Integrand size = 23, antiderivative size = 140 \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\frac {b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {\left (a^2-b^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {2 a b \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{d^2 f (2+n)} \]
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Time = 0.16 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3624, 3619, 3557, 371} \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\frac {\left (a^2-b^2\right ) (d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2},\frac {n+3}{2},-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac {2 a b (d \tan (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{2},\frac {n+4}{2},-\tan ^2(e+f x)\right )}{d^2 f (n+2)}+\frac {b^2 (d \tan (e+f x))^{n+1}}{d f (n+1)} \]
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Rule 371
Rule 3557
Rule 3619
Rule 3624
Rubi steps \begin{align*} \text {integral}& = \frac {b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\int (d \tan (e+f x))^n \left (a^2-b^2+2 a b \tan (e+f x)\right ) \, dx \\ & = \frac {b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\left (a^2-b^2\right ) \int (d \tan (e+f x))^n \, dx+\frac {(2 a b) \int (d \tan (e+f x))^{1+n} \, dx}{d} \\ & = \frac {b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {(2 a b) \text {Subst}\left (\int \frac {x^{1+n}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}+\frac {\left (\left (a^2-b^2\right ) d\right ) \text {Subst}\left (\int \frac {x^n}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f} \\ & = \frac {b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {\left (a^2-b^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {2 a b \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{d^2 f (2+n)} \\ \end{align*}
Time = 0.51 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.83 \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\frac {\tan (e+f x) (d \tan (e+f x))^n \left (\left (a^2-b^2\right ) (2+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},-\tan ^2(e+f x)\right )+b \left (b (2+n)+2 a (1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)\right )\right )}{f (1+n) (2+n)} \]
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\[\int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +b \tan \left (f x +e \right )\right )^{2}d x\]
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\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
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\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\int \left (d \tan {\left (e + f x \right )}\right )^{n} \left (a + b \tan {\left (e + f x \right )}\right )^{2}\, dx \]
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\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
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\[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]
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Timed out. \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx=\int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \]
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