\(\int \frac {(c+d x)^{-1+m} (e+f x)}{(a+b x)^3} \, dx\) [248]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 22, antiderivative size = 109 \[ \int \frac {(c+d x)^{-1+m} (e+f x)}{(a+b x)^3} \, dx=-\frac {f (c+d x)^m}{b d (2-m) (a+b x)^2}+\frac {d (b (2 c f-d e (2-m))-a d f m) (c+d x)^m \operatorname {Hypergeometric2F1}\left (3,m,1+m,\frac {b (c+d x)}{b c-a d}\right )}{b (b c-a d)^3 (2-m) m} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x)^{-1+m} (e+f x)}{(a+b x)^3} \, dx=\frac {(c+d x)^m \left (\frac {-b e+a f}{(a+b x)^2}+\frac {d (2 b c f+b d e (-2+m)-a d f m) \operatorname {Hypergeometric2F1}\left (2,m,1+m,\frac {b (c+d x)}{b c-a d}\right )}{(b c-a d)^2 m}\right )}{2 b (b c-a d)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.04, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {87, 78}

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.

\(\displaystyle \int \frac {(e+f x) (c+d x)^{m-1}}{(a+b x)^3} \, dx\)

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 78

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 78
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

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]

\[\int \frac {\left (x d +c \right )^{m -1} \left (f x +e \right )}{\left (b x +a \right )^{3}}d x\]

Input:

int((d*x+c)^(m-1)*(f*x+e)/(b*x+a)^3,x)
 

Output:

int((d*x+c)^(m-1)*(f*x+e)/(b*x+a)^3,x)
 

Fricas [F]

\[ \int \frac {(c+d x)^{-1+m} (e+f x)}{(a+b x)^3} \, dx=\int { \frac {{\left (f x + e\right )} {\left (d x + c\right )}^{m - 1}}{{\left (b x + a\right )}^{3}} \,d x } \] Input:

integrate((d*x+c)^(-1+m)*(f*x+e)/(b*x+a)^3,x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {(c+d x)^{-1+m} (e+f x)}{(a+b x)^3} \, dx=\int \frac {\left (c + d x\right )^{m - 1} \left (e + f x\right )}{\left (a + b x\right )^{3}}\, dx \] Input:

integrate((d*x+c)**(-1+m)*(f*x+e)/(b*x+a)**3,x)
 

Output:

Integral((c + d*x)**(m - 1)*(e + f*x)/(a + b*x)**3, x)
 

Maxima [F]

\[ \int \frac {(c+d x)^{-1+m} (e+f x)}{(a+b x)^3} \, dx=\int { \frac {{\left (f x + e\right )} {\left (d x + c\right )}^{m - 1}}{{\left (b x + a\right )}^{3}} \,d x } \] Input:

integrate((d*x+c)^(-1+m)*(f*x+e)/(b*x+a)^3,x, algorithm="maxima")
 

Output:

integrate((f*x + e)*(d*x + c)^(m - 1)/(b*x + a)^3, x)
 

Giac [F]

\[ \int \frac {(c+d x)^{-1+m} (e+f x)}{(a+b x)^3} \, dx=\int { \frac {{\left (f x + e\right )} {\left (d x + c\right )}^{m - 1}}{{\left (b x + a\right )}^{3}} \,d x } \] Input:

integrate((d*x+c)^(-1+m)*(f*x+e)/(b*x+a)^3,x, algorithm="giac")
 

Output:

integrate((f*x + e)*(d*x + c)^(m - 1)/(b*x + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{-1+m} (e+f x)}{(a+b x)^3} \, dx=\int \frac {\left (e+f\,x\right )\,{\left (c+d\,x\right )}^{m-1}}{{\left (a+b\,x\right )}^3} \,d x \] Input:

int(((e + f*x)*(c + d*x)^(m - 1))/(a + b*x)^3,x)
 

Output:

int(((e + f*x)*(c + d*x)^(m - 1))/(a + b*x)^3, x)
 

Reduce [F]

\[ \int \frac {(c+d x)^{-1+m} (e+f x)}{(a+b x)^3} \, dx=\text {too large to display} \] Input:

int((d*x+c)^(-1+m)*(f*x+e)/(b*x+a)^3,x)
 

Output:

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