Optimal. Leaf size=98 \[ \frac {2 A \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {b B \log (a+b \sin (x))}{a^2-b^2}-\frac {B \log (1-\sin (x))}{2 (a+b)}+\frac {B \log (\sin (x)+1)}{2 (a-b)} \]
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Rubi [A] time = 0.26, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4226, 4401, 2660, 618, 204, 2668, 706, 31, 633} \[ \frac {2 A \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {b B \log (a+b \sin (x))}{a^2-b^2}-\frac {B \log (1-\sin (x))}{2 (a+b)}+\frac {B \log (\sin (x)+1)}{2 (a-b)} \]
Antiderivative was successfully verified.
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Rule 31
Rule 204
Rule 618
Rule 633
Rule 706
Rule 2660
Rule 2668
Rule 4226
Rule 4401
Rubi steps
\begin {align*} \int \frac {A+B \sec (x)}{a+b \sin (x)} \, dx &=\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) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+(b B) \operatorname {Subst}\left (\int \frac {1}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (x)\right )\\ &=-\left ((4 A) \operatorname {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) \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \sin (x)\right )}{a^2-b^2}-\frac {(b B) \operatorname {Subst}\left (\int \frac {-a+x}{b^2-x^2} \, dx,x,b \sin (x)\right )}{a^2-b^2}\\ &=\frac {2 A \tan ^{-1}\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 \operatorname {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (x)\right )}{2 (a-b)}+\frac {B \operatorname {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (x)\right )}{2 (a+b)}\\ &=\frac {2 A \tan ^{-1}\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*}
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Mathematica [A] time = 0.24, size = 108, normalized size = 1.10 \[ \frac {2 A \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )+B \left (-b \log (a+b \sin (x))+(b-a) \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+(a+b) \log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )\right )}{(a-b) (a+b)} \]
Antiderivative was successfully verified.
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fricas [A] time = 6.33, size = 278, normalized size = 2.84 \[ \left [-\frac {B b \log \left (-b^{2} \cos \relax (x)^{2} + 2 \, a b \sin \relax (x) + a^{2} + b^{2}\right ) + \sqrt {-a^{2} + b^{2}} A \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2} + 2 \, {\left (a \cos \relax (x) \sin \relax (x) + b \cos \relax (x)\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) - {\left (B a + B b\right )} \log \left (\sin \relax (x) + 1\right ) + {\left (B a - B b\right )} \log \left (-\sin \relax (x) + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )}}, -\frac {B b \log \left (-b^{2} \cos \relax (x)^{2} + 2 \, a b \sin \relax (x) + a^{2} + b^{2}\right ) + 2 \, \sqrt {a^{2} - b^{2}} A \arctan \left (-\frac {a \sin \relax (x) + b}{\sqrt {a^{2} - b^{2}} \cos \relax (x)}\right ) - {\left (B a + B b\right )} \log \left (\sin \relax (x) + 1\right ) + {\left (B a - B b\right )} \log \left (-\sin \relax (x) + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 116, normalized size = 1.18 \[ -\frac {B b \log \left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}{a^{2} - b^{2}} + \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} A}{\sqrt {a^{2} - b^{2}}} + \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{a - b} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{a + b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.26, size = 182, normalized size = 1.86 \[ -\frac {2 B \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{2 a +2 b}+\frac {2 B \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{2 a -2 b}-\frac {B b \ln \left (\left (\tan ^{2}\left (\frac {x}{2}\right )\right ) a +2 \tan \left (\frac {x}{2}\right ) b +a \right )}{\left (a -b \right ) \left (a +b \right )}+\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) a^{2} A}{\left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}}-\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) A \,b^{2}}{\left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 20.28, size = 755, normalized size = 7.70 \[ \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} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \sec {\relax (x )}}{a + b \sin {\relax (x )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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