Integrand size = 27, antiderivative size = 199 \[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\frac {27 (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{1445 (1+m)}+\frac {3 \left (117+47 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{7514 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {3 \left (117-47 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{7514 \left (13+2 \sqrt {13}\right ) (1+m)}+\frac {12 (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{425 (1+m)} \]
27/1445*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],-3/5-12/5*x)/(1+m)+12/425*( 1+4*x)^(1+m)*hypergeom([2, 1+m],[2+m],-3/5-12/5*x)/(1+m)+3/7514*(1+4*x)^(1 +m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13+2*13^(1/2)))*(117-47*13^(1/2))/ (1+m)/(13+2*13^(1/2))+3/7514*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4 *x)/(13-2*13^(1/2)))*(117+47*13^(1/2))/(1+m)/(13-2*13^(1/2))
Time = 0.17 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.76 \[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\frac {(1+4 x)^{1+m} \left (10530 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )+25 \left (211+65 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )+5275 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )-1625 \sqrt {13} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )+15912 \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {3}{5} (1+4 x)\right )\right )}{563550 (1+m)} \]
((1 + 4*x)^(1 + m)*(10530*Hypergeometric2F1[1, 1 + m, 2 + m, (-3*(1 + 4*x) )/5] + 25*(211 + 65*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x )/(13 - 2*Sqrt[13])] + 5275*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/ (13 + 2*Sqrt[13])] - 1625*Sqrt[13]*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])] + 15912*Hypergeometric2F1[2, 1 + m, 2 + m, (-3*( 1 + 4*x))/5]))/(563550*(1 + m))
Time = 0.34 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1200, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(4 x+1)^m}{(3 x+2)^2 \left (3 x^2-5 x+1\right )} \, dx\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \int \left (\frac {(46-27 x) (4 x+1)^m}{289 \left (3 x^2-5 x+1\right )}+\frac {27 (4 x+1)^m}{289 (3 x+2)}+\frac {3 (4 x+1)^m}{17 (3 x+2)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {27 (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {3}{5} (4 x+1)\right )}{1445 (m+1)}+\frac {3 \left (117+47 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{7514 \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {3 \left (117-47 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{7514 \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {12 (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,-\frac {3}{5} (4 x+1)\right )}{425 (m+1)}\) |
(27*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (-3*(1 + 4*x))/5] )/(1445*(1 + m)) + (3*(117 + 47*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric 2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(7514*(13 - 2*Sqrt[ 13])*(1 + m)) + (3*(117 - 47*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1 [1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])/(7514*(13 + 2*Sqrt[13] )*(1 + m)) + (12*(1 + 4*x)^(1 + m)*Hypergeometric2F1[2, 1 + m, 2 + m, (-3* (1 + 4*x))/5])/(425*(1 + m))
3.10.36.3.1 Defintions of rubi rules used
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
\[\int \frac {\left (1+4 x \right )^{m}}{\left (2+3 x \right )^{2} \left (3 x^{2}-5 x +1\right )}d x\]
\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )} {\left (3 \, x + 2\right )}^{2}} \,d x } \]
\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\int \frac {\left (4 x + 1\right )^{m}}{\left (3 x + 2\right )^{2} \cdot \left (3 x^{2} - 5 x + 1\right )}\, dx \]
\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )} {\left (3 \, x + 2\right )}^{2}} \,d x } \]
\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )} {\left (3 \, x + 2\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )} \, dx=\int \frac {{\left (4\,x+1\right )}^m}{{\left (3\,x+2\right )}^2\,\left (3\,x^2-5\,x+1\right )} \,d x \]