\(\int \frac {(g x)^n (d+e x)^m}{a+b x+c x^2} \, dx\) [132]

Optimal result

Integrand size = 25, antiderivative size = 211 \[ \int \frac {(g x)^n (d+e x)^m}{a+b x+c x^2} \, dx=\frac {2 c (g x)^{1+n} (d+e x)^m \left (1+\frac {e x}{d}\right )^{-m} \operatorname {AppellF1}\left (1+n,-m,1,2+n,-\frac {e x}{d},-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) g (1+n)}-\frac {2 c (g x)^{1+n} (d+e x)^m \left (1+\frac {e x}{d}\right )^{-m} \operatorname {AppellF1}\left (1+n,-m,1,2+n,-\frac {e x}{d},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) g (1+n)} \] Output:

2*c*(g*x)^(1+n)*(e*x+d)^m*AppellF1(1+n,1,-m,2+n,-2*c*x/(b-(-4*a*c+b^2)^(1/ 
2)),-e*x/d)/(-4*a*c+b^2)^(1/2)/(b-(-4*a*c+b^2)^(1/2))/g/(1+n)/((1+e*x/d)^m 
)-2*c*(g*x)^(1+n)*(e*x+d)^m*AppellF1(1+n,1,-m,2+n,-2*c*x/(b+(-4*a*c+b^2)^( 
1/2)),-e*x/d)/(-4*a*c+b^2)^(1/2)/(b+(-4*a*c+b^2)^(1/2))/g/(1+n)/((1+e*x/d) 
^m)
 

Mathematica [F]

\[ \int \frac {(g x)^n (d+e x)^m}{a+b x+c x^2} \, dx=\int \frac {(g x)^n (d+e x)^m}{a+b x+c x^2} \, dx \] Input:

Integrate[((g*x)^n*(d + e*x)^m)/(a + b*x + c*x^2),x]
 

Output:

Integrate[((g*x)^n*(d + e*x)^m)/(a + b*x + c*x^2), x]
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1205, 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.

\(\displaystyle \int \frac {(g x)^n (d+e x)^m}{a+b x+c x^2} \, dx\)

\(\Big \downarrow \) 1205

\(\displaystyle \int \left (\frac {2 c (g x)^n (d+e x)^m}{\sqrt {b^2-4 a c} \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}-\frac {2 c (g x)^n (d+e x)^m}{\sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b+2 c x\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 c (g x)^{n+1} (d+e x)^m \left (\frac {e x}{d}+1\right )^{-m} \operatorname {AppellF1}\left (n+1,-m,1,n+2,-\frac {e x}{d},-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{g (n+1) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c (g x)^{n+1} (d+e x)^m \left (\frac {e x}{d}+1\right )^{-m} \operatorname {AppellF1}\left (n+1,-m,1,n+2,-\frac {e x}{d},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{g (n+1) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}\)

Input:

Int[((g*x)^n*(d + e*x)^m)/(a + b*x + c*x^2),x]
 

Output:

(2*c*(g*x)^(1 + n)*(d + e*x)^m*AppellF1[1 + n, -m, 1, 2 + n, -((e*x)/d), ( 
-2*c*x)/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c 
])*g*(1 + n)*(1 + (e*x)/d)^m) - (2*c*(g*x)^(1 + n)*(d + e*x)^m*AppellF1[1 
+ n, -m, 1, 2 + n, -((e*x)/d), (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(Sqrt[b^ 
2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*g*(1 + n)*(1 + (e*x)/d)^m)
 

Defintions of rubi rules used

rule 1205

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, 1/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && 
 !IntegerQ[m] &&  !IntegerQ[n]
 

rule 2009

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

Maple [F]

\[\int \frac {\left (g x \right )^{n} \left (e x +d \right )^{m}}{c \,x^{2}+b x +a}d x\]

Input:

int((g*x)^n*(e*x+d)^m/(c*x^2+b*x+a),x)
 

Output:

int((g*x)^n*(e*x+d)^m/(c*x^2+b*x+a),x)
 

Fricas [F]

\[ \int \frac {(g x)^n (d+e x)^m}{a+b x+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{m} \left (g x\right )^{n}}{c x^{2} + b x + a} \,d x } \] Input:

integrate((g*x)^n*(e*x+d)^m/(c*x^2+b*x+a),x, algorithm="fricas")
 

Output:

integral((e*x + d)^m*(g*x)^n/(c*x^2 + b*x + a), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(g x)^n (d+e x)^m}{a+b x+c x^2} \, dx=\text {Timed out} \] Input:

integrate((g*x)**n*(e*x+d)**m/(c*x**2+b*x+a),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(g x)^n (d+e x)^m}{a+b x+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{m} \left (g x\right )^{n}}{c x^{2} + b x + a} \,d x } \] Input:

integrate((g*x)^n*(e*x+d)^m/(c*x^2+b*x+a),x, algorithm="maxima")
 

Output:

integrate((e*x + d)^m*(g*x)^n/(c*x^2 + b*x + a), x)
 

Giac [F]

\[ \int \frac {(g x)^n (d+e x)^m}{a+b x+c x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{m} \left (g x\right )^{n}}{c x^{2} + b x + a} \,d x } \] Input:

integrate((g*x)^n*(e*x+d)^m/(c*x^2+b*x+a),x, algorithm="giac")
 

Output:

integrate((e*x + d)^m*(g*x)^n/(c*x^2 + b*x + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(g x)^n (d+e x)^m}{a+b x+c x^2} \, dx=\int \frac {{\left (g\,x\right )}^n\,{\left (d+e\,x\right )}^m}{c\,x^2+b\,x+a} \,d x \] Input:

int(((g*x)^n*(d + e*x)^m)/(a + b*x + c*x^2),x)
 

Output:

int(((g*x)^n*(d + e*x)^m)/(a + b*x + c*x^2), x)
 

Reduce [F]

\[ \int \frac {(g x)^n (d+e x)^m}{a+b x+c x^2} \, dx=g^{n} \left (\int \frac {x^{n} \left (e x +d \right )^{m}}{c \,x^{2}+b x +a}d x \right ) \] Input:

int((g*x)^n*(e*x+d)^m/(c*x^2+b*x+a),x)
 

Output:

g**n*int((x**n*(d + e*x)**m)/(a + b*x + c*x**2),x)