Optimal. Leaf size=105 \[ \frac {(d \tan (e+f x))^{m+1} \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} F_1\left (\frac {m+1}{2};1,-p;\frac {m+3}{2};-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right )}{d f (m+1)} \]
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Rubi [A] time = 0.20, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4141, 1975, 511, 510} \[ \frac {(d \tan (e+f x))^{m+1} \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} F_1\left (\frac {m+1}{2};1,-p;\frac {m+3}{2};-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right )}{d f (m+1)} \]
Antiderivative was successfully verified.
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Rule 510
Rule 511
Rule 1975
Rule 4141
Rubi steps
\begin {align*} \int \left (a+b \sec ^2(e+f x)\right )^p (d \tan (e+f x))^m \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(d x)^m \left (a+b \left (1+x^2\right )\right )^p}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(d x)^m \left (a+b+b x^2\right )^p}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\left (\left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}\right ) \operatorname {Subst}\left (\int \frac {(d x)^m \left (1+\frac {b x^2}{a+b}\right )^p}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {F_1\left (\frac {1+m}{2};1,-p;\frac {3+m}{2};-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right ) (d \tan (e+f x))^{1+m} \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}}{d f (1+m)}\\ \end {align*}
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Mathematica [B] time = 3.49, size = 259, normalized size = 2.47 \[ \frac {\sin (e+f x) \cos (e+f x) (d \tan (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p F_1\left (\frac {m+1}{2};-p,1;\frac {m+3}{2};-\frac {b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right )}{f (m+1) \left (\frac {2 \tan ^2(e+f x) \left (b p F_1\left (\frac {m+3}{2};1-p,1;\frac {m+5}{2};-\frac {b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right )-(a+b) F_1\left (\frac {m+3}{2};-p,2;\frac {m+5}{2};-\frac {b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right )\right )}{(m+3) (a+b)}+F_1\left (\frac {m+1}{2};-p,1;\frac {m+3}{2};-\frac {b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right )\right )} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \tan \left (f x + e\right )\right )^{m}, x\right ) \]
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 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \tan \left (f x + e\right )\right )^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 2.89, size = 0, normalized size = 0.00 \[ \int \left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )^{p} \left (d \tan \left (f x +e \right )\right )^{m}\, dx \]
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 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \tan \left (f x + e\right )\right )^{m}\,{d x} \]
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 \[ \int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^m\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \]
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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