b^2*C - 
 24*a^2*b^4*C + 8*b^6*C - 3*b*(-24*a^4*A*b + 3*a^2*A*b^3 - 4*A*b^5 + 8*a^5 
*B + 14*a^3*b^2*B + 3*a*b^4*B)*Sin[d + e*x] + 6*b^2*(9*a^3*A*b + a*A*b^3 - 
 2*a^4*B - 9*a^2*b^2*B + b^4*B)*Sin[2*(d + e*x)] + 11*a^2*A*b^4*Sin[3*(d + 
 e*x)] + 4*A*b^6*Sin[3*(d + e*x)] - 2*a^3*b^3*B*Sin[3*(d + e*x)] - 13*a*b^ 
5*B*Sin[3*(d + e*x)])/(b*(-a^2 + b^2)^3*(a + b*Cos[d + e*x])^3))/(24*e)
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.16, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.516, Rules used = {3042, 4877, 3042, 3147, 17, 3233, 25, 3042, 3233, 25, 3042, 3233, 27, 3042, 3138, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^4}dx\)

\(\Big \downarrow \) 4877

\(\displaystyle \int \frac {A+B \cos (d+e x)}{(a+b \cos (d+e x))^4}dx+C \int \frac {\sin (d+e x)}{(a+b \cos (d+e x))^4}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sin \left (d+e x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (d+e x+\frac {\pi }{2}\right )\right )^4}dx+C \int \frac {\cos \left (d+e x-\frac {\pi }{2}\right )}{\left (a-b \sin \left (d+e x-\frac {\pi }{2}\right )\right )^4}dx\)

\(\Big \downarrow \) 3147

\(\displaystyle \int \frac {A+B \sin \left (d+e x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (d+e x+\frac {\pi }{2}\right )\right )^4}dx-\frac {C \int \frac {1}{(a+b \cos (d+e x))^4}d(b \cos (d+e x))}{b e}\)

\(\Big \downarrow \) 17

\(\displaystyle \int \frac {A+B \sin \left (d+e x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (d+e x+\frac {\pi }{2}\right )\right )^4}dx+\frac {C}{3 b e (a+b \cos (d+e x))^3}\)

\(\Big \downarrow \) 3233

\(\displaystyle -\frac {\int -\frac {3 (a A-b B)-2 (A b-a B) \cos (d+e x)}{(a+b \cos (d+e x))^3}dx}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (d+e x)}{3 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^3}+\frac {C}{3 b e (a+b \cos (d+e x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {3 (a A-b B)-2 (A b-a B) \cos (d+e x)}{(a+b \cos (d+e x))^3}dx}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (d+e x)}{3 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^3}+\frac {C}{3 b e (a+b \cos (d+e x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {3 (a A-b B)-2 (A b-a B) \sin \left (d+e x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (d+e x+\frac {\pi }{2}\right )\right )^3}dx}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (d+e x)}{3 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^3}+\frac {C}{3 b e (a+b \cos (d+e x))^3}\)

\(\Big \downarrow \) 3233

\(\displaystyle \frac {-\frac {\int -\frac {2 \left (3 A a^2-5 b B a+2 A b^2\right )-\left (-2 B a^2+5 A b a-3 b^2 B\right ) \cos (d+e x)}{(a+b \cos (d+e x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \sin (d+e x)}{2 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (d+e x)}{3 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^3}+\frac {C}{3 b e (a+b \cos (d+e x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {2 \left (3 A a^2-5 b B a+2 A b^2\right )-\left (-2 B a^2+5 A b a-3 b^2 B\right ) \cos (d+e x)}{(a+b \cos (d+e x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \sin (d+e x)}{2 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (d+e x)}{3 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^3}+\frac {C}{3 b e (a+b \cos (d+e x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {2 \left (3 A a^2-5 b B a+2 A b^2\right )+\left (2 B a^2-5 A b a+3 b^2 B\right ) \sin \left (d+e x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (d+e x+\frac {\pi }{2}\right )\right )^2}dx}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \sin (d+e x)}{2 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (d+e x)}{3 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^3}+\frac {C}{3 b e (a+b \cos (d+e x))^3}\)

\(\Big \downarrow \) 3233

\(\displaystyle \frac {\frac {-\frac {\int -\frac {3 \left (2 A a^3-4 b B a^2+3 A b^2 a-b^3 B\right )}{a+b \cos (d+e x)}dx}{a^2-b^2}-\frac {\left (-2 a^3 B+11 a^2 A b-13 a b^2 B+4 A b^3\right ) \sin (d+e x)}{e \left (a^2-b^2\right ) (a+b \cos (d+e x))}}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \sin (d+e x)}{2 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (d+e x)}{3 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^3}+\frac {C}{3 b e (a+b \cos (d+e x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 \left (2 a^3 A-4 a^2 b B+3 a A b^2-b^3 B\right ) \int \frac {1}{a+b \cos (d+e x)}dx}{a^2-b^2}-\frac {\left (-2 a^3 B+11 a^2 A b-13 a b^2 B+4 A b^3\right ) \sin (d+e x)}{e \left (a^2-b^2\right ) (a+b \cos (d+e x))}}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \sin (d+e x)}{2 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (d+e x)}{3 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^3}+\frac {C}{3 b e (a+b \cos (d+e x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (2 a^3 A-4 a^2 b B+3 a A b^2-b^3 B\right ) \int \frac {1}{a+b \sin \left (d+e x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {\left (-2 a^3 B+11 a^2 A b-13 a b^2 B+4 A b^3\right ) \sin (d+e x)}{e \left (a^2-b^2\right ) (a+b \cos (d+e x))}}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \sin (d+e x)}{2 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (d+e x)}{3 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^3}+\frac {C}{3 b e (a+b \cos (d+e x))^3}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {\frac {6 \left (2 a^3 A-4 a^2 b B+3 a A b^2-b^3 B\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (d+e x)\right )+a+b}d\tan \left (\frac {1}{2} (d+e x)\right )}{e \left (a^2-b^2\right )}-\frac {\left (-2 a^3 B+11 a^2 A b-13 a b^2 B+4 A b^3\right ) \sin (d+e x)}{e \left (a^2-b^2\right ) (a+b \cos (d+e x))}}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \sin (d+e x)}{2 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^2}}{3 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (d+e x)}{3 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^3}+\frac {C}{3 b e (a+b \cos (d+e x))^3}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {(A b-a B) \sin (d+e x)}{3 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^3}+\frac {\frac {\frac {6 \left (2 a^3 A-4 a^2 b B+3 a A b^2-b^3 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a+b}}\right )}{e \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\left (-2 a^3 B+11 a^2 A b-13 a b^2 B+4 A b^3\right ) \sin (d+e x)}{e \left (a^2-b^2\right ) (a+b \cos (d+e x))}}{2 \left (a^2-b^2\right )}-\frac {\left (-2 a^2 B+5 a A b-3 b^2 B\right ) \sin (d+e x)}{2 e \left (a^2-b^2\right ) (a+b \cos (d+e x))^2}}{3 \left (a^2-b^2\right )}+\frac {C}{3 b e (a+b \cos (d+e x))^3}\)

Input:

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

Output:

C/(3*b*e*(a + b*Cos[d + e*x])^3) - ((A*b - a*B)*Sin[d + e*x])/(3*(a^2 - b^ 
2)*e*(a + b*Cos[d + e*x])^3) + (-1/2*((5*a*A*b - 2*a^2*B - 3*b^2*B)*Sin[d 
+ e*x])/((a^2 - b^2)*e*(a + b*Cos[d + e*x])^2) + ((6*(2*a^3*A + 3*a*A*b^2 
- 4*a^2*b*B - b^3*B)*ArcTan[(Sqrt[a - b]*Tan[(d + e*x)/2])/Sqrt[a + b]])/( 
Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*e) - ((11*a^2*A*b + 4*A*b^3 - 2*a^3*B 
- 13*a*b^2*B)*Sin[d + e*x])/((a^2 - b^2)*e*(a + b*Cos[d + e*x])))/(2*(a^2 
- b^2)))/(3*(a^2 - b^2))
 

Defintions of rubi rules used

rule 17

Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[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 3147

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) 
/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p 
 - 1)/2] && NeQ[a^2 - b^2, 0]
 

rule 3233

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] + Simp[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 4877

Int[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] : 
> With[{e = FreeFactors[Cos[c*(a + b*x)], x]}, Int[ActivateTrig[u*v], x] + 
Simp[d   Int[ActivateTrig[u]*Sin[c*(a + b*x)]^n, x], x] /; FunctionOfQ[Cos[ 
c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] &&  !FreeQ[v, x] && Intege 
rQ[(n - 1)/2] && NonsumQ[u] && (EqQ[F, Sin] || EqQ[F, sin])
 

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 456, normalized size of antiderivative = 1.75

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (6 A \,a^{2} b +3 A a \,b^{2}+2 A \,b^{3}-2 B \,a^{3}-2 B \,a^{2} b -6 B a \,b^{2}-B \,b^{3}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{5}}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 C \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{a -b}-\frac {4 \left (9 A \,a^{2} b +A \,b^{3}-3 B \,a^{3}-7 B a \,b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {4 C a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{a^{2}-2 a b +b^{2}}-\frac {\left (6 A \,a^{2} b -3 A a \,b^{2}+2 A \,b^{3}-2 B \,a^{3}+2 B \,a^{2} b -6 B a \,b^{2}+B \,b^{3}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {2 C \left (3 a^{2}+b^{2}\right )}{3 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{\left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+a +b \right )^{3}}+\frac {\left (2 A \,a^{3}+3 A a \,b^{2}-4 B \,a^{2} b -B \,b^{3}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{e}\)	\(456\)
default	\(\frac {\frac {-\frac {\left (6 A \,a^{2} b +3 A a \,b^{2}+2 A \,b^{3}-2 B \,a^{3}-2 B \,a^{2} b -6 B a \,b^{2}-B \,b^{3}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{5}}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 C \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{a -b}-\frac {4 \left (9 A \,a^{2} b +A \,b^{3}-3 B \,a^{3}-7 B a \,b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {4 C a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{a^{2}-2 a b +b^{2}}-\frac {\left (6 A \,a^{2} b -3 A a \,b^{2}+2 A \,b^{3}-2 B \,a^{3}+2 B \,a^{2} b -6 B a \,b^{2}+B \,b^{3}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {2 C \left (3 a^{2}+b^{2}\right )}{3 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{\left (a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}-b \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+a +b \right )^{3}}+\frac {\left (2 A \,a^{3}+3 A a \,b^{2}-4 B \,a^{2} b -B \,b^{3}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{e}\)	\(456\)
risch	\(\text {Expression too large to display}\)	\(1282\)

Input:

int((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^4,x,method=_RETURNVERBO 
SE)
 

Output:

1/e*(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)/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*e*x+1/2*d)^5-C/(a-b)*tan(1/2*e* 
x+1/2*d)^4-2/3*(9*A*a^2*b+A*b^3-3*B*a^3-7*B*a*b^2)/(a^2-2*a*b+b^2)/(a^2+2* 
a*b+b^2)*tan(1/2*e*x+1/2*d)^3-2*C*a/(a^2-2*a*b+b^2)*tan(1/2*e*x+1/2*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)/(a+b)/( 
a^3-3*a^2*b+3*a*b^2-b^3)*tan(1/2*e*x+1/2*d)-1/3*C*(3*a^2+b^2)/(a^3-3*a^2*b 
+3*a*b^2-b^3))/(a*tan(1/2*e*x+1/2*d)^2-b*tan(1/2*e*x+1/2*d)^2+a+b)^3+(2*A* 
a^3+3*A*a*b^2-4*B*a^2*b-B*b^3)/(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*e*x+1/2*d)/((a-b)*(a+b))^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 632 vs. \(2 (243) = 486\).

Time = 0.22 (sec) , antiderivative size = 1334, normalized size of antiderivative = 5.13 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^4} \, dx=\text {Too large to display} \] Input:

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^4,x, algorithm="f 
ricas")
 

Output:

[1/12*(4*C*a^8 - 16*C*a^6*b^2 + 24*C*a^4*b^4 - 16*C*a^2*b^6 + 4*C*b^8 - 3* 
(2*A*a^6*b - 4*B*a^5*b^2 + 3*A*a^4*b^3 - B*a^3*b^4 + (2*A*a^3*b^4 - 4*B*a^ 
2*b^5 + 3*A*a*b^6 - B*b^7)*cos(e*x + d)^3 + 3*(2*A*a^4*b^3 - 4*B*a^3*b^4 + 
 3*A*a^2*b^5 - B*a*b^6)*cos(e*x + d)^2 + 3*(2*A*a^5*b^2 - 4*B*a^4*b^3 + 3* 
A*a^3*b^4 - B*a^2*b^5)*cos(e*x + d))*sqrt(-a^2 + b^2)*log((2*a*b*cos(e*x + 
 d) + (2*a^2 - b^2)*cos(e*x + d)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(e*x + d) + 
b)*sin(e*x + d) - a^2 + 2*b^2)/(b^2*cos(e*x + d)^2 + 2*a*b*cos(e*x + d) + 
a^2)) + 2*(6*B*a^7*b - 18*A*a^6*b^2 + 4*B*a^5*b^3 + 23*A*a^4*b^4 - 11*B*a^ 
3*b^5 - 7*A*a^2*b^6 + B*a*b^7 + 2*A*b^8 + (2*B*a^5*b^3 - 11*A*a^4*b^4 + 11 
*B*a^3*b^5 + 7*A*a^2*b^6 - 13*B*a*b^7 + 4*A*b^8)*cos(e*x + d)^2 + 3*(2*B*a 
^6*b^2 - 9*A*a^5*b^3 + 7*B*a^4*b^4 + 8*A*a^3*b^5 - 10*B*a^2*b^6 + A*a*b^7 
+ B*b^8)*cos(e*x + d))*sin(e*x + d))/((a^8*b^4 - 4*a^6*b^6 + 6*a^4*b^8 - 4 
*a^2*b^10 + b^12)*e*cos(e*x + d)^3 + 3*(a^9*b^3 - 4*a^7*b^5 + 6*a^5*b^7 - 
4*a^3*b^9 + a*b^11)*e*cos(e*x + d)^2 + 3*(a^10*b^2 - 4*a^8*b^4 + 6*a^6*b^6 
 - 4*a^4*b^8 + a^2*b^10)*e*cos(e*x + d) + (a^11*b - 4*a^9*b^3 + 6*a^7*b^5 
- 4*a^5*b^7 + a^3*b^9)*e), 1/6*(2*C*a^8 - 8*C*a^6*b^2 + 12*C*a^4*b^4 - 8*C 
*a^2*b^6 + 2*C*b^8 + 3*(2*A*a^6*b - 4*B*a^5*b^2 + 3*A*a^4*b^3 - B*a^3*b^4 
+ (2*A*a^3*b^4 - 4*B*a^2*b^5 + 3*A*a*b^6 - B*b^7)*cos(e*x + d)^3 + 3*(2*A* 
a^4*b^3 - 4*B*a^3*b^4 + 3*A*a^2*b^5 - B*a*b^6)*cos(e*x + d)^2 + 3*(2*A*a^5 
*b^2 - 4*B*a^4*b^3 + 3*A*a^3*b^4 - B*a^2*b^5)*cos(e*x + d))*sqrt(a^2 - ...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^4,x, algorithm="m 
axima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 918 vs. \(2 (243) = 486\).

Time = 0.19 (sec) , antiderivative size = 918, normalized size of antiderivative = 3.53 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^4} \, dx=\text {Too large to display} \] Input:

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+b*cos(e*x+d))^4,x, algorithm="g 
iac")
 

Output:

-1/3*(3*(2*A*a^3 - 4*B*a^2*b + 3*A*a*b^2 - B*b^3)*(pi*floor(1/2*(e*x + d)/ 
pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*e*x + 1/2*d) - b*tan(1/2*e* 
x + 1/2*d))/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*e*x + 1/2*d)^5 - 18*A*a^4*b*tan(1/2*e*x + 1/2 
*d)^5 - 6*B*a^4*b*tan(1/2*e*x + 1/2*d)^5 + 27*A*a^3*b^2*tan(1/2*e*x + 1/2* 
d)^5 + 12*B*a^3*b^2*tan(1/2*e*x + 1/2*d)^5 - 6*A*a^2*b^3*tan(1/2*e*x + 1/2 
*d)^5 - 27*B*a^2*b^3*tan(1/2*e*x + 1/2*d)^5 + 3*A*a*b^4*tan(1/2*e*x + 1/2* 
d)^5 + 12*B*a*b^4*tan(1/2*e*x + 1/2*d)^5 - 6*A*b^5*tan(1/2*e*x + 1/2*d)^5 
+ 3*B*b^5*tan(1/2*e*x + 1/2*d)^5 - 6*C*a^5*tan(1/2*e*x + 1/2*d)^4 - 6*C*a^ 
4*b*tan(1/2*e*x + 1/2*d)^4 + 12*C*a^3*b^2*tan(1/2*e*x + 1/2*d)^4 + 12*C*a^ 
2*b^3*tan(1/2*e*x + 1/2*d)^4 - 6*C*a*b^4*tan(1/2*e*x + 1/2*d)^4 - 6*C*b^5* 
tan(1/2*e*x + 1/2*d)^4 + 12*B*a^5*tan(1/2*e*x + 1/2*d)^3 - 36*A*a^4*b*tan( 
1/2*e*x + 1/2*d)^3 + 16*B*a^3*b^2*tan(1/2*e*x + 1/2*d)^3 + 32*A*a^2*b^3*ta 
n(1/2*e*x + 1/2*d)^3 - 28*B*a*b^4*tan(1/2*e*x + 1/2*d)^3 + 4*A*b^5*tan(1/2 
*e*x + 1/2*d)^3 - 12*C*a^5*tan(1/2*e*x + 1/2*d)^2 - 24*C*a^4*b*tan(1/2*e*x 
 + 1/2*d)^2 + 24*C*a^2*b^3*tan(1/2*e*x + 1/2*d)^2 + 12*C*a*b^4*tan(1/2*e*x 
 + 1/2*d)^2 + 6*B*a^5*tan(1/2*e*x + 1/2*d) - 18*A*a^4*b*tan(1/2*e*x + 1/2* 
d) + 6*B*a^4*b*tan(1/2*e*x + 1/2*d) - 27*A*a^3*b^2*tan(1/2*e*x + 1/2*d) + 
12*B*a^3*b^2*tan(1/2*e*x + 1/2*d) - 6*A*a^2*b^3*tan(1/2*e*x + 1/2*d) + 27* 
B*a^2*b^3*tan(1/2*e*x + 1/2*d) - 3*A*a*b^4*tan(1/2*e*x + 1/2*d) + 12*B*...
 

Mupad [B] (verification not implemented)

Time = 2.88 (sec) , antiderivative size = 502, normalized size of antiderivative = 1.93 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^4} \, dx=\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (2\,A\,a^3-4\,B\,a^2\,b+3\,A\,a\,b^2-B\,b^3\right )}{e\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {2\,C\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4}{a-b}+\frac {2\,C\,\left (3\,a^2+b^2\right )}{3\,{\left (a-b\right )}^3}+\frac {4\,C\,a\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2}{{\left (a-b\right )}^2}+\frac {4\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (-3\,B\,a^3+9\,A\,a^2\,b-7\,B\,a\,b^2+A\,b^3\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}-\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^5\,\left (2\,B\,a^3-2\,A\,b^3+B\,b^3-3\,A\,a\,b^2-6\,A\,a^2\,b+6\,B\,a\,b^2+2\,B\,a^2\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,A\,b^3-2\,B\,a^3+B\,b^3-3\,A\,a\,b^2+6\,A\,a^2\,b-6\,B\,a\,b^2+2\,B\,a^2\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{e\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1239, normalized size of antiderivative = 4.77 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+b \cos (d+e x))^4} \, dx =\text {Too large to display} \] Input:

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

Output:

(12*sqrt(a**2 - b**2)*atan((tan((d + e*x)/2)*a - tan((d + e*x)/2)*b)/sqrt( 
a**2 - b**2))*cos(d + e*x)*sin(d + e*x)**2*a**2*b**3 + 6*sqrt(a**2 - b**2) 
*atan((tan((d + e*x)/2)*a - tan((d + e*x)/2)*b)/sqrt(a**2 - b**2))*cos(d + 
 e*x)*sin(d + e*x)**2*b**5 - 36*sqrt(a**2 - b**2)*atan((tan((d + e*x)/2)*a 
 - tan((d + e*x)/2)*b)/sqrt(a**2 - b**2))*cos(d + e*x)*a**4*b - 30*sqrt(a* 
*2 - b**2)*atan((tan((d + e*x)/2)*a - tan((d + e*x)/2)*b)/sqrt(a**2 - b**2 
))*cos(d + e*x)*a**2*b**3 - 6*sqrt(a**2 - b**2)*atan((tan((d + e*x)/2)*a - 
 tan((d + e*x)/2)*b)/sqrt(a**2 - b**2))*cos(d + e*x)*b**5 + 36*sqrt(a**2 - 
 b**2)*atan((tan((d + e*x)/2)*a - tan((d + e*x)/2)*b)/sqrt(a**2 - b**2))*s 
in(d + e*x)**2*a**3*b**2 + 18*sqrt(a**2 - b**2)*atan((tan((d + e*x)/2)*a - 
 tan((d + e*x)/2)*b)/sqrt(a**2 - b**2))*sin(d + e*x)**2*a*b**4 - 12*sqrt(a 
**2 - b**2)*atan((tan((d + e*x)/2)*a - tan((d + e*x)/2)*b)/sqrt(a**2 - b** 
2))*a**5 - 42*sqrt(a**2 - b**2)*atan((tan((d + e*x)/2)*a - tan((d + e*x)/2 
)*b)/sqrt(a**2 - b**2))*a**3*b**2 - 18*sqrt(a**2 - b**2)*atan((tan((d + e* 
x)/2)*a - tan((d + e*x)/2)*b)/sqrt(a**2 - b**2))*a*b**4 - 2*cos(d + e*x)*s 
in(d + e*x)**2*a**3*b**2*c + 6*cos(d + e*x)*sin(d + e*x)**2*a**2*b**3*c - 
6*cos(d + e*x)*sin(d + e*x)**2*a*b**4*c + 2*cos(d + e*x)*sin(d + e*x)**2*b 
**5*c + 21*cos(d + e*x)*sin(d + e*x)*a**4*b**2 - 24*cos(d + e*x)*sin(d + e 
*x)*a**2*b**4 + 3*cos(d + e*x)*sin(d + e*x)*b**6 + 6*cos(d + e*x)*a**5*c - 
 18*cos(d + e*x)*a**4*b*c + 20*cos(d + e*x)*a**3*b**2*c - 12*cos(d + e*...