\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx\) [980]

Optimal result

Integrand size = 33, antiderivative size = 97 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {(b B-a C) x}{b^2}+\frac {2 \left (A b^2-a (b B-a C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^2 \sqrt {a+b} d}+\frac {C \sin (c+d x)}{b d} \] Output:

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

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.95 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {(b B-a C) (c+d x)-\frac {2 \left (A b^2+a (-b B+a C)\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+b C \sin (c+d x)}{b^2 d} \] Input:

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

Output:

((b*B - a*C)*(c + d*x) - (2*(A*b^2 + a*(-(b*B) + a*C))*ArcTanh[((a - b)*Ta 
n[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + b*C*Sin[c + d*x])/(b 
^2*d)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\int \frac {A b+(b B-a C) \cos (c+d x)}{a+b \cos (c+d x)}dx}{b}+\frac {C \sin (c+d x)}{b d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(((b*B - a*C)*x)/b + (2*(A*b^2 - a*(b*B - a*C))*ArcTan[(Sqrt[a - b]*Tan[(c 
 + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d))/b + (C*Sin[c + d* 
x])/(b*d)
 

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 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 = 0.46 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(\frac {\frac {2 \left (A \,b^{2}-B a b +a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\frac {2 C b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 \left (B b -C a \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}}{d}\)	\(117\)
default	\(\frac {\frac {2 \left (A \,b^{2}-B a b +a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {\frac {2 C b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 \left (B b -C a \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}}{d}\)	\(117\)
risch	\(\frac {x B}{b}-\frac {a C x}{b^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 b d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 b d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B a}{\sqrt {-a^{2}+b^{2}}\, d b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2} C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B a}{\sqrt {-a^{2}+b^{2}}\, d b}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2} C}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}\)	\(491\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

1/d*(2*(A*b^2-B*a*b+C*a^2)/b^2/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d* 
x+1/2*c)/((a-b)*(a+b))^(1/2))+2/b^2*(C*b*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x 
+1/2*c)^2)+(B*b-C*a)*arctan(tan(1/2*d*x+1/2*c))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 331, normalized size of antiderivative = 3.41 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\left [-\frac {2 \, {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} d x + {\left (C a^{2} - B a b + A b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d}, -\frac {{\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} d x - {\left (C a^{2} - B a b + A b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{2} b^{2} - b^{4}\right )} d}\right ] \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="f 
ricas")
 

Output:

[-1/2*(2*(C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*d*x + (C*a^2 - B*a*b + A*b^2) 
*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*(C*a^2*b - C*b^3)*sin(d*x 
+ c))/((a^2*b^2 - b^4)*d), -((C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*d*x - (C* 
a^2 - B*a*b + A*b^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^ 
2 - b^2)*sin(d*x + c))) - (C*a^2*b - C*b^3)*sin(d*x + c))/((a^2*b^2 - b^4) 
*d)]
                                                                                    
                                                                                    
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="m 
axima")
 

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 = 0.13 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.52 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {\frac {{\left (C a - B b\right )} {\left (d x + c\right )}}{b^{2}} - \frac {2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b} + \frac {2 \, {\left (C a^{2} - B a b + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{2}}}{d} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="g 
iac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 2.77 (sec) , antiderivative size = 4410, normalized size of antiderivative = 45.46 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*B*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(d*(b^4 - a^2*b^2)) 
+ (2*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(d*(b^4 - a^2*b^2) 
) + (C*b^3*sin(c + d*x))/(d*(b^4 - a^2*b^2)) - (C*a^2*b*sin(c + d*x))/(d*( 
b^4 - a^2*b^2)) + (A*b^2*atan((C^2*a^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2 
)*2i - B^2*b^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i - A^2*b^7*sin(c/2 + 
 (d*x)/2)*(b^2 - a^2)^(1/2)*1i + C^2*a^7*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1 
/2)*2i + A^2*a*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2)*2i - A^2*a*b^6*sin 
(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i + B^2*a*b^6*sin(c/2 + (d*x)/2)*(b^2 - 
 a^2)^(1/2)*1i + A^2*a^2*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i + A^2 
*a^3*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i + B^2*a^2*b^5*sin(c/2 + ( 
d*x)/2)*(b^2 - a^2)^(1/2)*1i + B^2*a^3*b^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^ 
(3/2)*2i - B^2*a^3*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*3i + B^2*a^5*b 
^2*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*2i - C^2*a^2*b^5*sin(c/2 + (d*x)/2 
)*(b^2 - a^2)^(1/2)*1i + C^2*a^3*b^4*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)* 
1i + C^2*a^4*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*1i - C^2*a^5*b^2*sin 
(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*3i + A*B*a*b^6*sin(c/2 + (d*x)/2)*(b^2 - 
 a^2)^(1/2)*2i + B*C*a*b^6*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*2i - B*C*a 
^4*b*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2)*4i - B*C*a^6*b*sin(c/2 + (d*x)/2 
)*(b^2 - a^2)^(1/2)*4i - A*B*a^2*b^3*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(3/2)* 
4i + A*B*a^2*b^5*sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2)*2i - A*B*a^3*b^4*...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.33 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2} c +\sin \left (d x +c \right ) a^{2} b c -\sin \left (d x +c \right ) b^{3} c -a^{3} c d x +a^{2} b^{2} d x +a \,b^{2} c d x -b^{4} d x}{b^{2} d \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

(2*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*c + sin(c + d*x)*a**2*b*c - sin(c + d*x)*b**3*c - a**3*c 
*d*x + a**2*b**2*d*x + a*b**2*c*d*x - b**4*d*x)/(b**2*d*(a**2 - b**2))