\(\int \frac {(e+f x)^m}{(a c+(b c+a d) x+b d x^2)^2} \, dx\) [404]

Optimal result

Integrand size = 29, antiderivative size = 286 \[ \int \frac {(e+f x)^m}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=-\frac {(e+f x)^{1+m} \left (a b d^2 e-a^2 d^2 f+b^2 c (d e-c f)+b d (2 b d e-(b c+a d) f) x\right )}{(b c-a d)^2 (b e-a f) (d e-c f) \left (a c+(b c+a d) x+b d x^2\right )}-\frac {b^2 (a d f (2-m)-b (2 d e-c f m)) (e+f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {b (e+f x)}{b e-a f}\right )}{(b c-a d)^3 (b e-a f)^2 (1+m)}-\frac {d^2 (2 b d e-b c f (2-m)-a d f m) (e+f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {d (e+f x)}{d e-c f}\right )}{(b c-a d)^3 (d e-c f)^2 (1+m)} \] Output:

-(f*x+e)^(1+m)*(a*b*d^2*e-a^2*d^2*f+b^2*c*(-c*f+d*e)+b*d*(2*b*d*e-(a*d+b*c 
)*f)*x)/(-a*d+b*c)^2/(-a*f+b*e)/(-c*f+d*e)/(a*c+(a*d+b*c)*x+b*d*x^2)-b^2*( 
a*d*f*(2-m)-b*(-c*f*m+2*d*e))*(f*x+e)^(1+m)*hypergeom([1, 1+m],[2+m],b*(f* 
x+e)/(-a*f+b*e))/(-a*d+b*c)^3/(-a*f+b*e)^2/(1+m)-d^2*(2*b*d*e-b*c*f*(2-m)- 
a*d*f*m)*(f*x+e)^(1+m)*hypergeom([1, 1+m],[2+m],d*(f*x+e)/(-c*f+d*e))/(-a* 
d+b*c)^3/(-c*f+d*e)^2/(1+m)
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.88 \[ \int \frac {(e+f x)^m}{\left (a c+(b c+a d) x+b d x^2\right )^2} \, dx=\frac {(e+f x)^{1+m} \left (-\frac {d (-2 b d e+b c f+a d f)}{(b c-a d) (-d e+c f) (c+d x)}-\frac {b}{(a+b x) (c+d x)}+\frac {-\frac {b^2 (d e-c f) (a d f (-2+m)+b (2 d e-c f m)) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {b (e+f x)}{b e-a f}\right )}{b e-a f}+\frac {d^2 (b e-a f) (2 b d e+b c f (-2+m)-a d f m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {d (e+f x)}{d e-c f}\right )}{d e-c f}}{(b c-a d)^2 (-d e+c f) (1+m)}\right )}{(b c-a d) (b e-a f)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, 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.

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

\(\Big \downarrow \) 1165

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

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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