\(\int \frac {(a+b x)^m \sqrt {2+d x}}{\sqrt {2-d x}} \, dx\) [129]

Optimal result

Integrand size = 27, antiderivative size = 82 \[ \int \frac {(a+b x)^m \sqrt {2+d x}}{\sqrt {2-d x}} \, dx=-\frac {4 (a+b x)^m \left (\frac {d (a+b x)}{2 b+a d}\right )^{-m} \sqrt {2-d x} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-m,\frac {3}{2},\frac {1}{4} (2-d x),\frac {b (2-d x)}{2 b+a d}\right )}{d} \] Output:

-4*(b*x+a)^m*(-d*x+2)^(1/2)*AppellF1(1/2,-m,-1/2,3/2,b*(-d*x+2)/(a*d+2*b), 
-1/4*d*x+1/2)/d/((d*(b*x+a)/(a*d+2*b))^m)
 

Mathematica [A] (verified)

Time = 2.36 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.51 \[ \int \frac {(a+b x)^m \sqrt {2+d x}}{\sqrt {2-d x}} \, dx=\frac {(a+b x)^{1+m} \sqrt {-\frac {b (-2+d x)}{2 b+a d}} \sqrt {2+d x} \operatorname {AppellF1}\left (1+m,\frac {1}{2},-\frac {1}{2},2+m,\frac {d (a+b x)}{2 b+a d},\frac {d (a+b x)}{-2 b+a d}\right )}{b (1+m) \sqrt {2-d x} \sqrt {\frac {b (2+d x)}{2 b-a d}}} \] Input:

Integrate[((a + b*x)^m*Sqrt[2 + d*x])/Sqrt[2 - d*x],x]
 

Output:

((a + b*x)^(1 + m)*Sqrt[-((b*(-2 + d*x))/(2*b + a*d))]*Sqrt[2 + d*x]*Appel 
lF1[1 + m, 1/2, -1/2, 2 + m, (d*(a + b*x))/(2*b + a*d), (d*(a + b*x))/(-2* 
b + a*d)])/(b*(1 + m)*Sqrt[2 - d*x]*Sqrt[(b*(2 + d*x))/(2*b - a*d)])
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 82, 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.

\(\displaystyle \int \frac {\sqrt {d x+2} (a+b x)^m}{\sqrt {2-d x}} \, dx\)

\(\Big \downarrow \) 156

\(\displaystyle (a+b x)^m \left (\frac {d (a+b x)}{a d+2 b}\right )^{-m} \int \frac {\sqrt {d x+2} \left (\frac {b x d}{2 b+a d}+\frac {a d}{2 b+a d}\right )^m}{\sqrt {2-d x}}dx\)

\(\Big \downarrow \) 155

\(\displaystyle -\frac {4 \sqrt {2-d x} (a+b x)^m \left (\frac {d (a+b x)}{a d+2 b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-m,\frac {3}{2},\frac {1}{4} (2-d x),\frac {b (2-d x)}{2 b+a d}\right )}{d}\)

Input:

Int[((a + b*x)^m*Sqrt[2 + d*x])/Sqrt[2 - d*x],x]
 

Output:

(-4*(a + b*x)^m*Sqrt[2 - d*x]*AppellF1[1/2, -1/2, -m, 3/2, (2 - d*x)/4, (b 
*(2 - d*x))/(2*b + a*d)])/(d*((d*(a + b*x))/(2*b + a*d))^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]
 

Maple [F]

\[\int \frac {\left (b x +a \right )^{m} \sqrt {d x +2}}{\sqrt {-d x +2}}d x\]

Input:

int((b*x+a)^m*(d*x+2)^(1/2)/(-d*x+2)^(1/2),x)
 

Output:

int((b*x+a)^m*(d*x+2)^(1/2)/(-d*x+2)^(1/2),x)
 

Fricas [F]

\[ \int \frac {(a+b x)^m \sqrt {2+d x}}{\sqrt {2-d x}} \, dx=\int { \frac {\sqrt {d x + 2} {\left (b x + a\right )}^{m}}{\sqrt {-d x + 2}} \,d x } \] Input:

integrate((b*x+a)^m*(d*x+2)^(1/2)/(-d*x+2)^(1/2),x, algorithm="fricas")
 

Output:

integral(-sqrt(d*x + 2)*sqrt(-d*x + 2)*(b*x + a)^m/(d*x - 2), x)
 

Sympy [F]

\[ \int \frac {(a+b x)^m \sqrt {2+d x}}{\sqrt {2-d x}} \, dx=\int \frac {\left (a + b x\right )^{m} \sqrt {d x + 2}}{\sqrt {- d x + 2}}\, dx \] Input:

integrate((b*x+a)**m*(d*x+2)**(1/2)/(-d*x+2)**(1/2),x)
 

Output:

Integral((a + b*x)**m*sqrt(d*x + 2)/sqrt(-d*x + 2), x)
 

Maxima [F]

\[ \int \frac {(a+b x)^m \sqrt {2+d x}}{\sqrt {2-d x}} \, dx=\int { \frac {\sqrt {d x + 2} {\left (b x + a\right )}^{m}}{\sqrt {-d x + 2}} \,d x } \] Input:

integrate((b*x+a)^m*(d*x+2)^(1/2)/(-d*x+2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(d*x + 2)*(b*x + a)^m/sqrt(-d*x + 2), x)
 

Giac [F]

\[ \int \frac {(a+b x)^m \sqrt {2+d x}}{\sqrt {2-d x}} \, dx=\int { \frac {\sqrt {d x + 2} {\left (b x + a\right )}^{m}}{\sqrt {-d x + 2}} \,d x } \] Input:

integrate((b*x+a)^m*(d*x+2)^(1/2)/(-d*x+2)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(d*x + 2)*(b*x + a)^m/sqrt(-d*x + 2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^m \sqrt {2+d x}}{\sqrt {2-d x}} \, dx=\int \frac {\sqrt {d\,x+2}\,{\left (a+b\,x\right )}^m}{\sqrt {2-d\,x}} \,d x \] Input:

int(((d*x + 2)^(1/2)*(a + b*x)^m)/(2 - d*x)^(1/2),x)
 

Output:

int(((d*x + 2)^(1/2)*(a + b*x)^m)/(2 - d*x)^(1/2), x)
 

Reduce [F]

\[ \int \frac {(a+b x)^m \sqrt {2+d x}}{\sqrt {2-d x}} \, dx=\int \frac {\left (b x +a \right )^{m} \sqrt {d x +2}}{\sqrt {-d x +2}}d x \] Input:

int((b*x+a)^m*(d*x+2)^(1/2)/(-d*x+2)^(1/2),x)
 

Output:

int(((a + b*x)**m*sqrt(d*x + 2))/sqrt( - d*x + 2),x)