3.10.42 \(\int \frac {(1+4 x)^m}{(2+3 x) (1-5 x+3 x^2)^2} \, dx\) [942]

3.10.42.1 Optimal result

Integrand size = 27, antiderivative size = 340 \[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )^2} \, dx=\frac {(43-33 x) (1+4 x)^{1+m}}{663 \left (1-5 x+3 x^2\right )}+\frac {9 (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{1445 (1+m)}+\frac {9 \left (13+9 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{7514 \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {\left (81+\left (62+22 \sqrt {13}\right ) m\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{221 \sqrt {13} \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {9 \left (13-9 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{7514 \left (13+2 \sqrt {13}\right ) (1+m)}+\frac {\left (81+\left (62-22 \sqrt {13}\right ) m\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{221 \sqrt {13} \left (13+2 \sqrt {13}\right ) (1+m)} \]

output

1/663*(43-33*x)*(1+4*x)^(1+m)/(3*x^2-5*x+1)+9/1445*(1+4*x)^(1+m)*hypergeom 
([1, 1+m],[2+m],-3/5-12/5*x)/(1+m)+9/7514*(1+4*x)^(1+m)*hypergeom([1, 1+m] 
,[2+m],3*(1+4*x)/(13+2*13^(1/2)))*(13-9*13^(1/2))/(1+m)/(13+2*13^(1/2))+1/ 
2873*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13+2*13^(1/2)))*(81 
+m*(62-22*13^(1/2)))/(1+m)*13^(1/2)/(13+2*13^(1/2))+9/7514*(1+4*x)^(1+m)*h 
ypergeom([1, 1+m],[2+m],3*(1+4*x)/(13-2*13^(1/2)))*(13+9*13^(1/2))/(1+m)/( 
13-2*13^(1/2))-1/2873*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13 
-2*13^(1/2)))*(81+m*(62+22*13^(1/2)))/(1+m)/(13-2*13^(1/2))*13^(1/2)
 

3.10.42.2 Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.81 \[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )^2} \, dx=\frac {(1+4 x)^{1+m} \left (\frac {2210 (43-33 x)}{1-5 x+3 x^2}+\frac {9126 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{1+m}+\frac {1755 \left (13+9 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )}{\left (13-2 \sqrt {13}\right ) (1+m)}+\frac {1755 \left (13-9 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (1+m)}+\frac {510 \sqrt {13} \left (\frac {\left (81+\left (62+22 \sqrt {13}\right ) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )}{-13+2 \sqrt {13}}+\frac {\left (81+\left (62-22 \sqrt {13}\right ) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )}{13+2 \sqrt {13}}\right )}{1+m}\right )}{1465230} \]

input

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

output

((1 + 4*x)^(1 + m)*((2210*(43 - 33*x))/(1 - 5*x + 3*x^2) + (9126*Hypergeom 
etric2F1[1, 1 + m, 2 + m, (-3*(1 + 4*x))/5])/(1 + m) + (1755*(13 + 9*Sqrt[ 
13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 - 2*Sqrt[13])])/((1 
3 - 2*Sqrt[13])*(1 + m)) + (1755*(13 - 9*Sqrt[13])*Hypergeometric2F1[1, 1 
+ m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13])])/((13 + 2*Sqrt[13])*(1 + m)) + ( 
510*Sqrt[13]*(((81 + (62 + 22*Sqrt[13])*m)*Hypergeometric2F1[1, 1 + m, 2 + 
 m, (3 + 12*x)/(13 - 2*Sqrt[13])])/(-13 + 2*Sqrt[13]) + ((81 + (62 - 22*Sq 
rt[13])*m)*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(13 + 2*Sqrt[13]) 
])/(13 + 2*Sqrt[13])))/(1 + m)))/1465230
 

3.10.42.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 340, 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.

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

\(\Big \downarrow \) 1289

\(\displaystyle \int \left (-\frac {3 (3 x-7) (4 x+1)^m}{289 \left (3 x^2-5 x+1\right )}+\frac {(7-3 x) (4 x+1)^m}{17 \left (3 x^2-5 x+1\right )^2}+\frac {9 (4 x+1)^m}{289 (3 x+2)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {9 (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {3}{5} (4 x+1)\right )}{1445 (m+1)}-\frac {\left (\left (62+22 \sqrt {13}\right ) m+81\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{221 \sqrt {13} \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {9 \left (13+9 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{7514 \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {\left (\left (62-22 \sqrt {13}\right ) m+81\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{221 \sqrt {13} \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {9 \left (13-9 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{7514 \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {(43-33 x) (4 x+1)^{m+1}}{663 \left (3 x^2-5 x+1\right )}\)

input

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

output

((43 - 33*x)*(1 + 4*x)^(1 + m))/(663*(1 - 5*x + 3*x^2)) + (9*(1 + 4*x)^(1 
+ m)*Hypergeometric2F1[1, 1 + m, 2 + m, (-3*(1 + 4*x))/5])/(1445*(1 + m)) 
+ (9*(13 + 9*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)) - ((8 
1 + (62 + 22*Sqrt[13])*m)*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 
+ m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(221*Sqrt[13]*(13 - 2*Sqrt[13])*(1 
+ m)) + (9*(13 - 9*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)) 
 + ((81 + (62 - 22*Sqrt[13])*m)*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + 
 m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])/(221*Sqrt[13]*(13 + 2*Sqrt[13 
])*(1 + m))
 

3.10.42.3.1 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]
 

3.10.42.4 Maple [F]

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

input

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

output

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

3.10.42.5 Fricas [F]

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

input

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

output

integral((4*x + 1)^m/(27*x^5 - 72*x^4 + 33*x^3 + 32*x^2 - 17*x + 2), x)
 

3.10.42.6 Sympy [F]

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

input

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

output

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

3.10.42.7 Maxima [F]

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

input

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

3.10.42.8 Giac [F]

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

input

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

3.10.42.9 Mupad [F(-1)]

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

input

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

output

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