Optimal. Leaf size=156 \[ \frac{2 e (a+b \sin (c+d x))^{5/2} (e \cos (c+d x))^{p-1} \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{\frac{1-p}{2}} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{\frac{1-p}{2}} F_1\left (\frac{5}{2};\frac{1-p}{2},\frac{1-p}{2};\frac{7}{2};\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right )}{5 b d} \]
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Rubi [A] time = 0.115663, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2704, 138} \[ \frac{2 e (a+b \sin (c+d x))^{5/2} (e \cos (c+d x))^{p-1} \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{\frac{1-p}{2}} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{\frac{1-p}{2}} F_1\left (\frac{5}{2};\frac{1-p}{2},\frac{1-p}{2};\frac{7}{2};\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right )}{5 b d} \]
Antiderivative was successfully verified.
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Rule 2704
Rule 138
Rubi steps
\begin{align*} \int (e \cos (c+d x))^p (a+b \sin (c+d x))^{3/2} \, dx &=\frac{\left (e (e \cos (c+d x))^{-1+p} \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{\frac{1-p}{2}} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{\frac{1-p}{2}}\right ) \operatorname{Subst}\left (\int (a+b x)^{3/2} \left (-\frac{b}{a-b}-\frac{b x}{a-b}\right )^{\frac{1}{2} (-1+p)} \left (\frac{b}{a+b}-\frac{b x}{a+b}\right )^{\frac{1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{2 e F_1\left (\frac{5}{2};\frac{1-p}{2},\frac{1-p}{2};\frac{7}{2};\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right ) (e \cos (c+d x))^{-1+p} (a+b \sin (c+d x))^{5/2} \left (1-\frac{a+b \sin (c+d x)}{a-b}\right )^{\frac{1-p}{2}} \left (1-\frac{a+b \sin (c+d x)}{a+b}\right )^{\frac{1-p}{2}}}{5 b d}\\ \end{align*}
Mathematica [A] time = 0.820552, size = 187, normalized size = 1.2 \[ \frac{2 e (a+b \sin (c+d x))^{5/2} (e \cos (c+d x))^{p-1} \left (\frac{\sqrt{b^2}-b \sin (c+d x)}{a+\sqrt{b^2}}\right )^{\frac{1-p}{2}} \left (\frac{\sqrt{b^2}+b \sin (c+d x)}{\sqrt{b^2}-a}\right )^{\frac{1-p}{2}} F_1\left (\frac{5}{2};\frac{1-p}{2},\frac{1-p}{2};\frac{7}{2};\frac{a+b \sin (c+d x)}{a-\sqrt{b^2}},\frac{a+b \sin (c+d x)}{a+\sqrt{b^2}}\right )}{5 b d} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.132, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{p} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}\, dx \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 \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \left (e \cos \left (d x + c\right )\right )^{p}\,{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 \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \left (e \cos \left (d x + c\right )\right )^{p}, 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 \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \left (e \cos \left (d x + c\right )\right )^{p}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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