Optimal. Leaf size=302 \[ -\frac{\sqrt{e} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{\sqrt{b} d \sqrt [4]{b^2-a^2}}+\frac{\sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{\sqrt{b} d \sqrt [4]{b^2-a^2}}+\frac{a e \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{b d \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \sin (c+d x)}}+\frac{a e \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{b d \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \sin (c+d x)}} \]
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Rubi [A] time = 0.607455, antiderivative size = 302, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2701, 2807, 2805, 329, 298, 205, 208} \[ -\frac{\sqrt{e} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{\sqrt{b} d \sqrt [4]{b^2-a^2}}+\frac{\sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{\sqrt{b} d \sqrt [4]{b^2-a^2}}+\frac{a e \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{b d \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \sin (c+d x)}}+\frac{a e \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{b d \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \sin (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 2701
Rule 2807
Rule 2805
Rule 329
Rule 298
Rule 205
Rule 208
Rubi steps
\begin{align*} \int \frac{\sqrt{e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx &=-\frac{(a e) \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 b}+\frac{(a e) \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 b}-\frac{(b e) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{\left (a^2-b^2\right ) e^2+b^2 x^2} \, dx,x,e \sin (c+d x)\right )}{d}\\ &=-\frac{(2 b e) \operatorname{Subst}\left (\int \frac{x^2}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d}-\frac{\left (a e \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)} \left (\sqrt{-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 b \sqrt{e \sin (c+d x)}}+\frac{\left (a e \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)} \left (\sqrt{-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 b \sqrt{e \sin (c+d x)}}\\ &=\frac{a e \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{b \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{a e \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{b \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{e \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e-b x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d}-\frac{e \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e+b x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d}\\ &=-\frac{\sqrt{e} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{\sqrt{b} \sqrt [4]{-a^2+b^2} d}+\frac{\sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{\sqrt{b} \sqrt [4]{-a^2+b^2} d}+\frac{a e \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{b \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{a e \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{b \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}\\ \end{align*}
Mathematica [C] time = 1.85541, size = 361, normalized size = 1.2 \[ \frac{2 \cos (c+d x) \sqrt{e \sin (c+d x)} \left (a+b \sqrt{\cos ^2(c+d x)}\right ) \left (\frac{a \sin ^{\frac{3}{2}}(c+d x) F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right )}{3 \left (a^2-b^2\right )}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \left (-\log \left (-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}+i b \sin (c+d x)\right )+\log \left ((1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}+i b \sin (c+d x)\right )+2 \tan ^{-1}\left (1-\frac{(1+i) \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (1+\frac{(1+i) \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )\right )}{\sqrt{b} \sqrt [4]{b^2-a^2}}\right )}{d \sqrt{\sin (c+d x)} \sqrt{\cos ^2(c+d x)} (a+b \cos (c+d x))} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 4.738, size = 815, normalized size = 2.7 \begin{align*} \text{result too large to display} \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 \frac{\sqrt{e \sin \left (d x + c\right )}}{b \cos \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 \frac{\sqrt{e \sin{\left (c + d x \right )}}}{a + b \cos{\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 \frac{\sqrt{e \sin \left (d x + c\right )}}{b \cos \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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