\(\int \frac {(2+3 x)^m (3+5 x)^2}{1-2 x} \, dx\) [1679]

Optimal result

Integrand size = 22, antiderivative size = 72 \[ \int \frac {(2+3 x)^m (3+5 x)^2}{1-2 x} \, dx=-\frac {155 (2+3 x)^{1+m}}{36 (1+m)}-\frac {25 (2+3 x)^{2+m}}{18 (2+m)}+\frac {121 (2+3 x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2}{7} (2+3 x)\right )}{28 (1+m)} \] Output:

-155*(2+3*x)^(1+m)/(36+36*m)-25*(2+3*x)^(2+m)/(36+18*m)+121*(2+3*x)^(1+m)* 
hypergeom([1, 1+m],[2+m],4/7+6/7*x)/(28+28*m)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.83 \[ \int \frac {(2+3 x)^m (3+5 x)^2}{1-2 x} \, dx=\frac {(2+3 x)^{1+m} \left (-35 (82+30 x+m (51+30 x))+1089 (2+m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2}{7} (2+3 x)\right )\right )}{252 (1+m) (2+m)} \] Input:

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

Output:

((2 + 3*x)^(1 + m)*(-35*(82 + 30*x + m*(51 + 30*x)) + 1089*(2 + m)*Hyperge 
ometric2F1[1, 1 + m, 2 + m, (2*(2 + 3*x))/7]))/(252*(1 + m)*(2 + m))
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {99, 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.

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

\(\Big \downarrow \) 99

\(\displaystyle \int \left (\frac {121 (3 x+2)^m}{4 (1-2 x)}-\frac {155}{12} (3 x+2)^m-\frac {25}{6} (3 x+2)^{m+1}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {121 (3 x+2)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {2}{7} (3 x+2)\right )}{28 (m+1)}-\frac {155 (3 x+2)^{m+1}}{36 (m+1)}-\frac {25 (3 x+2)^{m+2}}{18 (m+2)}\)

Input:

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

Output:

(-155*(2 + 3*x)^(1 + m))/(36*(1 + m)) - (25*(2 + 3*x)^(2 + m))/(18*(2 + m) 
) + (121*(2 + 3*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (2*(2 + 3*x) 
)/7])/(28*(1 + m))
 

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(-(25*x^2 + 30*x + 9)*(3*x + 2)^m/(2*x - 1), x)
 

Sympy [F]

\[ \int \frac {(2+3 x)^m (3+5 x)^2}{1-2 x} \, dx=- \int \frac {9 \left (3 x + 2\right )^{m}}{2 x - 1}\, dx - \int \frac {30 x \left (3 x + 2\right )^{m}}{2 x - 1}\, dx - \int \frac {25 x^{2} \left (3 x + 2\right )^{m}}{2 x - 1}\, dx \] Input:

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

Output:

-Integral(9*(3*x + 2)**m/(2*x - 1), x) - Integral(30*x*(3*x + 2)**m/(2*x - 
 1), x) - Integral(25*x**2*(3*x + 2)**m/(2*x - 1), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(2+3 x)^m (3+5 x)^2}{1-2 x} \, dx=\frac {-450 \left (3 x +2\right )^{m} m^{2} x^{2}-1065 \left (3 x +2\right )^{m} m^{2} x +216 \left (3 x +2\right )^{m} m^{2}-450 \left (3 x +2\right )^{m} m \,x^{2}-1530 \left (3 x +2\right )^{m} m x +1358 \left (3 x +2\right )^{m} m +1452 \left (3 x +2\right )^{m}-7623 \left (\int \frac {\left (3 x +2\right )^{m} x}{6 x^{2}+x -2}d x \right ) m^{3}-22869 \left (\int \frac {\left (3 x +2\right )^{m} x}{6 x^{2}+x -2}d x \right ) m^{2}-15246 \left (\int \frac {\left (3 x +2\right )^{m} x}{6 x^{2}+x -2}d x \right ) m}{36 m \left (m^{2}+3 m +2\right )} \] Input:

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

Output:

( - 450*(3*x + 2)**m*m**2*x**2 - 1065*(3*x + 2)**m*m**2*x + 216*(3*x + 2)* 
*m*m**2 - 450*(3*x + 2)**m*m*x**2 - 1530*(3*x + 2)**m*m*x + 1358*(3*x + 2) 
**m*m + 1452*(3*x + 2)**m - 7623*int(((3*x + 2)**m*x)/(6*x**2 + x - 2),x)* 
m**3 - 22869*int(((3*x + 2)**m*x)/(6*x**2 + x - 2),x)*m**2 - 15246*int(((3 
*x + 2)**m*x)/(6*x**2 + x - 2),x)*m)/(36*m*(m**2 + 3*m + 2))