Integrand size = 22, antiderivative size = 158 \[ \int x^m (3-2 a x)^{-2+n} (6+4 a x)^n \, dx=\frac {2^n 9^{-1+n} x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},2-n,\frac {3+m}{2},\frac {4 a^2 x^2}{9}\right )}{1+m}+\frac {2^{2+n} 3^{-3+2 n} a x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {2+m}{2},2-n,\frac {4+m}{2},\frac {4 a^2 x^2}{9}\right )}{2+m}+\frac {2^{2+n} 9^{-2+n} a^2 x^{3+m} \operatorname {Hypergeometric2F1}\left (\frac {3+m}{2},2-n,\frac {5+m}{2},\frac {4 a^2 x^2}{9}\right )}{3+m} \]
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Time = 0.08 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {137, 126, 371} \[ \int x^m (3-2 a x)^{-2+n} (6+4 a x)^n \, dx=\frac {2^n 9^{n-1} x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},2-n,\frac {m+3}{2},\frac {4 a^2 x^2}{9}\right )}{m+1}+\frac {a 2^{n+2} 3^{2 n-3} x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {m+2}{2},2-n,\frac {m+4}{2},\frac {4 a^2 x^2}{9}\right )}{m+2}+\frac {a^2 2^{n+2} 9^{n-2} x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {m+3}{2},2-n,\frac {m+5}{2},\frac {4 a^2 x^2}{9}\right )}{m+3} \]
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Rule 126
Rule 137
Rule 371
Rubi steps \begin{align*} \text {integral}& = \int \left (36 x^m (3-2 a x)^{-2+n} (6+4 a x)^{-2+n}+48 a x^{1+m} (3-2 a x)^{-2+n} (6+4 a x)^{-2+n}+16 a^2 x^{2+m} (3-2 a x)^{-2+n} (6+4 a x)^{-2+n}\right ) \, dx \\ & = 36 \int x^m (3-2 a x)^{-2+n} (6+4 a x)^{-2+n} \, dx+(48 a) \int x^{1+m} (3-2 a x)^{-2+n} (6+4 a x)^{-2+n} \, dx+\left (16 a^2\right ) \int x^{2+m} (3-2 a x)^{-2+n} (6+4 a x)^{-2+n} \, dx \\ & = 36 \int x^m \left (18-8 a^2 x^2\right )^{-2+n} \, dx+(48 a) \int x^{1+m} \left (18-8 a^2 x^2\right )^{-2+n} \, dx+\left (16 a^2\right ) \int x^{2+m} \left (18-8 a^2 x^2\right )^{-2+n} \, dx \\ & = \frac {2^n 9^{-1+n} x^{1+m} \, _2F_1\left (\frac {1+m}{2},2-n;\frac {3+m}{2};\frac {4 a^2 x^2}{9}\right )}{1+m}+\frac {2^{2+n} 3^{-3+2 n} a x^{2+m} \, _2F_1\left (\frac {2+m}{2},2-n;\frac {4+m}{2};\frac {4 a^2 x^2}{9}\right )}{2+m}+\frac {2^{2+n} 9^{-2+n} a^2 x^{3+m} \, _2F_1\left (\frac {3+m}{2},2-n;\frac {5+m}{2};\frac {4 a^2 x^2}{9}\right )}{3+m} \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.09 \[ \int x^m (3-2 a x)^{-2+n} (6+4 a x)^n \, dx=\frac {9^{-2+n} x^{1+m} \left (36-16 a^2 x^2\right )^n \left (18-8 a^2 x^2\right )^{-n} \left (9 \left (6+5 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},2-n,\frac {3+m}{2},\frac {4 a^2 x^2}{9}\right )+4 a (1+m) x \left (3 (3+m) \operatorname {Hypergeometric2F1}\left (\frac {2+m}{2},2-n,\frac {4+m}{2},\frac {4 a^2 x^2}{9}\right )+a (2+m) x \operatorname {Hypergeometric2F1}\left (\frac {3+m}{2},2-n,\frac {5+m}{2},\frac {4 a^2 x^2}{9}\right )\right )\right )}{(1+m) (2+m) (3+m)} \]
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\[\int x^{m} \left (-2 a x +3\right )^{-2+n} \left (4 a x +6\right )^{n}d x\]
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\[ \int x^m (3-2 a x)^{-2+n} (6+4 a x)^n \, dx=\int { {\left (4 \, a x + 6\right )}^{n} {\left (-2 \, a x + 3\right )}^{n - 2} x^{m} \,d x } \]
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\[ \int x^m (3-2 a x)^{-2+n} (6+4 a x)^n \, dx=2^{n} \int x^{m} \left (- 2 a x + 3\right )^{n - 2} \left (2 a x + 3\right )^{n}\, dx \]
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\[ \int x^m (3-2 a x)^{-2+n} (6+4 a x)^n \, dx=\int { {\left (4 \, a x + 6\right )}^{n} {\left (-2 \, a x + 3\right )}^{n - 2} x^{m} \,d x } \]
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\[ \int x^m (3-2 a x)^{-2+n} (6+4 a x)^n \, dx=\int { {\left (4 \, a x + 6\right )}^{n} {\left (-2 \, a x + 3\right )}^{n - 2} x^{m} \,d x } \]
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Timed out. \[ \int x^m (3-2 a x)^{-2+n} (6+4 a x)^n \, dx=\int x^m\,{\left (3-2\,a\,x\right )}^{n-2}\,{\left (4\,a\,x+6\right )}^n \,d x \]
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