Integrand size = 25, antiderivative size = 179 \[ \int \frac {(2+3 x) (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\frac {(20-21 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {\left (81+2 \left (5+7 \sqrt {13}\right ) m\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{13 \sqrt {13} \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {\left (81+2 \left (5-7 \sqrt {13}\right ) m\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{13 \sqrt {13} \left (13+2 \sqrt {13}\right ) (1+m)} \]
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Time = 0.12 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {836, 844, 70} \[ \int \frac {(2+3 x) (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=-\frac {\left (2 \left (5+7 \sqrt {13}\right ) m+81\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{13 \sqrt {13} \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {\left (2 \left (5-7 \sqrt {13}\right ) m+81\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{13 \sqrt {13} \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {(20-21 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )} \]
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Rule 70
Rule 836
Rule 844
Rubi steps \begin{align*} \text {integral}& = \frac {(20-21 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {1}{507} \int \frac {(1+4 x)^m (13 (81+80 m)-1092 m x)}{1-5 x+3 x^2} \, dx \\ & = \frac {(20-21 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {1}{507} \int \left (\frac {\left (-1092 m+6 \sqrt {13} (81+10 m)\right ) (1+4 x)^m}{-5-\sqrt {13}+6 x}+\frac {\left (-1092 m-6 \sqrt {13} (81+10 m)\right ) (1+4 x)^m}{-5+\sqrt {13}+6 x}\right ) \, dx \\ & = \frac {(20-21 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}+\frac {1}{169} \left (2 \left (182 m+\sqrt {13} (81+10 m)\right )\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx-\frac {1}{507} \left (-1092 m+6 \sqrt {13} (81+10 m)\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx \\ & = \frac {(20-21 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {\left (182 m+\sqrt {13} (81+10 m)\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{169 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {\left (81+2 \left (5-7 \sqrt {13}\right ) m\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 \sqrt {13} \left (13+2 \sqrt {13}\right ) (1+m)} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.83 \[ \int \frac {(2+3 x) (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\frac {1}{507} (1+4 x)^{1+m} \left (\frac {260-273 x}{1-5 x+3 x^2}+\frac {3 \left (182 m+\sqrt {13} (81+10 m)\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )}{\left (-13+2 \sqrt {13}\right ) (1+m)}-\frac {3 \left (182 m-\sqrt {13} (81+10 m)\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (1+m)}\right ) \]
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\[\int \frac {\left (2+3 x \right ) \left (1+4 x \right )^{m}}{\left (3 x^{2}-5 x +1\right )^{2}}d x\]
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\[ \int \frac {(2+3 x) (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
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\[ \int \frac {(2+3 x) (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {\left (3 x + 2\right ) \left (4 x + 1\right )^{m}}{\left (3 x^{2} - 5 x + 1\right )^{2}}\, dx \]
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\[ \int \frac {(2+3 x) (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
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\[ \int \frac {(2+3 x) (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(2+3 x) (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {\left (3\,x+2\right )\,{\left (4\,x+1\right )}^m}{{\left (3\,x^2-5\,x+1\right )}^2} \,d x \]
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