Integrand size = 20, antiderivative size = 177 \[ \int \frac {(1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\frac {(7-6 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {2 \left (9+2 \left (2+\sqrt {13}\right ) m\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{13 \sqrt {13} \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {2 \left (9+2 \left (2-\sqrt {13}\right ) m\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{13 \sqrt {13} \left (13+2 \sqrt {13}\right ) (1+m)} \]
1/39*(7-6*x)*(1+4*x)^(1+m)/(3*x^2-5*x+1)+2/169*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13+2*13^(1/2)))*(9+2*m*(2-13^(1/2)))/(1+m)*13^(1/2) /(13+2*13^(1/2))-2/169*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(1 3-2*13^(1/2)))*(9+2*m*(2+13^(1/2)))/(1+m)/(13-2*13^(1/2))*13^(1/2)
Time = 0.17 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.85 \[ \int \frac {(1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\frac {1}{507} (1+4 x)^{1+m} \left (\frac {91-78 x}{1-5 x+3 x^2}+\frac {6 \left (26 m+\sqrt {13} (9+4 m)\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )}{\left (-13+2 \sqrt {13}\right ) (1+m)}+\frac {6 \sqrt {13} \left (9-2 \left (-2+\sqrt {13}\right ) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (1+m)}\right ) \]
((1 + 4*x)^(1 + m)*((91 - 78*x)/(1 - 5*x + 3*x^2) + (6*(26*m + Sqrt[13]*(9 + 4*m))*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt[13])]) /((-13 + 2*Sqrt[13])*(1 + m)) + (6*Sqrt[13]*(9 - 2*(-2 + Sqrt[13])*m)*Hype rgeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])])/((13 + 2*Sqr t[13])*(1 + m))))/507
Time = 0.35 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1165, 27, 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.
\(\displaystyle \int \frac {(4 x+1)^m}{\left (3 x^2-5 x+1\right )^2} \, dx\) |
\(\Big \downarrow \) 1165 |
\(\displaystyle \frac {(7-6 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )}-\frac {1}{507} \int \frac {26 (4 x+1)^m (-12 x m+14 m+9)}{3 x^2-5 x+1}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(7-6 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )}-\frac {2}{39} \int \frac {(4 x+1)^m (-12 x m+14 m+9)}{3 x^2-5 x+1}dx\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \frac {(7-6 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )}-\frac {2}{39} \int \left (\frac {\left (\frac {6 (4 m+9)}{\sqrt {13}}-12 m\right ) (4 x+1)^m}{6 x-\sqrt {13}-5}+\frac {\left (-12 m-\frac {6 (4 m+9)}{\sqrt {13}}\right ) (4 x+1)^m}{6 x+\sqrt {13}-5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(7-6 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )}-\frac {2}{39} \left (\frac {3 \left (2 \left (2+\sqrt {13}\right ) m+9\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{\sqrt {13} \left (13-2 \sqrt {13}\right ) (m+1)}-\frac {3 \left (2 \left (2-\sqrt {13}\right ) m+9\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{\sqrt {13} \left (13+2 \sqrt {13}\right ) (m+1)}\right )\) |
((7 - 6*x)*(1 + 4*x)^(1 + m))/(39*(1 - 5*x + 3*x^2)) - (2*((3*(9 + 2*(2 + Sqrt[13])*m)*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(Sqrt[13]*(13 - 2*Sqrt[13])*(1 + m)) - (3*(9 + 2 *(2 - Sqrt[13])*m)*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3 *(1 + 4*x))/(13 + 2*Sqrt[13])])/(Sqrt[13]*(13 + 2*Sqrt[13])*(1 + m))))/39
3.10.41.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e) *x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^ 2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m + 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]
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]
\[\int \frac {\left (1+4 x \right )^{m}}{\left (3 x^{2}-5 x +1\right )^{2}}d x\]
\[ \int \frac {(1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
\[ \int \frac {(1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {\left (4 x + 1\right )^{m}}{\left (3 x^{2} - 5 x + 1\right )^{2}}\, dx \]
\[ \int \frac {(1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
\[ \int \frac {(1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {{\left (4\,x+1\right )}^m}{{\left (3\,x^2-5\,x+1\right )}^2} \,d x \]