Integrand size = 33, antiderivative size = 178 \[ \int (a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\frac {C (a+b \tan (e+f x))^{1+m}}{b f (1+m)}+\frac {(A-i B-C) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (i a+b) f (1+m)}+\frac {(i A-B-i C) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a+i b) f (1+m)} \]
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Time = 0.22 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3711, 3620, 3618, 70} \[ \int (a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\frac {(A-i B-C) (a+b \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a)}+\frac {(i A-B-i C) (a+b \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)}+\frac {C (a+b \tan (e+f x))^{m+1}}{b f (m+1)} \]
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Rule 70
Rule 3618
Rule 3620
Rule 3711
Rubi steps \begin{align*} \text {integral}& = \frac {C (a+b \tan (e+f x))^{1+m}}{b f (1+m)}+\int (a+b \tan (e+f x))^m (A-C+B \tan (e+f x)) \, dx \\ & = \frac {C (a+b \tan (e+f x))^{1+m}}{b f (1+m)}+\frac {1}{2} (A-i B-C) \int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx+\frac {1}{2} (A+i B-C) \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx \\ & = \frac {C (a+b \tan (e+f x))^{1+m}}{b f (1+m)}+\frac {(i A+B-i C) \text {Subst}\left (\int \frac {(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 f}+\frac {(i (-A-i B+C)) \text {Subst}\left (\int \frac {(a+i b x)^m}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{2 f} \\ & = \frac {C (a+b \tan (e+f x))^{1+m}}{b f (1+m)}-\frac {(i A+B-i C) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a-i b) f (1+m)}+\frac {(i A-B-i C) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a+i b) f (1+m)} \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.76 \[ \int (a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\frac {\left (\frac {2 C}{b}-\frac {i (A-i B-C) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \tan (e+f x)}{a-i b}\right )}{a-i b}+\frac {i (A+i B-C) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \tan (e+f x)}{a+i b}\right )}{a+i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 f (1+m)} \]
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\[\int \left (a +b \tan \left (f x +e \right )\right )^{m} \left (A +B \tan \left (f x +e \right )+C \tan \left (f x +e \right )^{2}\right )d x\]
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\[ \int (a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\int { {\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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\[ \int (a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\int \left (a + b \tan {\left (e + f x \right )}\right )^{m} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )\, dx \]
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\[ \int (a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\int { {\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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\[ \int (a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\int { {\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m\,\left (C\,{\mathrm {tan}\left (e+f\,x\right )}^2+B\,\mathrm {tan}\left (e+f\,x\right )+A\right ) \,d x \]
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