Integrand size = 27, antiderivative size = 183 \[ \int \frac {(2+3 x)^4 (1+4 x)^m}{1-5 x+3 x^2} \, dx=\frac {3687 (1+4 x)^{1+m}}{64 (1+m)}+\frac {207 (1+4 x)^{2+m}}{32 (2+m)}+\frac {27 (1+4 x)^{3+m}}{64 (3+m)}-\frac {3 \left (5499-1631 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{26 \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {3 \left (5499+1631 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{26 \left (13+2 \sqrt {13}\right ) (1+m)} \]
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Time = 0.14 (sec) , antiderivative size = 183, 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} \[ \int \frac {(2+3 x)^4 (1+4 x)^m}{1-5 x+3 x^2} \, dx=-\frac {3 \left (5499-1631 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{26 \left (13-2 \sqrt {13}\right ) (m+1)}-\frac {3 \left (5499+1631 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{26 \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {3687 (4 x+1)^{m+1}}{64 (m+1)}+\frac {207 (4 x+1)^{m+2}}{32 (m+2)}+\frac {27 (4 x+1)^{m+3}}{64 (m+3)} \]
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Rule 70
Rule 1642
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3687}{16} (1+4 x)^m+\frac {207}{8} (1+4 x)^{1+m}+\frac {27}{16} (1+4 x)^{2+m}+\frac {\left (1269+\frac {4893}{\sqrt {13}}\right ) (1+4 x)^m}{-5-\sqrt {13}+6 x}+\frac {\left (1269-\frac {4893}{\sqrt {13}}\right ) (1+4 x)^m}{-5+\sqrt {13}+6 x}\right ) \, dx \\ & = \frac {3687 (1+4 x)^{1+m}}{64 (1+m)}+\frac {207 (1+4 x)^{2+m}}{32 (2+m)}+\frac {27 (1+4 x)^{3+m}}{64 (3+m)}+\frac {1}{13} \left (3 \left (5499-1631 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx+\frac {1}{13} \left (3 \left (5499+1631 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx \\ & = \frac {3687 (1+4 x)^{1+m}}{64 (1+m)}+\frac {207 (1+4 x)^{2+m}}{32 (2+m)}+\frac {27 (1+4 x)^{3+m}}{64 (3+m)}-\frac {3 \left (5499-1631 \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 (5499+1631 \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 = 0.29 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.83 \[ \int \frac {(2+3 x)^4 (1+4 x)^m}{1-5 x+3 x^2} \, dx=\frac {3}{832} (1+4 x)^{1+m} \left (\frac {15977}{1+m}+\frac {1794 (1+4 x)}{2+m}+\frac {117 (1+4 x)^2}{3+m}-\frac {32 \left (-5499+1631 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )}{\left (-13+2 \sqrt {13}\right ) (1+m)}-\frac {32 \left (5499+1631 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (1+m)}\right ) \]
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\[\int \frac {\left (2+3 x \right )^{4} \left (1+4 x \right )^{m}}{3 x^{2}-5 x +1}d x\]
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\[ \int \frac {(2+3 x)^4 (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{4}}{3 \, x^{2} - 5 \, x + 1} \,d x } \]
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\[ \int \frac {(2+3 x)^4 (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int \frac {\left (3 x + 2\right )^{4} \left (4 x + 1\right )^{m}}{3 x^{2} - 5 x + 1}\, dx \]
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\[ \int \frac {(2+3 x)^4 (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{4}}{3 \, x^{2} - 5 \, x + 1} \,d x } \]
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\[ \int \frac {(2+3 x)^4 (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{4}}{3 \, x^{2} - 5 \, x + 1} \,d x } \]
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Timed out. \[ \int \frac {(2+3 x)^4 (1+4 x)^m}{1-5 x+3 x^2} \, dx=\int \frac {{\left (3\,x+2\right )}^4\,{\left (4\,x+1\right )}^m}{3\,x^2-5\,x+1} \,d x \]
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