\(\int \frac {A+B \cos (c+d x)}{a+b \cos (c+d x)} \, dx\) [253]

Optimal result

Integrand size = 23, antiderivative size = 67 \[ \int \frac {A+B \cos (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {B x}{b}+\frac {2 (A b-a B) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b \sqrt {a+b} d} \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.01 \[ \int \frac {A+B \cos (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {B (c+d x)+\frac {2 (-A b+a B) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}}{b d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

\(\displaystyle \frac {(A b-a B) \int \frac {1}{a+b \cos (c+d x)}dx}{b}+\frac {B x}{b}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Maple [A] (verified)

Time = 1.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {\frac {2 \left (A b -B a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}}{d}\)	\(73\)
default	\(\frac {\frac {2 \left (A b -B a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 B \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}}{d}\)	\(73\)
risch	\(\frac {B x}{b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B a}{\sqrt {-a^{2}+b^{2}}\, d b}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) B a}{\sqrt {-a^{2}+b^{2}}\, d b}\)	\(290\)

Input:

int((A+B*cos(d*x+c))/(a+cos(d*x+c)*b),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 242, normalized size of antiderivative = 3.61 \[ \int \frac {A+B \cos (c+d x)}{a+b \cos (c+d x)} \, dx=\left [\frac {2 \, {\left (B a^{2} - B b^{2}\right )} d x + {\left (B a - A b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d}, \frac {{\left (B a^{2} - B b^{2}\right )} d x - {\left (B a - A b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right )}{{\left (a^{2} b - b^{3}\right )} d}\right ] \] Input:

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c)),x, algorithm="fricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 524 vs. \(2 (56) = 112\).

Time = 11.33 (sec) , antiderivative size = 524, normalized size of antiderivative = 7.82 \[ \int \frac {A+B \cos (c+d x)}{a+b \cos (c+d x)} \, dx=\begin {cases} \frac {\tilde {\infty } x \left (A + B \cos {\left (c \right )}\right )}{\cos {\left (c \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{b d} + \frac {B x}{b} - \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{b d} & \text {for}\: a = b \\\frac {A}{b d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}} + \frac {B x}{b} + \frac {B}{b d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}} & \text {for}\: a = - b \\\frac {A x + \frac {B \sin {\left (c + d x \right )}}{d}}{a} & \text {for}\: b = 0 \\\frac {x \left (A + B \cos {\left (c \right )}\right )}{a + b \cos {\left (c \right )}} & \text {for}\: d = 0 \\\frac {A b \log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )}}{a b d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - b^{2} d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {A b \log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )}}{a b d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - b^{2} d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} + \frac {B a d x \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}}{a b d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - b^{2} d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {B a \log {\left (- \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )}}{a b d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - b^{2} d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} + \frac {B a \log {\left (\sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} + \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )}}{a b d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - b^{2} d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} - \frac {B b d x \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}}{a b d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}} - b^{2} d \sqrt {- \frac {a}{a - b} - \frac {b}{a - b}}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((zoo*x*(A + B*cos(c))/cos(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ( 
A*tan(c/2 + d*x/2)/(b*d) + B*x/b - B*tan(c/2 + d*x/2)/(b*d), Eq(a, b)), (A 
/(b*d*tan(c/2 + d*x/2)) + B*x/b + B/(b*d*tan(c/2 + d*x/2)), Eq(a, -b)), (( 
A*x + B*sin(c + d*x)/d)/a, Eq(b, 0)), (x*(A + B*cos(c))/(a + b*cos(c)), Eq 
(d, 0)), (A*b*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(a*b*d 
*sqrt(-a/(a - b) - b/(a - b)) - b**2*d*sqrt(-a/(a - b) - b/(a - b))) - A*b 
*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(a*b*d*sqrt(-a/(a - 
b) - b/(a - b)) - b**2*d*sqrt(-a/(a - b) - b/(a - b))) + B*a*d*x*sqrt(-a/( 
a - b) - b/(a - b))/(a*b*d*sqrt(-a/(a - b) - b/(a - b)) - b**2*d*sqrt(-a/( 
a - b) - b/(a - b))) - B*a*log(-sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d 
*x/2))/(a*b*d*sqrt(-a/(a - b) - b/(a - b)) - b**2*d*sqrt(-a/(a - b) - b/(a 
 - b))) + B*a*log(sqrt(-a/(a - b) - b/(a - b)) + tan(c/2 + d*x/2))/(a*b*d* 
sqrt(-a/(a - b) - b/(a - b)) - b**2*d*sqrt(-a/(a - b) - b/(a - b))) - B*b* 
d*x*sqrt(-a/(a - b) - b/(a - b))/(a*b*d*sqrt(-a/(a - b) - b/(a - b)) - b** 
2*d*sqrt(-a/(a - b) - b/(a - b))), True))
 

Maxima [F(-2)]

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

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c)),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. 296 vs. \(2 (58) = 116\).

Time = 0.20 (sec) , antiderivative size = 296, normalized size of antiderivative = 4.42 \[ \int \frac {A+B \cos (c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {\frac {{\left (\sqrt {a^{2} - b^{2}} B {\left (2 \, a - b\right )} {\left | a - b \right |} - \sqrt {a^{2} - b^{2}} A b {\left | a - b \right |} - \sqrt {a^{2} - b^{2}} A {\left | a - b \right |} {\left | b \right |} + \sqrt {a^{2} - b^{2}} B {\left | a - b \right |} {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {2 \, a + \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{a - b}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} b^{2} + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} {\left | b \right |}} + \frac {{\left (2 \, B a - A b - B b + A {\left | b \right |} - B {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {\frac {2 \, a - \sqrt {-4 \, {\left (a + b\right )} {\left (a - b\right )} + 4 \, a^{2}}}{a - b}}}\right )\right )}}{b^{2} - a {\left | b \right |}}}{d} \] Input:

integrate((A+B*cos(d*x+c))/(a+b*cos(d*x+c)),x, algorithm="giac")
 

Output:

-((sqrt(a^2 - b^2)*B*(2*a - b)*abs(a - b) - sqrt(a^2 - b^2)*A*b*abs(a - b) 
 - sqrt(a^2 - b^2)*A*abs(a - b)*abs(b) + sqrt(a^2 - b^2)*B*abs(a - b)*abs( 
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 + sqrt(-4*(a + b)*(a - b) + 4*a^2))/(a - b))))/((a^2 - 2*a 
*b + b^2)*b^2 + (a^3 - 2*a^2*b + a*b^2)*abs(b)) + (2*B*a - A*b - B*b + A*a 
bs(b) - B*abs(b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(2*sqrt(1/2)*t 
an(1/2*d*x + 1/2*c)/sqrt((2*a - sqrt(-4*(a + b)*(a - b) + 4*a^2))/(a - b)) 
))/(b^2 - a*abs(b)))/d
 

Mupad [B] (verification not implemented)

Time = 42.83 (sec) , antiderivative size = 344, normalized size of antiderivative = 5.13 \[ \int \frac {A+B \cos (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {a\,\left (B\,\ln \left (\frac {b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}-B\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}\right )-A\,b\,\ln \left (\frac {b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}+A\,b\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}}{b\,d\,\left (a^2-b^2\right )}+\frac {2\,B\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d} \] Input:

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

Output:

(a*(B*log((b*sin(c/2 + (d*x)/2) - a*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2 
)*(b^2 - a^2)^(1/2))/cos(c/2 + (d*x)/2))*(-(a + b)*(a - b))^(1/2) - B*log( 
(a*sin(c/2 + (d*x)/2) - b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(b^2 - a 
^2)^(1/2))/cos(c/2 + (d*x)/2))*(b^2 - a^2)^(1/2)) - A*b*log((b*sin(c/2 + ( 
d*x)/2) - a*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/cos 
(c/2 + (d*x)/2))*(-(a + b)*(a - b))^(1/2) + A*b*log((a*sin(c/2 + (d*x)/2) 
- b*sin(c/2 + (d*x)/2) + cos(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/cos(c/2 + ( 
d*x)/2))*(b^2 - a^2)^(1/2))/(b*d*(a^2 - b^2)) + (2*B*atan(sin(c/2 + (d*x)/ 
2)/cos(c/2 + (d*x)/2)))/(b*d)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.01 \[ \int \frac {A+B \cos (c+d x)}{a+b \cos (c+d x)} \, dx=x \] Input:

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

Output:

x