Integrand size = 15, antiderivative size = 98 \[ \int \frac {A+B \sec (x)}{a+b \sin (x)} \, dx=\frac {2 A \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {B \log (1-\sin (x))}{2 (a+b)}+\frac {B \log (1+\sin (x))}{2 (a-b)}-\frac {b B \log (a+b \sin (x))}{a^2-b^2} \]
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Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4311, 4486, 2739, 632, 210, 2747, 720, 31, 647} \[ \int \frac {A+B \sec (x)}{a+b \sin (x)} \, dx=\frac {2 A \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {b B \log (a+b \sin (x))}{a^2-b^2}-\frac {B \log (1-\sin (x))}{2 (a+b)}+\frac {B \log (\sin (x)+1)}{2 (a-b)} \]
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Rule 31
Rule 210
Rule 632
Rule 647
Rule 720
Rule 2739
Rule 2747
Rule 4311
Rule 4486
Rubi steps \begin{align*} \text {integral}& = \int \frac {(B+A \cos (x)) \sec (x)}{a+b \sin (x)} \, dx \\ & = \int \left (\frac {A}{a+b \sin (x)}+\frac {B \sec (x)}{a+b \sin (x)}\right ) \, dx \\ & = A \int \frac {1}{a+b \sin (x)} \, dx+B \int \frac {\sec (x)}{a+b \sin (x)} \, dx \\ & = (2 A) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+(b B) \text {Subst}\left (\int \frac {1}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (x)\right ) \\ & = -\left ((4 A) \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 )\right )-\frac {(b B) \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \sin (x)\right )}{a^2-b^2}-\frac {(b B) \text {Subst}\left (\int \frac {-a+x}{b^2-x^2} \, dx,x,b \sin (x)\right )}{a^2-b^2} \\ & = \frac {2 A \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {b B \log (a+b \sin (x))}{a^2-b^2}-\frac {B \text {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (x)\right )}{2 (a-b)}+\frac {B \text {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (x)\right )}{2 (a+b)} \\ & = \frac {2 A \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {B \log (1-\sin (x))}{2 (a+b)}+\frac {B \log (1+\sin (x))}{2 (a-b)}-\frac {b B \log (a+b \sin (x))}{a^2-b^2} \\ \end{align*}
Time = 0.36 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.10 \[ \int \frac {A+B \sec (x)}{a+b \sin (x)} \, dx=\frac {2 A \sqrt {a^2-b^2} \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )+B \left ((-a+b) \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+(a+b) \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )-b \log (a+b \sin (x))\right )}{(a-b) (a+b)} \]
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Time = 0.80 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.99
method | result | size |
parts | \(\frac {2 A \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}+B \left (\frac {\ln \left (1+\sin \left (x \right )\right )}{2 a -2 b}-\frac {\ln \left (\sin \left (x \right )-1\right )}{2 a +2 b}-\frac {b \ln \left (a +b \sin \left (x \right )\right )}{\left (a -b \right ) \left (a +b \right )}\right )\) | \(97\) |
default | \(\frac {-B b \ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a \right )+\frac {2 \left (A \,a^{2}-A \,b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}}{\left (a -b \right ) \left (a +b \right )}-\frac {2 B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{2 a +2 b}+\frac {2 B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{2 a -2 b}\) | \(126\) |
risch | \(\frac {2 i x B \,a^{2} b}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {2 i x B \,b^{3}}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {i x B}{a -b}+\frac {i x B}{a +b}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A +\sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) B b}{\left (a +b \right ) \left (a -b \right )}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A +\sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) \sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{\left (a +b \right ) \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A -\sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) B b}{\left (a +b \right ) \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i a A -\sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) \sqrt {-A^{2} a^{2}+A^{2} b^{2}}}{\left (a +b \right ) \left (a -b \right )}+\frac {B \ln \left ({\mathrm e}^{i x}+i\right )}{a -b}-\frac {B \ln \left ({\mathrm e}^{i x}-i\right )}{a +b}\) | \(361\) |
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Time = 3.50 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.84 \[ \int \frac {A+B \sec (x)}{a+b \sin (x)} \, dx=\left [-\frac {B b \log \left (-b^{2} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2}\right ) + \sqrt {-a^{2} + b^{2}} A \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - {\left (B a + B b\right )} \log \left (\sin \left (x\right ) + 1\right ) + {\left (B a - B b\right )} \log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )}}, -\frac {B b \log \left (-b^{2} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2}\right ) + 2 \, \sqrt {a^{2} - b^{2}} A \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) - {\left (B a + B b\right )} \log \left (\sin \left (x\right ) + 1\right ) + {\left (B a - B b\right )} \log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )}}\right ] \]
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\[ \int \frac {A+B \sec (x)}{a+b \sin (x)} \, dx=\int \frac {A + B \sec {\left (x \right )}}{a + b \sin {\left (x \right )}}\, dx \]
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Exception generated. \[ \int \frac {A+B \sec (x)}{a+b \sin (x)} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.33 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.18 \[ \int \frac {A+B \sec (x)}{a+b \sin (x)} \, dx=-\frac {B b \log \left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}{a^{2} - b^{2}} + \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} A}{\sqrt {a^{2} - b^{2}}} + \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{a - b} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{a + b} \]
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Time = 20.04 (sec) , antiderivative size = 755, normalized size of antiderivative = 7.70 \[ \int \frac {A+B \sec (x)}{a+b \sin (x)} \, dx=\frac {B\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}{a-b}-\frac {B\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )}{a+b}-\frac {\ln \left (32\,A\,B^2\,a^2-32\,A^2\,B\,a^2+\frac {\left (A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,b^3+B\,a^2\,b\right )\,\left (32\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (A^2\,a^2+A^2\,b^2-4\,A\,B\,b^2+B^2\,a^2+3\,B^2\,b^2\right )+64\,A^2\,a^2\,b+32\,B^2\,a^2\,b-\frac {\left (A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,b^3+B\,a^2\,b\right )\,\left (32\,A\,a^4+32\,B\,a^4-32\,A\,a^2\,b^2+64\,B\,a^2\,b^2+32\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,A\,a^2-2\,A\,b^2+4\,B\,a^2-B\,b^2\right )-\frac {96\,a\,b\,\left (a+b\,\mathrm {tan}\left (\frac {x}{2}\right )\right )\,\left (A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,b^3+B\,a^2\,b\right )}{a^2-b^2}\right )}{{\left (a^2-b^2\right )}^2}-64\,A\,B\,a^2\,b\right )}{{\left (a^2-b^2\right )}^2}-32\,B\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,{\left (A-B\right )}^2\right )\,\left (A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,b^3+B\,a^2\,b\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {\ln \left (32\,A\,B^2\,a^2-32\,A^2\,B\,a^2-\frac {\left (B\,b^3+A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^2\,b\right )\,\left (32\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (A^2\,a^2+A^2\,b^2-4\,A\,B\,b^2+B^2\,a^2+3\,B^2\,b^2\right )+64\,A^2\,a^2\,b+32\,B^2\,a^2\,b+\frac {\left (B\,b^3+A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^2\,b\right )\,\left (32\,A\,a^4+32\,B\,a^4-32\,A\,a^2\,b^2+64\,B\,a^2\,b^2+32\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,A\,a^2-2\,A\,b^2+4\,B\,a^2-B\,b^2\right )+\frac {96\,a\,b\,\left (a+b\,\mathrm {tan}\left (\frac {x}{2}\right )\right )\,\left (B\,b^3+A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^2\,b\right )}{a^2-b^2}\right )}{{\left (a^2-b^2\right )}^2}-64\,A\,B\,a^2\,b\right )}{{\left (a^2-b^2\right )}^2}-32\,B\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,{\left (A-B\right )}^2\right )\,\left (B\,b^3+A\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}-B\,a^2\,b\right )}{a^4-2\,a^2\,b^2+b^4} \]
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