Integrand size = 27, antiderivative size = 181 \[ \int \frac {(2+3 x)^3 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\frac {(209-426 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {\left (1521+\sqrt {13} \left (1701-1168 m+568 \sqrt {13} m\right )\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{338 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {\left (\sqrt {13} (1701-1168 m)-13 (117+568 m)\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{338 \left (13+2 \sqrt {13}\right ) (1+m)} \]
[Out]
Time = 0.16 (sec) , antiderivative size = 181, 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, 844, 70} \[ \int \frac {(2+3 x)^3 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=-\frac {\left (\sqrt {13} \left (568 \sqrt {13} m-1168 m+1701\right )+1521\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{338 \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {\left (\sqrt {13} (1701-1168 m)-13 (568 m+117)\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{338 \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {(209-426 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )} \]
[In]
[Out]
Rule 70
Rule 844
Rule 1662
Rubi steps \begin{align*} \text {integral}& = \frac {(209-426 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {1}{507} \int \frac {(1+4 x)^m (13 (1143+836 m)-39 (117+568 m) x)}{1-5 x+3 x^2} \, dx \\ & = \frac {(209-426 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {1}{507} \int \left (\frac {\left (-39 (117+568 m)-3 \sqrt {13} (-1701+1168 m)\right ) (1+4 x)^m}{-5-\sqrt {13}+6 x}+\frac {\left (-39 (117+568 m)+3 \sqrt {13} (-1701+1168 m)\right ) (1+4 x)^m}{-5+\sqrt {13}+6 x}\right ) \, dx \\ & = \frac {(209-426 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {1}{169} \left (\sqrt {13} (1701-1168 m)-13 (117+568 m)\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx+\frac {1}{169} \left (\sqrt {13} (1701-1168 m)+13 (117+568 m)\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx \\ & = \frac {(209-426 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {\left (\sqrt {13} (1701-1168 m)+13 (117+568 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 )}{338 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {\left (\sqrt {13} (1701-1168 m)-13 (117+568 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 )}{338 \left (13+2 \sqrt {13}\right ) (1+m)} \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.39 \[ \int \frac {(2+3 x)^3 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\frac {(1+4 x)^{1+m} \left (\frac {5434-11076 x}{1-5 x+3 x^2}-\frac {351 \left (-13+27 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )}{\left (-13+2 \sqrt {13}\right ) (1+m)}-\frac {12 \left (\sqrt {13} (1215-292 m)+1846 m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )}{\left (13-2 \sqrt {13}\right ) (1+m)}-\frac {351 \left (13+27 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (1+m)}+\frac {12 \left (\sqrt {13} (1215-292 m)-1846 m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (1+m)}\right )}{1014} \]
[In]
[Out]
\[\int \frac {\left (2+3 x \right )^{3} \left (1+4 x \right )^{m}}{\left (3 x^{2}-5 x +1\right )^{2}}d x\]
[In]
[Out]
\[ \int \frac {(2+3 x)^3 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{3}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
[In]
[Out]
\[ \int \frac {(2+3 x)^3 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {\left (3 x + 2\right )^{3} \left (4 x + 1\right )^{m}}{\left (3 x^{2} - 5 x + 1\right )^{2}}\, dx \]
[In]
[Out]
\[ \int \frac {(2+3 x)^3 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{3}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
[In]
[Out]
\[ \int \frac {(2+3 x)^3 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{3}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(2+3 x)^3 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {{\left (3\,x+2\right )}^3\,{\left (4\,x+1\right )}^m}{{\left (3\,x^2-5\,x+1\right )}^2} \,d x \]
[In]
[Out]