\(\int \frac {(d+e x)^m}{a+b x+c x^2} \, dx\) [765]

Optimal result

Integrand size = 20, antiderivative size = 191 \[ \int \frac {(d+e x)^m}{a+b x+c x^2} \, dx=-\frac {2 c (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (1+m)}+\frac {2 c (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+m)} \] Output:

-2*c*(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)+ 
2*c*(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 = 0.44 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.85 \[ \int \frac {(d+e x)^m}{a+b x+c x^2} \, dx=\frac {2 c (d+e x)^{1+m} \left (-\frac {\operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}+\frac {\operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1150, 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.

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

\(\Big \downarrow \) 1150

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-2*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)])/(Sqrt[b^2 - 4*a*c]*(2*c*d - (b - Sq 
rt[b^2 - 4*a*c])*e)*(1 + m)) + (2*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)])/(Sqrt 
[b^2 - 4*a*c]*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(1 + m))
 

Defintions of rubi rules used

rule 1150

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol 
] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + b*x + c*x^2), x], x] /; FreeQ[ 
{a, b, c, d, e, m}, x] &&  !IntegerQ[2*m]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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