\(\int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx\) [1012]

Optimal result

Integrand size = 48, antiderivative size = 175 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {\left (2 a^2 b B+b^3 B-2 a^3 C-4 a b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {b (b B-2 a C) \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {b \left (3 a b B-4 a^2 C-2 b^2 C\right ) \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

(2*B*a^2*b+B*b^3-2*C*a^3-4*C*a*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/ 
(a+b)^(1/2))/(a-b)^(5/2)/(a+b)^(5/2)/d-1/2*b*(B*b-2*C*a)*sin(d*x+c)/(a^2-b 
^2)/d/(a+b*cos(d*x+c))^2-1/2*b*(3*B*a*b-4*C*a^2-2*C*b^2)*sin(d*x+c)/(a^2-b 
^2)^2/d/(a+b*cos(d*x+c))
 

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.98 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {\frac {2 \left (-2 a^2 b B-b^3 B+2 a^3 C+4 a b^2 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+\frac {b (-b B+2 a C) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}+\frac {b \left (-3 a b B+4 a^2 C+2 b^2 C\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}}{2 d} \] Input:

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

Output:

((2*(-2*a^2*b*B - b^3*B + 2*a^3*C + 4*a*b^2*C)*ArcTanh[((a - b)*Tan[(c + d 
*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + (b*(-(b*B) + 2*a*C)*Sin[c 
+ d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])^2) + (b*(-3*a*b*B + 4*a^2*C 
+ 2*b^2*C)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])))/(2*d)
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.21, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {2014, 3042, 3233, 25, 3042, 3233, 25, 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.

\(\displaystyle \int \frac {a^2 (-C)+a b B+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx\)

\(\Big \downarrow \) 2014

\(\displaystyle \frac {\int \frac {C \cos (c+d x) b^3+(b B-a C) b^2}{(a+b \cos (c+d x))^3}dx}{b^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {\frac {2 b^2 \left (-2 a^3 C+2 a^2 b B-4 a b^2 C+b^3 B\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {b^3 \left (-4 a^2 C+3 a b B-2 b^2 C\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{b^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {2 b^2 \left (-2 a^3 C+2 a^2 b B-4 a b^2 C+b^3 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {b^3 \left (-4 a^2 C+3 a b B-2 b^2 C\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{b^2}\)

Input:

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

Output:

(-1/2*(b^3*(b*B - 2*a*C)*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x]) 
^2) + ((2*b^2*(2*a^2*b*B + b^3*B - 2*a^3*C - 4*a*b^2*C)*ArcTan[(Sqrt[a - b 
]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*d) 
- (b^3*(3*a*b*B - 4*a^2*C - 2*b^2*C)*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*C 
os[c + d*x])))/(2*(a^2 - b^2)))/b^2
 

Defintions of rubi rules used

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 2014

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_S 
ymbol] :> Simp[1/b^2   Int[u*(a + b*v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], 
x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] && LeQ 
[m, -1]
 

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 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]
 

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.39

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (4 B a b +B \,b^{2}-6 a^{2} C -2 a b C -2 C \,b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (4 B a b -B \,b^{2}-6 a^{2} C +2 a b C -2 C \,b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (2 B \,a^{2} b +B \,b^{3}-2 a^{3} C -4 C a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(244\)
default	\(\frac {\frac {-\frac {\left (4 B a b +B \,b^{2}-6 a^{2} C -2 a b C -2 C \,b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (4 B a b -B \,b^{2}-6 a^{2} C +2 a b C -2 C \,b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (2 B \,a^{2} b +B \,b^{3}-2 a^{3} C -4 C a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(244\)
risch	\(\frac {i \left (-2 B \,a^{2} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-B \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}+2 C \,a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}+4 C a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-6 B \,a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+8 C \,a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+8 C \,a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 C \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-10 B \,a^{2} b^{2} {\mathrm e}^{i \left (d x +c \right )}+B \,b^{4} {\mathrm e}^{i \left (d x +c \right )}+14 C \,a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}+4 C a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}-3 B a \,b^{3}+4 C \,a^{2} b^{2}+2 C \,b^{4}\right )}{\left (a^{2}-b^{2}\right )^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )} b +2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B \,a^{2} b}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B \,b^{3}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{3} C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C a \,b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B \,a^{2} b}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B \,b^{3}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{3} C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C a \,b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\)	\(952\)

Input:

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

Output:

1/d*(2*(-1/2*(4*B*a*b+B*b^2-6*C*a^2-2*C*a*b-2*C*b^2)*b/(a-b)/(a^2+2*a*b+b^ 
2)*tan(1/2*d*x+1/2*c)^3-1/2*(4*B*a*b-B*b^2-6*C*a^2+2*C*a*b-2*C*b^2)*b/(a+b 
)/(a^2-2*a*b+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)^2+(2*B*a^2*b+B*b^3-2*C*a^3-4*C*a*b^2)/(a^4-2*a^2*b^2+b^4)/ 
((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. 367 vs. \(2 (165) = 330\).

Time = 0.16 (sec) , antiderivative size = 803, normalized size of antiderivative = 4.59 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/4*((2*C*a^5 - 2*B*a^4*b + 4*C*a^3*b^2 - B*a^2*b^3 + (2*C*a^3*b^2 - 2*B* 
a^2*b^3 + 4*C*a*b^4 - B*b^5)*cos(d*x + c)^2 + 2*(2*C*a^4*b - 2*B*a^3*b^2 + 
 4*C*a^2*b^3 - B*a*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*C*a^5*b - 4*B*a^4*b^2 - 6*C*a^3*b^3 + 5*B*a^2*b^4 - B*b^6 + 
(4*C*a^4*b^2 - 3*B*a^3*b^3 - 2*C*a^2*b^4 + 3*B*a*b^5 - 2*C*b^6)*cos(d*x + 
c))*sin(d*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^ 
2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^8 - 3*a^ 
6*b^2 + 3*a^4*b^4 - a^2*b^6)*d), -1/2*((2*C*a^5 - 2*B*a^4*b + 4*C*a^3*b^2 
- B*a^2*b^3 + (2*C*a^3*b^2 - 2*B*a^2*b^3 + 4*C*a*b^4 - B*b^5)*cos(d*x + c) 
^2 + 2*(2*C*a^4*b - 2*B*a^3*b^2 + 4*C*a^2*b^3 - B*a*b^4)*cos(d*x + c))*sqr 
t(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) 
- (6*C*a^5*b - 4*B*a^4*b^2 - 6*C*a^3*b^3 + 5*B*a^2*b^4 - B*b^6 + (4*C*a^4* 
b^2 - 3*B*a^3*b^3 - 2*C*a^2*b^4 + 3*B*a*b^5 - 2*C*b^6)*cos(d*x + c))*sin(d 
*x + c))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d*cos(d*x + c)^2 + 2*(a^ 
7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^8 - 3*a^6*b^2 + 3 
*a^4*b^4 - a^2*b^6)*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

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. 381 vs. \(2 (165) = 330\).

Time = 0.18 (sec) , antiderivative size = 381, normalized size of antiderivative = 2.18 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=-\frac {\frac {{\left (2 \, C a^{3} - 2 \, B a^{2} b + 4 \, C a b^{2} - B b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {6 \, C a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, B a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}}}{d} \] Input:

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

Output:

-((2*C*a^3 - 2*B*a^2*b + 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^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) - (6*C*a^3 
*b*tan(1/2*d*x + 1/2*c)^3 - 4*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 4*C*a^2*b 
^2*tan(1/2*d*x + 1/2*c)^3 + 3*B*a*b^3*tan(1/2*d*x + 1/2*c)^3 + B*b^4*tan(1 
/2*d*x + 1/2*c)^3 - 2*C*b^4*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^3*b*tan(1/2*d*x 
 + 1/2*c) - 4*B*a^2*b^2*tan(1/2*d*x + 1/2*c) + 4*C*a^2*b^2*tan(1/2*d*x + 1 
/2*c) - 3*B*a*b^3*tan(1/2*d*x + 1/2*c) + B*b^4*tan(1/2*d*x + 1/2*c) + 2*C* 
b^4*tan(1/2*d*x + 1/2*c))/((a^4 - 2*a^2*b^2 + b^4)*(a*tan(1/2*d*x + 1/2*c) 
^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2))/d
 

Mupad [B] (verification not implemented)

Time = 2.68 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.53 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,C\,b^3-B\,b^3-4\,B\,a\,b^2+2\,C\,a\,b^2+6\,C\,a^2\,b\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B\,b^3+2\,C\,b^3-4\,B\,a\,b^2-2\,C\,a\,b^2+6\,C\,a^2\,b\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (2\,a\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}}\right )\,\left (-2\,C\,a^3+2\,B\,a^2\,b-4\,C\,a\,b^2+B\,b^3\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 1140, normalized size of antiderivative = 6.51 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t(a**2 - b**2))*cos(c + d*x)*a**4*b*c + 8*sqrt(a**2 - b**2)*atan((tan((c + 
 d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a**3*b**3 
 - 16*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t(a**2 - b**2))*cos(c + d*x)*a**2*b**3*c + 4*sqrt(a**2 - b**2)*atan((tan(( 
c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d*x)*a*b**5 
 + 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt 
(a**2 - b**2))*sin(c + d*x)**2*a**3*b**2*c - 4*sqrt(a**2 - b**2)*atan((tan 
((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a 
**2*b**4 + 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2) 
*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a*b**4*c - 2*sqrt(a**2 - b**2)*atan 
((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x) 
**2*b**6 - 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2) 
*b)/sqrt(a**2 - b**2))*a**5*c + 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2) 
*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**4*b**2 - 12*sqrt(a**2 - b** 
2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**3* 
b**2*c + 6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b 
)/sqrt(a**2 - b**2))*a**2*b**4 - 8*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2 
)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a*b**4*c + 2*sqrt(a**2 - b**2 
)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*b**...