Optimal. Leaf size=232 \[ \frac{\cos (c+d x) (e \sin (c+d x))^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )}{a d e (m+1) \sqrt{\cos ^2(c+d x)}}-\frac{b e (e \sin (c+d x))^{m-1} \left (-\frac{a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac{1-m}{2}} \left (\frac{a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac{1-m}{2}} F_1\left (1-m;\frac{1-m}{2},\frac{1-m}{2};2-m;-\frac{a-b}{b+a \cos (c+d x)},\frac{a+b}{b+a \cos (c+d x)}\right )}{a^2 d (1-m)} \]
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Rubi [A] time = 0.258745, antiderivative size = 232, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3872, 2867, 2643, 2703} \[ \frac{\cos (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{a d e (m+1) \sqrt{\cos ^2(c+d x)}}-\frac{b e (e \sin (c+d x))^{m-1} \left (-\frac{a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac{1-m}{2}} \left (\frac{a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac{1-m}{2}} F_1\left (1-m;\frac{1-m}{2},\frac{1-m}{2};2-m;-\frac{a-b}{b+a \cos (c+d x)},\frac{a+b}{b+a \cos (c+d x)}\right )}{a^2 d (1-m)} \]
Antiderivative was successfully verified.
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Rule 3872
Rule 2867
Rule 2643
Rule 2703
Rubi steps
\begin{align*} \int \frac{(e \sin (c+d x))^m}{a+b \sec (c+d x)} \, dx &=-\int \frac{\cos (c+d x) (e \sin (c+d x))^m}{-b-a \cos (c+d x)} \, dx\\ &=\frac{\int (e \sin (c+d x))^m \, dx}{a}+\frac{b \int \frac{(e \sin (c+d x))^m}{-b-a \cos (c+d x)} \, dx}{a}\\ &=-\frac{b e F_1\left (1-m;\frac{1-m}{2},\frac{1-m}{2};2-m;-\frac{a-b}{b+a \cos (c+d x)},\frac{a+b}{b+a \cos (c+d x)}\right ) \left (-\frac{a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac{1-m}{2}} \left (\frac{a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac{1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^2 d (1-m)}+\frac{\cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{a d e (1+m) \sqrt{\cos ^2(c+d x)}}\\ \end{align*}
Mathematica [B] time = 5.62394, size = 687, normalized size = 2.96 \[ \frac{2 \tan \left (\frac{1}{2} (c+d x)\right ) (e \sin (c+d x))^m \left ((a+b) \text{Hypergeometric2F1}\left (\frac{m+1}{2},m+1,\frac{m+3}{2},-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-b F_1\left (\frac{m+1}{2};m,1;\frac{m+3}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),\frac{(a-b) \tan ^2\left (\frac{1}{2} (c+d x)\right )}{a+b}\right )\right )}{d (a+b \sec (c+d x)) \left (2 m \tan ^2\left (\frac{1}{2} (c+d x)\right ) \left ((a+b) \text{Hypergeometric2F1}\left (\frac{m+1}{2},m+1,\frac{m+3}{2},-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-b F_1\left (\frac{m+1}{2};m,1;\frac{m+3}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),\frac{(a-b) \tan ^2\left (\frac{1}{2} (c+d x)\right )}{a+b}\right )\right )+2 m \tan \left (\frac{1}{2} (c+d x)\right ) \cot (c+d x) \left ((a+b) \text{Hypergeometric2F1}\left (\frac{m+1}{2},m+1,\frac{m+3}{2},-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-b F_1\left (\frac{m+1}{2};m,1;\frac{m+3}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),\frac{(a-b) \tan ^2\left (\frac{1}{2} (c+d x)\right )}{a+b}\right )\right )+\sec ^2\left (\frac{1}{2} (c+d x)\right ) \left ((a+b) \text{Hypergeometric2F1}\left (\frac{m+1}{2},m+1,\frac{m+3}{2},-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-b F_1\left (\frac{m+1}{2};m,1;\frac{m+3}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),\frac{(a-b) \tan ^2\left (\frac{1}{2} (c+d x)\right )}{a+b}\right )\right )+\frac{(m+1) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (\frac{2 b \tan ^2\left (\frac{1}{2} (c+d x)\right ) \left ((b-a) F_1\left (\frac{m+3}{2};m,2;\frac{m+5}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),\frac{(a-b) \tan ^2\left (\frac{1}{2} (c+d x)\right )}{a+b}\right )+m (a+b) F_1\left (\frac{m+3}{2};m+1,1;\frac{m+5}{2};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),\frac{(a-b) \tan ^2\left (\frac{1}{2} (c+d x)\right )}{a+b}\right )\right )}{m+3}-(a+b)^2 \left (\text{Hypergeometric2F1}\left (\frac{m+1}{2},m+1,\frac{m+3}{2},-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-\sec ^2\left (\frac{1}{2} (c+d x)\right )^{-m-1}\right )\right )}{a+b}\right )} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.575, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( e\sin \left ( dx+c \right ) \right ) ^{m}}{a+b\sec \left ( dx+c \right ) }}\, 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 \frac{\left (e \sin \left (d x + c\right )\right )^{m}}{b \sec \left (d x + c\right ) + a}\,{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 (\frac{\left (e \sin \left (d x + c\right )\right )^{m}}{b \sec \left (d x + c\right ) + a}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sin{\left (c + d x \right )}\right )^{m}}{a + b \sec{\left (c + d x \right )}}\, dx \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 \frac{\left (e \sin \left (d x + c\right )\right )^{m}}{b \sec \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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