\(\int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx\) [254]

Optimal result

Integrand size = 29, antiderivative size = 76 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 (A b-a B) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b} d}+\frac {A \text {arctanh}(\sin (c+d x))}{a d} \] Output:

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

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.47 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {2 (A b-a B) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+A \left (-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{a d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3042, 3480, 3042, 3138, 218, 4257}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

(-2*(A*b - a*B)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqr 
t[a - b]*Sqrt[a + b]*d) + (A*ArcTanh[Sin[c + d*x]])/(a*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 3480

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b 
- a*B)/(b*c - a*d)   Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ 
(b*c - a*d)   Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f 
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 1.72 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(\frac {-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a}+\frac {2 \left (-A b +B a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a}}{d}\)	\(92\)
default	\(\frac {-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a}+\frac {2 \left (-A b +B a \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a}}{d}\)	\(92\)
risch	\(-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A b}{\sqrt {-a^{2}+b^{2}}\, d a}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) A b}{\sqrt {-a^{2}+b^{2}}\, d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a d}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a d}\)	\(331\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 304, normalized size of antiderivative = 4.00 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\left [\frac {{\left (B a - A b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) + {\left (A a^{2} - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} - A b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d}, \frac {2 \, {\left (B a - A b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) + {\left (A a^{2} - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{2} - A b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{3} - a b^{2}\right )} d}\right ] \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

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

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

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 = 127, normalized size of antiderivative = 1.67 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} + \frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (B a - A b\right )}}{\sqrt {a^{2} - b^{2}} a}}{d} \] Input:

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

Output:

(A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - A*log(abs(tan(1/2*d*x + 1/2*c) - 
 1))/a + 2*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*ta 
n(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))*(B*a - A*b) 
/(sqrt(a^2 - b^2)*a))/d
 

Mupad [B] (verification not implemented)

Time = 42.94 (sec) , antiderivative size = 342, normalized size of antiderivative = 4.50 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2\,A\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}+\frac {b\,\left (A\,\ln \left (\frac {a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}-A\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}\right )-B\,a\,\ln \left (\frac {a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}+B\,a\,\ln \left (\frac {a\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sqrt {b^2-a^2}}{a\,d\,\left (a^2-b^2\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.41 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d} \] Input:

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

Output:

( - log(tan((c + d*x)/2) - 1) + log(tan((c + d*x)/2) + 1))/d