Integrand size = 27, antiderivative size = 376 \[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\frac {(268-195 x) (1+4 x)^{1+m}}{11271 \left (1-5 x+3 x^2\right )}+\frac {162 (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{24565 (1+m)}+\frac {9 \left (117+64 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{63869 \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {\left (423+2 \left (211+65 \sqrt {13}\right ) m\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{3757 \sqrt {13} \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {9 \left (117-64 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{63869 \left (13+2 \sqrt {13}\right ) (1+m)}+\frac {\left (423+\left (422-130 \sqrt {13}\right ) m\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{3757 \sqrt {13} \left (13+2 \sqrt {13}\right ) (1+m)}+\frac {36 (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{7225 (1+m)} \]
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Time = 0.27 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {974, 70, 836, 844} \[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\frac {162 (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {3}{5} (4 x+1)\right )}{24565 (m+1)}-\frac {\left (2 \left (211+65 \sqrt {13}\right ) m+423\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{3757 \sqrt {13} \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {9 \left (117+64 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{63869 \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {\left (\left (422-130 \sqrt {13}\right ) m+423\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{3757 \sqrt {13} \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {9 \left (117-64 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{63869 \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {36 (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,-\frac {3}{5} (4 x+1)\right )}{7225 (m+1)}+\frac {(268-195 x) (4 x+1)^{m+1}}{11271 \left (3 x^2-5 x+1\right )} \]
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Rule 70
Rule 836
Rule 844
Rule 974
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {9 (1+4 x)^m}{289 (2+3 x)^2}+\frac {162 (1+4 x)^m}{4913 (2+3 x)}+\frac {(46-27 x) (1+4 x)^m}{289 \left (1-5 x+3 x^2\right )^2}-\frac {3 (1+4 x)^m (-109+54 x)}{4913 \left (1-5 x+3 x^2\right )}\right ) \, dx \\ & = -\frac {3 \int \frac {(1+4 x)^m (-109+54 x)}{1-5 x+3 x^2} \, dx}{4913}+\frac {1}{289} \int \frac {(46-27 x) (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx+\frac {9}{289} \int \frac {(1+4 x)^m}{(2+3 x)^2} \, dx+\frac {162 \int \frac {(1+4 x)^m}{2+3 x} \, dx}{4913} \\ & = \frac {(268-195 x) (1+4 x)^{1+m}}{11271 \left (1-5 x+3 x^2\right )}+\frac {162 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{24565 (1+m)}+\frac {36 (1+4 x)^{1+m} \, _2F_1\left (2,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{7225 (1+m)}-\frac {\int \frac {(1+4 x)^m (13 (423+1072 m)-10140 m x)}{1-5 x+3 x^2} \, dx}{146523}-\frac {3 \int \left (\frac {\left (54-\frac {384}{\sqrt {13}}\right ) (1+4 x)^m}{-5-\sqrt {13}+6 x}+\frac {\left (54+\frac {384}{\sqrt {13}}\right ) (1+4 x)^m}{-5+\sqrt {13}+6 x}\right ) \, dx}{4913} \\ & = \frac {(268-195 x) (1+4 x)^{1+m}}{11271 \left (1-5 x+3 x^2\right )}+\frac {162 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{24565 (1+m)}+\frac {36 (1+4 x)^{1+m} \, _2F_1\left (2,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{7225 (1+m)}-\frac {\int \left (\frac {\left (-10140 m+6 \sqrt {13} (423+422 m)\right ) (1+4 x)^m}{-5-\sqrt {13}+6 x}+\frac {\left (-10140 m-6 \sqrt {13} (423+422 m)\right ) (1+4 x)^m}{-5+\sqrt {13}+6 x}\right ) \, dx}{146523}-\frac {\left (18 \left (117-64 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx}{63869}-\frac {\left (18 \left (117+64 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx}{63869} \\ & = \frac {(268-195 x) (1+4 x)^{1+m}}{11271 \left (1-5 x+3 x^2\right )}+\frac {162 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{24565 (1+m)}+\frac {9 \left (117+64 \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 )}{63869 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {9 \left (117-64 \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 )}{63869 \left (13+2 \sqrt {13}\right ) (1+m)}+\frac {36 (1+4 x)^{1+m} \, _2F_1\left (2,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{7225 (1+m)}-\frac {\left (2 \left (423+\left (422-130 \sqrt {13}\right ) m\right )\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx}{3757 \sqrt {13}}+\frac {\left (2 \left (1690 m+\sqrt {13} (423+422 m)\right )\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx}{48841} \\ & = \frac {(268-195 x) (1+4 x)^{1+m}}{11271 \left (1-5 x+3 x^2\right )}+\frac {162 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{24565 (1+m)}+\frac {9 \left (117+64 \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 )}{63869 \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {\left (1690 m+\sqrt {13} (423+422 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 )}{48841 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {9 \left (117-64 \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 )}{63869 \left (13+2 \sqrt {13}\right ) (1+m)}+\frac {\left (423+\left (422-130 \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 )}{3757 \sqrt {13} \left (13+2 \sqrt {13}\right ) (1+m)}+\frac {36 (1+4 x)^{1+m} \, _2F_1\left (2,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{7225 (1+m)} \\ \end{align*}
Time = 0.52 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.76 \[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\frac {(1+4 x)^{1+m} \left (\frac {16575 (268-195 x)}{1-5 x+3 x^2}+\frac {1232010 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{1+m}+\frac {26325 \left (117+64 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )}{\left (13-2 \sqrt {13}\right ) (1+m)}+\frac {26325 \left (117-64 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (1+m)}-\frac {425 \left (\left (423 \left (2+\sqrt {13}\right )+\left (2534+682 \sqrt {13}\right ) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )+\left (-423 \left (-2+\sqrt {13}\right )+\left (2534-682 \sqrt {13}\right ) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )\right )}{1+m}+\frac {930852 \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{1+m}\right )}{186816825} \]
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\[\int \frac {\left (1+4 x \right )^{m}}{\left (2+3 x \right )^{2} \left (3 x^{2}-5 x +1\right )^{2}}d x\]
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\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2} {\left (3 \, x + 2\right )}^{2}} \,d x } \]
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\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {\left (4 x + 1\right )^{m}}{\left (3 x + 2\right )^{2} \left (3 x^{2} - 5 x + 1\right )^{2}}\, dx \]
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\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2} {\left (3 \, x + 2\right )}^{2}} \,d x } \]
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\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2} {\left (3 \, x + 2\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {{\left (4\,x+1\right )}^m}{{\left (3\,x+2\right )}^2\,{\left (3\,x^2-5\,x+1\right )}^2} \,d x \]
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