Integrand size = 27, antiderivative size = 340 \[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )^2} \, dx=\frac {(43-33 x) (1+4 x)^{1+m}}{663 \left (1-5 x+3 x^2\right )}+\frac {9 (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{1445 (1+m)}+\frac {9 \left (13+9 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{7514 \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {\left (81+\left (62+22 \sqrt {13}\right ) m\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{221 \sqrt {13} \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {9 \left (13-9 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{7514 \left (13+2 \sqrt {13}\right ) (1+m)}+\frac {\left (81+\left (62-22 \sqrt {13}\right ) m\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{221 \sqrt {13} \left (13+2 \sqrt {13}\right ) (1+m)} \]
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Time = 0.26 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {974, 70, 836, 844} \[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )^2} \, dx=\frac {9 (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {3}{5} (4 x+1)\right )}{1445 (m+1)}-\frac {\left (\left (62+22 \sqrt {13}\right ) m+81\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{221 \sqrt {13} \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {9 \left (13+9 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{7514 \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {\left (\left (62-22 \sqrt {13}\right ) m+81\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{221 \sqrt {13} \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {9 \left (13-9 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{7514 \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {(43-33 x) (4 x+1)^{m+1}}{663 \left (3 x^2-5 x+1\right )} \]
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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)}+\frac {(7-3 x) (1+4 x)^m}{17 \left (1-5 x+3 x^2\right )^2}-\frac {3 (-7+3 x) (1+4 x)^m}{289 \left (1-5 x+3 x^2\right )}\right ) \, dx \\ & = -\left (\frac {3}{289} \int \frac {(-7+3 x) (1+4 x)^m}{1-5 x+3 x^2} \, dx\right )+\frac {9}{289} \int \frac {(1+4 x)^m}{2+3 x} \, dx+\frac {1}{17} \int \frac {(7-3 x) (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx \\ & = \frac {(43-33 x) (1+4 x)^{1+m}}{663 \left (1-5 x+3 x^2\right )}+\frac {9 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{1445 (1+m)}-\frac {\int \frac {(1+4 x)^m (13 (81+172 m)-1716 m x)}{1-5 x+3 x^2} \, dx}{8619}-\frac {3}{289} \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 {(43-33 x) (1+4 x)^{1+m}}{663 \left (1-5 x+3 x^2\right )}+\frac {9 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{1445 (1+m)}-\frac {\int \left (\frac {\left (-1716 m+6 \sqrt {13} (81+62 m)\right ) (1+4 x)^m}{-5-\sqrt {13}+6 x}+\frac {\left (-1716 m-6 \sqrt {13} (81+62 m)\right ) (1+4 x)^m}{-5+\sqrt {13}+6 x}\right ) \, dx}{8619}-\frac {\left (9 \left (13-9 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx}{3757}-\frac {\left (9 \left (13+9 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx}{3757} \\ & = \frac {(43-33 x) (1+4 x)^{1+m}}{663 \left (1-5 x+3 x^2\right )}+\frac {9 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{1445 (1+m)}+\frac {9 \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 )}{7514 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {9 \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 )}{7514 \left (13+2 \sqrt {13}\right ) (1+m)}-\frac {\left (2 \left (81+\left (62-22 \sqrt {13}\right ) m\right )\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx}{221 \sqrt {13}}+\frac {\left (2 \left (81+\left (62+22 \sqrt {13}\right ) m\right )\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx}{221 \sqrt {13}} \\ & = \frac {(43-33 x) (1+4 x)^{1+m}}{663 \left (1-5 x+3 x^2\right )}+\frac {9 (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {3}{5} (1+4 x)\right )}{1445 (1+m)}+\frac {9 \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 )}{7514 \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {\left (81+\left (62+22 \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 )}{221 \sqrt {13} \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {9 \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 )}{7514 \left (13+2 \sqrt {13}\right ) (1+m)}+\frac {\left (81+\left (62-22 \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 )}{221 \sqrt {13} \left (13+2 \sqrt {13}\right ) (1+m)} \\ \end{align*}
Time = 0.54 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.81 \[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )^2} \, dx=\frac {(1+4 x)^{1+m} \left (\frac {2210 (43-33 x)}{1-5 x+3 x^2}+\frac {9126 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{1+m}+\frac {1755 \left (13+9 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )}{\left (13-2 \sqrt {13}\right ) (1+m)}+\frac {1755 \left (13-9 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (1+m)}+\frac {510 \sqrt {13} \left (\frac {\left (81+\left (62+22 \sqrt {13}\right ) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )}{-13+2 \sqrt {13}}+\frac {\left (81+\left (62-22 \sqrt {13}\right ) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{13+2 \sqrt {13}}\right )}{1+m}\right )}{1465230} \]
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\[\int \frac {\left (1+4 x \right )^{m}}{\left (2+3 x \right ) \left (3 x^{2}-5 x +1\right )^{2}}d x\]
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\[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2} {\left (3 \, x + 2\right )}} \,d x } \]
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\[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {\left (4 x + 1\right )^{m}}{\left (3 x + 2\right ) \left (3 x^{2} - 5 x + 1\right )^{2}}\, dx \]
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\[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2} {\left (3 \, x + 2\right )}} \,d x } \]
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\[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2} {\left (3 \, x + 2\right )}} \,d x } \]
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Timed out. \[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {{\left (4\,x+1\right )}^m}{\left (3\,x+2\right )\,{\left (3\,x^2-5\,x+1\right )}^2} \,d x \]
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