\(\int \frac {(d+e x)^{1+m} (f+g x)}{a+b x+c x^2} \, dx\) [1117]

Optimal result

Integrand size = 27, antiderivative size = 234 \[ \int \frac {(d+e x)^{1+m} (f+g x)}{a+b x+c x^2} \, dx=-\frac {\left (2 c f-\left (b-\sqrt {b^2-4 a c}\right ) g\right ) (d+e x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (2+m)}+\frac {\left (2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g\right ) (d+e x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (2+m)} \] Output:

-(2*c*f-(b-(-4*a*c+b^2)^(1/2))*g)*(e*x+d)^(2+m)*hypergeom([1, 2+m],[3+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)/(2+m)+(2*c*f-(b+(-4*a*c+b^2)^(1/2))*g)*(e*x+d)^(2+m 
)*hypergeom([1, 2+m],[3+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)/(2+m)
 

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.84 \[ \int \frac {(d+e x)^{1+m} (f+g x)}{a+b x+c x^2} \, dx=\frac {(d+e x)^{2+m} \left (-\frac {\left (g+\frac {2 c f-b g}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}-\frac {\left (g+\frac {-2 c f+b g}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,2+m,3+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2+m} \] Input:

Integrate[((d + e*x)^(1 + m)*(f + g*x))/(a + b*x + c*x^2),x]
 

Output:

((d + e*x)^(2 + m)*(-(((g + (2*c*f - b*g)/Sqrt[b^2 - 4*a*c])*Hypergeometri 
c2F1[1, 2 + m, 3 + 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)) - ((g + (-2*c*f + b*g)/Sqrt[b^2 - 
 4*a*c])*Hypergeometric2F1[1, 2 + m, 3 + 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)))/(2 + m)
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.91, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, 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.

\(\displaystyle \int \frac {(f+g x) (d+e x)^{m+1}}{a+b x+c x^2} \, dx\)

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {(d+e x)^{m+2} \left (\frac {2 c f-b g}{\sqrt {b^2-4 a c}}+g\right ) \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{(m+2) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {(d+e x)^{m+2} \left (g-\frac {2 c f-b g}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,m+2,m+3,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{(m+2) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}\)

Input:

Int[((d + e*x)^(1 + m)*(f + g*x))/(a + b*x + c*x^2),x]
 

Output:

-(((g + (2*c*f - b*g)/Sqrt[b^2 - 4*a*c])*(d + e*x)^(2 + m)*Hypergeometric2 
F1[1, 2 + m, 3 + 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)*(2 + m))) - ((g - (2*c*f - b*g)/Sqrt[ 
b^2 - 4*a*c])*(d + e*x)^(2 + m)*Hypergeometric2F1[1, 2 + m, 3 + 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)*(2 + 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]

\[\int \frac {\left (e x +d \right )^{1+m} \left (g x +f \right )}{c \,x^{2}+b x +a}d x\]

Input:

int((e*x+d)^(1+m)*(g*x+f)/(c*x^2+b*x+a),x)
 

Output:

int((e*x+d)^(1+m)*(g*x+f)/(c*x^2+b*x+a),x)
 

Fricas [F]

\[ \int \frac {(d+e x)^{1+m} (f+g x)}{a+b x+c x^2} \, dx=\int { \frac {{\left (g x + f\right )} {\left (e x + d\right )}^{m + 1}}{c x^{2} + b x + a} \,d x } \] Input:

integrate((e*x+d)^(1+m)*(g*x+f)/(c*x^2+b*x+a),x, algorithm="fricas")
 

Output:

integral((g*x + f)*(e*x + d)^(m + 1)/(c*x^2 + b*x + a), x)
 

Sympy [F]

\[ \int \frac {(d+e x)^{1+m} (f+g x)}{a+b x+c x^2} \, dx=\int \frac {\left (d + e x\right )^{m + 1} \left (f + g x\right )}{a + b x + c x^{2}}\, dx \] Input:

integrate((e*x+d)**(1+m)*(g*x+f)/(c*x**2+b*x+a),x)
 

Output:

Integral((d + e*x)**(m + 1)*(f + g*x)/(a + b*x + c*x**2), x)
 

Maxima [F]

\[ \int \frac {(d+e x)^{1+m} (f+g x)}{a+b x+c x^2} \, dx=\int { \frac {{\left (g x + f\right )} {\left (e x + d\right )}^{m + 1}}{c x^{2} + b x + a} \,d x } \] Input:

integrate((e*x+d)^(1+m)*(g*x+f)/(c*x^2+b*x+a),x, algorithm="maxima")
 

Output:

integrate((g*x + f)*(e*x + d)^(m + 1)/(c*x^2 + b*x + a), x)
 

Giac [F]

\[ \int \frac {(d+e x)^{1+m} (f+g x)}{a+b x+c x^2} \, dx=\int { \frac {{\left (g x + f\right )} {\left (e x + d\right )}^{m + 1}}{c x^{2} + b x + a} \,d x } \] Input:

integrate((e*x+d)^(1+m)*(g*x+f)/(c*x^2+b*x+a),x, algorithm="giac")
 

Output:

integrate((g*x + f)*(e*x + d)^(m + 1)/(c*x^2 + b*x + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{1+m} (f+g x)}{a+b x+c x^2} \, dx=\int \frac {\left (f+g\,x\right )\,{\left (d+e\,x\right )}^{m+1}}{c\,x^2+b\,x+a} \,d x \] Input:

int(((f + g*x)*(d + e*x)^(m + 1))/(a + b*x + c*x^2),x)
 

Output:

int(((f + g*x)*(d + e*x)^(m + 1))/(a + b*x + c*x^2), x)
 

Reduce [F]

\[ \int \frac {(d+e x)^{1+m} (f+g x)}{a+b x+c x^2} \, dx =\text {Too large to display} \] Input:

int((e*x+d)^(1+m)*(g*x+f)/(c*x^2+b*x+a),x)
 

Output:

( - (d + e*x)**m*a*e**2*g*m - (d + e*x)**m*a*e**2*g - (d + e*x)**m*b*d*e*g 
 + (d + e*x)**m*b*e**2*g*m*x + (d + e*x)**m*c*d**2*g*m + (d + e*x)**m*c*d* 
*2*g + 2*(d + e*x)**m*c*d*e*f*m + 2*(d + e*x)**m*c*d*e*f + 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**2*e**3*g*m* 
*2 + 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**2*e**3*g*m - 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*c*d**2*e*g*m**2 - 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*c*d**2*e*g*m - 2*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*c*d*e**2 
*f*m**2 - 2*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*c*d*e**2*f*m + 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)*b*c*d**2*e*f*m**2 + 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)*b*c*d**2*e*f*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)*a*c*e**3*g*m**2 + 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)*a*c*e**3*g*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**2*e**3*g*m**2 - 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**2*e**3*g*m + 2*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*c*d*e**2*g*m**2 + 2*int(((d + e*x)**m*...