\(\int \frac {(d+e x)^m}{x (a+b x+c x^2)} \, dx\) [129]

Optimal result

Integrand size = 23, antiderivative size = 264 \[ \int \frac {(d+e x)^m}{x \left (a+b x+c x^2\right )} \, dx=\frac {c \left (b+\sqrt {b^2-4 a c}\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{a \sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (1+m)}-\frac {c \left (b-\sqrt {b^2-4 a c}\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{a \sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+m)}-\frac {(d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,1+\frac {e x}{d}\right )}{a d (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.78 \[ \int \frac {(d+e x)^m}{x \left (a+b x+c x^2\right )} \, dx=\frac {(d+e x)^{1+m} \left (\frac {c \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}+\frac {c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+m,2+m,1+\frac {e x}{d}\right )}{d}\right )}{a (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, 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.

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

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((e*x + d)^m/(c*x^3 + b*x^2 + a*x), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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