\(\int \frac {(e x)^m (c+d x)}{(a x^2+b x^3)^3} \, dx\) [361]

Optimal result

Integrand size = 24, antiderivative size = 85 \[ \int \frac {(e x)^m (c+d x)}{\left (a x^2+b x^3\right )^3} \, dx=-\frac {d e^5 (e x)^{-5+m}}{b (7-m) (a+b x)^2}-\frac {e^5 \left (\frac {c}{5-m}-\frac {a d}{7 b-b m}\right ) (e x)^{-5+m} \operatorname {Hypergeometric2F1}\left (3,-5+m,-4+m,-\frac {b x}{a}\right )}{a^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.87 \[ \int \frac {(e x)^m (c+d x)}{\left (a x^2+b x^3\right )^3} \, dx=\frac {(e x)^m \left (\frac {a^2 (b c-a d)}{(a+b x)^2}-\frac {(b c (-7+m)-a d (-5+m)) \operatorname {Hypergeometric2F1}\left (2,-5+m,-4+m,-\frac {b x}{a}\right )}{-5+m}\right )}{2 a^3 b x^5} \] Input:

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

Output:

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

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, 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.

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

\(\Big \downarrow \) 9

\(\displaystyle e^6 \int \frac {(e x)^{m-6} (c+d x)}{(a+b x)^3}dx\)

\(\Big \downarrow \) 87

\(\displaystyle e^6 \left (\frac {(e x)^{m-5} (b c-a d)}{2 a b e (a+b x)^2}-\frac {(a d (5-m)-b c (7-m)) \int \frac {(e x)^{m-6}}{(a+b x)^2}dx}{2 a b}\right )\)

\(\Big \downarrow \) 74

\(\displaystyle e^6 \left (\frac {(e x)^{m-5} (a d (5-m)-b c (7-m)) \operatorname {Hypergeometric2F1}\left (2,m-5,m-4,-\frac {b x}{a}\right )}{2 a^3 b e (5-m)}+\frac {(e x)^{m-5} (b c-a d)}{2 a b e (a+b x)^2}\right )\)

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d*x + c)*(e*x)^m/(b^3*x^9 + 3*a*b^2*x^8 + 3*a^2*b*x^7 + a^3*x^6) 
, x)
 

Sympy [F]

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

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

Output:

Integral((e*x)**m*(c + d*x)/(x**6*(a + b*x)**3), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(e**m*(x**m*d - int(x**m/(a**3*m*x**6 - 7*a**3*x**6 + 3*a**2*b*m*x**7 - 21 
*a**2*b*x**7 + 3*a*b**2*m*x**8 - 21*a*b**2*x**8 + b**3*m*x**9 - 7*b**3*x** 
9),x)*a**3*d*m**2*x**5 + 12*int(x**m/(a**3*m*x**6 - 7*a**3*x**6 + 3*a**2*b 
*m*x**7 - 21*a**2*b*x**7 + 3*a*b**2*m*x**8 - 21*a*b**2*x**8 + b**3*m*x**9 
- 7*b**3*x**9),x)*a**3*d*m*x**5 - 35*int(x**m/(a**3*m*x**6 - 7*a**3*x**6 + 
 3*a**2*b*m*x**7 - 21*a**2*b*x**7 + 3*a*b**2*m*x**8 - 21*a*b**2*x**8 + b** 
3*m*x**9 - 7*b**3*x**9),x)*a**3*d*x**5 + int(x**m/(a**3*m*x**6 - 7*a**3*x* 
*6 + 3*a**2*b*m*x**7 - 21*a**2*b*x**7 + 3*a*b**2*m*x**8 - 21*a*b**2*x**8 + 
 b**3*m*x**9 - 7*b**3*x**9),x)*a**2*b*c*m**2*x**5 - 14*int(x**m/(a**3*m*x* 
*6 - 7*a**3*x**6 + 3*a**2*b*m*x**7 - 21*a**2*b*x**7 + 3*a*b**2*m*x**8 - 21 
*a*b**2*x**8 + b**3*m*x**9 - 7*b**3*x**9),x)*a**2*b*c*m*x**5 + 49*int(x**m 
/(a**3*m*x**6 - 7*a**3*x**6 + 3*a**2*b*m*x**7 - 21*a**2*b*x**7 + 3*a*b**2* 
m*x**8 - 21*a*b**2*x**8 + b**3*m*x**9 - 7*b**3*x**9),x)*a**2*b*c*x**5 - 2* 
int(x**m/(a**3*m*x**6 - 7*a**3*x**6 + 3*a**2*b*m*x**7 - 21*a**2*b*x**7 + 3 
*a*b**2*m*x**8 - 21*a*b**2*x**8 + b**3*m*x**9 - 7*b**3*x**9),x)*a**2*b*d*m 
**2*x**6 + 24*int(x**m/(a**3*m*x**6 - 7*a**3*x**6 + 3*a**2*b*m*x**7 - 21*a 
**2*b*x**7 + 3*a*b**2*m*x**8 - 21*a*b**2*x**8 + b**3*m*x**9 - 7*b**3*x**9) 
,x)*a**2*b*d*m*x**6 - 70*int(x**m/(a**3*m*x**6 - 7*a**3*x**6 + 3*a**2*b*m* 
x**7 - 21*a**2*b*x**7 + 3*a*b**2*m*x**8 - 21*a*b**2*x**8 + b**3*m*x**9 - 7 
*b**3*x**9),x)*a**2*b*d*x**6 + 2*int(x**m/(a**3*m*x**6 - 7*a**3*x**6 + ...