\(\int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx\) [231]

Optimal result

Integrand size = 23, antiderivative size = 196 \[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=-\frac {\left (b B-2 A c-B \sqrt {b^2-4 a c}\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) e (1+m)}-\frac {\left (2 A c-B \left (b+\sqrt {b^2-4 a c}\right )\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) e (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.69 \[ \int \frac {(e x)^m (A+B x)}{a+b x+c x^2} \, dx=\frac {x (e x)^m \left (\left (-2 a B+A \left (b+\sqrt {b^2-4 a c}\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+\left (2 a B+A \left (-b+\sqrt {b^2-4 a c}\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )\right )}{2 a \sqrt {b^2-4 a c} (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.88, 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.

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

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((B - (b*B - 2*A*c)/Sqrt[b^2 - 4*a*c])*(e*x)^(1 + m)*Hypergeometric2F1[1, 
1 + m, 2 + m, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c])])/((b - Sqrt[b^2 - 4*a*c])* 
e*(1 + m)) + ((B + (b*B - 2*A*c)/Sqrt[b^2 - 4*a*c])*(e*x)^(1 + m)*Hypergeo 
metric2F1[1, 1 + m, 2 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/((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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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