\(\int \frac {(a+b x)^m (c+d x)^n}{(e+f x)^3} \, dx\) [1801]

Optimal result

Integrand size = 22, antiderivative size = 103 \[ \int \frac {(a+b x)^m (c+d x)^n}{(e+f x)^3} \, dx=\frac {b^2 (a+b x)^{1+m} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \operatorname {AppellF1}\left (1+m,-n,3,2+m,-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{(b e-a f)^3 (1+m)} \] Output:

b^2*(b*x+a)^(1+m)*(d*x+c)^n*AppellF1(1+m,-n,3,2+m,-d*(b*x+a)/(-a*d+b*c),-f 
*(b*x+a)/(-a*f+b*e))/(-a*f+b*e)^3/(1+m)/((b*(d*x+c)/(-a*d+b*c))^n)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^m (c+d x)^n}{(e+f x)^3} \, dx=\frac {b^2 (a+b x)^{1+m} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \operatorname {AppellF1}\left (1+m,-n,3,2+m,\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )}{(b e-a f)^3 (1+m)} \] Input:

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

Output:

(b^2*(a + b*x)^(1 + m)*(c + d*x)^n*AppellF1[1 + m, -n, 3, 2 + m, (d*(a + b 
*x))/(-(b*c) + a*d), (f*(a + b*x))/(-(b*e) + a*f)])/((b*e - a*f)^3*(1 + m) 
*((b*(c + d*x))/(b*c - a*d))^n)
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {154, 153}

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.

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

\(\Big \downarrow \) 154

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

\(\Big \downarrow \) 153

\(\displaystyle \frac {b^2 (a+b x)^{m+1} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \operatorname {AppellF1}\left (m+1,-n,3,m+2,-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^3}\)

Input:

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

Output:

(b^2*(a + b*x)^(1 + m)*(c + d*x)^n*AppellF1[1 + m, -n, 3, 2 + m, -((d*(a + 
 b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/((b*e - a*f)^3*(1 + m) 
*((b*(c + d*x))/(b*c - a*d))^n)
 

Defintions of rubi rules used

rule 153

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(b*e - a*f)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*Simp 
lify[b/(b*c - a*d)]^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c 
 - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, 
n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[p] && GtQ[Simplify[b/( 
b*c - a*d)], 0] &&  !(GtQ[Simplify[d/(d*a - c*b)], 0] && SimplerQ[c + d*x, 
a + b*x])
 

rule 154

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n 
]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c 
 - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[p] &&  !G 
tQ[Simplify[b/(b*c - a*d)], 0] &&  !SimplerQ[c + d*x, a + b*x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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