Optimal. Leaf size=49 \[ -\frac {\sqrt {\frac {2}{5}} \tan ^{-1}\left (\frac {\sin \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}{\sqrt {2} \sqrt {\cos \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )-1}}\right )}{e} \]
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Rubi [A] time = 0.06, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3115, 2649, 204} \[ -\frac {\sqrt {\frac {2}{5}} \tan ^{-1}\left (\frac {\sin \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}{\sqrt {2} \sqrt {\cos \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )-1}}\right )}{e} \]
Antiderivative was successfully verified.
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Rule 204
Rule 2649
Rule 3115
Rubi steps
\begin {align*} \int \frac {1}{\sqrt {-5+4 \cos (d+e x)+3 \sin (d+e x)}} \, dx &=\int \frac {1}{\sqrt {-5+5 \cos \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}} \, dx\\ &=-\frac {2 \operatorname {Subst}\left (\int \frac {1}{-10-x^2} \, dx,x,-\frac {5 \sin \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}{\sqrt {-5+5 \cos \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}}\right )}{e}\\ &=-\frac {\sqrt {\frac {2}{5}} \tan ^{-1}\left (\frac {\sin \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}{\sqrt {2} \sqrt {-1+\cos \left (d+e x-\tan ^{-1}\left (\frac {3}{4}\right )\right )}}\right )}{e}\\ \end {align*}
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Mathematica [C] time = 0.09, size = 99, normalized size = 2.02 \[ \frac {\left (\frac {2}{5}+\frac {6 i}{5}\right ) \sqrt {-\frac {4}{5}-\frac {3 i}{5}} \left (\cos \left (\frac {1}{2} (d+e x)\right )-3 \sin \left (\frac {1}{2} (d+e x)\right )\right ) \tanh ^{-1}\left (\left (\frac {1}{10}+\frac {3 i}{10}\right ) \sqrt {-\frac {4}{5}-\frac {3 i}{5}} \left (\tan \left (\frac {1}{4} (d+e x)\right )+3\right )\right )}{e \sqrt {3 \sin (d+e x)+4 \cos (d+e x)-5}} \]
Antiderivative was successfully verified.
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fricas [B] time = 2.11, size = 88, normalized size = 1.80 \[ \frac {\sqrt {5} \sqrt {2} \arctan \left (-\frac {{\left (3 \, \sqrt {5} \sqrt {2} \cos \left (e x + d\right ) + \sqrt {5} \sqrt {2} \sin \left (e x + d\right ) + 3 \, \sqrt {5} \sqrt {2}\right )} \sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}{10 \, {\left (\cos \left (e x + d\right ) - 3 \, \sin \left (e x + d\right ) + 1\right )}}\right )}{5 \, e} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.33, size = 77, normalized size = 1.57 \[ \frac {\left (\sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-1\right ) \sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}\, \arctan \left (\frac {\sqrt {-5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )-5}\, \sqrt {10}}{10}\right )}{5 \cos \left (e x +d +\arctan \left (\frac {4}{3}\right )\right ) \sqrt {-5+5 \sin \left (e x +d +\arctan \left (\frac {4}{3}\right )\right )}\, e} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {4 \, \cos \left (e x + d\right ) + 3 \, \sin \left (e x + d\right ) - 5}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {4\,\cos \left (d+e\,x\right )+3\,\sin \left (d+e\,x\right )-5}} \,d x \]
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 \[ \int \frac {1}{\sqrt {3 \sin {\left (d + e x \right )} + 4 \cos {\left (d + e x \right )} - 5}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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