Integrand size = 15, antiderivative size = 65 \[ \int \frac {A+B \tan (x)}{a+b \cos (x)} \, dx=\frac {2 A \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}-\frac {B \log (\cos (x))}{a}+\frac {B \log (a+b \cos (x))}{a} \] Output:
2*A*arctan((a-b)^(1/2)*tan(1/2*x)/(a+b)^(1/2))/(a-b)^(1/2)/(a+b)^(1/2)-B*l n(cos(x))/a+B*ln(a+b*cos(x))/a
Time = 0.35 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \frac {A+B \tan (x)}{a+b \cos (x)} \, dx=-\frac {2 A \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+\frac {B (-\log (\cos (x))+\log (a+b \cos (x)))}{a} \] Input:
Integrate[(A + B*Tan[x])/(a + b*Cos[x]),x]
Output:
(-2*A*ArcTanh[((a - b)*Tan[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + (B* (-Log[Cos[x]] + Log[a + b*Cos[x]]))/a
Time = 0.36 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4901, 2009}
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.
\(\displaystyle \int \frac {A+B \tan (x)}{a+b \cos (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \tan (x)}{a+b \cos (x)}dx\) |
\(\Big \downarrow \) 4901 |
\(\displaystyle \int \left (\frac {A}{a+b \cos (x)}+\frac {B \tan (x)}{a+b \cos (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 A \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+\frac {B \log (a+b \cos (x))}{a}-\frac {B \log (\cos (x))}{a}\) |
Input:
Int[(A + B*Tan[x])/(a + b*Cos[x]),x]
Output:
(2*A*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]) - (B*Log[Cos[x]])/a + (B*Log[a + b*Cos[x]])/a
Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /; !InertTrigFreeQ[u]
Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.66
method | result | size |
default | \(\frac {\frac {2 \left (B a -B b \right ) \ln \left (a \tan \left (\frac {x}{2}\right )^{2}-b \tan \left (\frac {x}{2}\right )^{2}+a +b \right )}{2 a -2 b}+\frac {2 A a \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}}{a}-\frac {B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{a}-\frac {B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{a}\) | \(108\) |
risch | \(-\frac {2 i x B \,a^{3}}{a^{4}-b^{2} a^{2}}+\frac {2 i x B a \,b^{2}}{a^{4}-b^{2} a^{2}}+\frac {2 i x B}{a}+\frac {a \ln \left ({\mathrm e}^{i x}+\frac {A \,a^{2}-i \sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) B}{a^{2}-b^{2}}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {A \,a^{2}-i \sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) B \,b^{2}}{\left (a^{2}-b^{2}\right ) a}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {A \,a^{2}-i \sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) \sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{\left (a^{2}-b^{2}\right ) a}+\frac {a \ln \left ({\mathrm e}^{i x}+\frac {A \,a^{2}+i \sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) B}{a^{2}-b^{2}}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {A \,a^{2}+i \sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) B \,b^{2}}{\left (a^{2}-b^{2}\right ) a}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {A \,a^{2}+i \sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) \sqrt {-A^{2} a^{4}+A^{2} a^{2} b^{2}}}{\left (a^{2}-b^{2}\right ) a}-\frac {B \ln \left ({\mathrm e}^{2 i x}+1\right )}{a}\) | \(492\) |
Input:
int((A+B*tan(x))/(a+b*cos(x)),x,method=_RETURNVERBOSE)
Output:
2/a*(1/2*(B*a-B*b)/(a-b)*ln(a*tan(1/2*x)^2-b*tan(1/2*x)^2+a+b)+A*a/((a-b)* (a+b))^(1/2)*arctan((a-b)*tan(1/2*x)/((a-b)*(a+b))^(1/2)))-B/a*ln(tan(1/2* x)-1)-B/a*ln(tan(1/2*x)+1)
Time = 0.16 (sec) , antiderivative size = 263, normalized size of antiderivative = 4.05 \[ \int \frac {A+B \tan (x)}{a+b \cos (x)} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} A a \log \left (\frac {2 \, a b \cos \left (x\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}\right ) + 2 \, {\left (B a^{2} - B b^{2}\right )} \log \left (-\cos \left (x\right )\right )}{2 \, {\left (a^{3} - a b^{2}\right )}}, \frac {2 \, \sqrt {a^{2} - b^{2}} A a \arctan \left (-\frac {a \cos \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (x\right )}\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}\right ) - 2 \, {\left (B a^{2} - B b^{2}\right )} \log \left (-\cos \left (x\right )\right )}{2 \, {\left (a^{3} - a b^{2}\right )}}\right ] \] Input:
integrate((A+B*tan(x))/(a+b*cos(x)),x, algorithm="fricas")
Output:
[-1/2*(sqrt(-a^2 + b^2)*A*a*log((2*a*b*cos(x) + (2*a^2 - b^2)*cos(x)^2 + 2 *sqrt(-a^2 + b^2)*(a*cos(x) + b)*sin(x) - a^2 + 2*b^2)/(b^2*cos(x)^2 + 2*a *b*cos(x) + a^2)) - (B*a^2 - B*b^2)*log(b^2*cos(x)^2 + 2*a*b*cos(x) + a^2) + 2*(B*a^2 - B*b^2)*log(-cos(x)))/(a^3 - a*b^2), 1/2*(2*sqrt(a^2 - b^2)*A *a*arctan(-(a*cos(x) + b)/(sqrt(a^2 - b^2)*sin(x))) + (B*a^2 - B*b^2)*log( b^2*cos(x)^2 + 2*a*b*cos(x) + a^2) - 2*(B*a^2 - B*b^2)*log(-cos(x)))/(a^3 - a*b^2)]
\[ \int \frac {A+B \tan (x)}{a+b \cos (x)} \, dx=\int \frac {A + B \tan {\left (x \right )}}{a + b \cos {\left (x \right )}}\, dx \] Input:
integrate((A+B*tan(x))/(a+b*cos(x)),x)
Output:
Integral((A + B*tan(x))/(a + b*cos(x)), x)
Exception generated. \[ \int \frac {A+B \tan (x)}{a+b \cos (x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((A+B*tan(x))/(a+b*cos(x)),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
Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (55) = 110\).
Time = 0.13 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.86 \[ \int \frac {A+B \tan (x)}{a+b \cos (x)} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} A}{\sqrt {a^{2} - b^{2}}} + \frac {B \log \left (-a \tan \left (\frac {1}{2} \, x\right )^{2} + b \tan \left (\frac {1}{2} \, x\right )^{2} - a - b\right )}{a} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{a} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{a} \] Input:
integrate((A+B*tan(x))/(a+b*cos(x)),x, algorithm="giac")
Output:
-2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*t an(1/2*x))/sqrt(a^2 - b^2)))*A/sqrt(a^2 - b^2) + B*log(-a*tan(1/2*x)^2 + b *tan(1/2*x)^2 - a - b)/a - B*log(abs(tan(1/2*x) + 1))/a - B*log(abs(tan(1/ 2*x) - 1))/a
Time = 3.01 (sec) , antiderivative size = 1540, normalized size of antiderivative = 23.69 \[ \int \frac {A+B \tan (x)}{a+b \cos (x)} \, dx=\text {Too large to display} \] Input:
int((A + B*tan(x))/(a + b*cos(x)),x)
Output:
(log((a + b*cos(x))/(cos(x) + 1))*(2*B*a^3 - 2*B*a*b^2))/(2*(a^4 - a^2*b^2 )) + (2*A*atan(((a^2 - b^2)*((A*(64*A*B*a^3 + ((2*B*a^3 - 2*B*a*b^2)*(32*A *a^4 + 32*A*a^2*b^2 - 64*A*a^3*b))/(2*(a^4 - a^2*b^2)) + 64*A*B*a*b^2 - 12 8*A*B*a^2*b))/(a^2 - b^2)^(1/2) + (A*(2*B*a^3 - 2*B*a*b^2)*(32*A*a^4 + 32* A*a^2*b^2 - 64*A*a^3*b))/(2*(a^4 - a^2*b^2)*(a^2 - b^2)^(1/2)))*(A^2*a^2 - 4*B^2*a^2 + 4*B^2*b^2))/((32*A*a - 32*A*b)*(a - b)*(A^2*a^2 + 4*B^2*a^2 - 4*B^2*b^2)^2) - (tan(x/2)*(a^2 - b^2)^(3/2)*(((((2*B*a^3 - 2*B*a*b^2)*((A *(64*B*a^4 + 64*B*a^2*b^2 - 128*B*a^3*b - ((2*B*a^3 - 2*B*a*b^2)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2))/(2*(a^4 - a^2*b^2))))/(a^2 - b^2)^(1/2) - (A *(2*B*a^3 - 2*B*a*b^2)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2))/(2*(a^4 - a^ 2*b^2)*(a^2 - b^2)^(1/2))))/(2*(a^4 - a^2*b^2)) + (A^3*(64*a^4*b + 64*a^2* b^3 - 128*a^3*b^2))/(a^2 - b^2)^(3/2) + (A*(64*B^2*b^3 - 32*A^2*a^3 + 32*A ^2*a^2*b - 128*B^2*a*b^2 + 64*B^2*a^2*b + ((2*B*a^3 - 2*B*a*b^2)*(64*B*a^4 + 64*B*a^2*b^2 - 128*B*a^3*b - ((2*B*a^3 - 2*B*a*b^2)*(64*a^4*b + 64*a^2* b^3 - 128*a^3*b^2))/(2*(a^4 - a^2*b^2))))/(2*(a^4 - a^2*b^2))))/(a^2 - b^2 )^(1/2))*(A^2*a^2 - 4*B^2*a^2 + 4*B^2*b^2))/((a^2 - b^2)^(1/2)*(a - b)*(A^ 2*a^2 + 4*B^2*a^2 - 4*B^2*b^2)^2) - (4*A*B*a*(64*B^3*a^2 + 64*B^3*b^2 + 32 *A^2*B*a^2 + (A*((A*(64*B*a^4 + 64*B*a^2*b^2 - 128*B*a^3*b - ((2*B*a^3 - 2 *B*a*b^2)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2))/(2*(a^4 - a^2*b^2))))/(a^ 2 - b^2)^(1/2) - (A*(2*B*a^3 - 2*B*a*b^2)*(64*a^4*b + 64*a^2*b^3 - 128*...
Time = 0.25 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.45 \[ \int \frac {A+B \tan (x)}{a+b \cos (x)} \, dx=\frac {2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a -\tan \left (\frac {x}{2}\right ) b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2}-\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) a^{2} b +\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) b^{3}-\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) a^{2} b +\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) b^{3}+\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2} a -\tan \left (\frac {x}{2}\right )^{2} b +a +b \right ) a^{2} b -\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2} a -\tan \left (\frac {x}{2}\right )^{2} b +a +b \right ) b^{3}}{a \left (a^{2}-b^{2}\right )} \] Input:
int((A+B*tan(x))/(a+b*cos(x)),x)
Output:
(2*sqrt(a**2 - b**2)*atan((tan(x/2)*a - tan(x/2)*b)/sqrt(a**2 - b**2))*a** 2 - log(tan(x/2) - 1)*a**2*b + log(tan(x/2) - 1)*b**3 - log(tan(x/2) + 1)* a**2*b + log(tan(x/2) + 1)*b**3 + log(tan(x/2)**2*a - tan(x/2)**2*b + a + b)*a**2*b - log(tan(x/2)**2*a - tan(x/2)**2*b + a + b)*b**3)/(a*(a**2 - b* *2))