\(\int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx\) [461]

Optimal result

Integrand size = 21, antiderivative size = 155 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 a x}{b^3}+\frac {2 a^2 \left (2 a^2-3 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^3 (a+b)^{3/2} d}+\frac {\left (2 a^2-b^2\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d}-\frac {a^2 \cos (c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 1.00 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {-2 a (c+d x)+\frac {2 a^2 \left (2 a^2-3 b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+\left (b+\frac {a^3 b}{(a-b) (a+b) (a+b \cos (c+d x))}\right ) \sin (c+d x)}{b^3 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.14, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3271, 3042, 3502, 3042, 3214, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3271

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Defintions of rubi rules used

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 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3271

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co 
s[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f* 
(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin 
[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^ 
2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*d*(a^2 
+ b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - 
 d^2) - m*(b*c - a*d)^2 + d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x] 
, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - 
b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || 
IntegersQ[2*m, 2*n])
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

Maple [A] (verified)

Time = 1.23 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{3}}+\frac {2 a^{2} \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (2 a^{2}-3 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3}}}{d}\)	\(178\)
default	\(\frac {-\frac {2 \left (-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{3}}+\frac {2 a^{2} \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (2 a^{2}-3 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3}}}{d}\)	\(178\)
risch	\(-\frac {2 a x}{b^{3}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 d \,b^{2}}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d \,b^{2}}+\frac {2 i a^{3} \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b^{3} \left (a^{2}-b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {2 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}+\frac {2 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}\)	\(456\)

Input:

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

Output:

1/d*(-2/b^3*(-b*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+2*a*arctan(tan 
(1/2*d*x+1/2*c)))+2*a^2/b^3*(a*b/(a^2-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*a^2-3*b^2)/(a-b)/(a+b)/((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 = 554, normalized size of antiderivative = 3.57 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\left [-\frac {4 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d x \cos \left (d x + c\right ) + 4 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d x + {\left (2 \, a^{5} - 3 \, a^{3} b^{2} + {\left (2 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d\right )}}, -\frac {2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d x - {\left (2 \, a^{5} - 3 \, a^{3} b^{2} + {\left (2 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (2 \, a^{5} b - 3 \, a^{3} b^{3} + a b^{5} + {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{3} - 2 \, a^{3} b^{5} + a b^{7}\right )} d}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(cos(d*x+c)^3/(a+b*cos(d*x+c))^2,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. 847 vs. \(2 (146) = 292\).

Time = 0.58 (sec) , antiderivative size = 847, normalized size of antiderivative = 5.46 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((4*a^6*b^2 - 2*a^5*b^3 - 9*a^4*b^4 + 4*a^3*b^5 + 5*a^2*b^6 - 2*a*b^7 + 2* 
a^3*abs(-a^2*b^3 + b^5) - a^2*b*abs(-a^2*b^3 + b^5) - 2*a*b^2*abs(-a^2*b^3 
 + b^5))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2)*tan(1/2*d* 
x + 1/2*c)/sqrt((2*a^3*b^2 - 2*a*b^4 + sqrt(-4*(a^3*b^2 + a^2*b^3 - a*b^4 
- b^5)*(a^3*b^2 - a^2*b^3 - a*b^4 + b^5) + 4*(a^3*b^2 - a*b^4)^2))/(a^3*b^ 
2 - a^2*b^3 - a*b^4 + b^5))))/(a^3*b^2*abs(-a^2*b^3 + b^5) - a*b^4*abs(-a^ 
2*b^3 + b^5) + (a^2*b^3 - b^5)^2) - ((2*a^3 - a^2*b - 2*a*b^2)*sqrt(a^2 - 
b^2)*abs(-a^2*b^3 + b^5)*abs(-a + b) - (4*a^6*b^2 - 2*a^5*b^3 - 9*a^4*b^4 
+ 4*a^3*b^5 + 5*a^2*b^6 - 2*a*b^7)*sqrt(a^2 - b^2)*abs(-a + b))*(pi*floor( 
1/2*(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2)*tan(1/2*d*x + 1/2*c)/sqrt((2* 
a^3*b^2 - 2*a*b^4 - sqrt(-4*(a^3*b^2 + a^2*b^3 - a*b^4 - b^5)*(a^3*b^2 - a 
^2*b^3 - a*b^4 + b^5) + 4*(a^3*b^2 - a*b^4)^2))/(a^3*b^2 - a^2*b^3 - a*b^4 
 + b^5))))/((a^2*b^3 - b^5)^2*(a^2 - 2*a*b + b^2) - (a^5*b^2 - 2*a^4*b^3 + 
 2*a^2*b^5 - a*b^6)*abs(-a^2*b^3 + b^5)) + 2*(2*a^3*tan(1/2*d*x + 1/2*c)^3 
 - a^2*b*tan(1/2*d*x + 1/2*c)^3 - a*b^2*tan(1/2*d*x + 1/2*c)^3 + b^3*tan(1 
/2*d*x + 1/2*c)^3 + 2*a^3*tan(1/2*d*x + 1/2*c) + a^2*b*tan(1/2*d*x + 1/2*c 
) - a*b^2*tan(1/2*d*x + 1/2*c) - b^3*tan(1/2*d*x + 1/2*c))/((a*tan(1/2*d*x 
 + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 + 2*a*tan(1/2*d*x + 1/2*c)^2 + a + 
b)*(a^2*b^2 - b^4)))/d
 

Mupad [B] (verification not implemented)

Time = 47.93 (sec) , antiderivative size = 3180, normalized size of antiderivative = 20.52 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

- ((2*tan(c/2 + (d*x)/2)^3*(a*b^2 + a^2*b - 2*a^3 - b^3))/(b^2*(a + b)*(a 
- b)) + (2*tan(c/2 + (d*x)/2)*(a*b^2 - a^2*b - 2*a^3 + b^3))/(b^2*(a + b)* 
(a - b)))/(d*(a + b + tan(c/2 + (d*x)/2)^4*(a - b) + 2*a*tan(c/2 + (d*x)/2 
)^2)) - (4*a*atan(((2*a*((32*tan(c/2 + (d*x)/2)*(8*a^8 - 8*a^7*b + 4*a^2*b 
^6 - 8*a^3*b^5 + 5*a^4*b^4 + 16*a^5*b^3 - 16*a^6*b^2))/(a*b^6 + b^7 - a^2* 
b^5 - a^3*b^4) + (a*((32*(2*a*b^11 - 3*a^2*b^10 - 3*a^3*b^9 + 5*a^4*b^8 + 
a^5*b^7 - 2*a^6*b^6))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (a*tan(c/2 + (d* 
x)/2)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b 
^6)*64i)/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4)))*2i)/b^3))/b^3 + (2*a*((3 
2*tan(c/2 + (d*x)/2)*(8*a^8 - 8*a^7*b + 4*a^2*b^6 - 8*a^3*b^5 + 5*a^4*b^4 
+ 16*a^5*b^3 - 16*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) - (a*((32*(2 
*a*b^11 - 3*a^2*b^10 - 3*a^3*b^9 + 5*a^4*b^8 + a^5*b^7 - 2*a^6*b^6))/(a*b^ 
8 + b^9 - a^2*b^7 - a^3*b^6) + (a*tan(c/2 + (d*x)/2)*(2*a*b^11 - 2*a^2*b^1 
0 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6)*64i)/(b^3*(a*b^6 + b^7 
- a^2*b^5 - a^3*b^4)))*2i)/b^3))/b^3)/((64*(8*a^8 - 4*a^7*b + 12*a^4*b^4 + 
 6*a^5*b^3 - 20*a^6*b^2))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (a*((32*tan( 
c/2 + (d*x)/2)*(8*a^8 - 8*a^7*b + 4*a^2*b^6 - 8*a^3*b^5 + 5*a^4*b^4 + 16*a 
^5*b^3 - 16*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) + (a*((32*(2*a*b^1 
1 - 3*a^2*b^10 - 3*a^3*b^9 + 5*a^4*b^8 + a^5*b^7 - 2*a^6*b^6))/(a*b^8 + b^ 
9 - a^2*b^7 - a^3*b^6) - (a*tan(c/2 + (d*x)/2)*(2*a*b^11 - 2*a^2*b^10 -...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 521, normalized size of antiderivative = 3.36 \[ \int \frac {\cos ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) a^{4} b -6 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) \cos \left (d x +c \right ) a^{2} b^{3}+4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{5}-6 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{3} b^{2}+\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4} b^{2}-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{4}+\cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{6}-2 \cos \left (d x +c \right ) a^{5} b c -2 \cos \left (d x +c \right ) a^{5} b d x +4 \cos \left (d x +c \right ) a^{3} b^{3} c +4 \cos \left (d x +c \right ) a^{3} b^{3} d x -2 \cos \left (d x +c \right ) a \,b^{5} c -2 \cos \left (d x +c \right ) a \,b^{5} d x +2 \sin \left (d x +c \right ) a^{5} b -3 \sin \left (d x +c \right ) a^{3} b^{3}+\sin \left (d x +c \right ) a \,b^{5}-2 a^{6} c -2 a^{6} d x +4 a^{4} b^{2} c +4 a^{4} b^{2} d x -2 a^{2} b^{4} c -2 a^{2} b^{4} d x}{b^{3} d \left (\cos \left (d x +c \right ) a^{4} b -2 \cos \left (d x +c \right ) a^{2} b^{3}+\cos \left (d x +c \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:

int(cos(d*x+c)^3/(a+b*cos(d*x+c))^2,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**4*b - 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)*a**2*b**3 + 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 - 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**3*b**2 + cos(c + d*x)*sin(c + d*x)*a**4*b* 
*2 - 2*cos(c + d*x)*sin(c + d*x)*a**2*b**4 + cos(c + d*x)*sin(c + d*x)*b** 
6 - 2*cos(c + d*x)*a**5*b*c - 2*cos(c + d*x)*a**5*b*d*x + 4*cos(c + d*x)*a 
**3*b**3*c + 4*cos(c + d*x)*a**3*b**3*d*x - 2*cos(c + d*x)*a*b**5*c - 2*co 
s(c + d*x)*a*b**5*d*x + 2*sin(c + d*x)*a**5*b - 3*sin(c + d*x)*a**3*b**3 + 
 sin(c + d*x)*a*b**5 - 2*a**6*c - 2*a**6*d*x + 4*a**4*b**2*c + 4*a**4*b**2 
*d*x - 2*a**2*b**4*c - 2*a**2*b**4*d*x)/(b**3*d*(cos(c + d*x)*a**4*b - 2*c 
os(c + d*x)*a**2*b**3 + cos(c + d*x)*b**5 + a**5 - 2*a**3*b**2 + a*b**4))