Integrand size = 27, antiderivative size = 164 \[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )} \, dx=\frac {3 (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{85 (1+m)}+\frac {3 \left (13+9 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{442 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {3 \left (13-9 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{442 \left (13+2 \sqrt {13}\right ) (1+m)} \]
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Time = 0.10 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {974, 70, 844} \[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )} \, dx=\frac {3 (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {3}{5} (4 x+1)\right )}{85 (m+1)}+\frac {3 \left (13+9 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{442 \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {3 \left (13-9 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{442 \left (13+2 \sqrt {13}\right ) (m+1)} \]
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Rule 70
Rule 844
Rule 974
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3 (1+4 x)^m}{17 (2+3 x)}+\frac {(7-3 x) (1+4 x)^m}{17 \left (1-5 x+3 x^2\right )}\right ) \, dx \\ & = \frac {1}{17} \int \frac {(7-3 x) (1+4 x)^m}{1-5 x+3 x^2} \, dx+\frac {3}{17} \int \frac {(1+4 x)^m}{2+3 x} \, dx \\ & = \frac {3 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{85 (1+m)}+\frac {1}{17} \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 {3 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{85 (1+m)}-\frac {1}{221} \left (3 \left (13-9 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx-\frac {1}{221} \left (3 \left (13+9 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx \\ & = \frac {3 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{85 (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 )}{442 \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 )}{442 \left (13+2 \sqrt {13}\right ) (1+m)} \\ \end{align*}
Time = 0.36 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.67 \[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )} \, dx=\frac {(1+4 x)^{1+m} \left (234 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )+5 \left (31+11 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )+5 \left (31-11 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )\right )}{6630 (1+m)} \]
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\[\int \frac {\left (1+4 x \right )^{m}}{\left (2+3 x \right ) \left (3 x^{2}-5 x +1\right )}d x\]
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\[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )} {\left (3 \, x + 2\right )}} \,d x } \]
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\[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )} \, dx=\int \frac {\left (4 x + 1\right )^{m}}{\left (3 x + 2\right ) \left (3 x^{2} - 5 x + 1\right )}\, dx \]
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\[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )} {\left (3 \, x + 2\right )}} \,d x } \]
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\[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )} {\left (3 \, x + 2\right )}} \,d x } \]
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Timed out. \[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )} \, dx=\int \frac {{\left (4\,x+1\right )}^m}{\left (3\,x+2\right )\,\left (3\,x^2-5\,x+1\right )} \,d x \]
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