Integrand size = 25, antiderivative size = 121 \[ \int (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1+m}{2},\frac {2+m}{2},-p,\frac {3+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right ) \sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a}\right )^{-p}}{f (1+m)} \]
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Time = 0.18 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3748, 525, 524} \[ \int (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\frac {\tan (e+f x) \sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \left (\frac {b \tan ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {m+1}{2},\frac {m+2}{2},-p,\frac {m+3}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )}{f (m+1)} \]
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Rule 524
Rule 525
Rule 3748
Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \tan ^{-m}(e+f x)\right ) \text {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 ) \text {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 {\operatorname {AppellF1}\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*}
Leaf count is larger than twice the leaf count of optimal. \(275\) vs. \(2(121)=242\).
Time = 4.00 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.27 \[ \int (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\frac {a (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {2+m}{2},-p,\frac {3+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right ) \cos (e+f x) \sin (e+f x) (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p}{f (1+m) \left (a (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {2+m}{2},-p,\frac {3+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )+\left (2 b p \operatorname {AppellF1}\left (\frac {3+m}{2},\frac {2+m}{2},1-p,\frac {5+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )-a (2+m) \operatorname {AppellF1}\left (\frac {3+m}{2},\frac {4+m}{2},-p,\frac {5+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )\right ) \tan ^2(e+f x)\right )} \]
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\[\int \left (d \sin \left (f x +e \right )\right )^{m} \left (a +b \tan \left (f x +e \right )^{2}\right )^{p}d x\]
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\[ \int (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]
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Timed out. \[ \int (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
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\[ \int (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]
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\[ \int (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m} \,d x } \]
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Timed out. \[ \int (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^m\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p \,d x \]
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