\(\int \frac {(d+e x)^m}{(b x+c x^2)^3} \, dx\) [243]

Optimal result

Integrand size = 19, antiderivative size = 378 \[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^3} \, dx=\frac {c \left (6 c^2 d^2-b^2 e^2 (1-m)-b c d e (4+m)\right ) (d+e x)^{1+m}}{2 b^3 d^2 (c d-b e) (b+c x)^2}-\frac {(d+e x)^{1+m}}{2 b d x^2 (b+c x)^2}+\frac {(4 c d+b e (1-m)) (d+e x)^{1+m}}{2 b^2 d^2 x (b+c x)^2}+\frac {c (2 c d-b e) \left (6 c^2 d^2-6 b c d e-b^2 e^2 (1-m)\right ) (d+e x)^{1+m}}{2 b^4 d^2 (c d-b e)^2 (b+c x)}+\frac {c^3 \left (12 c^2 d^2-6 b c d e (4-m)+b^2 e^2 \left (12-7 m+m^2\right )\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {c (d+e x)}{c d-b e}\right )}{2 b^5 (c d-b e)^3 (1+m)}-\frac {\left (12 c^2 d^2-6 b c d e m-b^2 e^2 (1-m) m\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,1+\frac {e x}{d}\right )}{2 b^5 d^3 (1+m)} \] Output:

1/2*c*(6*c^2*d^2-b^2*e^2*(1-m)-b*c*d*e*(4+m))*(e*x+d)^(1+m)/b^3/d^2/(-b*e+ 
c*d)/(c*x+b)^2-1/2*(e*x+d)^(1+m)/b/d/x^2/(c*x+b)^2+1/2*(4*c*d+b*e*(1-m))*( 
e*x+d)^(1+m)/b^2/d^2/x/(c*x+b)^2+1/2*c*(-b*e+2*c*d)*(6*c^2*d^2-6*b*c*d*e-b 
^2*e^2*(1-m))*(e*x+d)^(1+m)/b^4/d^2/(-b*e+c*d)^2/(c*x+b)+1/2*c^3*(12*c^2*d 
^2-6*b*c*d*e*(4-m)+b^2*e^2*(m^2-7*m+12))*(e*x+d)^(1+m)*hypergeom([1, 1+m], 
[2+m],c*(e*x+d)/(-b*e+c*d))/b^5/(-b*e+c*d)^3/(1+m)-1/2*(12*c^2*d^2-6*b*c*d 
*e*m-b^2*e^2*(1-m)*m)*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+m],1+e*x/d)/b^5/ 
d^3/(1+m)
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 337, normalized size of antiderivative = 0.89 \[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^3} \, dx=-\frac {(d+e x)^{1+m} \left (2 b^4 d^2 (c d-b e)^3 (1+m)-2 b^3 d (c d-b e)^3 (4 c d-b e (-1+m)) (1+m) x+x^2 \left (-2 b^2 c d (c d-b e)^2 (1+m) \left (6 c^2 d^2+b^2 e^2 (-1+m)-b c d e (4+m)\right )-(b+c x) \left (-2 b c d (2 c d-b e) (-c d+b e) \left (6 c^2 d^2-6 b c d e+b^2 e^2 (-1+m)\right ) (1+m)+(b+c x) \left (2 c^3 d^3 \left (12 c^2 d^2+6 b c d e (-4+m)+b^2 e^2 \left (12-7 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {c (d+e x)}{c d-b e}\right )-2 (c d-b e)^3 \left (12 c^2 d^2-6 b c d e m+b^2 e^2 (-1+m) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,1+\frac {e x}{d}\right )\right )\right )\right )\right )}{4 b^5 d^3 (c d-b e)^3 (1+m) x^2 (b+c x)^2} \] Input:

Integrate[(d + e*x)^m/(b*x + c*x^2)^3,x]
 

Output:

-1/4*((d + e*x)^(1 + m)*(2*b^4*d^2*(c*d - b*e)^3*(1 + m) - 2*b^3*d*(c*d - 
b*e)^3*(4*c*d - b*e*(-1 + m))*(1 + m)*x + x^2*(-2*b^2*c*d*(c*d - b*e)^2*(1 
 + m)*(6*c^2*d^2 + b^2*e^2*(-1 + m) - b*c*d*e*(4 + m)) - (b + c*x)*(-2*b*c 
*d*(2*c*d - b*e)*(-(c*d) + b*e)*(6*c^2*d^2 - 6*b*c*d*e + b^2*e^2*(-1 + m)) 
*(1 + m) + (b + c*x)*(2*c^3*d^3*(12*c^2*d^2 + 6*b*c*d*e*(-4 + m) + b^2*e^2 
*(12 - 7*m + m^2))*Hypergeometric2F1[1, 1 + m, 2 + m, (c*(d + e*x))/(c*d - 
 b*e)] - 2*(c*d - b*e)^3*(12*c^2*d^2 - 6*b*c*d*e*m + b^2*e^2*(-1 + m)*m)*H 
ypergeometric2F1[1, 1 + m, 2 + m, 1 + (e*x)/d])))))/(b^5*d^3*(c*d - b*e)^3 
*(1 + m)*x^2*(b + c*x)^2)
 

Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1165, 1235, 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.

\(\displaystyle \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^3} \, dx\)

\(\Big \downarrow \) 1165

\(\displaystyle -\frac {\int \frac {(d+e x)^m \left (6 c^2 d^2-b c e (m+4) d-b^2 e^2 (1-m)+c e (2 c d-b e) (2-m) x\right )}{\left (c x^2+b x\right )^2}dx}{2 b^2 d (c d-b e)}-\frac {(d+e x)^{m+1} (c x (2 c d-b e)+b (c d-b e))}{2 b^2 d \left (b x+c x^2\right )^2 (c d-b e)}\)

\(\Big \downarrow \) 1235

\(\displaystyle -\frac {-\frac {\int \frac {(d+e x)^m \left ((c d-b e)^2 \left (12 c^2 d^2-6 b c e m d-b^2 e^2 (1-m) m\right )-c e (2 c d-b e) \left (6 c^2 d^2-6 b c e d-b^2 e^2 (1-m)\right ) m x\right )}{c x^2+b x}dx}{b^2 d (c d-b e)}-\frac {(d+e x)^{m+1} \left (c x (2 c d-b e) \left (-b^2 e^2 (1-m)-6 b c d e+6 c^2 d^2\right )+b (c d-b e) \left (-b^2 e^2 (1-m)-b c d e (m+4)+6 c^2 d^2\right )\right )}{b^2 d \left (b x+c x^2\right ) (c d-b e)}}{2 b^2 d (c d-b e)}-\frac {(d+e x)^{m+1} (c x (2 c d-b e)+b (c d-b e))}{2 b^2 d \left (b x+c x^2\right )^2 (c d-b e)}\)

\(\Big \downarrow \) 1200

\(\displaystyle -\frac {-\frac {\int \left (\frac {(c d-b e)^2 \left (12 c^2 d^2-6 b c e m d-b^2 e^2 (1-m) m\right ) (d+e x)^m}{b x}+\frac {c^3 d^2 \left (-12 c^2 d^2+6 b c e (4-m) d-b^2 e^2 \left (m^2-7 m+12\right )\right ) (d+e x)^m}{b (b+c x)}\right )dx}{b^2 d (c d-b e)}-\frac {(d+e x)^{m+1} \left (c x (2 c d-b e) \left (-b^2 e^2 (1-m)-6 b c d e+6 c^2 d^2\right )+b (c d-b e) \left (-b^2 e^2 (1-m)-b c d e (m+4)+6 c^2 d^2\right )\right )}{b^2 d \left (b x+c x^2\right ) (c d-b e)}}{2 b^2 d (c d-b e)}-\frac {(d+e x)^{m+1} (c x (2 c d-b e)+b (c d-b e))}{2 b^2 d \left (b x+c x^2\right )^2 (c d-b e)}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {-\frac {(d+e x)^{m+1} \left (c x (2 c d-b e) \left (-b^2 e^2 (1-m)-6 b c d e+6 c^2 d^2\right )+b (c d-b e) \left (-b^2 e^2 (1-m)-b c d e (m+4)+6 c^2 d^2\right )\right )}{b^2 d \left (b x+c x^2\right ) (c d-b e)}-\frac {\frac {c^3 d^2 (d+e x)^{m+1} \left (b^2 e^2 \left (m^2-7 m+12\right )-6 b c d e (4-m)+12 c^2 d^2\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {c (d+e x)}{c d-b e}\right )}{b (m+1) (c d-b e)}-\frac {(c d-b e)^2 (d+e x)^{m+1} \left (-b^2 e^2 (1-m) m-6 b c d e m+12 c^2 d^2\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {e x}{d}+1\right )}{b d (m+1)}}{b^2 d (c d-b e)}}{2 b^2 d (c d-b e)}-\frac {(d+e x)^{m+1} (c x (2 c d-b e)+b (c d-b e))}{2 b^2 d \left (b x+c x^2\right )^2 (c d-b e)}\)

Input:

Int[(d + e*x)^m/(b*x + c*x^2)^3,x]
 

Output:

-1/2*((d + e*x)^(1 + m)*(b*(c*d - b*e) + c*(2*c*d - b*e)*x))/(b^2*d*(c*d - 
 b*e)*(b*x + c*x^2)^2) - (-(((d + e*x)^(1 + m)*(b*(c*d - b*e)*(6*c^2*d^2 - 
 b^2*e^2*(1 - m) - b*c*d*e*(4 + m)) + c*(2*c*d - b*e)*(6*c^2*d^2 - 6*b*c*d 
*e - b^2*e^2*(1 - m))*x))/(b^2*d*(c*d - b*e)*(b*x + c*x^2))) - ((c^3*d^2*( 
12*c^2*d^2 - 6*b*c*d*e*(4 - m) + b^2*e^2*(12 - 7*m + m^2))*(d + e*x)^(1 + 
m)*Hypergeometric2F1[1, 1 + m, 2 + m, (c*(d + e*x))/(c*d - b*e)])/(b*(c*d 
- b*e)*(1 + m)) - ((c*d - b*e)^2*(12*c^2*d^2 - 6*b*c*d*e*m - b^2*e^2*(1 - 
m)*m)*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, 1 + (e*x)/d])/( 
b*d*(1 + m)))/(b^2*d*(c*d - b*e)))/(2*b^2*d*(c*d - b*e))
 

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 1235

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2 
*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*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 
*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 
 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* 
m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - 
f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m}, x] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p] 
)
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((e*x+d)^m/(c*x^2+b*x)^3,x, algorithm="fricas")
 

Output:

integral((e*x + d)^m/(c^3*x^6 + 3*b*c^2*x^5 + 3*b^2*c*x^4 + b^3*x^3), x)
 

Sympy [F]

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

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

Output:

Integral((d + e*x)**m/(x**3*(b + c*x)**3), x)
 

Maxima [F]

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

integrate((e*x+d)^m/(c*x^2+b*x)^3,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((e*x+d)^m/(c*x^2+b*x)^3,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^3} \, dx=\int \frac {\left (e x +d \right )^{m}}{c^{3} x^{6}+3 b \,c^{2} x^{5}+3 b^{2} c \,x^{4}+b^{3} x^{3}}d x \] Input:

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

Output:

int((d + e*x)**m/(b**3*x**3 + 3*b**2*c*x**4 + 3*b*c**2*x**5 + c**3*x**6),x 
)