\(\int \frac {(e+f x)^n \sqrt {1-d^2 x^2}}{1-d x} \, dx\) [124]

Optimal result

Integrand size = 30, antiderivative size = 83 \[ \int \frac {(e+f x)^n \sqrt {1-d^2 x^2}}{1-d x} \, dx=-\frac {2 \sqrt {2} \sqrt {1-d x} (e+f x)^n \left (\frac {d (e+f x)}{d e+f}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-d x),\frac {f (1-d x)}{d e+f}\right )}{d} \] Output:

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

Mathematica [F]

\[ \int \frac {(e+f x)^n \sqrt {1-d^2 x^2}}{1-d x} \, dx=\int \frac {(e+f x)^n \sqrt {1-d^2 x^2}}{1-d x} \, dx \] Input:

Integrate[((e + f*x)^n*Sqrt[1 - d^2*x^2])/(1 - d*x),x]
 

Output:

Integrate[((e + f*x)^n*Sqrt[1 - d^2*x^2])/(1 - d*x), x]
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {667, 717, 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.

\(\displaystyle \int \frac {\sqrt {1-d^2 x^2} (e+f x)^n}{1-d x} \, dx\)

\(\Big \downarrow \) 667

\(\displaystyle \int \frac {(d x+1) (e+f x)^n}{\sqrt {1-d^2 x^2}}dx\)

\(\Big \downarrow \) 717

\(\displaystyle \int \frac {\sqrt {d x+1} (e+f x)^n}{\sqrt {1-d x}}dx\)

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

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

Input:

Int[((e + f*x)^n*Sqrt[1 - d^2*x^2])/(1 - d*x),x]
 

Output:

(-2*Sqrt[2]*Sqrt[1 - d*x]*(e + f*x)^n*AppellF1[1/2, -1/2, -n, 3/2, (1 - d* 
x)/2, (f*(1 - d*x))/(d*e + f)])/(d*((d*(e + f*x))/(d*e + f))^n)
 

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 667

Int[(((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*( 
x_)), x_Symbol] :> Int[(a/d + c*(x/e))*(f + g*x)^n*(a + c*x^2)^(p - 1), x] 
/; FreeQ[{a, c, d, e, f, g, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0]
 

rule 717

Int[((d_) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_) 
^2)^(p_), x_Symbol] :> Int[(d + e*x)^(m + p)*(f + g*x)^n*(a/d + (c/e)*x)^p, 
 x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a 
, 0] && GtQ[d, 0]
 

Maple [F]

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

Input:

int((f*x+e)^n*(-d^2*x^2+1)^(1/2)/(-d*x+1),x)
 

Output:

int((f*x+e)^n*(-d^2*x^2+1)^(1/2)/(-d*x+1),x)
 

Fricas [F]

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

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

Output:

integral(-sqrt(-d^2*x^2 + 1)*(f*x + e)^n/(d*x - 1), x)
 

Sympy [F]

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

integrate((f*x+e)**n*(-d**2*x**2+1)**(1/2)/(-d*x+1),x)
 

Output:

-Integral((e + f*x)**n*sqrt(-d**2*x**2 + 1)/(d*x - 1), x)
 

Maxima [F]

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

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

Output:

-integrate(sqrt(-d^2*x^2 + 1)*(f*x + e)^n/(d*x - 1), x)
 

Giac [F]

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

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

Output:

integrate(-sqrt(-d^2*x^2 + 1)*(f*x + e)^n/(d*x - 1), x)
 

Mupad [F(-1)]

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

int(-((e + f*x)^n*(1 - d^2*x^2)^(1/2))/(d*x - 1),x)
 

Output:

-int(((e + f*x)^n*(1 - d^2*x^2)^(1/2))/(d*x - 1), x)
 

Reduce [F]

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

int((f*x+e)^n*(-d^2*x^2+1)^(1/2)/(-d*x+1),x)
 

Output:

 - int(((e + f*x)**n*sqrt( - d**2*x**2 + 1))/(d*x - 1),x)