3.10.43 \(\int \frac {(1+4 x)^m}{(2+3 x)^2 (1-5 x+3 x^2)^2} \, dx\) [943]

3.10.43.1 Optimal result

Integrand size = 27, antiderivative size = 376 \[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\frac {(268-195 x) (1+4 x)^{1+m}}{11271 \left (1-5 x+3 x^2\right )}+\frac {162 (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{24565 (1+m)}+\frac {9 \left (117+64 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{63869 \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {\left (423+2 \left (211+65 \sqrt {13}\right ) m\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{3757 \sqrt {13} \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {9 \left (117-64 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{63869 \left (13+2 \sqrt {13}\right ) (1+m)}+\frac {\left (423+\left (422-130 \sqrt {13}\right ) m\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{3757 \sqrt {13} \left (13+2 \sqrt {13}\right ) (1+m)}+\frac {36 (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{7225 (1+m)} \]

output

1/11271*(268-195*x)*(1+4*x)^(1+m)/(3*x^2-5*x+1)+162/24565*(1+4*x)^(1+m)*hy 
pergeom([1, 1+m],[2+m],-3/5-12/5*x)/(1+m)+36/7225*(1+4*x)^(1+m)*hypergeom( 
[2, 1+m],[2+m],-3/5-12/5*x)/(1+m)+9/63869*(1+4*x)^(1+m)*hypergeom([1, 1+m] 
,[2+m],3*(1+4*x)/(13+2*13^(1/2)))*(117-64*13^(1/2))/(1+m)/(13+2*13^(1/2))+ 
1/48841*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13+2*13^(1/2)))* 
(423+m*(422-130*13^(1/2)))/(1+m)*13^(1/2)/(13+2*13^(1/2))+9/63869*(1+4*x)^ 
(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13-2*13^(1/2)))*(117+64*13^(1/2) 
)/(1+m)/(13-2*13^(1/2))-1/48841*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*( 
1+4*x)/(13-2*13^(1/2)))*(423+2*m*(211+65*13^(1/2)))/(1+m)/(13-2*13^(1/2))* 
13^(1/2)
 

3.10.43.2 Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.76 \[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\frac {(1+4 x)^{1+m} \left (\frac {16575 (268-195 x)}{1-5 x+3 x^2}+\frac {1232010 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{1+m}+\frac {26325 \left (117+64 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )}{\left (13-2 \sqrt {13}\right ) (1+m)}+\frac {26325 \left (117-64 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (1+m)}-\frac {425 \left (\left (423 \left (2+\sqrt {13}\right )+\left (2534+682 \sqrt {13}\right ) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )+\left (-423 \left (-2+\sqrt {13}\right )+\left (2534-682 \sqrt {13}\right ) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )\right )}{1+m}+\frac {930852 \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{1+m}\right )}{186816825} \]

input

Integrate[(1 + 4*x)^m/((2 + 3*x)^2*(1 - 5*x + 3*x^2)^2),x]
 

output

((1 + 4*x)^(1 + m)*((16575*(268 - 195*x))/(1 - 5*x + 3*x^2) + (1232010*Hyp 
ergeometric2F1[1, 1 + m, 2 + m, (-3*(1 + 4*x))/5])/(1 + m) + (26325*(117 + 
 64*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt[1 
3])])/((13 - 2*Sqrt[13])*(1 + m)) + (26325*(117 - 64*Sqrt[13])*Hypergeomet 
ric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])])/((13 + 2*Sqrt[13])* 
(1 + m)) - (425*((423*(2 + Sqrt[13]) + (2534 + 682*Sqrt[13])*m)*Hypergeome 
tric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt[13])] + (-423*(-2 + Sqrt[ 
13]) + (2534 - 682*Sqrt[13])*m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12 
*x)/(13 + 2*Sqrt[13])]))/(1 + m) + (930852*Hypergeometric2F1[2, 1 + m, 2 + 
 m, (-3*(1 + 4*x))/5])/(1 + m)))/186816825
 

3.10.43.3 Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 376, 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 = {1289, 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.

\(\displaystyle \int \frac {(4 x+1)^m}{(3 x+2)^2 \left (3 x^2-5 x+1\right )^2} \, dx\)

\(\Big \downarrow \) 1289

\(\displaystyle \int \left (-\frac {3 (54 x-109) (4 x+1)^m}{4913 \left (3 x^2-5 x+1\right )}+\frac {(46-27 x) (4 x+1)^m}{289 \left (3 x^2-5 x+1\right )^2}+\frac {162 (4 x+1)^m}{4913 (3 x+2)}+\frac {9 (4 x+1)^m}{289 (3 x+2)^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {162 (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {3}{5} (4 x+1)\right )}{24565 (m+1)}-\frac {\left (2 \left (211+65 \sqrt {13}\right ) m+423\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{3757 \sqrt {13} \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {9 \left (117+64 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{63869 \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {\left (\left (422-130 \sqrt {13}\right ) m+423\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{3757 \sqrt {13} \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {9 \left (117-64 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{63869 \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {36 (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,-\frac {3}{5} (4 x+1)\right )}{7225 (m+1)}+\frac {(268-195 x) (4 x+1)^{m+1}}{11271 \left (3 x^2-5 x+1\right )}\)

input

Int[(1 + 4*x)^m/((2 + 3*x)^2*(1 - 5*x + 3*x^2)^2),x]
 

output

((268 - 195*x)*(1 + 4*x)^(1 + m))/(11271*(1 - 5*x + 3*x^2)) + (162*(1 + 4* 
x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (-3*(1 + 4*x))/5])/(24565*(1 
 + m)) + (9*(117 + 64*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + 
 m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(63869*(13 - 2*Sqrt[13])*(1 + 
 m)) - ((423 + 2*(211 + 65*Sqrt[13])*m)*(1 + 4*x)^(1 + m)*Hypergeometric2F 
1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(3757*Sqrt[13]*(13 - 
2*Sqrt[13])*(1 + m)) + (9*(117 - 64*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeome 
tric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])/(63869*(13 + 2* 
Sqrt[13])*(1 + m)) + ((423 + (422 - 130*Sqrt[13])*m)*(1 + 4*x)^(1 + m)*Hyp 
ergeometric2F1[1, 1 + m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])/(3757*Sq 
rt[13]*(13 + 2*Sqrt[13])*(1 + m)) + (36*(1 + 4*x)^(1 + m)*Hypergeometric2F 
1[2, 1 + m, 2 + m, (-3*(1 + 4*x))/5])/(7225*(1 + m))
 

3.10.43.3.1 Defintions of rubi rules used

rule 1289

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x 
_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + 
 g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && ( 
IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0]))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

3.10.43.4 Maple [F]

\[\int \frac {\left (1+4 x \right )^{m}}{\left (2+3 x \right )^{2} \left (3 x^{2}-5 x +1\right )^{2}}d x\]

input

int((1+4*x)^m/(2+3*x)^2/(3*x^2-5*x+1)^2,x)
 

output

int((1+4*x)^m/(2+3*x)^2/(3*x^2-5*x+1)^2,x)
 

3.10.43.5 Fricas [F]

\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2} {\left (3 \, x + 2\right )}^{2}} \,d x } \]

input

integrate((1+4*x)^m/(2+3*x)^2/(3*x^2-5*x+1)^2,x, algorithm="fricas")
 

output

integral((4*x + 1)^m/(81*x^6 - 162*x^5 - 45*x^4 + 162*x^3 + 13*x^2 - 28*x 
+ 4), x)
 

3.10.43.6 Sympy [F]

\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {\left (4 x + 1\right )^{m}}{\left (3 x + 2\right )^{2} \left (3 x^{2} - 5 x + 1\right )^{2}}\, dx \]

input

integrate((1+4*x)**m/(2+3*x)**2/(3*x**2-5*x+1)**2,x)
 

output

Integral((4*x + 1)**m/((3*x + 2)**2*(3*x**2 - 5*x + 1)**2), x)
 

3.10.43.7 Maxima [F]

\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2} {\left (3 \, x + 2\right )}^{2}} \,d x } \]

input

integrate((1+4*x)^m/(2+3*x)^2/(3*x^2-5*x+1)^2,x, algorithm="maxima")
 

output

integrate((4*x + 1)^m/((3*x^2 - 5*x + 1)^2*(3*x + 2)^2), x)
 

3.10.43.8 Giac [F]

\[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\int { \frac {{\left (4 \, x + 1\right )}^{m}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2} {\left (3 \, x + 2\right )}^{2}} \,d x } \]

input

integrate((1+4*x)^m/(2+3*x)^2/(3*x^2-5*x+1)^2,x, algorithm="giac")
 

output

integrate((4*x + 1)^m/((3*x^2 - 5*x + 1)^2*(3*x + 2)^2), x)
 

3.10.43.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(1+4 x)^m}{(2+3 x)^2 \left (1-5 x+3 x^2\right )^2} \, dx=\int \frac {{\left (4\,x+1\right )}^m}{{\left (3\,x+2\right )}^2\,{\left (3\,x^2-5\,x+1\right )}^2} \,d x \]

input

int((4*x + 1)^m/((3*x + 2)^2*(3*x^2 - 5*x + 1)^2),x)
 

output

int((4*x + 1)^m/((3*x + 2)^2*(3*x^2 - 5*x + 1)^2), x)