\(\int (a+b x)^m (c+d x)^{1-m} (e+f x)^p \, dx\) [1822]

Optimal result

Integrand size = 26, antiderivative size = 131 \[ \int (a+b x)^m (c+d x)^{1-m} (e+f x)^p \, dx=\frac {(b c-a d) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (1+m,-1+m,-p,2+m,-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^2 (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95 \[ \int (a+b x)^m (c+d x)^{1-m} (e+f x)^p \, dx=\frac {(a+b x)^{1+m} (c+d x)^{1-m} \left (\frac {b (c+d x)}{b c-a d}\right )^{-1+m} (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (1+m,-1+m,-p,2+m,\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )}{b (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {157, 156, 155}

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.

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

\(\Big \downarrow \) 157

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

\(\Big \downarrow \) 156

\(\displaystyle \frac {(b c-a d) (c+d x)^{-m} (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{1-m} \left (\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}\right )^pdx}{b}\)

\(\Big \downarrow \) 155

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

Input:

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

Output:

((b*c - a*d)*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*(e + f*x)^p*A 
ppellF1[1 + m, -1 + m, -p, 2 + m, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + 
b*x))/(b*e - a*f))])/(b^2*(1 + m)*(c + d*x)^m*((b*(e + f*x))/(b*e - a*f))^ 
p)
 

Defintions of rubi rules used

rule 155

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

rule 156

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

rule 157

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, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
&  !GtQ[Simplify[b/(b*c - a*d)], 0] &&  !SimplerQ[c + d*x, a + b*x] &&  !Si 
mplerQ[e + f*x, a + b*x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(((e + f*x)**p*(a + b*x)**m*x)/(c + d*x)**m,x)*d + int(((e + f*x)**p*(a 
 + b*x)**m)/(c + d*x)**m,x)*c