\(\int \frac {(e x)^m (c+d x)}{(a x+b x^2)^3} \, dx\) [234]

Optimal result

Integrand size = 22, antiderivative size = 85 \[ \int \frac {(e x)^m (c+d x)}{\left (a x+b x^2\right )^3} \, dx=-\frac {d e^2 (e x)^{-2+m}}{b (4-m) (a+b x)^2}-\frac {e^2 \left (\frac {c}{2-m}-\frac {a d}{4 b-b m}\right ) (e x)^{-2+m} \operatorname {Hypergeometric2F1}\left (3,-2+m,-1+m,-\frac {b x}{a}\right )}{a^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.87 \[ \int \frac {(e x)^m (c+d x)}{\left (a x+b x^2\right )^3} \, dx=\frac {(e x)^m \left (\frac {a^2 (b c-a d)}{(a+b x)^2}-\frac {(b c (-4+m)-a d (-2+m)) \operatorname {Hypergeometric2F1}\left (2,-2+m,-1+m,-\frac {b x}{a}\right )}{-2+m}\right )}{2 a^3 b x^2} \] Input:

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

Output:

((e*x)^m*((a^2*(b*c - a*d))/(a + b*x)^2 - ((b*c*(-4 + m) - a*d*(-2 + m))*H 
ypergeometric2F1[2, -2 + m, -1 + m, -((b*x)/a)])/(-2 + m)))/(2*a^3*b*x^2)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.18, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {9, 87, 74}

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.

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

\(\Big \downarrow \) 9

\(\displaystyle e^3 \int \frac {(e x)^{m-3} (c+d x)}{(a+b x)^3}dx\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 74

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 9

Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, 
x, Min]}, Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, 
x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && 
  !MonomialQ[Px, x]
 

rule 74

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x 
)^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] 
/; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] 
 &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(e**m*(x**m*d - int(x**m/(a**3*m*x**3 - 4*a**3*x**3 + 3*a**2*b*m*x**4 - 12 
*a**2*b*x**4 + 3*a*b**2*m*x**5 - 12*a*b**2*x**5 + b**3*m*x**6 - 4*b**3*x** 
6),x)*a**3*d*m**2*x**2 + 6*int(x**m/(a**3*m*x**3 - 4*a**3*x**3 + 3*a**2*b* 
m*x**4 - 12*a**2*b*x**4 + 3*a*b**2*m*x**5 - 12*a*b**2*x**5 + b**3*m*x**6 - 
 4*b**3*x**6),x)*a**3*d*m*x**2 - 8*int(x**m/(a**3*m*x**3 - 4*a**3*x**3 + 3 
*a**2*b*m*x**4 - 12*a**2*b*x**4 + 3*a*b**2*m*x**5 - 12*a*b**2*x**5 + b**3* 
m*x**6 - 4*b**3*x**6),x)*a**3*d*x**2 + int(x**m/(a**3*m*x**3 - 4*a**3*x**3 
 + 3*a**2*b*m*x**4 - 12*a**2*b*x**4 + 3*a*b**2*m*x**5 - 12*a*b**2*x**5 + b 
**3*m*x**6 - 4*b**3*x**6),x)*a**2*b*c*m**2*x**2 - 8*int(x**m/(a**3*m*x**3 
- 4*a**3*x**3 + 3*a**2*b*m*x**4 - 12*a**2*b*x**4 + 3*a*b**2*m*x**5 - 12*a* 
b**2*x**5 + b**3*m*x**6 - 4*b**3*x**6),x)*a**2*b*c*m*x**2 + 16*int(x**m/(a 
**3*m*x**3 - 4*a**3*x**3 + 3*a**2*b*m*x**4 - 12*a**2*b*x**4 + 3*a*b**2*m*x 
**5 - 12*a*b**2*x**5 + b**3*m*x**6 - 4*b**3*x**6),x)*a**2*b*c*x**2 - 2*int 
(x**m/(a**3*m*x**3 - 4*a**3*x**3 + 3*a**2*b*m*x**4 - 12*a**2*b*x**4 + 3*a* 
b**2*m*x**5 - 12*a*b**2*x**5 + b**3*m*x**6 - 4*b**3*x**6),x)*a**2*b*d*m**2 
*x**3 + 12*int(x**m/(a**3*m*x**3 - 4*a**3*x**3 + 3*a**2*b*m*x**4 - 12*a**2 
*b*x**4 + 3*a*b**2*m*x**5 - 12*a*b**2*x**5 + b**3*m*x**6 - 4*b**3*x**6),x) 
*a**2*b*d*m*x**3 - 16*int(x**m/(a**3*m*x**3 - 4*a**3*x**3 + 3*a**2*b*m*x** 
4 - 12*a**2*b*x**4 + 3*a*b**2*m*x**5 - 12*a*b**2*x**5 + b**3*m*x**6 - 4*b* 
*3*x**6),x)*a**2*b*d*x**3 + 2*int(x**m/(a**3*m*x**3 - 4*a**3*x**3 + 3*a...