\(\int \frac {(1+4 x)^m}{(2+3 x) (1-5 x+3 x^2)} \, dx\) [709]

Optimal result

Integrand size = 27, antiderivative size = 164 \[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )} \, dx=\frac {3 (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )}{85 (1+m)}+\frac {3 \left (13+9 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{442 \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {3 \left (13-9 \sqrt {13}\right ) (1+4 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{442 \left (13+2 \sqrt {13}\right ) (1+m)} \] Output:

3*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],-3/5-12/5*x)/(85+85*m)+3/442*(13+ 
9*13^(1/2))*(1+4*x)^(1+m)*hypergeom([1, 1+m],[2+m],3*(1+4*x)/(13-2*13^(1/2 
)))/(13-2*13^(1/2))/(1+m)+3/442*(13-9*13^(1/2))*(1+4*x)^(1+m)*hypergeom([1 
, 1+m],[2+m],3*(1+4*x)/(13+2*13^(1/2)))/(13+2*13^(1/2))/(1+m)
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.67 \[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )} \, dx=\frac {(1+4 x)^{1+m} \left (234 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {3}{5} (1+4 x)\right )+5 \left (31+11 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13-2 \sqrt {13}}\right )+5 \left (31-11 \sqrt {13}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {3+12 x}{13+2 \sqrt {13}}\right )\right )}{6630 (1+m)} \] Input:

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

Output:

((1 + 4*x)^(1 + m)*(234*Hypergeometric2F1[1, 1 + m, 2 + m, (-3*(1 + 4*x))/ 
5] + 5*(31 + 11*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, (3 + 12*x)/(1 
3 - 2*Sqrt[13])] + 5*(31 - 11*Sqrt[13])*Hypergeometric2F1[1, 1 + m, 2 + m, 
 (3 + 12*x)/(13 + 2*Sqrt[13])]))/(6630*(1 + m))
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 164, 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.

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

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {3}{5} (4 x+1)\right )}{85 (m+1)}+\frac {3 \left (13+9 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{442 \left (13-2 \sqrt {13}\right ) (m+1)}+\frac {3 \left (13-9 \sqrt {13}\right ) (4 x+1)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{442 \left (13+2 \sqrt {13}\right ) (m+1)}\)

Input:

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

Output:

(3*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (-3*(1 + 4*x))/5]) 
/(85*(1 + m)) + (3*(13 + 9*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1 
, 1 + m, 2 + m, (3*(1 + 4*x))/(13 - 2*Sqrt[13])])/(442*(13 - 2*Sqrt[13])*( 
1 + m)) + (3*(13 - 9*Sqrt[13])*(1 + 4*x)^(1 + m)*Hypergeometric2F1[1, 1 + 
m, 2 + m, (3*(1 + 4*x))/(13 + 2*Sqrt[13])])/(442*(13 + 2*Sqrt[13])*(1 + m) 
)
 

Defintions of rubi rules used

rule 1200

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]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((4*x + 1)^m/(9*x^3 - 9*x^2 - 7*x + 2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(1+4 x)^m}{(2+3 x) \left (1-5 x+3 x^2\right )} \, dx=\int \frac {\left (4 x +1\right )^{m}}{9 x^{3}-9 x^{2}-7 x +2}d x \] Input:

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

Output:

int((4*x + 1)**m/(9*x**3 - 9*x**2 - 7*x + 2),x)