Optimal. Leaf size=121 \[ \frac{\tan (e+f x) \sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \left (\frac{b \tan ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{m+1}{2};\frac{m+2}{2},-p;\frac{m+3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )}{f (m+1)} \]
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Rubi [A] time = 0.141187, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3667, 511, 510} \[ \frac{\tan (e+f x) \sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \left (\frac{b \tan ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{m+1}{2};\frac{m+2}{2},-p;\frac{m+3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )}{f (m+1)} \]
Antiderivative was successfully verified.
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Rule 3667
Rule 511
Rule 510
Rubi steps
\begin{align*} \int (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\frac{\left (\sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \tan ^{-m}(e+f x)\right ) \operatorname{Subst}\left (\int x^m \left (1+x^2\right )^{-1-\frac{m}{2}} \left (a+b x^2\right )^p \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\left (\sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \tan ^{-m}(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int x^m \left (1+x^2\right )^{-1-\frac{m}{2}} \left (1+\frac{b x^2}{a}\right )^p \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{F_1\left (\frac{1+m}{2};\frac{2+m}{2},-p;\frac{3+m}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right ) \sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a}\right )^{-p}}{f (1+m)}\\ \end{align*}
Mathematica [B] time = 2.3441, size = 275, normalized size = 2.27 \[ \frac{a (m+3) \sin (e+f x) \cos (e+f x) (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p F_1\left (\frac{m+1}{2};\frac{m+2}{2},-p;\frac{m+3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )}{f (m+1) \left (\tan ^2(e+f x) \left (2 b p F_1\left (\frac{m+3}{2};\frac{m+2}{2},1-p;\frac{m+5}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )-a (m+2) F_1\left (\frac{m+3}{2};\frac{m+4}{2},-p;\frac{m+5}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )\right )+a (m+3) F_1\left (\frac{m+1}{2};\frac{m+2}{2},-p;\frac{m+3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )\right )} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.658, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sin \left ( fx+e \right ) \right ) ^{m} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, 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{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m}\,{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 ({\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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