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 = 0.14, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 510
Rule 511
Rule 3667
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*}
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Mathematica [B] time = 2.43, 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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fricas [F] time = 1.38, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m}, x\right ) \]
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 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 2.14, size = 0, normalized size = 0.00 \[ \int \left (d \sin \left (f x +e \right )\right )^{m} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]
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 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m}\,{d x} \]
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 \[ \int {\left (d\,\sin \left (e+f\,x\right )\right )}^m\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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