Integrand size = 27, antiderivative size = 165 \[ \int \frac {(2+3 x)^3 (1+4 x)^m}{1-5 x+3 x^2} \, dx=\frac {123 (1+4 x)^{1+m}}{16 (1+m)}+\frac {9 (1+4 x)^{2+m}}{16 (2+m)}-\frac {3 \left (416-135 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{13 \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {3 \left (416+135 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{13 \left (13+2 \sqrt {13}\right ) (1+m)} \]
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Time = 0.08 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1642, 70} \[ \int \frac {(2+3 x)^3 (1+4 x)^m}{1-5 x+3 x^2} \, dx=-\frac {3 \left (416-135 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{13 \left (13-2 \sqrt {13}\right ) (m+1)}-\frac {3 \left (416+135 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{13 \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {123 (4 x+1)^{m+1}}{16 (m+1)}+\frac {9 (4 x+1)^{m+2}}{16 (m+2)} \]
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Rule 70
Rule 1642
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {123}{4} (1+4 x)^m+\frac {9}{4} (1+4 x)^{1+m}+\frac {\left (192+\frac {810}{\sqrt {13}}\right ) (1+4 x)^m}{-5-\sqrt {13}+6 x}+\frac {\left (192-\frac {810}{\sqrt {13}}\right ) (1+4 x)^m}{-5+\sqrt {13}+6 x}\right ) \, dx \\ & = \frac {123 (1+4 x)^{1+m}}{16 (1+m)}+\frac {9 (1+4 x)^{2+m}}{16 (2+m)}+\frac {1}{13} \left (6 \left (416-135 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx+\frac {1}{13} \left (6 \left (416+135 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx \\ & = \frac {123 (1+4 x)^{1+m}}{16 (1+m)}+\frac {9 (1+4 x)^{2+m}}{16 (2+m)}-\frac {3 \left (416-135 \sqrt {13}\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{13 \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {3 \left (416+135 \sqrt {13}\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{13 \left (13+2 \sqrt {13}\right ) (1+m)} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.71 \[ \int \frac {(2+3 x)^3 (1+4 x)^m}{1-5 x+3 x^2} \, dx=\frac {(1+4 x)^{1+m} \left (117 (85+12 x+4 m (11+3 x))+16 \left (-146+71 \sqrt {13}\right ) (2+m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )-16 \left (146+71 \sqrt {13}\right ) (2+m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )\right )}{624 \left (2+3 m+m^2\right )} \]
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\[\int \frac {\left (2+3 x \right )^{3} \left (1+4 x \right )^{m}}{3 x^{2}-5 x +1}d x\]
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\[ \int \frac {(2+3 x)^3 (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{3}}{3 \, x^{2} - 5 \, x + 1} \,d x } \]
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\[ \int \frac {(2+3 x)^3 (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int \frac {\left (3 x + 2\right )^{3} \left (4 x + 1\right )^{m}}{3 x^{2} - 5 x + 1}\, dx \]
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\[ \int \frac {(2+3 x)^3 (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{3}}{3 \, x^{2} - 5 \, x + 1} \,d x } \]
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\[ \int \frac {(2+3 x)^3 (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{3}}{3 \, x^{2} - 5 \, x + 1} \,d x } \]
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Timed out. \[ \int \frac {(2+3 x)^3 (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int \frac {{\left (3\,x+2\right )}^3\,{\left (4\,x+1\right )}^m}{3\,x^2-5\,x+1} \,d x \]
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