Integrand size = 27, antiderivative size = 202 \[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\frac {9 (1+4 x)^{1+m}}{4 (1+m)}+\frac {(844-2355 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {\left (13689-\sqrt {13} \left (297+4474 m-1570 \sqrt {13} m\right )\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{169 \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {\left (13689+\sqrt {13} \left (297+4474 m+1570 \sqrt {13} m\right )\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{169 \left (13+2 \sqrt {13}\right ) (1+m)} \]
9/4*(1+4*x)^(1+m)/(1+m)+1/39*(844-2355*x)*(1+4*x)^(1+m)/(3*x^2-5*x+1)-1/16 9*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13-2*13^(1/2)))*(13689 -13^(1/2)*(297+4474*m-1570*m*13^(1/2)))/(1+m)/(13-2*13^(1/2))-1/169*(1+4*x )^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13+2*13^(1/2)))*(13689+13^(1/2 )*(297+4474*m+1570*m*13^(1/2)))/(1+m)/(13+2*13^(1/2))
Time = 0.32 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.24 \[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\frac {(1+4 x)^{1+m} \left (\frac {13689}{4+4 m}+\frac {39 (844-2355 x)}{1-5 x+3 x^2}-\frac {1053 \left (-117+128 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )}{\left (-13+2 \sqrt {13}\right ) (1+m)}-\frac {1053 \left (117+128 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (1+m)}-\frac {-\left (\left (-14679 \left (2+\sqrt {13}\right )+2 \left (-5731+667 \sqrt {13}\right ) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )\right )+\left (-14679 \left (-2+\sqrt {13}\right )+2 \left (5731+667 \sqrt {13}\right ) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{1+m}\right )}{1521} \]
((1 + 4*x)^(1 + m)*(13689/(4 + 4*m) + (39*(844 - 2355*x))/(1 - 5*x + 3*x^2 ) - (1053*(-117 + 128*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12 *x)/(13 - 2*Sqrt[13])])/((-13 + 2*Sqrt[13])*(1 + m)) - (1053*(117 + 128*Sq rt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])])/ ((13 + 2*Sqrt[13])*(1 + m)) - (-((-14679*(2 + Sqrt[13]) + 2*(-5731 + 667*S qrt[13])*m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt[13] )]) + (-14679*(-2 + Sqrt[13]) + 2*(5731 + 667*Sqrt[13])*m)*Hypergeometric2 F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])])/(1 + m)))/1521
Time = 0.45 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1265, 27, 2159, 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.
| \(\displaystyle \int \frac {(3 x+2)^4 (4 x+1)^m}{\left (3 x^2-5 x+1\right )^2} \, dx\) |
| \(\Big \downarrow \) 1265 |
| \(\displaystyle \frac {(844-2355 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )}-\frac {1}{507} \int \frac {13 (4 x+1)^m \left (-1053 x^2-3 (3140 m+1521) x+3376 m+4617\right )}{3 x^2-5 x+1}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(844-2355 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )}-\frac {1}{39} \int \frac {(4 x+1)^m \left (-1053 x^2-3 (3140 m+1521) x+3376 m+4617\right )}{3 x^2-5 x+1}dx\) |
| \(\Big \downarrow \) 2159 |
| \(\displaystyle \frac {(844-2355 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )}-\frac {1}{39} \int \left (\frac {\left (-9420 m-\frac {6 (4474 m+297)}{\sqrt {13}}-6318\right ) (4 x+1)^m}{6 x-\sqrt {13}-5}+\frac {\left (-9420 m+\frac {6 (4474 m+297)}{\sqrt {13}}-6318\right ) (4 x+1)^m}{6 x+\sqrt {13}-5}-351 (4 x+1)^m\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{39} \left (-\frac {3 \left (13689-\sqrt {13} \left (-1570 \sqrt {13} m+4474 m+297\right )\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{13 \left (13-2 \sqrt {13}\right ) (m+1)}-\frac {3 \left (\sqrt {13} \left (1570 \sqrt {13} m+4474 m+297\right )+13689\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{13 \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {351 (4 x+1)^{m+1}}{4 (m+1)}\right )+\frac {(844-2355 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )}\) |
((844 - 2355*x)*(1 + 4*x)^(1 + m))/(39*(1 - 5*x + 3*x^2)) + ((351*(1 + 4*x )^(1 + m))/(4*(1 + m)) - (3*(13689 - Sqrt[13]*(297 + 4474*m - 1570*Sqrt[13 ]*m))*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/( 13 - 2*Sqrt[13])])/(13*(13 - 2*Sqrt[13])*(1 + m)) - (3*(13689 + Sqrt[13]*( 297 + 4474*m + 1570*Sqrt[13]*m))*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])/(13*(13 + 2*Sqrt[13])*(1 + m )))/39
3.10.37.3.1 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_.)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(f + g*x) ^n, a + b*x + c*x^2, x], R = Coeff[PolynomialRemainder[(f + g*x)^n, a + b*x + c*x^2, x], x, 0], S = Coeff[PolynomialRemainder[(f + g*x)^n, a + b*x + c *x^2, x], x, 1]}, Simp[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1)*((R*(b*c *d - b^2*e + 2*a*c*e) - a*S*(2*c*d - b*e) + c*(R*(2*c*d - b*e) - S*(b*d - 2 *a*e))*x)/((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*(a + b*x + c *x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*Q + R*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a *c*e^2*(m + 2*p + 3)) - S*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(S*(b*d - 2*a*e) - R*(2*c*d - b*e))*(m + 2*p + 4)* x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && IGtQ[n, 1] && LtQ[p , -1] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
\[\int \frac {\left (2+3 x \right )^{4} \left (1+4 x \right )^{m}}{\left (3 x^{2}-5 x +1\right )^{2}}d x\]
\[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{4}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
integral((81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*(4*x + 1)^m/(9*x^4 - 30* x^3 + 31*x^2 - 10*x + 1), x)
\[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {\left (3 x + 2\right )^{4} \left (4 x + 1\right )^{m}}{\left (3 x^{2} - 5 x + 1\right )^{2}}\, dx \]
\[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{4}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
\[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{4}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(2+3 x)^4 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {{\left (3\,x+2\right )}^4\,{\left (4\,x+1\right )}^m}{{\left (3\,x^2-5\,x+1\right )}^2} \,d x \]