Optimal. Leaf size=498 \[ -\frac{7 e^{9/2} \left (5 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{9/2} d \sqrt [4]{b^2-a^2}}+\frac{7 e^{9/2} \left (5 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{9/2} d \sqrt [4]{b^2-a^2}}+\frac{7 a e^5 \left (5 a^2-2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b^5 d \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \sin (c+d x)}}+\frac{7 a e^5 \left (5 a^2-2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b^5 d \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \sin (c+d x)}}+\frac{7 e^3 (e \sin (c+d x))^{3/2} (5 a+2 b \cos (c+d x))}{12 b^3 d (a+b \cos (c+d x))}-\frac{35 a e^4 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{4 b^4 d \sqrt{\sin (c+d x)}}+\frac{e (e \sin (c+d x))^{7/2}}{2 b d (a+b \cos (c+d x))^2} \]
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Rubi [A] time = 1.11482, antiderivative size = 498, 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 = {2693, 2863, 2867, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ -\frac{7 e^{9/2} \left (5 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{9/2} d \sqrt [4]{b^2-a^2}}+\frac{7 e^{9/2} \left (5 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{9/2} d \sqrt [4]{b^2-a^2}}+\frac{7 a e^5 \left (5 a^2-2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b^5 d \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \sin (c+d x)}}+\frac{7 a e^5 \left (5 a^2-2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b^5 d \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \sin (c+d x)}}+\frac{7 e^3 (e \sin (c+d x))^{3/2} (5 a+2 b \cos (c+d x))}{12 b^3 d (a+b \cos (c+d x))}-\frac{35 a e^4 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{4 b^4 d \sqrt{\sin (c+d x)}}+\frac{e (e \sin (c+d x))^{7/2}}{2 b d (a+b \cos (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 2693
Rule 2863
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))^3} \, dx &=\frac{e (e \sin (c+d x))^{7/2}}{2 b d (a+b \cos (c+d x))^2}-\frac{\left (7 e^2\right ) \int \frac{\cos (c+d x) (e \sin (c+d x))^{5/2}}{(a+b \cos (c+d x))^2} \, dx}{4 b}\\ &=\frac{7 e^3 (5 a+2 b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{12 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{7/2}}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (7 e^4\right ) \int \frac{\left (-b-\frac{5}{2} a \cos (c+d x)\right ) \sqrt{e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{4 b^3}\\ &=\frac{7 e^3 (5 a+2 b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{12 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{7/2}}{2 b d (a+b \cos (c+d x))^2}-\frac{\left (35 a e^4\right ) \int \sqrt{e \sin (c+d x)} \, dx}{8 b^4}+\frac{\left (7 \left (5 a^2-2 b^2\right ) e^4\right ) \int \frac{\sqrt{e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{8 b^4}\\ &=\frac{7 e^3 (5 a+2 b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{12 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{7/2}}{2 b d (a+b \cos (c+d x))^2}-\frac{\left (7 a \left (5 a^2-2 b^2\right ) 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}{16 b^5}+\frac{\left (7 a \left (5 a^2-2 b^2\right ) 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}{16 b^5}-\frac{\left (7 \left (5 a^2-2 b^2\right ) 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 )}{8 b^3 d}-\frac{\left (35 a e^4 \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{8 b^4 \sqrt{\sin (c+d x)}}\\ &=-\frac{35 a 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)}}{4 b^4 d \sqrt{\sin (c+d x)}}+\frac{7 e^3 (5 a+2 b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{12 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{7/2}}{2 b d (a+b \cos (c+d x))^2}-\frac{\left (7 \left (5 a^2-2 b^2\right ) 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 )}{4 b^3 d}-\frac{\left (7 a \left (5 a^2-2 b^2\right ) 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}{16 b^5 \sqrt{e \sin (c+d x)}}+\frac{\left (7 a \left (5 a^2-2 b^2\right ) 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}{16 b^5 \sqrt{e \sin (c+d x)}}\\ &=\frac{7 a \left (5 a^2-2 b^2\right ) 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)}}{8 b^5 \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{7 a \left (5 a^2-2 b^2\right ) 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)}}{8 b^5 \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{35 a 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)}}{4 b^4 d \sqrt{\sin (c+d x)}}+\frac{7 e^3 (5 a+2 b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{12 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{7/2}}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (7 \left (5 a^2-2 b^2\right ) 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 )}{8 b^4 d}-\frac{\left (7 \left (5 a^2-2 b^2\right ) 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 )}{8 b^4 d}\\ &=-\frac{7 \left (5 a^2-2 b^2\right ) e^{9/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{8 b^{9/2} \sqrt [4]{-a^2+b^2} d}+\frac{7 \left (5 a^2-2 b^2\right ) e^{9/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{8 b^{9/2} \sqrt [4]{-a^2+b^2} d}+\frac{7 a \left (5 a^2-2 b^2\right ) 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)}}{8 b^5 \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{7 a \left (5 a^2-2 b^2\right ) 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)}}{8 b^5 \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{35 a 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)}}{4 b^4 d \sqrt{\sin (c+d x)}}+\frac{7 e^3 (5 a+2 b \cos (c+d x)) (e \sin (c+d x))^{3/2}}{12 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{7/2}}{2 b d (a+b \cos (c+d x))^2}\\ \end{align*}
Mathematica [C] time = 14.6169, size = 837, normalized size = 1.68 \[ \frac{\csc ^4(c+d x) (e \sin (c+d x))^{9/2} \left (\frac{11 a \sin (c+d x)}{4 b^3 (a+b \cos (c+d x))}+\frac{2 \sin (c+d x)}{3 b^3}+\frac{b^2 \sin (c+d x)-a^2 \sin (c+d x)}{2 b^3 (a+b \cos (c+d x))^2}\right )}{d}-\frac{7 (e \sin (c+d x))^{9/2} \left (\frac{5 a \left (8 F_1\left (\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right ) \sin ^{\frac{3}{2}}(c+d x) b^{5/2}+3 \sqrt{2} a \left (a^2-b^2\right )^{3/4} \left (2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )-\log \left (b \sin (c+d x)-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\sin (c+d x)}+\sqrt{a^2-b^2}\right )+\log \left (b \sin (c+d x)+\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\sin (c+d x)}+\sqrt{a^2-b^2}\right )\right )\right ) \left (a+b \sqrt{1-\sin ^2(c+d x)}\right ) \cos ^2(c+d x)}{12 b^{3/2} \left (b^2-a^2\right ) (a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac{4 b \left (\frac{a F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right ) \sin ^{\frac{3}{2}}(c+d x)}{3 \left (a^2-b^2\right )}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \left (2 \tan ^{-1}\left (1-\frac{(1+i) \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac{(1+i) \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )-\log \left (i b \sin (c+d x)-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}\right )+\log \left (i b \sin (c+d x)+(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}\right )\right )}{\sqrt{b} \sqrt [4]{b^2-a^2}}\right ) \left (a+b \sqrt{1-\sin ^2(c+d x)}\right ) \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt{1-\sin ^2(c+d x)}}\right )}{8 b^3 d \sin ^{\frac{9}{2}}(c+d x)} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 13.824, size = 5791, normalized size = 11.6 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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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(-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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