Optimal. Leaf size=102 \[ -\frac {2^{\frac {p}{2}-1} (\sin (c+d x)+1)^{1-\frac {p}{2}} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac {4-p}{2},\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d e (p+1) (a \sin (c+d x)+a)^{3/2}} \]
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Rubi [A] time = 0.12, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2689, 70, 69} \[ -\frac {2^{\frac {p}{2}-1} (\sin (c+d x)+1)^{1-\frac {p}{2}} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac {4-p}{2},\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d e (p+1) (a \sin (c+d x)+a)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 69
Rule 70
Rule 2689
Rubi steps
\begin {align*} \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac {\left (a^2 (e \cos (c+d x))^{1+p} (a-a \sin (c+d x))^{\frac {1}{2} (-1-p)} (a+a \sin (c+d x))^{\frac {1}{2} (-1-p)}\right ) \operatorname {Subst}\left (\int (a-a x)^{\frac {1}{2} (-1+p)} (a+a x)^{-\frac {3}{2}+\frac {1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac {\left (2^{-2+\frac {p}{2}} a (e \cos (c+d x))^{1+p} (a-a \sin (c+d x))^{\frac {1}{2} (-1-p)} (a+a \sin (c+d x))^{-1+\frac {1}{2} (-1-p)+\frac {p}{2}} \left (\frac {a+a \sin (c+d x)}{a}\right )^{1-\frac {p}{2}}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {3}{2}+\frac {1}{2} (-1+p)} (a-a x)^{\frac {1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=-\frac {2^{-1+\frac {p}{2}} (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac {4-p}{2},\frac {1+p}{2};\frac {3+p}{2};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{1-\frac {p}{2}}}{d e (1+p) (a+a \sin (c+d x))^{3/2}}\\ \end {align*}
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Mathematica [A] time = 0.17, size = 101, normalized size = 0.99 \[ -\frac {2^{\frac {p}{2}-1} \cos (c+d x) (\sin (c+d x)+1)^{1-\frac {p}{2}} (e \cos (c+d x))^p \, _2F_1\left (2-\frac {p}{2},\frac {p+1}{2};\frac {p+3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d (p+1) (a (\sin (c+d x)+1))^{3/2}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {a \sin \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{p}}{a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}}, x\right ) \]
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 \[ \int \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.17, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{\left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}\, dx \]
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 \[ \int \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]
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 \[ \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^p}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]
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 \[ \int \frac {\left (e \cos {\left (c + d x \right )}\right )^{p}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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