\(\int \frac {(1+d x)^{3/2} (e+f x)^n}{\sqrt {1-d x}} \, dx\) [128]

Optimal result

Integrand size = 27, antiderivative size = 83 \[ \int \frac {(1+d x)^{3/2} (e+f x)^n}{\sqrt {1-d x}} \, dx=-\frac {4 \sqrt {2} \sqrt {1-d x} (e+f x)^n \left (\frac {d (e+f x)}{d e+f}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{2},-n,\frac {3}{2},\frac {1}{2} (1-d x),\frac {f (1-d x)}{d e+f}\right )}{d} \] Output:

-4*2^(1/2)*(-d*x+1)^(1/2)*(f*x+e)^n*AppellF1(1/2,-n,-3/2,3/2,f*(-d*x+1)/(d 
*e+f),-1/2*d*x+1/2)/d/((d*(f*x+e)/(d*e+f))^n)
 

Mathematica [A] (verified)

Time = 1.40 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.53 \[ \int \frac {(1+d x)^{3/2} (e+f x)^n}{\sqrt {1-d x}} \, dx=-\frac {2 \sqrt {2-2 d x} (e+f x)^n \left (\frac {d (e+f x)}{d e+f}\right )^{-n} \left (6 \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-n,\frac {3}{2},\frac {1}{2}-\frac {d x}{2},\frac {f-d f x}{d e+f}\right )+(-1+d x) \operatorname {AppellF1}\left (\frac {3}{2},-\frac {1}{2},-n,\frac {5}{2},\frac {1}{2}-\frac {d x}{2},\frac {f-d f x}{d e+f}\right )\right )}{3 d} \] Input:

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

Output:

(-2*Sqrt[2 - 2*d*x]*(e + f*x)^n*(6*AppellF1[1/2, -1/2, -n, 3/2, 1/2 - (d*x 
)/2, (f - d*f*x)/(d*e + f)] + (-1 + d*x)*AppellF1[3/2, -1/2, -n, 5/2, 1/2 
- (d*x)/2, (f - d*f*x)/(d*e + f)]))/(3*d*((d*(e + f*x))/(d*e + f))^n)
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {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.

\(\displaystyle \int \frac {(d x+1)^{3/2} (e+f x)^n}{\sqrt {1-d x}} \, dx\)

\(\Big \downarrow \) 156

\(\displaystyle (e+f x)^n \left (\frac {d (e+f x)}{d e+f}\right )^{-n} \int \frac {(d x+1)^{3/2} \left (\frac {d e}{d e+f}+\frac {d f x}{d e+f}\right )^n}{\sqrt {1-d x}}dx\)

\(\Big \downarrow \) 155

\(\displaystyle -\frac {4 \sqrt {2} \sqrt {1-d x} (e+f x)^n \left (\frac {d (e+f x)}{d e+f}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{2},-n,\frac {3}{2},\frac {1}{2} (1-d x),\frac {f (1-d x)}{d e+f}\right )}{d}\)

Input:

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

Output:

(-4*Sqrt[2]*Sqrt[1 - d*x]*(e + f*x)^n*AppellF1[1/2, -3/2, -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]
 

Maple [F]

\[\int \frac {\left (d x +1\right )^{\frac {3}{2}} \left (f x +e \right )^{n}}{\sqrt {-d x +1}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {(1+d x)^{3/2} (e+f x)^n}{\sqrt {1-d x}} \, dx=\int { \frac {{\left (d x + 1\right )}^{\frac {3}{2}} {\left (f x + e\right )}^{n}}{\sqrt {-d x + 1}} \,d x } \] Input:

integrate((d*x+1)^(3/2)*(f*x+e)^n/(-d*x+1)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {(1+d x)^{3/2} (e+f x)^n}{\sqrt {1-d x}} \, dx=\int \frac {\left (e + f x\right )^{n} \left (d x + 1\right )^{\frac {3}{2}}}{\sqrt {- d x + 1}}\, dx \] Input:

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

Output:

Integral((e + f*x)**n*(d*x + 1)**(3/2)/sqrt(-d*x + 1), x)
 

Maxima [F]

\[ \int \frac {(1+d x)^{3/2} (e+f x)^n}{\sqrt {1-d x}} \, dx=\int { \frac {{\left (d x + 1\right )}^{\frac {3}{2}} {\left (f x + e\right )}^{n}}{\sqrt {-d x + 1}} \,d x } \] Input:

integrate((d*x+1)^(3/2)*(f*x+e)^n/(-d*x+1)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {(1+d x)^{3/2} (e+f x)^n}{\sqrt {1-d x}} \, dx=\int { \frac {{\left (d x + 1\right )}^{\frac {3}{2}} {\left (f x + e\right )}^{n}}{\sqrt {-d x + 1}} \,d x } \] Input:

integrate((d*x+1)^(3/2)*(f*x+e)^n/(-d*x+1)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(1+d x)^{3/2} (e+f x)^n}{\sqrt {1-d x}} \, dx=\int \frac {{\left (e+f\,x\right )}^n\,{\left (d\,x+1\right )}^{3/2}}{\sqrt {1-d\,x}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(1+d x)^{3/2} (e+f x)^n}{\sqrt {1-d x}} \, dx=\left (\int \frac {\left (f x +e \right )^{n} \sqrt {d x +1}\, x}{\sqrt {-d x +1}}d x \right ) d +\int \frac {\left (f x +e \right )^{n} \sqrt {d x +1}}{\sqrt {-d x +1}}d x \] Input:

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

Output:

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