Optimal. Leaf size=75 \[ \frac{2 \sqrt{a \cos (c+d x)+b \sin (c+d x)} E\left (\left .\frac{1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{d \sqrt{\frac{a \cos (c+d x)+b \sin (c+d x)}{\sqrt{a^2+b^2}}}} \]
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Rubi [A] time = 0.0296799, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3078, 2639} \[ \frac{2 \sqrt{a \cos (c+d x)+b \sin (c+d x)} E\left (\left .\frac{1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{d \sqrt{\frac{a \cos (c+d x)+b \sin (c+d x)}{\sqrt{a^2+b^2}}}} \]
Antiderivative was successfully verified.
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Rule 3078
Rule 2639
Rubi steps
\begin{align*} \int \sqrt{a \cos (c+d x)+b \sin (c+d x)} \, dx &=\frac{\sqrt{a \cos (c+d x)+b \sin (c+d x)} \int \sqrt{\cos \left (c+d x-\tan ^{-1}(a,b)\right )} \, dx}{\sqrt{\frac{a \cos (c+d x)+b \sin (c+d x)}{\sqrt{a^2+b^2}}}}\\ &=\frac{2 E\left (\left .\frac{1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right ) \sqrt{a \cos (c+d x)+b \sin (c+d x)}}{d \sqrt{\frac{a \cos (c+d x)+b \sin (c+d x)}{\sqrt{a^2+b^2}}}}\\ \end{align*}
Mathematica [C] time = 1.14195, size = 268, normalized size = 3.57 \[ \frac{\cos \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right ) \left (\sqrt{\sin ^2\left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )} \left (b \left (a^2+b^2\right ) \sin \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )-2 a \left (a^2+b^2\right ) \cos \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )+2 a^2 \sqrt{\frac{b^2}{a^2}+1} \sqrt{a \cos (c+d x)+b \sin (c+d x)} \sqrt{a \sqrt{\frac{b^2}{a^2}+1} \cos \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )}\right )-b \left (a^2+b^2\right ) \sin \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )\right )\right )}{b d \sqrt{\sin ^2\left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )} \left (a \sqrt{\frac{b^2}{a^2}+1} \cos \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )\right )^{3/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 1.204, size = 159, normalized size = 2.1 \begin{align*} -{\frac{1}{\cos \left ( dx+c-\arctan \left ( -a,b \right ) \right ) d}\sqrt{{a}^{2}+{b}^{2}}\sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) }\sqrt{-2\,\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) +2}\sqrt{-\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) } \left ( 2\,{\it EllipticE} \left ( \sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) },1/2\,\sqrt{2} \right ) -{\it EllipticF} \left ( \sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) },{\frac{\sqrt{2}}{2}} \right ) \right ){\frac{1}{\sqrt{\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \sqrt{{a}^{2}+{b}^{2}}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \cos{\left (c + d x \right )} + b \sin{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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