Optimal. Leaf size=604 \[ -\frac{9 e^{11/2} \left (-9 a^2 b^2+7 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{11/2} d \left (b^2-a^2\right )^{3/4}}-\frac{9 e^{11/2} \left (-9 a^2 b^2+7 a^4+2 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{11/2} d \left (b^2-a^2\right )^{3/4}}-\frac{3 e^5 \sqrt{e \sin (c+d x)} \left (3 \left (7 a^2-2 b^2\right )-7 a b \cos (c+d x)\right )}{4 b^5 d}+\frac{3 a e^6 \left (21 a^2-13 b^2\right ) \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{4 b^6 d \sqrt{e \sin (c+d x)}}-\frac{9 a e^6 \left (-9 a^2 b^2+7 a^4+2 b^4\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b^6 d \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \sin (c+d x)}}-\frac{9 a e^6 \left (-9 a^2 b^2+7 a^4+2 b^4\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b^6 d \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \sin (c+d x)}}+\frac{9 e^3 (e \sin (c+d x))^{5/2} (7 a+2 b \cos (c+d x))}{20 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{9/2}}{2 b d (a+b \cos (c+d x))^2} \]
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Rubi [A] time = 1.58172, antiderivative size = 604, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 13, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.52, Rules used = {2693, 2863, 2865, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ -\frac{9 e^{11/2} \left (-9 a^2 b^2+7 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{11/2} d \left (b^2-a^2\right )^{3/4}}-\frac{9 e^{11/2} \left (-9 a^2 b^2+7 a^4+2 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{11/2} d \left (b^2-a^2\right )^{3/4}}-\frac{3 e^5 \sqrt{e \sin (c+d x)} \left (3 \left (7 a^2-2 b^2\right )-7 a b \cos (c+d x)\right )}{4 b^5 d}+\frac{3 a e^6 \left (21 a^2-13 b^2\right ) \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{4 b^6 d \sqrt{e \sin (c+d x)}}-\frac{9 a e^6 \left (-9 a^2 b^2+7 a^4+2 b^4\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b^6 d \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \sin (c+d x)}}-\frac{9 a e^6 \left (-9 a^2 b^2+7 a^4+2 b^4\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b^6 d \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \sin (c+d x)}}+\frac{9 e^3 (e \sin (c+d x))^{5/2} (7 a+2 b \cos (c+d x))}{20 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{9/2}}{2 b d (a+b \cos (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 2693
Rule 2863
Rule 2865
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))^{11/2}}{(a+b \cos (c+d x))^3} \, dx &=\frac{e (e \sin (c+d x))^{9/2}}{2 b d (a+b \cos (c+d x))^2}-\frac{\left (9 e^2\right ) \int \frac{\cos (c+d x) (e \sin (c+d x))^{7/2}}{(a+b \cos (c+d x))^2} \, dx}{4 b}\\ &=\frac{9 e^3 (7 a+2 b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{20 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{9/2}}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (9 e^4\right ) \int \frac{\left (-b-\frac{7}{2} a \cos (c+d x)\right ) (e \sin (c+d x))^{3/2}}{a+b \cos (c+d x)} \, dx}{4 b^3}\\ &=-\frac{3 e^5 \left (3 \left (7 a^2-2 b^2\right )-7 a b \cos (c+d x)\right ) \sqrt{e \sin (c+d x)}}{4 b^5 d}+\frac{9 e^3 (7 a+2 b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{20 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{9/2}}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (3 e^6\right ) \int \frac{\frac{1}{2} b \left (7 a^2-3 b^2\right )+\frac{1}{4} a \left (21 a^2-13 b^2\right ) \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}} \, dx}{2 b^5}\\ &=-\frac{3 e^5 \left (3 \left (7 a^2-2 b^2\right )-7 a b \cos (c+d x)\right ) \sqrt{e \sin (c+d x)}}{4 b^5 d}+\frac{9 e^3 (7 a+2 b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{20 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{9/2}}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (3 a \left (21 a^2-13 b^2\right ) e^6\right ) \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx}{8 b^6}-\frac{\left (9 \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^6\right ) \int \frac{1}{(a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}} \, dx}{8 b^6}\\ &=-\frac{3 e^5 \left (3 \left (7 a^2-2 b^2\right )-7 a b \cos (c+d x)\right ) \sqrt{e \sin (c+d x)}}{4 b^5 d}+\frac{9 e^3 (7 a+2 b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{20 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{9/2}}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (9 a \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^6\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^6 \sqrt{-a^2+b^2}}+\frac{\left (9 a \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^6\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^6 \sqrt{-a^2+b^2}}+\frac{\left (9 \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^7\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 )}{8 b^5 d}+\frac{\left (3 a \left (21 a^2-13 b^2\right ) e^6 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{8 b^6 \sqrt{e \sin (c+d x)}}\\ &=\frac{3 a \left (21 a^2-13 b^2\right ) e^6 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{4 b^6 d \sqrt{e \sin (c+d x)}}-\frac{3 e^5 \left (3 \left (7 a^2-2 b^2\right )-7 a b \cos (c+d x)\right ) \sqrt{e \sin (c+d x)}}{4 b^5 d}+\frac{9 e^3 (7 a+2 b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{20 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{9/2}}{2 b d (a+b \cos (c+d x))^2}+\frac{\left (9 \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^7\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 )}{4 b^5 d}+\frac{\left (9 a \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^6 \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^6 \sqrt{-a^2+b^2} \sqrt{e \sin (c+d x)}}+\frac{\left (9 a \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^6 \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^6 \sqrt{-a^2+b^2} \sqrt{e \sin (c+d x)}}\\ &=\frac{3 a \left (21 a^2-13 b^2\right ) e^6 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{4 b^6 d \sqrt{e \sin (c+d x)}}-\frac{9 a \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^6 \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^6 \left (a^2-b \left (b-\sqrt{-a^2+b^2}\right )\right ) d \sqrt{e \sin (c+d x)}}+\frac{9 a \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^6 \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^6 \sqrt{-a^2+b^2} \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{3 e^5 \left (3 \left (7 a^2-2 b^2\right )-7 a b \cos (c+d x)\right ) \sqrt{e \sin (c+d x)}}{4 b^5 d}+\frac{9 e^3 (7 a+2 b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{20 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{9/2}}{2 b d (a+b \cos (c+d x))^2}-\frac{\left (9 \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^6\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^5 \sqrt{-a^2+b^2} d}-\frac{\left (9 \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^6\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^5 \sqrt{-a^2+b^2} d}\\ &=-\frac{9 \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^{11/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{8 b^{11/2} \left (-a^2+b^2\right )^{3/4} d}-\frac{9 \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^{11/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{8 b^{11/2} \left (-a^2+b^2\right )^{3/4} d}+\frac{3 a \left (21 a^2-13 b^2\right ) e^6 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{4 b^6 d \sqrt{e \sin (c+d x)}}-\frac{9 a \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^6 \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^6 \left (a^2-b \left (b-\sqrt{-a^2+b^2}\right )\right ) d \sqrt{e \sin (c+d x)}}+\frac{9 a \left (7 a^4-9 a^2 b^2+2 b^4\right ) e^6 \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^6 \sqrt{-a^2+b^2} \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{3 e^5 \left (3 \left (7 a^2-2 b^2\right )-7 a b \cos (c+d x)\right ) \sqrt{e \sin (c+d x)}}{4 b^5 d}+\frac{9 e^3 (7 a+2 b \cos (c+d x)) (e \sin (c+d x))^{5/2}}{20 b^3 d (a+b \cos (c+d x))}+\frac{e (e \sin (c+d x))^{9/2}}{2 b d (a+b \cos (c+d x))^2}\\ \end{align*}
Mathematica [C] time = 16.6971, size = 2024, normalized size = 3.35 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 15.757, size = 7238, normalized size = 12. \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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