\(\int (c+d x)^n (e+f x)^{-1+n} (a (d e+c f)+2 a d f x)^m \, dx\) [1814]

Optimal result

Integrand size = 35, antiderivative size = 268 \[ \int (c+d x)^n (e+f x)^{-1+n} (a (d e+c f)+2 a d f x)^m \, dx=\frac {(c+d x)^n (e+f x)^n (a (d e+c f)+2 a d f x)^{1+m} \left (1-\frac {(d e+c f+2 d f x)^2}{(d e-c f)^2}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},1-n,\frac {3+m}{2},\frac {(d e+c f+2 d f x)^2}{(d e-c f)^2}\right )}{a f (d e-c f) (1+m)}-\frac {(c+d x)^n (e+f x)^n (a (d e+c f)+2 a d f x)^{2+m} \left (1-\frac {(d e+c f+2 d f x)^2}{(d e-c f)^2}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {2+m}{2},1-n,\frac {4+m}{2},\frac {(d e+c f+2 d f x)^2}{(d e-c f)^2}\right )}{a^2 f (d e-c f)^2 (2+m)} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.47 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.53 \[ \int (c+d x)^n (e+f x)^{-1+n} (a (d e+c f)+2 a d f x)^m \, dx=\frac {(c+d x)^{1+n} (e+f x)^{-1+n} \left (\frac {d (e+f x)}{d e-c f}\right )^{1-n} \left (\frac {d e+c f+2 d f x}{d e-c f}\right )^{-m} (a (c f+d (e+2 f x)))^m \operatorname {AppellF1}\left (1+n,1-n,-m,2+n,\frac {f (c+d x)}{-d e+c f},\frac {2 f (c+d x)}{-d e+c f}\right )}{d (1+n)} \] Input:

Integrate[(c + d*x)^n*(e + f*x)^(-1 + n)*(a*(d*e + c*f) + 2*a*d*f*x)^m,x]
 

Output:

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

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.33 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.56, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, 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.

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

\(\Big \downarrow \) 157

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

\(\Big \downarrow \) 156

\(\displaystyle \frac {d (e+f x)^n \left (\frac {c f+d e+2 d f x}{d e-c f}\right )^{-m} \left (\frac {d (e+f x)}{d e-c f}\right )^{-n} (a (c f+d e)+2 a d f x)^m \int (c+d x)^n \left (\frac {d e}{d e-c f}+\frac {d f x}{d e-c f}\right )^{n-1} \left (\frac {d e+c f}{d e-c f}+\frac {2 d f x}{d e-c f}\right )^mdx}{d e-c f}\)

\(\Big \downarrow \) 155

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

Input:

Int[(c + d*x)^n*(e + f*x)^(-1 + n)*(a*(d*e + c*f) + 2*a*d*f*x)^m,x]
 

Output:

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

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]

\[\int \left (x d +c \right )^{n} \left (f x +e \right )^{-1+n} \left (a \left (c f +d e \right )+2 a d f x \right )^{m}d x\]

Input:

int((d*x+c)^n*(f*x+e)^(-1+n)*(a*(c*f+d*e)+2*a*d*f*x)^m,x)
 

Output:

int((d*x+c)^n*(f*x+e)^(-1+n)*(a*(c*f+d*e)+2*a*d*f*x)^m,x)
 

Fricas [F]

\[ \int (c+d x)^n (e+f x)^{-1+n} (a (d e+c f)+2 a d f x)^m \, dx=\int { {\left (2 \, a d f x + {\left (d e + c f\right )} a\right )}^{m} {\left (d x + c\right )}^{n} {\left (f x + e\right )}^{n - 1} \,d x } \] Input:

integrate((d*x+c)^n*(f*x+e)^(-1+n)*(a*(c*f+d*e)+2*a*d*f*x)^m,x, algorithm= 
"fricas")
 

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int (c+d x)^n (e+f x)^{-1+n} (a (d e+c f)+2 a d f x)^m \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

integrate((d*x+c)**n*(f*x+e)**(-1+n)*(a*(c*f+d*e)+2*a*d*f*x)**m,x)
 

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

\[ \int (c+d x)^n (e+f x)^{-1+n} (a (d e+c f)+2 a d f x)^m \, dx=\int { {\left (2 \, a d f x + {\left (d e + c f\right )} a\right )}^{m} {\left (d x + c\right )}^{n} {\left (f x + e\right )}^{n - 1} \,d x } \] Input:

integrate((d*x+c)^n*(f*x+e)^(-1+n)*(a*(c*f+d*e)+2*a*d*f*x)^m,x, algorithm= 
"maxima")
 

Output:

integrate((2*a*d*f*x + (d*e + c*f)*a)^m*(d*x + c)^n*(f*x + e)^(n - 1), x)
 

Giac [F]

\[ \int (c+d x)^n (e+f x)^{-1+n} (a (d e+c f)+2 a d f x)^m \, dx=\int { {\left (2 \, a d f x + {\left (d e + c f\right )} a\right )}^{m} {\left (d x + c\right )}^{n} {\left (f x + e\right )}^{n - 1} \,d x } \] Input:

integrate((d*x+c)^n*(f*x+e)^(-1+n)*(a*(c*f+d*e)+2*a*d*f*x)^m,x, algorithm= 
"giac")
 

Output:

integrate((2*a*d*f*x + (d*e + c*f)*a)^m*(d*x + c)^n*(f*x + e)^(n - 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^n (e+f x)^{-1+n} (a (d e+c f)+2 a d f x)^m \, dx=\int {\left (e+f\,x\right )}^{n-1}\,{\left (a\,\left (c\,f+d\,e\right )+2\,a\,d\,f\,x\right )}^m\,{\left (c+d\,x\right )}^n \,d x \] Input:

int((e + f*x)^(n - 1)*(a*(c*f + d*e) + 2*a*d*f*x)^m*(c + d*x)^n,x)
 

Output:

int((e + f*x)^(n - 1)*(a*(c*f + d*e) + 2*a*d*f*x)^m*(c + d*x)^n, x)
 

Reduce [F]

\[ \int (c+d x)^n (e+f x)^{-1+n} (a (d e+c f)+2 a d f x)^m \, dx=\text {too large to display} \] Input:

int((d*x+c)^n*(f*x+e)^(-1+n)*(a*(c*f+d*e)+2*a*d*f*x)^m,x)
 

Output:

(3*(e + f*x)**n*(c + d*x)**n*(a*c*f + a*d*e + 2*a*d*f*x)**m*c*f + (e + f*x 
)**n*(c + d*x)**n*(a*c*f + a*d*e + 2*a*d*f*x)**m*d*e - 2*int(((e + f*x)**n 
*(c + d*x)**n*(a*c*f + a*d*e + 2*a*d*f*x)**m*x**2)/(c**3*e*f**2*m + 2*c**3 
*e*f**2*n + c**3*f**3*m*x + 2*c**3*f**3*n*x + 2*c**2*d*e**2*f*m + 4*c**2*d 
*e**2*f*n + 5*c**2*d*e*f**2*m*x + 10*c**2*d*e*f**2*n*x + 3*c**2*d*f**3*m*x 
**2 + 6*c**2*d*f**3*n*x**2 + c*d**2*e**3*m + 2*c*d**2*e**3*n + 5*c*d**2*e* 
*2*f*m*x + 10*c*d**2*e**2*f*n*x + 6*c*d**2*e*f**2*m*x**2 + 12*c*d**2*e*f** 
2*n*x**2 + 2*c*d**2*f**3*m*x**3 + 4*c*d**2*f**3*n*x**3 + d**3*e**3*m*x + 2 
*d**3*e**3*n*x + 3*d**3*e**2*f*m*x**2 + 6*d**3*e**2*f*n*x**2 + 2*d**3*e*f* 
*2*m*x**3 + 4*d**3*e*f**2*n*x**3),x)*c**2*d**2*f**4*m**2 - 8*int(((e + f*x 
)**n*(c + d*x)**n*(a*c*f + a*d*e + 2*a*d*f*x)**m*x**2)/(c**3*e*f**2*m + 2* 
c**3*e*f**2*n + c**3*f**3*m*x + 2*c**3*f**3*n*x + 2*c**2*d*e**2*f*m + 4*c* 
*2*d*e**2*f*n + 5*c**2*d*e*f**2*m*x + 10*c**2*d*e*f**2*n*x + 3*c**2*d*f**3 
*m*x**2 + 6*c**2*d*f**3*n*x**2 + c*d**2*e**3*m + 2*c*d**2*e**3*n + 5*c*d** 
2*e**2*f*m*x + 10*c*d**2*e**2*f*n*x + 6*c*d**2*e*f**2*m*x**2 + 12*c*d**2*e 
*f**2*n*x**2 + 2*c*d**2*f**3*m*x**3 + 4*c*d**2*f**3*n*x**3 + d**3*e**3*m*x 
 + 2*d**3*e**3*n*x + 3*d**3*e**2*f*m*x**2 + 6*d**3*e**2*f*n*x**2 + 2*d**3* 
e*f**2*m*x**3 + 4*d**3*e*f**2*n*x**3),x)*c**2*d**2*f**4*m*n - 8*int(((e + 
f*x)**n*(c + d*x)**n*(a*c*f + a*d*e + 2*a*d*f*x)**m*x**2)/(c**3*e*f**2*m + 
 2*c**3*e*f**2*n + c**3*f**3*m*x + 2*c**3*f**3*n*x + 2*c**2*d*e**2*f*m ...