Optimal. Leaf size=48 \[ \frac {x (a \text {a1}+b \text {b1})}{a^2+b^2}-\frac {(\text {a1} b-a \text {b1}) \log (a \sin (x)+b \cos (x))}{a^2+b^2} \]
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Rubi [A] time = 0.03, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {3133} \[ \frac {x (a \text {a1}+b \text {b1})}{a^2+b^2}-\frac {(\text {a1} b-a \text {b1}) \log (a \sin (x)+b \cos (x))}{a^2+b^2} \]
Antiderivative was successfully verified.
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Rule 3133
Rubi steps
\begin {align*} \int \frac {\text {b1} \cos (x)+\text {a1} \sin (x)}{b \cos (x)+a \sin (x)} \, dx &=\frac {(a \text {a1}+b \text {b1}) x}{a^2+b^2}-\frac {(\text {a1} b-a \text {b1}) \log (b \cos (x)+a \sin (x))}{a^2+b^2}\\ \end {align*}
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Mathematica [A] time = 0.13, size = 39, normalized size = 0.81 \[ \frac {x (a \text {a1}+b \text {b1})+(a \text {b1}-\text {a1} b) \log (a \sin (x)+b \cos (x))}{a^2+b^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 60, normalized size = 1.25 \[ \frac {2 \, {\left (a a_{1} + b b_{1}\right )} x - {\left (a_{1} b - a b_{1}\right )} \log \left (2 \, a b \cos \relax (x) \sin \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + a^{2}\right )}{2 \, {\left (a^{2} + b^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.05, size = 77, normalized size = 1.60 \[ \frac {{\left (a a_{1} + b b_{1}\right )} x}{a^{2} + b^{2}} + \frac {{\left (a_{1} b - a b_{1}\right )} \log \left (\tan \relax (x)^{2} + 1\right )}{2 \, {\left (a^{2} + b^{2}\right )}} - \frac {{\left (a a_{1} b - a^{2} b_{1}\right )} \log \left ({\left | a \tan \relax (x) + b \right |}\right )}{a^{3} + a b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.13, size = 111, normalized size = 2.31 \[ \frac {a \mathit {a1} \arctan \left (\tan \relax (x )\right )}{a^{2}+b^{2}}-\frac {a \mathit {b1} \ln \left (\tan ^{2}\relax (x )+1\right )}{2 \left (a^{2}+b^{2}\right )}+\frac {a \mathit {b1} \ln \left (a \tan \relax (x )+b \right )}{a^{2}+b^{2}}+\frac {\mathit {a1} b \ln \left (\tan ^{2}\relax (x )+1\right )}{2 a^{2}+2 b^{2}}-\frac {\mathit {a1} b \ln \left (a \tan \relax (x )+b \right )}{a^{2}+b^{2}}+\frac {b \mathit {b1} \arctan \left (\tan \relax (x )\right )}{a^{2}+b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 1.30, size = 181, normalized size = 3.77 \[ a_{1} {\left (\frac {2 \, a \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{2} + b^{2}} - \frac {b \log \left (-b - \frac {2 \, a \sin \relax (x)}{\cos \relax (x) + 1} + \frac {b \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )}{a^{2} + b^{2}} + \frac {b \log \left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )}{a^{2} + b^{2}}\right )} + b_{1} {\left (\frac {2 \, b \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{2} + b^{2}} + \frac {a \log \left (-b - \frac {2 \, a \sin \relax (x)}{\cos \relax (x) + 1} + \frac {b \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )}{a^{2} + b^{2}} - \frac {a \log \left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )}{a^{2} + b^{2}}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 10.61, size = 2034, normalized size = 42.38 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.00, size = 360, normalized size = 7.50 \[ \begin {cases} \tilde {\infty } \left (- a_{1} \log {\left (\cos {\relax (x )} \right )} + b_{1} x\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {- a_{1} \log {\left (\cos {\relax (x )} \right )} + b_{1} x}{b} & \text {for}\: a = 0 \\\frac {i a_{1} x \sin {\relax (x )}}{2 b \sin {\relax (x )} + 2 i b \cos {\relax (x )}} - \frac {a_{1} x \cos {\relax (x )}}{2 b \sin {\relax (x )} + 2 i b \cos {\relax (x )}} - \frac {i a_{1} \cos {\relax (x )}}{2 b \sin {\relax (x )} + 2 i b \cos {\relax (x )}} + \frac {b_{1} x \sin {\relax (x )}}{2 b \sin {\relax (x )} + 2 i b \cos {\relax (x )}} + \frac {i b_{1} x \cos {\relax (x )}}{2 b \sin {\relax (x )} + 2 i b \cos {\relax (x )}} + \frac {b_{1} \cos {\relax (x )}}{2 b \sin {\relax (x )} + 2 i b \cos {\relax (x )}} & \text {for}\: a = - i b \\- \frac {i a_{1} x \sin {\relax (x )}}{2 b \sin {\relax (x )} - 2 i b \cos {\relax (x )}} - \frac {a_{1} x \cos {\relax (x )}}{2 b \sin {\relax (x )} - 2 i b \cos {\relax (x )}} + \frac {i a_{1} \cos {\relax (x )}}{2 b \sin {\relax (x )} - 2 i b \cos {\relax (x )}} + \frac {b_{1} x \sin {\relax (x )}}{2 b \sin {\relax (x )} - 2 i b \cos {\relax (x )}} - \frac {i b_{1} x \cos {\relax (x )}}{2 b \sin {\relax (x )} - 2 i b \cos {\relax (x )}} + \frac {b_{1} \cos {\relax (x )}}{2 b \sin {\relax (x )} - 2 i b \cos {\relax (x )}} & \text {for}\: a = i b \\\frac {a a_{1} x}{a^{2} + b^{2}} + \frac {a b_{1} \log {\left (\sin {\relax (x )} + \frac {b \cos {\relax (x )}}{a} \right )}}{a^{2} + b^{2}} - \frac {a_{1} b \log {\left (\sin {\relax (x )} + \frac {b \cos {\relax (x )}}{a} \right )}}{a^{2} + b^{2}} + \frac {b b_{1} x}{a^{2} + b^{2}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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