Integrand size = 27, antiderivative size = 202 \[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\frac {9 (1+4 x)^{1+m}}{4 (1+m)}+\frac {(844-2355 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {\left (13689-\sqrt {13} \left (297+4474 m-1570 \sqrt {13} m\right )\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{169 \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {\left (13689+\sqrt {13} \left (297+4474 m+1570 \sqrt {13} m\right )\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{169 \left (13+2 \sqrt {13}\right ) (1+m)} \]
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Time = 0.16 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1662, 1642, 70} \[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=-\frac {\left (13689-\sqrt {13} \left (-1570 \sqrt {13} m+4474 m+297\right )\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{169 \left (13-2 \sqrt {13}\right ) (m+1)}-\frac {\left (\sqrt {13} \left (1570 \sqrt {13} m+4474 m+297\right )+13689\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{169 \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {(844-2355 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )}+\frac {9 (4 x+1)^{m+1}}{4 (m+1)} \]
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Rule 70
Rule 1642
Rule 1662
Rubi steps \begin{align*} \text {integral}& = \frac {(844-2355 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {1}{507} \int \frac {(1+4 x)^m \left (13 (4617+3376 m)-39 (1521+3140 m) x-13689 x^2\right )}{1-5 x+3 x^2} \, dx \\ & = \frac {(844-2355 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {1}{507} \int \left (-4563 (1+4 x)^m+\frac {\left (-82134-122460 m-6 \sqrt {13} (297+4474 m)\right ) (1+4 x)^m}{-5-\sqrt {13}+6 x}+\frac {\left (-82134-122460 m+6 \sqrt {13} (297+4474 m)\right ) (1+4 x)^m}{-5+\sqrt {13}+6 x}\right ) \, dx \\ & = \frac {9 (1+4 x)^{1+m}}{4 (1+m)}+\frac {(844-2355 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {1}{507} \left (-82134-122460 m+6 \sqrt {13} (297+4474 m)\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx+\frac {1}{507} \left (82134+122460 m+6 \sqrt {13} (297+4474 m)\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx \\ & = \frac {9 (1+4 x)^{1+m}}{4 (1+m)}+\frac {(844-2355 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {\left (13689-\sqrt {13} \left (297+4474 m-1570 \sqrt {13} m\right )\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 (13689+\sqrt {13} \left (297+4474 m+1570 \sqrt {13} m\right )\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)} \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.24 \[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\frac {(1+4 x)^{1+m} \left (\frac {13689}{4+4 m}+\frac {39 (844-2355 x)}{1-5 x+3 x^2}-\frac {1053 \left (-117+128 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )}{\left (-13+2 \sqrt {13}\right ) (1+m)}-\frac {1053 \left (117+128 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (1+m)}-\frac {-\left (\left (-14679 \left (2+\sqrt {13}\right )+2 \left (-5731+667 \sqrt {13}\right ) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )\right )+\left (-14679 \left (-2+\sqrt {13}\right )+2 \left (5731+667 \sqrt {13}\right ) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{1+m}\right )}{1521} \]
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\[\int \frac {\left (2+3 x \right )^{4} \left (1+4 x \right )^{m}}{\left (3 x^{2}-5 x +1\right )^{2}}d x\]
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\[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{4}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
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\[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {\left (3 x + 2\right )^{4} \left (4 x + 1\right )^{m}}{\left (3 x^{2} - 5 x + 1\right )^{2}}\, dx \]
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\[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{4}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
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\[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{4}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {{\left (3\,x+2\right )}^4\,{\left (4\,x+1\right )}^m}{{\left (3\,x^2-5\,x+1\right )}^2} \,d x \]
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