\(\int \frac {\cos ^3(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [289]

Optimal result

Integrand size = 34, antiderivative size = 114 \[ \int \frac {\cos ^3(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {\left (2 a^2+b^2\right ) B x}{2 b^3}-\frac {2 a^3 B \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^3 \sqrt {a+b} d}-\frac {a B \sin (c+d x)}{b^2 d}+\frac {B \cos (c+d x) \sin (c+d x)}{2 b d} \] Output:

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

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^3(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {B \left (2 \left (2 a^2+b^2\right ) (c+d x)+\frac {8 a^3 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-4 a b \sin (c+d x)+b^2 \sin (2 (c+d x))\right )}{4 b^3 d} \] Input:

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

Output:

(B*(2*(2*a^2 + b^2)*(c + d*x) + (8*a^3*ArcTanh[((a - b)*Tan[(c + d*x)/2])/ 
Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] - 4*a*b*Sin[c + d*x] + b^2*Sin[2*(c + 
d*x)]))/(4*b^3*d)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2011, 3042, 3272, 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.

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3272

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

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 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

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 3272

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

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 = 2.84 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.22

method	result	size
derivativedivides	\(\frac {2 B \left (\frac {\frac {\left (-a b -\frac {1}{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a b +\frac {1}{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (2 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}}{b^{3}}-\frac {a^{3} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d}\)	\(139\)
default	\(\frac {2 B \left (\frac {\frac {\left (-a b -\frac {1}{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a b +\frac {1}{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {\left (2 a^{2}+b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}}{b^{3}}-\frac {a^{3} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d}\)	\(139\)
risch	\(\frac {B x \,a^{2}}{b^{3}}+\frac {B x}{2 b}+\frac {i B a \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}-\frac {i B a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}-\frac {a^{3} B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {a^{3} B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{3}}+\frac {B \sin \left (2 d x +2 c \right )}{4 b d}\)	\(227\)

Input:

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

Output:

2/d*B*(1/b^3*(((-a*b-1/2*b^2)*tan(1/2*d*x+1/2*c)^3+(-a*b+1/2*b^2)*tan(1/2* 
d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^2+1/2*(2*a^2+b^2)*arctan(tan(1/2*d*x+ 
1/2*c)))-a^3/b^3/((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 = 0.11 (sec) , antiderivative size = 350, normalized size of antiderivative = 3.07 \[ \int \frac {\cos ^3(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} B a^{3} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (2 \, B a^{4} - B a^{2} b^{2} - B b^{4}\right )} d x + {\left (2 \, B a^{3} b - 2 \, B a b^{3} - {\left (B a^{2} b^{2} - B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}, -\frac {2 \, \sqrt {a^{2} - b^{2}} B a^{3} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (2 \, B a^{4} - B a^{2} b^{2} - B b^{4}\right )} d x + {\left (2 \, B a^{3} b - 2 \, B a b^{3} - {\left (B a^{2} b^{2} - B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}\right ] \] Input:

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

Output:

[-1/2*(sqrt(-a^2 + b^2)*B*a^3*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*B*a^4 - B*a^2 
*b^2 - B*b^4)*d*x + (2*B*a^3*b - 2*B*a*b^3 - (B*a^2*b^2 - B*b^4)*cos(d*x + 
 c))*sin(d*x + c))/((a^2*b^3 - b^5)*d), -1/2*(2*sqrt(a^2 - b^2)*B*a^3*arct 
an(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (2*B*a^4 - B*a^ 
2*b^2 - B*b^4)*d*x + (2*B*a^3*b - 2*B*a*b^3 - (B*a^2*b^2 - B*b^4)*cos(d*x 
+ c))*sin(d*x + c))/((a^2*b^3 - b^5)*d)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(cos(d*x+c)^3*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algorith 
m="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 = 0.18 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.62 \[ \int \frac {\cos ^3(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} B a^{3}}{\sqrt {a^{2} - b^{2}} b^{3}} - \frac {{\left (2 \, B a^{2} + B b^{2}\right )} {\left (d x + c\right )}}{b^{3}} + \frac {2 \, {\left (2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 25.53 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.52 \[ \int \frac {\cos ^3(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {B\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d}+\frac {B\,\sin \left (2\,c+2\,d\,x\right )}{4\,b\,d}+\frac {2\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^3\,d}-\frac {B\,a\,\sin \left (c+d\,x\right )}{b^2\,d}-\frac {B\,a^3\,\mathrm {atan}\left (\frac {\left (a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}\right )\,2{}\mathrm {i}}{b^3\,d\,\sqrt {b^2-a^2}} \] Input:

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

Output:

(B*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b*d) + (B*sin(2*c + 2*d*x 
))/(4*b*d) + (2*B*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^3*d) 
 - (B*a*sin(c + d*x))/(b^2*d) - (B*a^3*atan(((a*sin(c/2 + (d*x)/2) - b*sin 
(c/2 + (d*x)/2))*1i)/(cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)))*2i)/(b^3*d*(b 
^2 - a^2)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.88 \[ \int \frac {\cos ^3(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {-4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{3}+\cos \left (d x +c \right )^{2} a^{2} b^{2} d x -\cos \left (d x +c \right )^{2} b^{4} d x +\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{2}-\cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4}+\sin \left (d x +c \right )^{2} a^{2} b^{2} d x -\sin \left (d x +c \right )^{2} b^{4} d x -2 \sin \left (d x +c \right ) a^{3} b +2 \sin \left (d x +c \right ) a \,b^{3}+2 a^{4} d x -2 a^{2} b^{2} d x}{2 b^{2} d \left (a^{2}-b^{2}\right )} \] Input:

int(cos(d*x+c)^3*(a*B+b*B*cos(d*x+c))/(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)/sqr 
t(a**2 - b**2))*a**3 + cos(c + d*x)**2*a**2*b**2*d*x - cos(c + d*x)**2*b** 
4*d*x + cos(c + d*x)*sin(c + d*x)*a**2*b**2 - cos(c + d*x)*sin(c + d*x)*b* 
*4 + sin(c + d*x)**2*a**2*b**2*d*x - sin(c + d*x)**2*b**4*d*x - 2*sin(c + 
d*x)*a**3*b + 2*sin(c + d*x)*a*b**3 + 2*a**4*d*x - 2*a**2*b**2*d*x)/(2*b** 
2*d*(a**2 - b**2))