\(\int \frac {(d+e x)^m}{(a+b x+c x^2)^2} \, dx\) [766]

Optimal result

Integrand size = 20, antiderivative size = 432 \[ \int \frac {(d+e x)^m}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {(d+e x)^{1+m} \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}+\frac {c \left (4 c^2 d^2+b \left (b+\sqrt {b^2-4 a c}\right ) e^2 m-2 c e \left (2 b d-2 a e (1-m)+\sqrt {b^2-4 a c} d m\right )\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{\left (b^2-4 a c\right )^{3/2} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right ) (1+m)}-\frac {c \left (4 c^2 d^2+b \left (b-\sqrt {b^2-4 a c}\right ) e^2 m-2 c e \left (2 b d-2 a e (1-m)-\sqrt {b^2-4 a c} d m\right )\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\left (b^2-4 a c\right )^{3/2} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right ) (1+m)} \] Output:

-(e*x+d)^(1+m)*(b*c*d-b^2*e+2*a*c*e+c*(-b*e+2*c*d)*x)/(-4*a*c+b^2)/(a*e^2- 
b*d*e+c*d^2)/(c*x^2+b*x+a)+c*(4*c^2*d^2+b*(b+(-4*a*c+b^2)^(1/2))*e^2*m-2*c 
*e*(2*b*d-2*a*e*(1-m)+(-4*a*c+b^2)^(1/2)*d*m))*(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)^(3/ 
2)/(2*c*d-(b-(-4*a*c+b^2)^(1/2))*e)/(a*e^2-b*d*e+c*d^2)/(1+m)-c*(4*c^2*d^2 
+b*(b-(-4*a*c+b^2)^(1/2))*e^2*m-2*c*e*(2*b*d-2*a*e*(1-m)-(-4*a*c+b^2)^(1/2 
)*d*m))*(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)^(3/2)/(2*c*d-(b+(-4*a*c+b^2)^(1/2))*e)/(a* 
e^2-b*d*e+c*d^2)/(1+m)
 

Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 339, normalized size of antiderivative = 0.78 \[ \int \frac {(d+e x)^m}{\left (a+b x+c x^2\right )^2} \, dx=\frac {(d+e x)^{1+m} \left (\frac {b^2 e-2 c (a e+c d x)+b c (-d+e x)}{a+x (b+c x)}-\frac {c \left (e (2 c d-b e) m+\frac {-4 c^2 d^2+4 c e (b d+a e (-1+m))-b^2 e^2 m}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{\left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) (1+m)}-\frac {c \left (e (2 c d-b e) m+\frac {4 c^2 d^2-4 c e (b d+a e (-1+m))+b^2 e^2 m}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+m)}\right )}{\left (b^2-4 a c\right ) \left (c d^2+e (-b d+a e)\right )} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 396, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1165, 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.

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

\(\Big \downarrow \) 1165

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

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 1165

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e) 
*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^ 
2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))   Int[(d 
+ e*x)^m*Simp[b*c*d*e*(2*p - m + 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p 
+ 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x, x]*(a + 
 b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p, -1] 
 && IntQuadraticQ[a, b, c, d, e, m, p, x]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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