\(\int \frac {x (d+e x)^m}{a+b x+c x^2} \, dx\) [127]

Optimal result

Integrand size = 21, antiderivative size = 218 \[ \int \frac {x (d+e x)^m}{a+b x+c x^2} \, dx=\frac {\left (b-\sqrt {b^2-4 a c}\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (1+m)}-\frac {\left (b+\sqrt {b^2-4 a c}\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.84 \[ \int \frac {x (d+e x)^m}{a+b x+c x^2} \, dx=\frac {(d+e x)^{1+m} \left (-\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}-\frac {\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{1+m} \] Input:

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

Output:

((d + e*x)^(1 + m)*(-(((1 - b/Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + 
m, 2 + m, (2*c*(d + e*x))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)])/(2*c*d + 
(-b + Sqrt[b^2 - 4*a*c])*e)) - ((1 + b/Sqrt[b^2 - 4*a*c])*Hypergeometric2F 
1[1, 1 + m, 2 + m, (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/( 
2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)))/(1 + m)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.91, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.

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

\(\Big \downarrow \) 1200

\(\displaystyle \int \left (\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) (d+e x)^m}{-\sqrt {b^2-4 a c}+b+2 c x}+\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) (d+e x)^m}{\sqrt {b^2-4 a c}+b+2 c x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) (d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) (d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}\)

Input:

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

Output:

-(((1 - b/Sqrt[b^2 - 4*a*c])*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 
 2 + m, (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)])/((2*c*d - (b 
 - Sqrt[b^2 - 4*a*c])*e)*(1 + m))) - ((1 + b/Sqrt[b^2 - 4*a*c])*(d + e*x)^ 
(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (2*c*(d + e*x))/(2*c*d - (b + S 
qrt[b^2 - 4*a*c])*e)])/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x*(d + e*x)**m/(a + b*x + c*x**2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

((d + e*x)**m*d - int((d + e*x)**m/(a*d + a*e*x + b*d*x + b*e*x**2 + c*d*x 
**2 + c*e*x**3),x)*a*d*e*m + int(((d + e*x)**m*x**2)/(a*d + a*e*x + b*d*x 
+ b*e*x**2 + c*d*x**2 + c*e*x**3),x)*b*e**2*m - int(((d + e*x)**m*x**2)/(a 
*d + a*e*x + b*d*x + b*e*x**2 + c*d*x**2 + c*e*x**3),x)*c*d*e*m)/(b*e*m)