Optimal. Leaf size=154 \[ \frac {2 e F_1\left (\frac {1}{2};\frac {1-p}{2},\frac {1-p}{2};\frac {3}{2};\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right ) (e \cos (c+d x))^{-1+p} \sqrt {a+b \sin (c+d x)} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}}}{b d} \]
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Rubi [A]
time = 0.08, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps
used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2783, 143}
\begin {gather*} \frac {2 e \sqrt {a+b \sin (c+d x)} (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 {1}{2};\frac {1-p}{2},\frac {1-p}{2};\frac {3}{2};\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right )}{b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 143
Rule 2783
Rubi steps
\begin {align*} \int \frac {(e \cos (c+d x))^p}{\sqrt {a+b \sin (c+d x)}} \, 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 ) \text {Subst}\left (\int \frac {\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)}}{\sqrt {a+b x}} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {2 e F_1\left (\frac {1}{2};\frac {1-p}{2},\frac {1-p}{2};\frac {3}{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} \sqrt {a+b \sin (c+d x)} \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}}}{b d}\\ \end {align*}
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Mathematica [A]
time = 2.05, size = 185, normalized size = 1.20 \begin {gather*} \frac {2 e F_1\left (\frac {1}{2};\frac {1-p}{2},\frac {1-p}{2};\frac {3}{2};\frac {a+b \sin (c+d x)}{a-\sqrt {b^2}},\frac {a+b \sin (c+d x)}{a+\sqrt {b^2}}\right ) (e \cos (c+d x))^{-1+p} \left (\frac {\sqrt {b^2}-b \sin (c+d x)}{a+\sqrt {b^2}}\right )^{\frac {1-p}{2}} \sqrt {a+b \sin (c+d x)} \left (\frac {\sqrt {b^2}+b \sin (c+d x)}{-a+\sqrt {b^2}}\right )^{\frac {1-p}{2}}}{b d} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{\sqrt {a +b \sin \left (d x +c \right )}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e \cos {\left (c + d x \right )}\right )^{p}}{\sqrt {a + b \sin {\left (c + d x \right )}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^p}{\sqrt {a+b\,\sin \left (c+d\,x\right )}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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