Integrand size = 25, antiderivative size = 129 \[ \int \frac {(2+3 x) (1+4 x)^m}{1-5 x+3 x^2} \, dx=-\frac {3 \left (13-9 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{26 \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {3 \left (13+9 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{26 \left (13+2 \sqrt {13}\right ) (1+m)} \]
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Time = 0.06 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {844, 70} \[ \int \frac {(2+3 x) (1+4 x)^m}{1-5 x+3 x^2} \, dx=-\frac {3 \left (13-9 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{26 \left (13-2 \sqrt {13}\right ) (m+1)}-\frac {3 \left (13+9 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{26 \left (13+2 \sqrt {13}\right ) (m+1)} \]
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Rule 70
Rule 844
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (3+\frac {27}{\sqrt {13}}\right ) (1+4 x)^m}{-5-\sqrt {13}+6 x}+\frac {\left (3-\frac {27}{\sqrt {13}}\right ) (1+4 x)^m}{-5+\sqrt {13}+6 x}\right ) \, dx \\ & = \frac {1}{13} \left (3 \left (13-9 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx+\frac {1}{13} \left (3 \left (13+9 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx \\ & = -\frac {3 \left (13-9 \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 )}{26 \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {3 \left (13+9 \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 )}{26 \left (13+2 \sqrt {13}\right ) (1+m)} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.69 \[ \int \frac {(2+3 x) (1+4 x)^m}{1-5 x+3 x^2} \, dx=\frac {(1+4 x)^{1+m} \left (\left (5+7 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )+\left (5-7 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )\right )}{78 (1+m)} \]
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\[\int \frac {\left (2+3 x \right ) \left (1+4 x \right )^{m}}{3 x^{2}-5 x +1}d x\]
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\[ \int \frac {(2+3 x) (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}}{3 \, x^{2} - 5 \, x + 1} \,d x } \]
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\[ \int \frac {(2+3 x) (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int \frac {\left (3 x + 2\right ) \left (4 x + 1\right )^{m}}{3 x^{2} - 5 x + 1}\, dx \]
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\[ \int \frac {(2+3 x) (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}}{3 \, x^{2} - 5 \, x + 1} \,d x } \]
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\[ \int \frac {(2+3 x) (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}}{3 \, x^{2} - 5 \, x + 1} \,d x } \]
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Timed out. \[ \int \frac {(2+3 x) (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int \frac {\left (3\,x+2\right )\,{\left (4\,x+1\right )}^m}{3\,x^2-5\,x+1} \,d x \]
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