Optimal. Leaf size=149 \[ \frac{2 e^2 \left (7 a^2+2 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{e \cos (c+d x)}}+\frac{2 e \left (7 a^2+2 b^2\right ) \sin (c+d x) \sqrt{e \cos (c+d x)}}{21 d}-\frac{18 a b (e \cos (c+d x))^{5/2}}{35 d e}-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{7 d e} \]
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Rubi [A] time = 0.164014, antiderivative size = 149, 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, 2642, 2641} \[ \frac{2 e^2 \left (7 a^2+2 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{e \cos (c+d x)}}+\frac{2 e \left (7 a^2+2 b^2\right ) \sin (c+d x) \sqrt{e \cos (c+d x)}}{21 d}-\frac{18 a b (e \cos (c+d x))^{5/2}}{35 d e}-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{7 d e} \]
Antiderivative was successfully verified.
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Rule 2692
Rule 2669
Rule 2635
Rule 2642
Rule 2641
Rubi steps
\begin{align*} \int (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2 \, dx &=-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{7 d e}+\frac{2}{7} \int (e \cos (c+d x))^{3/2} \left (\frac{7 a^2}{2}+b^2+\frac{9}{2} a b \sin (c+d x)\right ) \, dx\\ &=-\frac{18 a b (e \cos (c+d x))^{5/2}}{35 d e}-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{7 d e}+\frac{1}{7} \left (7 a^2+2 b^2\right ) \int (e \cos (c+d x))^{3/2} \, dx\\ &=-\frac{18 a b (e \cos (c+d x))^{5/2}}{35 d e}+\frac{2 \left (7 a^2+2 b^2\right ) e \sqrt{e \cos (c+d x)} \sin (c+d x)}{21 d}-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{7 d e}+\frac{1}{21} \left (\left (7 a^2+2 b^2\right ) e^2\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx\\ &=-\frac{18 a b (e \cos (c+d x))^{5/2}}{35 d e}+\frac{2 \left (7 a^2+2 b^2\right ) e \sqrt{e \cos (c+d x)} \sin (c+d x)}{21 d}-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{7 d e}+\frac{\left (\left (7 a^2+2 b^2\right ) e^2 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{21 \sqrt{e \cos (c+d x)}}\\ &=-\frac{18 a b (e \cos (c+d x))^{5/2}}{35 d e}+\frac{2 \left (7 a^2+2 b^2\right ) e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{e \cos (c+d x)}}+\frac{2 \left (7 a^2+2 b^2\right ) e \sqrt{e \cos (c+d x)} \sin (c+d x)}{21 d}-\frac{2 b (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))}{7 d e}\\ \end{align*}
Mathematica [A] time = 1.14132, size = 115, normalized size = 0.77 \[ \frac{(e \cos (c+d x))^{3/2} \left (20 \left (7 a^2+2 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\sqrt{\cos (c+d x)} \left (5 \left (28 a^2+5 b^2\right ) \sin (c+d x)-3 b (28 a+5 b \sin (3 (c+d x)))-84 a b \cos (2 (c+d x))\right )\right )}{210 d \cos ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Maple [B] time = 1.028, size = 343, normalized size = 2.3 \begin{align*} -{\frac{2\,{e}^{2}}{105\,d} \left ( -240\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-336\,ab \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}+360\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+140\,{a}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+504\,ab \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-140\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+35\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){a}^{2}+10\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){b}^{2}-70\,{a}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-252\,ab \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+10\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+42\,\sin \left ( 1/2\,dx+c/2 \right ) ab \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \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 (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}{\left (b \sin \left (d x + c\right ) + a\right )}^{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 \cos \left (d x + c\right )^{3} - 2 \, a b e \cos \left (d x + c\right ) \sin \left (d x + c\right ) -{\left (a^{2} + b^{2}\right )} e \cos \left (d x + c\right )\right )} \sqrt{e \cos \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 (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}{\left (b \sin \left (d x + c\right ) + a\right )}^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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