Optimal. Leaf size=154 \[ \frac{2 e^2 \left (9 a^2+2 b^2\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{15 d \sqrt{\sin (c+d x)}}-\frac{2 e \left (9 a^2+2 b^2\right ) \cos (c+d x) (e \sin (c+d x))^{3/2}}{45 d}+\frac{22 a b (e \sin (c+d x))^{7/2}}{63 d e}+\frac{2 b (e \sin (c+d x))^{7/2} (a+b \cos (c+d x))}{9 d e} \]
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Rubi [A] time = 0.160937, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2692, 2669, 2635, 2640, 2639} \[ \frac{2 e^2 \left (9 a^2+2 b^2\right ) E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{15 d \sqrt{\sin (c+d x)}}-\frac{2 e \left (9 a^2+2 b^2\right ) \cos (c+d x) (e \sin (c+d x))^{3/2}}{45 d}+\frac{22 a b (e \sin (c+d x))^{7/2}}{63 d e}+\frac{2 b (e \sin (c+d x))^{7/2} (a+b \cos (c+d x))}{9 d e} \]
Antiderivative was successfully verified.
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Rule 2692
Rule 2669
Rule 2635
Rule 2640
Rule 2639
Rubi steps
\begin{align*} \int (a+b \cos (c+d x))^2 (e \sin (c+d x))^{5/2} \, dx &=\frac{2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{9 d e}+\frac{2}{9} \int \left (\frac{9 a^2}{2}+b^2+\frac{11}{2} a b \cos (c+d x)\right ) (e \sin (c+d x))^{5/2} \, dx\\ &=\frac{22 a b (e \sin (c+d x))^{7/2}}{63 d e}+\frac{2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{9 d e}+\frac{1}{9} \left (9 a^2+2 b^2\right ) \int (e \sin (c+d x))^{5/2} \, dx\\ &=-\frac{2 \left (9 a^2+2 b^2\right ) e \cos (c+d x) (e \sin (c+d x))^{3/2}}{45 d}+\frac{22 a b (e \sin (c+d x))^{7/2}}{63 d e}+\frac{2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{9 d e}+\frac{1}{15} \left (\left (9 a^2+2 b^2\right ) e^2\right ) \int \sqrt{e \sin (c+d x)} \, dx\\ &=-\frac{2 \left (9 a^2+2 b^2\right ) e \cos (c+d x) (e \sin (c+d x))^{3/2}}{45 d}+\frac{22 a b (e \sin (c+d x))^{7/2}}{63 d e}+\frac{2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{9 d e}+\frac{\left (\left (9 a^2+2 b^2\right ) e^2 \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{15 \sqrt{\sin (c+d x)}}\\ &=\frac{2 \left (9 a^2+2 b^2\right ) e^2 E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{15 d \sqrt{\sin (c+d x)}}-\frac{2 \left (9 a^2+2 b^2\right ) e \cos (c+d x) (e \sin (c+d x))^{3/2}}{45 d}+\frac{22 a b (e \sin (c+d x))^{7/2}}{63 d e}+\frac{2 b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{9 d e}\\ \end{align*}
Mathematica [A] time = 0.801063, size = 116, normalized size = 0.75 \[ -\frac{(e \sin (c+d x))^{5/2} \left (84 \left (9 a^2+2 b^2\right ) E\left (\left .\frac{1}{4} (-2 c-2 d x+\pi )\right |2\right )+\sin ^{\frac{3}{2}}(c+d x) \left (21 \left (12 a^2+b^2\right ) \cos (c+d x)+5 b (36 a \cos (2 (c+d x))-36 a+7 b \cos (3 (c+d x)))\right )\right )}{630 d \sin ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Maple [A] time = 1.938, size = 332, normalized size = 2.2 \begin{align*}{\frac{1}{d} \left ({\frac{4\,ab}{7\,e} \left ( e\sin \left ( dx+c \right ) \right ) ^{{\frac{7}{2}}}}-{\frac{{e}^{3}}{45\,\cos \left ( dx+c \right ) } \left ( 10\,{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}+54\,\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticE} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ){a}^{2}+12\,\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticE} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ){b}^{2}-27\,\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ){a}^{2}-6\,\sqrt{1-\sin \left ( dx+c \right ) }\sqrt{2+2\,\sin \left ( dx+c \right ) }\sqrt{\sin \left ( dx+c \right ) }{\it EllipticF} \left ( \sqrt{1-\sin \left ( dx+c \right ) },1/2\,\sqrt{2} \right ){b}^{2}-18\,{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}-14\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}{b}^{2}+18\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}{a}^{2}+4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{2} \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}} \right ) } \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{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (b^{2} e^{2} \cos \left (d x + c\right )^{4} + 2 \, a b e^{2} \cos \left (d x + c\right )^{3} - 2 \, a b e^{2} \cos \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} e^{2} \cos \left (d x + c\right )^{2} - a^{2} e^{2}\right )} \sqrt{e \sin \left (d x + c\right )}, x\right ) \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{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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