Integrand size = 15, antiderivative size = 61 \[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\frac {2 (a A-b B) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}-\frac {B \text {arctanh}(\cos (x))}{a} \]
[Out]
Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2907, 3080, 3855, 2739, 632, 210} \[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\frac {2 (a A-b B) \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2}}-\frac {B \text {arctanh}(\cos (x))}{a} \]
[In]
[Out]
Rule 210
Rule 632
Rule 2739
Rule 2907
Rule 3080
Rule 3855
Rubi steps \begin{align*} \text {integral}& = \int \frac {\csc (x) (B+A \sin (x))}{a+b \sin (x)} \, dx \\ & = \frac {B \int \csc (x) \, dx}{a}+\frac {(a A-b B) \int \frac {1}{a+b \sin (x)} \, dx}{a} \\ & = -\frac {B \text {arctanh}(\cos (x))}{a}+\frac {(2 (a A-b B)) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{a} \\ & = -\frac {B \text {arctanh}(\cos (x))}{a}-\frac {(4 (a A-b B)) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{a} \\ & = \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} \\ \end{align*}
Time = 0.55 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.18 \[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\frac {\frac {2 (a A-b B) \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+B \left (-\log \left (\cos \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )\right )\right )}{a} \]
[In]
[Out]
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 | \(-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) B b}{\sqrt {-a^{2}+b^{2}}\, a}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) B b}{\sqrt {-a^{2}+b^{2}}\, a}-\frac {B \ln \left ({\mathrm e}^{i x}+1\right )}{a}+\frac {B \ln \left ({\mathrm e}^{i x}-1\right )}{a}\) | \(275\) |
[In]
[Out]
none
Time = 0.59 (sec) , antiderivative size = 265, normalized size of antiderivative = 4.34 \[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\left [\frac {{\left (A a - B b\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )}}, -\frac {2 \, {\left (A a - B b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) + {\left (B a^{2} - B b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - {\left (B a^{2} - B b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{2 \, {\left (a^{3} - a b^{2}\right )}}\right ] \]
[In]
[Out]
\[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\int \frac {A + B \csc {\left (x \right )}}{a + b \sin {\left (x \right )}}\, dx \]
[In]
[Out]
Exception generated. \[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\text {Exception raised: ValueError} \]
[In]
[Out]
none
Time = 0.33 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.16 \[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a} + \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (A a - B b\right )}}{\sqrt {a^{2} - b^{2}} a} \]
[In]
[Out]
Time = 15.47 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.20 \[ \int \frac {A+B \csc (x)}{a+b \sin (x)} \, dx=\frac {B\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a}-\frac {\ln \left (b+a\,\mathrm {tan}\left (\frac {x}{2}\right )+\sqrt {b^2-a^2}\right )\,\sqrt {-\left (a+b\right )\,\left (a-b\right )}\,\left (A\,a-B\,b\right )}{a\,b^2-a^3}-\frac {\ln \left (b+a\,\mathrm {tan}\left (\frac {x}{2}\right )-\sqrt {b^2-a^2}\right )\,\left (A\,a\,\sqrt {b^2-a^2}-B\,b\,\sqrt {b^2-a^2}\right )}{a\,\left (a^2-b^2\right )} \]
[In]
[Out]