Optimal. Leaf size=138 \[ -\frac{2 \sec ^3(e+f x) \cos ^2(e+f x)^{\frac{p+5}{2}} (g \tan (e+f x))^{p+2} \, _2F_1\left (\frac{p+2}{2},\frac{p+5}{2};\frac{p+4}{2};\sin ^2(e+f x)\right )}{a^2 f g^2 (p+2)}+\frac{2 (g \tan (e+f x))^{p+3}}{a^2 f g^3 (p+3)}+\frac{(g \tan (e+f x))^{p+1}}{a^2 f g (p+1)} \]
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Rubi [A] time = 0.274754, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2711, 2607, 14, 16, 2617, 32} \[ -\frac{2 \sec ^3(e+f x) \cos ^2(e+f x)^{\frac{p+5}{2}} (g \tan (e+f x))^{p+2} \, _2F_1\left (\frac{p+2}{2},\frac{p+5}{2};\frac{p+4}{2};\sin ^2(e+f x)\right )}{a^2 f g^2 (p+2)}+\frac{2 (g \tan (e+f x))^{p+3}}{a^2 f g^3 (p+3)}+\frac{(g \tan (e+f x))^{p+1}}{a^2 f g (p+1)} \]
Antiderivative was successfully verified.
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Rule 2711
Rule 2607
Rule 14
Rule 16
Rule 2617
Rule 32
Rubi steps
\begin{align*} \int \frac{(g \tan (e+f x))^p}{(a+a \sin (e+f x))^2} \, dx &=\frac{\int \left (a^2 \sec ^4(e+f x) (g \tan (e+f x))^p-2 a^2 \sec ^3(e+f x) \tan (e+f x) (g \tan (e+f x))^p+a^2 \sec ^2(e+f x) \tan ^2(e+f x) (g \tan (e+f x))^p\right ) \, dx}{a^4}\\ &=\frac{\int \sec ^4(e+f x) (g \tan (e+f x))^p \, dx}{a^2}+\frac{\int \sec ^2(e+f x) \tan ^2(e+f x) (g \tan (e+f x))^p \, dx}{a^2}-\frac{2 \int \sec ^3(e+f x) \tan (e+f x) (g \tan (e+f x))^p \, dx}{a^2}\\ &=\frac{\operatorname{Subst}\left (\int (g x)^p \left (1+x^2\right ) \, dx,x,\tan (e+f x)\right )}{a^2 f}+\frac{\int \sec ^2(e+f x) (g \tan (e+f x))^{2+p} \, dx}{a^2 g^2}-\frac{2 \int \sec ^3(e+f x) (g \tan (e+f x))^{1+p} \, dx}{a^2 g}\\ &=-\frac{2 \cos ^2(e+f x)^{\frac{5+p}{2}} \, _2F_1\left (\frac{2+p}{2},\frac{5+p}{2};\frac{4+p}{2};\sin ^2(e+f x)\right ) \sec ^3(e+f x) (g \tan (e+f x))^{2+p}}{a^2 f g^2 (2+p)}+\frac{\operatorname{Subst}\left (\int \left ((g x)^p+\frac{(g x)^{2+p}}{g^2}\right ) \, dx,x,\tan (e+f x)\right )}{a^2 f}+\frac{\operatorname{Subst}\left (\int (g x)^{2+p} \, dx,x,\tan (e+f x)\right )}{a^2 f g^2}\\ &=\frac{(g \tan (e+f x))^{1+p}}{a^2 f g (1+p)}-\frac{2 \cos ^2(e+f x)^{\frac{5+p}{2}} \, _2F_1\left (\frac{2+p}{2},\frac{5+p}{2};\frac{4+p}{2};\sin ^2(e+f x)\right ) \sec ^3(e+f x) (g \tan (e+f x))^{2+p}}{a^2 f g^2 (2+p)}+\frac{2 (g \tan (e+f x))^{3+p}}{a^2 f g^3 (3+p)}\\ \end{align*}
Mathematica [B] time = 14.1776, size = 667, normalized size = 4.83 \[ \frac{2^{p+1} \tan \left (\frac{1}{2} (e+f x)\right ) \left (1-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )^p \left (-\frac{\tan \left (\frac{1}{2} (e+f x)\right )}{\tan ^2\left (\frac{1}{2} (e+f x)\right )-1}\right )^p \tan ^{-p}(e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 (g \tan (e+f x))^p \left (\frac{2 \tan ^4\left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (p+4,\frac{p+5}{2};\frac{p+7}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{p+5}+\frac{2 \tan ^3\left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (p+3,\frac{p+4}{2};\frac{p+6}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{p+4}-\frac{8 \tan ^3\left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (\frac{p+4}{2},p+4;\frac{p+6}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{p+4}+\frac{\tan ^2\left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (p+2,\frac{p+3}{2};\frac{p+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{p+3}-\frac{6 \tan ^2\left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (\frac{p+3}{2},p+3;\frac{p+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{p+3}+\frac{12 \tan ^2\left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (\frac{p+3}{2},p+4;\frac{p+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{p+3}-\frac{2 \tan \left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (\frac{p+2}{2},p+2;\frac{p+4}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{p+2}+\frac{6 \tan \left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (\frac{p+2}{2},p+3;\frac{p+4}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{p+2}-\frac{8 \tan \left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (\frac{p+2}{2},p+4;\frac{p+4}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{p+2}+\frac{\, _2F_1\left (\frac{p+1}{2},p+2;\frac{p+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{p+1}-\frac{2 \, _2F_1\left (\frac{p+1}{2},p+3;\frac{p+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{p+1}+\frac{2 \, _2F_1\left (\frac{p+1}{2},p+4;\frac{p+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{p+1}\right )}{f (a \sin (e+f x)+a)^2} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.41, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( g\tan \left ( fx+e \right ) \right ) ^{p}}{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{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 \frac{\left (g \tan \left (f x + e\right )\right )^{p}}{{\left (a \sin \left (f x + e\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 (-\frac{\left (g \tan \left (f x + e\right )\right )^{p}}{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}, 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*} \frac{\int \frac{\left (g \tan{\left (e + f x \right )}\right )^{p}}{\sin ^{2}{\left (e + f x \right )} + 2 \sin{\left (e + f x \right )} + 1}\, dx}{a^{2}} \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 (g \tan \left (f x + e\right )\right )^{p}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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