Optimal. Leaf size=410 \[ \frac{e^{3/2} \sqrt [4]{b^2-a^2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}+\frac{e^{3/2} \sqrt [4]{b^2-a^2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}-\frac{a e^2 \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{b^2 d \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \sin (c+d x)}}-\frac{a e^2 \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{b^2 d \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \sin (c+d x)}}+\frac{2 a e^2 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{b^2 d \sqrt{e \sin (c+d x)}}-\frac{2 e \sqrt{e \sin (c+d x)}}{b d} \]
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Rubi [A] time = 0.904185, antiderivative size = 410, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {2695, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ \frac{e^{3/2} \sqrt [4]{b^2-a^2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}+\frac{e^{3/2} \sqrt [4]{b^2-a^2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} d}-\frac{a e^2 \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{b^2 d \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \sin (c+d x)}}-\frac{a e^2 \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{b^2 d \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \sin (c+d x)}}+\frac{2 a e^2 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{b^2 d \sqrt{e \sin (c+d x)}}-\frac{2 e \sqrt{e \sin (c+d x)}}{b d} \]
Antiderivative was successfully verified.
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Rule 2695
Rule 2867
Rule 2642
Rule 2641
Rule 2702
Rule 2807
Rule 2805
Rule 329
Rule 212
Rule 208
Rule 205
Rubi steps
\begin{align*} \int \frac{(e \sin (c+d x))^{3/2}}{a+b \cos (c+d x)} \, dx &=-\frac{2 e \sqrt{e \sin (c+d x)}}{b d}-\frac{e^2 \int \frac{-b-a \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}} \, dx}{b}\\ &=-\frac{2 e \sqrt{e \sin (c+d x)}}{b d}+\frac{\left (a e^2\right ) \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx}{b^2}+\frac{\left (\left (-a^2+b^2\right ) e^2\right ) \int \frac{1}{(a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}} \, dx}{b^2}\\ &=-\frac{2 e \sqrt{e \sin (c+d x)}}{b d}-\frac{\left (a \sqrt{-a^2+b^2} e^2\right ) \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 b^2}-\frac{\left (a \sqrt{-a^2+b^2} e^2\right ) \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 b^2}+\frac{\left (\left (a^2-b^2\right ) e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (\left (a^2-b^2\right ) e^2+b^2 x^2\right )} \, dx,x,e \sin (c+d x)\right )}{b d}+\frac{\left (a e^2 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{b^2 \sqrt{e \sin (c+d x)}}\\ &=\frac{2 a e^2 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{b^2 d \sqrt{e \sin (c+d x)}}-\frac{2 e \sqrt{e \sin (c+d x)}}{b d}+\frac{\left (2 \left (a^2-b^2\right ) e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{b d}-\frac{\left (a \sqrt{-a^2+b^2} e^2 \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^2 \sqrt{e \sin (c+d x)}}-\frac{\left (a \sqrt{-a^2+b^2} e^2 \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^2 \sqrt{e \sin (c+d x)}}\\ &=\frac{2 a e^2 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{b^2 d \sqrt{e \sin (c+d x)}}+\frac{a \sqrt{-a^2+b^2} e^2 \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^2 \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{a \sqrt{-a^2+b^2} e^2 \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^2 \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{2 e \sqrt{e \sin (c+d x)}}{b d}+\frac{\left (\sqrt{-a^2+b^2} e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e-b x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{b d}+\frac{\left (\sqrt{-a^2+b^2} e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e+b x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{b d}\\ &=\frac{\sqrt [4]{-a^2+b^2} e^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{b^{3/2} d}+\frac{\sqrt [4]{-a^2+b^2} e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{b^{3/2} d}+\frac{2 a e^2 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{b^2 d \sqrt{e \sin (c+d x)}}+\frac{a \sqrt{-a^2+b^2} e^2 \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^2 \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{a \sqrt{-a^2+b^2} e^2 \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^2 \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{2 e \sqrt{e \sin (c+d x)}}{b d}\\ \end{align*}
Mathematica [C] time = 5.83121, size = 434, normalized size = 1.06 \[ -\frac{\left (\frac{1}{20}-\frac{i}{20}\right ) \cos (c+d x) (e \sin (c+d x))^{3/2} \left (a+b \sqrt{\cos ^2(c+d x)}\right ) \left ((4+4 i) a b^{3/2} \sin ^{\frac{5}{2}}(c+d x) F_1\left (\frac{5}{4};\frac{1}{2},1;\frac{9}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right )-5 \left (a^2-b^2\right ) \left (\sqrt [4]{b^2-a^2} \log \left (-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}+i b \sin (c+d x)\right )-\sqrt [4]{b^2-a^2} \log \left ((1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}+i b \sin (c+d x)\right )+2 \sqrt [4]{b^2-a^2} \tan ^{-1}\left (1-\frac{(1+i) \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \sqrt [4]{b^2-a^2} \tan ^{-1}\left (1+\frac{(1+i) \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )+(4+4 i) \sqrt{b} \sqrt{\sin (c+d x)}\right )\right )}{b^{3/2} d \left (b^2-a^2\right ) \sin ^{\frac{3}{2}}(c+d x) \sqrt{\cos ^2(c+d x)} (a+b \cos (c+d x))} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 5.977, size = 1314, normalized size = 3.2 \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{\left (e \sin \left (d x + c\right )\right )^{\frac{3}{2}}}{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(-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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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sin \left (d x + c\right )\right )^{\frac{3}{2}}}{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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