\(\int (1-x)^{\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-1-m)} (a+b x)^m \, dx\) [1807]

Optimal result

Integrand size = 36, antiderivative size = 128 \[ \int (1-x)^{\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-1-m)} (a+b x)^m \, dx=\frac {2^{\frac {1}{2}-\frac {m}{2}} (1-x)^{\frac {1}{2} (-1-m)} (1+x)^{\frac {1}{2} (-1-m)} \left (\frac {(a+b) (1+x)}{a+b x}\right )^{\frac {1+m}{2}} (a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {1+m}{2},\frac {1-m}{2},\frac {(a-b) (1-x)}{2 (a+b x)}\right )}{(a+b) (1+m)} \] Output:

2^(1/2-1/2*m)*(1-x)^(-1/2-1/2*m)*(1+x)^(-1/2-1/2*m)*((a+b)*(1+x)/(b*x+a))^ 
(1/2+1/2*m)*(b*x+a)^(1+m)*hypergeom([-1/2-1/2*m, 1/2+1/2*m],[1/2-1/2*m],(a 
-b)*(1-x)/(2*b*x+2*a))/(a+b)/(1+m)
 

Mathematica [A] (verified)

Time = 10.06 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.90 \[ \int (1-x)^{\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-1-m)} (a+b x)^m \, dx=\frac {2^{\frac {1}{2}-\frac {m}{2}} \left (\frac {(a+b) (1+x)}{a+b x}\right )^{\frac {1+m}{2}} (a+b x)^{1+m} \left (1-x^2\right )^{\frac {1}{2} (-1-m)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {1+m}{2},\frac {1-m}{2},-\frac {(a-b) (-1+x)}{2 (a+b x)}\right )}{(a+b) (1+m)} \] Input:

Integrate[(1 - x)^((-3 - m)/2)*(1 + x)^((-1 - m)/2)*(a + b*x)^m,x]
 

Output:

(2^(1/2 - m/2)*(((a + b)*(1 + x))/(a + b*x))^((1 + m)/2)*(a + b*x)^(1 + m) 
*(1 - x^2)^((-1 - m)/2)*Hypergeometric2F1[(-1 - m)/2, (1 + m)/2, (1 - m)/2 
, -1/2*((a - b)*(-1 + x))/(a + b*x)])/((a + b)*(1 + m))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {142}

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.

\(\displaystyle \int (1-x)^{\frac {1}{2} (-m-3)} (x+1)^{\frac {1}{2} (-m-1)} (a+b x)^m \, dx\)

\(\Big \downarrow \) 142

\(\displaystyle \frac {2^{\frac {1}{2}-\frac {m}{2}} (1-x)^{\frac {1}{2} (-m-1)} (x+1)^{\frac {1}{2} (-m-1)} \left (\frac {(x+1) (a+b)}{a+b x}\right )^{\frac {m+1}{2}} (a+b x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-m-1),\frac {m+1}{2},\frac {1-m}{2},\frac {(a-b) (1-x)}{2 (a+b x)}\right )}{(m+1) (a+b)}\)

Input:

Int[(1 - x)^((-3 - m)/2)*(1 + x)^((-1 - m)/2)*(a + b*x)^m,x]
 

Output:

(2^(1/2 - m/2)*(1 - x)^((-1 - m)/2)*(1 + x)^((-1 - m)/2)*(((a + b)*(1 + x) 
)/(a + b*x))^((1 + m)/2)*(a + b*x)^(1 + m)*Hypergeometric2F1[(-1 - m)/2, ( 
1 + m)/2, (1 - m)/2, ((a - b)*(1 - x))/(2*(a + b*x))])/((a + b)*(1 + m))
 

Defintions of rubi rules used

rule 142

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((b*e 
 - a*f)*(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f))*((a + 
b*x)/((b*c - a*d)*(e + f*x)))])/((b*e - a*f)*((c + d*x)/((b*c - a*d)*(e + f 
*x))))^n, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 
 0] &&  !IntegerQ[n]
 

Maple [F]

\[\int \left (1-x \right )^{-\frac {3}{2}-\frac {m}{2}} \left (1+x \right )^{-\frac {1}{2}-\frac {m}{2}} \left (b x +a \right )^{m}d x\]

Input:

int((1-x)^(-3/2-1/2*m)*(1+x)^(-1/2-1/2*m)*(b*x+a)^m,x)
 

Output:

int((1-x)^(-3/2-1/2*m)*(1+x)^(-1/2-1/2*m)*(b*x+a)^m,x)
 

Fricas [F]

\[ \int (1-x)^{\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-1-m)} (a+b x)^m \, dx=\int { {\left (b x + a\right )}^{m} {\left (x + 1\right )}^{-\frac {1}{2} \, m - \frac {1}{2}} {\left (-x + 1\right )}^{-\frac {1}{2} \, m - \frac {3}{2}} \,d x } \] Input:

integrate((1-x)^(-3/2-1/2*m)*(1+x)^(-1/2-1/2*m)*(b*x+a)^m,x, algorithm="fr 
icas")
 

Output:

integral((b*x + a)^m*(x + 1)^(-1/2*m - 1/2)*(-x + 1)^(-1/2*m - 3/2), x)
 

Sympy [F]

\[ \int (1-x)^{\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-1-m)} (a+b x)^m \, dx=\int \left (1 - x\right )^{- \frac {m}{2} - \frac {3}{2}} \left (a + b x\right )^{m} \left (x + 1\right )^{- \frac {m}{2} - \frac {1}{2}}\, dx \] Input:

integrate((1-x)**(-3/2-1/2*m)*(1+x)**(-1/2-1/2*m)*(b*x+a)**m,x)
 

Output:

Integral((1 - x)**(-m/2 - 3/2)*(a + b*x)**m*(x + 1)**(-m/2 - 1/2), x)
 

Maxima [F]

\[ \int (1-x)^{\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-1-m)} (a+b x)^m \, dx=\int { {\left (b x + a\right )}^{m} {\left (x + 1\right )}^{-\frac {1}{2} \, m - \frac {1}{2}} {\left (-x + 1\right )}^{-\frac {1}{2} \, m - \frac {3}{2}} \,d x } \] Input:

integrate((1-x)^(-3/2-1/2*m)*(1+x)^(-1/2-1/2*m)*(b*x+a)^m,x, algorithm="ma 
xima")
 

Output:

integrate((b*x + a)^m*(x + 1)^(-1/2*m - 1/2)*(-x + 1)^(-1/2*m - 3/2), x)
 

Giac [F]

\[ \int (1-x)^{\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-1-m)} (a+b x)^m \, dx=\int { {\left (b x + a\right )}^{m} {\left (x + 1\right )}^{-\frac {1}{2} \, m - \frac {1}{2}} {\left (-x + 1\right )}^{-\frac {1}{2} \, m - \frac {3}{2}} \,d x } \] Input:

integrate((1-x)^(-3/2-1/2*m)*(1+x)^(-1/2-1/2*m)*(b*x+a)^m,x, algorithm="gi 
ac")
 

Output:

integrate((b*x + a)^m*(x + 1)^(-1/2*m - 1/2)*(-x + 1)^(-1/2*m - 3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (1-x)^{\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-1-m)} (a+b x)^m \, dx=\int \frac {{\left (a+b\,x\right )}^m}{{\left (1-x\right )}^{\frac {m}{2}+\frac {3}{2}}\,{\left (x+1\right )}^{\frac {m}{2}+\frac {1}{2}}} \,d x \] Input:

int((a + b*x)^m/((1 - x)^(m/2 + 3/2)*(x + 1)^(m/2 + 1/2)),x)
 

Output:

int((a + b*x)^m/((1 - x)^(m/2 + 3/2)*(x + 1)^(m/2 + 1/2)), x)
 

Reduce [F]

\[ \int (1-x)^{\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-1-m)} (a+b x)^m \, dx=-\left (\int \frac {\left (b x +a \right )^{m}}{\left (x +1\right )^{\frac {m}{2}+\frac {1}{2}} \left (1-x \right )^{\frac {m}{2}+\frac {1}{2}} x -\left (x +1\right )^{\frac {m}{2}+\frac {1}{2}} \left (1-x \right )^{\frac {m}{2}+\frac {1}{2}}}d x \right ) \] Input:

int((1-x)^(-3/2-1/2*m)*(1+x)^(-1/2-1/2*m)*(b*x+a)^m,x)
 

Output:

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