\(\int \frac {(d+e x)^m}{x^2 (a+b x+c x^2)} \, dx\) [130]

Optimal result

Integrand size = 23, antiderivative size = 307 \[ \int \frac {(d+e x)^m}{x^2 \left (a+b x+c x^2\right )} \, dx=-\frac {(d+e x)^{1+m}}{a d x}-\frac {c \left (b^2-2 a c+b \sqrt {b^2-4 a c}\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{a^2 \sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (1+m)}+\frac {c \left (b^2-2 a c-b \sqrt {b^2-4 a c}\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{a^2 \sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+m)}+\frac {(b d-a e m) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,1+\frac {e x}{d}\right )}{a^2 d^2 (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.80 \[ \int \frac {(d+e x)^m}{x^2 \left (a+b x+c x^2\right )} \, dx=\frac {(d+e x)^{1+m} \left (-\frac {c \left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}-\frac {c \left (b+\frac {-b^2+2 a c}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}+\frac {b \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,1+\frac {e x}{d}\right )}{d}+\frac {a e \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,1+\frac {e x}{d}\right )}{d^2}\right )}{a^2 (1+m)} \] Input:

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

Output:

((d + e*x)^(1 + m)*(-((c*(b + (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*Hypergeomet 
ric2F1[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*(b + (-b^2 + 2*a*c)/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) + (b*Hyper 
geometric2F1[1, 1 + m, 2 + m, 1 + (e*x)/d])/d + (a*e*Hypergeometric2F1[2, 
1 + m, 2 + m, 1 + (e*x)/d])/d^2))/(a^2*(1 + m))
 

Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 296, normalized size of antiderivative = 0.96, 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.

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

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((c*(b + (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*(d + e*x)^(1 + m)*Hypergeometri 
c2F1[1, 1 + m, 2 + m, (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)] 
)/(a^2*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(1 + m))) - (c*(b - (b^2 - 2*a* 
c)/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*(2*c*d - (b + 
Sqrt[b^2 - 4*a*c])*e)*(1 + m)) + (b*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 
 1 + m, 2 + m, 1 + (e*x)/d])/(a^2*d*(1 + m)) + (e*(d + e*x)^(1 + m)*Hyperg 
eometric2F1[2, 1 + m, 2 + m, 1 + (e*x)/d])/(a*d^2*(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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((e*x + d)^m/(c*x^4 + b*x^3 + a*x^2), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(d+e x)^m}{x^2 \left (a+b x+c x^2\right )} \, dx=\int \frac {\left (e x +d \right )^{m}}{c \,x^{4}+b \,x^{3}+a \,x^{2}}d x \] Input:

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

Output:

int((d + e*x)**m/(a*x**2 + b*x**3 + c*x**4),x)