\(\int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [472]

Optimal result

Integrand size = 21, antiderivative size = 149 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\left (a^2+2 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {a^2 \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {a \left (a^2-4 b^2\right ) \sin (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {-\frac {2 \left (a^2+2 b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+\frac {a \left (-3 a b+\left (a^2-4 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))^2}}{2 d} \] Input:

Integrate[Cos[c + d*x]^2/(a + b*Cos[c + d*x])^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3269, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3269

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

Int[Cos[c + d*x]^2/(a + b*Cos[c + d*x])^3,x]
 

Output:

-1/2*(a^2*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + ((2*b*( 
a^2 + 2*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*(a^2 - 4*b^2)*Sin[c + d*x])/((a^2 - b^ 
2)*d*(a + b*Cos[c + d*x])))/(2*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 = 1.14 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (4 b +a \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (a -4 b \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (a^{2}+2 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(180\)
default	\(\frac {\frac {-\frac {\left (4 b +a \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{\left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (a -4 b \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (a^{2}+2 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\)	\(180\)
risch	\(\frac {i a \left (2 a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}-5 a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+2 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-7 a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-4 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+2 a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-11 b^{3} a \,{\mathrm e}^{i \left (d x +c \right )}+a^{2} b^{2}-4 b^{4}\right )}{b^{2} \left (a^{2}-b^{2}\right )^{2} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\)	\(503\)

Input:

int(cos(d*x+c)^2/(a+cos(d*x+c)*b)^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(2*(-1/2*(4*b+a)*a/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+1/2*(a-4 
*b)*a/(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-ta 
n(1/2*d*x+1/2*c)^2*b+a+b)^2+(a^2+2*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 [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 587, normalized size of antiderivative = 3.94 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\left [-\frac {{\left (a^{4} + 2 \, a^{2} b^{2} + {\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + 2 \, {\left (3 \, a^{4} b - 3 \, a^{2} b^{3} - {\left (a^{5} - 5 \, a^{3} b^{2} + 4 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d\right )}}, \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + {\left (a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (3 \, a^{4} b - 3 \, a^{2} b^{3} - {\left (a^{5} - 5 \, a^{3} b^{2} + 4 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d\right )}}\right ] \] Input:

integrate(cos(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
 

Output:

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

integrate(cos(d*x+c)**2/(a+b*cos(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cos(d*x+c)^2/(a+b*cos(d*x+c))^3,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 [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.68 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (a^{2} + 2 \, b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}}}{d} \] Input:

integrate(cos(d*x+c)^2/(a+b*cos(d*x+c))^3,x, algorithm="giac")
 

Output:

-((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^2 + 2*b^2)/((a^ 
4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (a^3*tan(1/2*d*x + 1/2*c)^3 + 3*a^ 
2*b*tan(1/2*d*x + 1/2*c)^3 - 4*a*b^2*tan(1/2*d*x + 1/2*c)^3 - a^3*tan(1/2* 
d*x + 1/2*c) + 3*a^2*b*tan(1/2*d*x + 1/2*c) + 4*a*b^2*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 = 43.67 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.36 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}}\right )\,\left (a^2+2\,b^2\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^2+4\,b\,a\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,a\,b-a^2\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (2\,a\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )} \] Input:

int(cos(c + d*x)^2/(a + b*cos(c + d*x))^3,x)
 

Output:

(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)))*(a^2 + 2*b^2))/(d*(a + b)^(5/2)*(a - b)^(5/2)) - ((tan(c 
/2 + (d*x)/2)^3*(4*a*b + a^2))/((a + b)^2*(a - b)) + (tan(c/2 + (d*x)/2)*( 
4*a*b - a^2))/((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 
))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 632, normalized size of antiderivative = 4.24 \[ \int \frac {\cos ^2(c+d x)}{(a+b \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(cos(d*x+c)^2/(a+b*cos(d*x+c))^3,x)
 

Output:

(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**3*b + 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*b**3 - 2*sqr 
t(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**2 - 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*b**4 + 2*sq 
rt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - 
 b**2))*a**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**2*b**2 + 4*sqrt(a**2 - b**2)*atan((tan((c + 
d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*b**4 + cos(c + d*x)*sin 
(c + d*x)*a**5 - 5*cos(c + d*x)*sin(c + d*x)*a**3*b**2 + 4*cos(c + d*x)*si 
n(c + d*x)*a*b**4 - 3*sin(c + d*x)*a**4*b + 3*sin(c + d*x)*a**2*b**3)/(2*d 
*(2*cos(c + d*x)*a**7*b - 6*cos(c + d*x)*a**5*b**3 + 6*cos(c + d*x)*a**3*b 
**5 - 2*cos(c + d*x)*a*b**7 - sin(c + d*x)**2*a**6*b**2 + 3*sin(c + d*x)** 
2*a**4*b**4 - 3*sin(c + d*x)**2*a**2*b**6 + sin(c + d*x)**2*b**8 + a**8 - 
2*a**6*b**2 + 2*a**2*b**6 - b**8))