\(\int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx\) [481]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F(-1)]
Maxima [F(-2)]
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 21, antiderivative size = 206 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {a \left (a^2+4 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{6 b \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \] Output:

a*(a^2+4*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^2*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^3+1/6*a 
*(a^2-6*b^2)*sin(d*x+c)/b/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^2+1/6*(a^4-10*a^2 
*b^2-6*b^4)*sin(d*x+c)/b/(a^2-b^2)^3/d/(a+b*cos(d*x+c))
 

Mathematica [A] (verified)

Time = 1.68 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.79 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {\frac {6 a \left (a^2+4 b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{7/2}}+\frac {\left (-13 a^4 b-2 a^2 b^3+3 a \left (a^4-9 a^2 b^2-2 b^4\right ) \cos (c+d x)+b \left (a^4-10 a^2 b^2-6 b^4\right ) \cos ^2(c+d x)\right ) \sin (c+d x)}{(a-b)^3 (a+b)^3 (a+b \cos (c+d x))^3}}{6 d} \] Input:

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

Output:

((6*a*(a^2 + 4*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/ 
(-a^2 + b^2)^(7/2) + ((-13*a^4*b - 2*a^2*b^3 + 3*a*(a^4 - 9*a^2*b^2 - 2*b^ 
4)*Cos[c + d*x] + b*(a^4 - 10*a^2*b^2 - 6*b^4)*Cos[c + d*x]^2)*Sin[c + d*x 
])/((a - b)^3*(a + b)^3*(a + b*Cos[c + d*x])^3))/(6*d)
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.19, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 3269, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4}dx\)

\(\Big \downarrow \) 3269

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

\(\displaystyle \frac {\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\int -\frac {2 b \left (2 a^2+3 b^2\right )+a \left (a^2-6 b^2\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2}dx}{2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {2 b \left (2 a^2+3 b^2\right )+a \left (a^2-6 b^2\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2}dx}{2 \left (a^2-b^2\right )}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {2 b \left (2 a^2+3 b^2\right )+a \left (a^2-6 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 \left (a^2-b^2\right )}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3233

\(\displaystyle \frac {\frac {\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int -\frac {3 a b \left (a^2+4 b^2\right )}{a+b \cos (c+d x)}dx}{a^2-b^2}}{2 \left (a^2-b^2\right )}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 a b \left (a^2+4 b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a^2-b^2}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 a b \left (a^2+4 b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\frac {\frac {6 a b \left (a^2+4 b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}+\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {a \left (a^2-6 b^2\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {\frac {6 a b \left (a^2+4 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}+\frac {\left (a^4-10 a^2 b^2-6 b^4\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\)

Input:

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

Output:

-1/3*(a^2*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^3) + ((a*(a^ 
2 - 6*b^2)*Sin[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + ((6*a* 
b*(a^2 + 4*b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[ 
a - b]*Sqrt[a + b]*(a^2 - b^2)*d) + ((a^4 - 10*a^2*b^2 - 6*b^4)*Sin[c + d* 
x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x])))/(2*(a^2 - b^2)))/(3*b*(a^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 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]
 

rule 3269
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^2, x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Cos[e 
+ f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] - Simp[ 
1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1) 
*(2*b*c*d - a*(c^2 + d^2)) + (a^2*d^2 - 2*a*b*c*d*(m + 2) + b^2*(d^2*(m + 1 
) + c^2*(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]
 
Maple [A] (verified)

Time = 1.49 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.41

method result size
derivativedivides \(\frac {\frac {-\frac {\left (a^{3}+6 a^{2} b +2 b^{2} a +2 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {4 \left (7 a^{2}+3 b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (a^{3}-6 a^{2} b +2 b^{2} a -2 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}+\frac {a \left (a^{2}+4 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) \(290\)
default \(\frac {\frac {-\frac {\left (a^{3}+6 a^{2} b +2 b^{2} a +2 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {4 \left (7 a^{2}+3 b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (a^{3}-6 a^{2} b +2 b^{2} a -2 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}+\frac {a \left (a^{2}+4 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) \(290\)
risch \(\frac {i \left (-3 a^{3} b^{4} {\mathrm e}^{5 i \left (d x +c \right )}-12 a \,b^{6} {\mathrm e}^{5 i \left (d x +c \right )}+6 a^{6} b \,{\mathrm e}^{4 i \left (d x +c \right )}-33 a^{4} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-42 a^{2} b^{5} {\mathrm e}^{4 i \left (d x +c \right )}-6 b^{7} {\mathrm e}^{4 i \left (d x +c \right )}+4 a^{7} {\mathrm e}^{3 i \left (d x +c \right )}-34 a^{5} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-84 a^{3} b^{4} {\mathrm e}^{3 i \left (d x +c \right )}-36 a \,b^{6} {\mathrm e}^{3 i \left (d x +c \right )}+6 a^{6} b \,{\mathrm e}^{2 i \left (d x +c \right )}-84 a^{4} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-60 a^{2} b^{5} {\mathrm e}^{2 i \left (d x +c \right )}-12 b^{7} {\mathrm e}^{2 i \left (d x +c \right )}+6 a^{5} b^{2} {\mathrm e}^{i \left (d x +c \right )}-57 a^{3} b^{4} {\mathrm e}^{i \left (d x +c \right )}-24 a \,b^{6} {\mathrm e}^{i \left (d x +c \right )}+a^{4} b^{3}-10 a^{2} b^{5}-6 b^{7}\right )}{3 b^{2} \left (a^{2}-b^{2}\right )^{3} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {2 a \,b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\) \(674\)

Input:

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

Output:

1/d*(2*(-1/2*(a^3+6*a^2*b+2*a*b^2+2*b^3)/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*t 
an(1/2*d*x+1/2*c)^5-2/3*(7*a^2+3*b^2)*b/(a^2-2*a*b+b^2)/(a^2+2*a*b+b^2)*ta 
n(1/2*d*x+1/2*c)^3+1/2*(a^3-6*a^2*b+2*a*b^2-2*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+a*(a^2+4*b^2)/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(1/2)*a 
rctan((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. 412 vs. \(2 (191) = 382\).

Time = 0.13 (sec) , antiderivative size = 893, normalized size of antiderivative = 4.33 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/12*(3*(a^6 + 4*a^4*b^2 + (a^3*b^3 + 4*a*b^5)*cos(d*x + c)^3 + 3*(a^4*b^ 
2 + 4*a^2*b^4)*cos(d*x + c)^2 + 3*(a^5*b + 4*a^3*b^3)*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*(13*a^6*b - 11*a^4*b^3 - 2*a^2*b^ 
5 - (a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + 6*b^7)*cos(d*x + c)^2 - 3*(a^7 - 10* 
a^5*b^2 + 7*a^3*b^4 + 2*a*b^6)*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*(a^6 + 4*a^4*b^2 + (a^3* 
b^3 + 4*a*b^5)*cos(d*x + c)^3 + 3*(a^4*b^2 + 4*a^2*b^4)*cos(d*x + c)^2 + 3 
*(a^5*b + 4*a^3*b^3)*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))) - (13*a^6*b - 11*a^4*b^3 - 2*a^2*b^5 
 - (a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + 6*b^7)*cos(d*x + c)^2 - 3*(a^7 - 10*a 
^5*b^2 + 7*a^3*b^4 + 2*a*b^6)*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. \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(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. 427 vs. \(2 (191) = 382\).

Time = 0.60 (sec) , antiderivative size = 427, normalized size of antiderivative = 2.07 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=-\frac {\frac {3 \, {\left (a^{3} + 4 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 28 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{3}}}{3 \, d} \] Input:

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

Output:

-1/3*(3*(a^3 + 4*a*b^2)*(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)) + (3*a^5*tan(1/2 
*d*x + 1/2*c)^5 + 12*a^4*b*tan(1/2*d*x + 1/2*c)^5 - 27*a^3*b^2*tan(1/2*d*x 
 + 1/2*c)^5 + 12*a^2*b^3*tan(1/2*d*x + 1/2*c)^5 - 6*a*b^4*tan(1/2*d*x + 1/ 
2*c)^5 + 6*b^5*tan(1/2*d*x + 1/2*c)^5 + 28*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 
16*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 12*b^5*tan(1/2*d*x + 1/2*c)^3 - 3*a^5* 
tan(1/2*d*x + 1/2*c) + 12*a^4*b*tan(1/2*d*x + 1/2*c) + 27*a^3*b^2*tan(1/2* 
d*x + 1/2*c) + 12*a^2*b^3*tan(1/2*d*x + 1/2*c) + 6*a*b^4*tan(1/2*d*x + 1/2 
*c) + 6*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 = 46.58 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.85 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx=\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+4\,b^2\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}\,\left (a^3+4\,a\,b^2\right )}\right )\,\left (a^2+4\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (a^3+6\,a^2\,b+2\,a\,b^2+2\,b^3\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,a^2\,b+3\,b^3\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^3-6\,a^2\,b+2\,a\,b^2-2\,b^3\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 1095, normalized size of antiderivative = 5.32 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:

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

Output:

(6*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)*sin(c + d*x)**2*a**3*b**3 + 24*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)*sin(c + d*x)**2*a*b**5 - 18*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**5*b - 78*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 - 24*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 + 18*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**4*b**2 + 72*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 - 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**6 - 42*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 - 72*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 - 3*cos( 
c + d*x)*sin(c + d*x)*a**7 + 30*cos(c + d*x)*sin(c + d*x)*a**5*b**2 - 21*c 
os(c + d*x)*sin(c + d*x)*a**3*b**4 - 6*cos(c + d*x)*sin(c + d*x)*a*b**6 + 
sin(c + d*x)**3*a**6*b - 11*sin(c + d*x)**3*a**4*b**3 + 4*sin(c + d*x)**3* 
a**2*b**5 + 6*sin(c + d*x)**3*b**7 + 12*sin(c + d*x)*a**6*b - 6*sin(c + d* 
x)*a**2*b**5 - 6*sin(c + d*x)*b**7)/(6*d*(cos(c + d*x)*sin(c + d*x)**2*...