Integrand size = 15, antiderivative size = 61 \[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\frac {2 (a A-b B) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}-\frac {B \text {arctanh}(\cos (x))}{a} \]
Time = 0.55 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.18 \[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\frac {\frac {2 (a A-b B) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+B \left (-\log \left (\cos \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )\right )\right )}{a} \]
((2*(a*A - b*B)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + B*(-Log[Cos[x/2]] + Log[Sin[x/2]]))/a
Time = 0.43 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 3307, 3042, 3480, 3042, 3139, 1083, 217, 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.
\(\displaystyle \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \csc (x)}{a+b \sin (x)}dx\) |
\(\Big \downarrow \) 3307 |
\(\displaystyle \int \frac {\csc (x) (A \sin (x)+B)}{a+b \sin (x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin (x)+B}{\sin (x) (a+b \sin (x))}dx\) |
\(\Big \downarrow \) 3480 |
\(\displaystyle \frac {(a A-b B) \int \frac {1}{a+b \sin (x)}dx}{a}+\frac {B \int \csc (x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(a A-b B) \int \frac {1}{a+b \sin (x)}dx}{a}+\frac {B \int \csc (x)dx}{a}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle \frac {2 (a A-b B) \int \frac {1}{a \tan ^2\left (\frac {x}{2}\right )+2 b \tan \left (\frac {x}{2}\right )+a}d\tan \left (\frac {x}{2}\right )}{a}+\frac {B \int \csc (x)dx}{a}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {B \int \csc (x)dx}{a}-\frac {4 (a A-b B) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{a}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {B \int \csc (x)dx}{a}+\frac {2 (a A-b B) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {2 (a A-b B) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}-\frac {B \text {arctanh}(\cos (x))}{a}\) |
(2*(a*A - b*B)*ArcTan[(2*b + 2*a*Tan[x/2])/(2*Sqrt[a^2 - b^2])])/(a*Sqrt[a ^2 - b^2]) - (B*ArcTanh[Cos[x]])/a
3.1.9.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*((d + c*Sin[e + f*x])^n/Sin[e + f*x]^n), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IntegerQ [n]
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]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.74 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {\left (2 a A -2 B b \right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a \sqrt {a^{2}-b^{2}}}+\frac {B \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a}\) | \(61\) |
parts | \(\frac {2 A \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}-\frac {2 B b \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a \sqrt {a^{2}-b^{2}}}+\frac {B \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a}\) | \(94\) |
risch | \(-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) B b}{\sqrt {-a^{2}+b^{2}}\, a}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) B b}{\sqrt {-a^{2}+b^{2}}\, a}-\frac {B \ln \left ({\mathrm e}^{i x}+1\right )}{a}+\frac {B \ln \left ({\mathrm e}^{i x}-1\right )}{a}\) | \(275\) |
(2*A*a-2*B*b)/a/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^ (1/2))+B/a*ln(tan(1/2*x))
Time = 0.59 (sec) , antiderivative size = 265, normalized size of antiderivative = 4.34 \[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\left [\frac {{\left (A a - B b\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )}}, -\frac {2 \, {\left (A a - B b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )}}\right ] \]
[1/2*((A*a - B*b)*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(x)^2 - 2*a*b*si n(x) - a^2 - b^2 - 2*(a*cos(x)*sin(x) + b*cos(x))*sqrt(-a^2 + b^2))/(b^2*c os(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) - (B*a^2 - B*b^2)*log(1/2*cos(x) + 1/ 2) + (B*a^2 - B*b^2)*log(-1/2*cos(x) + 1/2))/(a^3 - a*b^2), -1/2*(2*(A*a - B*b)*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x))) + ( B*a^2 - B*b^2)*log(1/2*cos(x) + 1/2) - (B*a^2 - B*b^2)*log(-1/2*cos(x) + 1 /2))/(a^3 - a*b^2)]
\[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\int \frac {A + B \csc {\left (x \right )}}{a + b \sin {\left (x \right )}}\, dx \]
Exception generated. \[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.33 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.16 \[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a} + \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (A a - B b\right )}}{\sqrt {a^{2} - b^{2}} a} \]
B*log(abs(tan(1/2*x)))/a + 2*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a* tan(1/2*x) + b)/sqrt(a^2 - b^2)))*(A*a - B*b)/(sqrt(a^2 - b^2)*a)
Time = 15.47 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.20 \[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\frac {B\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a}-\frac {\ln \left (b+a\,\mathrm {tan}\left (\frac {x}{2}\right )+\sqrt {b^2-a^2}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (A\,a-B\,b\right )}{a\,b^2-a^3}-\frac {\ln \left (b+a\,\mathrm {tan}\left (\frac {x}{2}\right )-\sqrt {b^2-a^2}\right )\,\left (A\,a\,\sqrt {b^2-a^2}-B\,b\,\sqrt {b^2-a^2}\right )}{a\,\left (a^2-b^2\right )} \]