Optimal. Leaf size=138 \[ \frac {(g \tan (e+f x))^{1+p}}{a^2 f g (1+p)}-\frac {2 \cos ^2(e+f x)^{\frac {5+p}{2}} \, _2F_1\left (\frac {2+p}{2},\frac {5+p}{2};\frac {4+p}{2};\sin ^2(e+f x)\right ) \sec ^3(e+f x) (g \tan (e+f x))^{2+p}}{a^2 f g^2 (2+p)}+\frac {2 (g \tan (e+f x))^{3+p}}{a^2 f g^3 (3+p)} \]
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Rubi [A]
time = 0.19, antiderivative size = 138, normalized size of antiderivative = 1.00, 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 = {2790, 2687,
14, 16, 2697, 32} \begin {gather*} \frac {2 (g \tan (e+f x))^{p+3}}{a^2 f g^3 (p+3)}-\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 {(g \tan (e+f x))^{p+1}}{a^2 f g (p+1)} \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 16
Rule 32
Rule 2687
Rule 2697
Rule 2790
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 {\text {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 {\text {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 {\text {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*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in
optimal.
time = 3.04, size = 710, normalized size = 5.14 \begin {gather*} \frac {(2+p) \left (F_1\left (1+p;p,2+p;2+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right )-2 F_1\left (1+p;p,3+p;2+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right )+2 F_1\left (1+p;p,4+p;2+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x) (g \tan (e+f x))^p}{a^2 f (1+p) (1+\sin (e+f x))^2 \left ((2+p) F_1\left (1+p;p,2+p;2+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right )-2 (2+p) F_1\left (1+p;p,3+p;2+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right )+4 F_1\left (1+p;p,4+p;2+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right )+2 p F_1\left (1+p;p,4+p;2+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right )-2 F_1\left (2+p;p,3+p;3+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )-p F_1\left (2+p;p,3+p;3+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )+6 F_1\left (2+p;p,4+p;3+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )+2 p F_1\left (2+p;p,4+p;3+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )-8 F_1\left (2+p;p,5+p;3+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )-2 p F_1\left (2+p;p,5+p;3+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )+p F_1\left (2+p;1+p,2+p;3+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )-2 p F_1\left (2+p;1+p,3+p;3+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )+2 p F_1\left (2+p;1+p,4+p;3+p;\tan \left (\frac {1}{2} (e+f x)\right ),-\tan \left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 1.35, size = 0, normalized size = 0.00 \[\int \frac {\left (g \tan \left (f x +e \right )\right )^{p}}{\left (a +a \sin \left (f x +e \right )\right )^{2}}\, dx\]
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 \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*} \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 {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 (g\,\mathrm {tan}\left (e+f\,x\right )\right )}^p}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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