Optimal. Leaf size=497 \[ -\frac{5 e^{7/2} \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{7/2} d \left (b^2-a^2\right )^{3/4}}-\frac{5 e^{7/2} \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{7/2} d \left (b^2-a^2\right )^{3/4}}+\frac{5 a e^4 \left (3 a^2-2 b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^4 d \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \cos (c+d x)}}+\frac{5 a e^4 \left (3 a^2-2 b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^4 d \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \cos (c+d x)}}-\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a+2 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))}-\frac{15 a e^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d \sqrt{e \cos (c+d x)}}-\frac{e (e \cos (c+d x))^{5/2}}{2 b d (a+b \sin (c+d x))^2} \]
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Rubi [A] time = 1.07856, antiderivative size = 497, 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, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ -\frac{5 e^{7/2} \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{7/2} d \left (b^2-a^2\right )^{3/4}}-\frac{5 e^{7/2} \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 b^{7/2} d \left (b^2-a^2\right )^{3/4}}+\frac{5 a e^4 \left (3 a^2-2 b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^4 d \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \cos (c+d x)}}+\frac{5 a e^4 \left (3 a^2-2 b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^4 d \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \cos (c+d x)}}-\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a+2 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))}-\frac{15 a e^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d \sqrt{e \cos (c+d x)}}-\frac{e (e \cos (c+d x))^{5/2}}{2 b d (a+b \sin (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 2693
Rule 2863
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 \cos (c+d x))^{7/2}}{(a+b \sin (c+d x))^3} \, dx &=-\frac{e (e \cos (c+d x))^{5/2}}{2 b d (a+b \sin (c+d x))^2}-\frac{\left (5 e^2\right ) \int \frac{(e \cos (c+d x))^{3/2} \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{4 b}\\ &=-\frac{e (e \cos (c+d x))^{5/2}}{2 b d (a+b \sin (c+d x))^2}-\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a+2 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))}+\frac{\left (5 e^4\right ) \int \frac{-b-\frac{3}{2} a \sin (c+d x)}{\sqrt{e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{4 b^3}\\ &=-\frac{e (e \cos (c+d x))^{5/2}}{2 b d (a+b \sin (c+d x))^2}-\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a+2 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))}-\frac{\left (15 a e^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx}{8 b^4}+\frac{\left (5 \left (3 a^2-2 b^2\right ) e^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)} (a+b \sin (c+d x))} \, dx}{8 b^4}\\ &=-\frac{e (e \cos (c+d x))^{5/2}}{2 b d (a+b \sin (c+d x))^2}-\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a+2 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))}-\frac{\left (5 a \left (3 a^2-2 b^2\right ) e^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)} \left (\sqrt{-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{16 b^4 \sqrt{-a^2+b^2}}-\frac{\left (5 a \left (3 a^2-2 b^2\right ) e^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)} \left (\sqrt{-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{16 b^4 \sqrt{-a^2+b^2}}+\frac{\left (5 \left (3 a^2-2 b^2\right ) e^5\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 \cos (c+d x)\right )}{8 b^3 d}-\frac{\left (15 a e^4 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{8 b^4 \sqrt{e \cos (c+d x)}}\\ &=-\frac{15 a e^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d \sqrt{e \cos (c+d x)}}-\frac{e (e \cos (c+d x))^{5/2}}{2 b d (a+b \sin (c+d x))^2}-\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a+2 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))}+\frac{\left (5 \left (3 a^2-2 b^2\right ) e^5\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt{e \cos (c+d x)}\right )}{4 b^3 d}-\frac{\left (5 a \left (3 a^2-2 b^2\right ) e^4 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \left (\sqrt{-a^2+b^2}-b \cos (c+d x)\right )} \, dx}{16 b^4 \sqrt{-a^2+b^2} \sqrt{e \cos (c+d x)}}-\frac{\left (5 a \left (3 a^2-2 b^2\right ) e^4 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \left (\sqrt{-a^2+b^2}+b \cos (c+d x)\right )} \, dx}{16 b^4 \sqrt{-a^2+b^2} \sqrt{e \cos (c+d x)}}\\ &=-\frac{15 a e^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d \sqrt{e \cos (c+d x)}}+\frac{5 a \left (3 a^2-2 b^2\right ) e^4 \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^4 \left (a^2-b \left (b-\sqrt{-a^2+b^2}\right )\right ) d \sqrt{e \cos (c+d x)}}-\frac{5 a \left (3 a^2-2 b^2\right ) e^4 \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^4 \sqrt{-a^2+b^2} \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}-\frac{e (e \cos (c+d x))^{5/2}}{2 b d (a+b \sin (c+d x))^2}-\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a+2 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))}-\frac{\left (5 \left (3 a^2-2 b^2\right ) e^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e-b x^2} \, dx,x,\sqrt{e \cos (c+d x)}\right )}{8 b^3 \sqrt{-a^2+b^2} d}-\frac{\left (5 \left (3 a^2-2 b^2\right ) e^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e+b x^2} \, dx,x,\sqrt{e \cos (c+d x)}\right )}{8 b^3 \sqrt{-a^2+b^2} d}\\ &=-\frac{5 \left (3 a^2-2 b^2\right ) e^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}-\frac{5 \left (3 a^2-2 b^2\right ) e^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \cos (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{8 b^{7/2} \left (-a^2+b^2\right )^{3/4} d}-\frac{15 a e^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 b^4 d \sqrt{e \cos (c+d x)}}+\frac{5 a \left (3 a^2-2 b^2\right ) e^4 \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^4 \left (a^2-b \left (b-\sqrt{-a^2+b^2}\right )\right ) d \sqrt{e \cos (c+d x)}}-\frac{5 a \left (3 a^2-2 b^2\right ) e^4 \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} (c+d x)\right |2\right )}{8 b^4 \sqrt{-a^2+b^2} \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \cos (c+d x)}}-\frac{e (e \cos (c+d x))^{5/2}}{2 b d (a+b \sin (c+d x))^2}-\frac{5 e^3 \sqrt{e \cos (c+d x)} (3 a+2 b \sin (c+d x))}{4 b^3 d (a+b \sin (c+d x))}\\ \end{align*}
Mathematica [C] time = 26.4244, size = 1954, normalized size = 3.93 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 24.696, size = 65216, normalized size = 131.2 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{7}{2}}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}\,{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(-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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