Optimal. Leaf size=461 \[ -\frac{e^{9/2} \left (b^2-a^2\right )^{7/4} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} d}+\frac{e^{9/2} \left (b^2-a^2\right )^{7/4} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} d}+\frac{2 e^3 (e \sin (c+d x))^{3/2} \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right )}{15 b^3 d}-\frac{2 a e^4 \left (5 a^2-8 b^2\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 b^4 d \sqrt{\sin (c+d x)}}+\frac{a e^5 \left (a^2-b^2\right )^2 \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{b^5 d \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \sin (c+d x)}}+\frac{a e^5 \left (a^2-b^2\right )^2 \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{b^5 d \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \sin (c+d x)}}-\frac{2 e (e \sin (c+d x))^{7/2}}{7 b d} \]
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Rubi [A] time = 1.33608, antiderivative size = 461, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {2695, 2865, 2867, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ -\frac{e^{9/2} \left (b^2-a^2\right )^{7/4} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} d}+\frac{e^{9/2} \left (b^2-a^2\right )^{7/4} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} d}+\frac{2 e^3 (e \sin (c+d x))^{3/2} \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right )}{15 b^3 d}-\frac{2 a e^4 \left (5 a^2-8 b^2\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 b^4 d \sqrt{\sin (c+d x)}}+\frac{a e^5 \left (a^2-b^2\right )^2 \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{b^5 d \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \sin (c+d x)}}+\frac{a e^5 \left (a^2-b^2\right )^2 \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{b^5 d \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \sin (c+d x)}}-\frac{2 e (e \sin (c+d x))^{7/2}}{7 b d} \]
Antiderivative was successfully verified.
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Rule 2695
Rule 2865
Rule 2867
Rule 2640
Rule 2639
Rule 2701
Rule 2807
Rule 2805
Rule 329
Rule 298
Rule 205
Rule 208
Rubi steps
\begin{align*} \int \frac{(e \sin (c+d x))^{9/2}}{a+b \cos (c+d x)} \, dx &=-\frac{2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac{e^2 \int \frac{(-b-a \cos (c+d x)) (e \sin (c+d x))^{5/2}}{a+b \cos (c+d x)} \, dx}{b}\\ &=\frac{2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac{2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac{\left (2 e^4\right ) \int \frac{\left (\frac{1}{2} b \left (2 a^2-5 b^2\right )+\frac{1}{2} a \left (5 a^2-8 b^2\right ) \cos (c+d x)\right ) \sqrt{e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{5 b^3}\\ &=\frac{2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac{2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac{\left (a \left (5 a^2-8 b^2\right ) e^4\right ) \int \sqrt{e \sin (c+d x)} \, dx}{5 b^4}+\frac{\left (\left (a^2-b^2\right )^2 e^4\right ) \int \frac{\sqrt{e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{b^4}\\ &=\frac{2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac{2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac{\left (a \left (a^2-b^2\right )^2 e^5\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^5}+\frac{\left (a \left (a^2-b^2\right )^2 e^5\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^5}-\frac{\left (\left (a^2-b^2\right )^2 e^5\right ) \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 )}{b^3 d}-\frac{\left (a \left (5 a^2-8 b^2\right ) e^4 \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{5 b^4 \sqrt{\sin (c+d x)}}\\ &=-\frac{2 a \left (5 a^2-8 b^2\right ) e^4 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 b^4 d \sqrt{\sin (c+d x)}}+\frac{2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac{2 e (e \sin (c+d x))^{7/2}}{7 b d}-\frac{\left (2 \left (a^2-b^2\right )^2 e^5\right ) \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 )}{b^3 d}-\frac{\left (a \left (a^2-b^2\right )^2 e^5 \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^5 \sqrt{e \sin (c+d x)}}+\frac{\left (a \left (a^2-b^2\right )^2 e^5 \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^5 \sqrt{e \sin (c+d x)}}\\ &=\frac{a \left (a^2-b^2\right )^2 e^5 \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^5 \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{a \left (a^2-b^2\right )^2 e^5 \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^5 \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{2 a \left (5 a^2-8 b^2\right ) e^4 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 b^4 d \sqrt{\sin (c+d x)}}+\frac{2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac{2 e (e \sin (c+d x))^{7/2}}{7 b d}+\frac{\left (\left (a^2-b^2\right )^2 e^5\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^4 d}-\frac{\left (\left (a^2-b^2\right )^2 e^5\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^4 d}\\ &=-\frac{\left (-a^2+b^2\right )^{7/4} e^{9/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{b^{9/2} d}+\frac{\left (-a^2+b^2\right )^{7/4} e^{9/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{b^{9/2} d}+\frac{a \left (a^2-b^2\right )^2 e^5 \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^5 \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{a \left (a^2-b^2\right )^2 e^5 \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^5 \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{2 a \left (5 a^2-8 b^2\right ) e^4 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 b^4 d \sqrt{\sin (c+d x)}}+\frac{2 e^3 \left (5 \left (a^2-b^2\right )-3 a b \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{15 b^3 d}-\frac{2 e (e \sin (c+d x))^{7/2}}{7 b d}\\ \end{align*}
Mathematica [C] time = 15.0076, size = 834, normalized size = 1.81 \[ \frac{\csc ^4(c+d x) (e \sin (c+d x))^{9/2} \left (-\frac{\left (37 b^2-28 a^2\right ) \sin (c+d x)}{42 b^3}-\frac{a \sin (2 (c+d x))}{5 b^2}+\frac{\sin (3 (c+d x))}{14 b}\right )}{d}-\frac{(e \sin (c+d x))^{9/2} \left (\frac{\left (5 a^3-8 a b^2\right ) \left (8 F_1\left (\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right ) \sin ^{\frac{3}{2}}(c+d x) b^{5/2}+3 \sqrt{2} a \left (a^2-b^2\right )^{3/4} \left (2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )-\log \left (b \sin (c+d x)-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\sin (c+d x)}+\sqrt{a^2-b^2}\right )+\log \left (b \sin (c+d x)+\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\sin (c+d x)}+\sqrt{a^2-b^2}\right )\right )\right ) \left (a+b \sqrt{1-\sin ^2(c+d x)}\right ) \cos ^2(c+d x)}{12 b^{3/2} \left (b^2-a^2\right ) (a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac{2 \left (2 a^2 b-5 b^3\right ) \left (\frac{a F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right ) \sin ^{\frac{3}{2}}(c+d x)}{3 \left (a^2-b^2\right )}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \left (2 \tan ^{-1}\left (1-\frac{(1+i) \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac{(1+i) \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )-\log \left (i b \sin (c+d x)-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}\right )+\log \left (i b \sin (c+d x)+(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}\right )\right )}{\sqrt{b} \sqrt [4]{b^2-a^2}}\right ) \left (a+b \sqrt{1-\sin ^2(c+d x)}\right ) \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt{1-\sin ^2(c+d x)}}\right )}{5 b^3 d \sin ^{\frac{9}{2}}(c+d x)} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 8.355, size = 2051, normalized size = 4.5 \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{9}{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{9}{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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