\(\int (1-x)^{\frac {1}{2} (-5-m)} (1+x)^{\frac {1}{2} (-3-m)} (a+b x)^m \, dx\) [1833]

Optimal result

Integrand size = 36, antiderivative size = 244 \[ \int (1-x)^{\frac {1}{2} (-5-m)} (1+x)^{\frac {1}{2} (-3-m)} (a+b x)^m \, dx=\frac {(1-x)^{\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-1-m)} (a+b x)^{1+m}}{(a+b) (3+m)}-\frac {b (a+2 b+a m) (a+b x)^{1+m} \left (1-x^2\right )^{\frac {1}{2} (-1-m)}}{(a-b) (a+b)^2 (1+m) (3+m)}+\frac {\left (2 a b-b^2+a^2 (2+m)\right ) (1-x) \left (-\frac {(a+b) (1+x)}{(a-b) (1-x)}\right )^{\frac {3+m}{2}} (a+b x)^{2+m} \left (1-x^2\right )^{\frac {1}{2} (-3-m)} \operatorname {Hypergeometric2F1}\left (2+m,\frac {3+m}{2},3+m,\frac {2 (a+b x)}{(a-b) (1-x)}\right )}{(a-b) (a+b)^3 (2+m) (3+m)} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.47 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.77 \[ \int (1-x)^{\frac {1}{2} (-5-m)} (1+x)^{\frac {1}{2} (-3-m)} (a+b x)^m \, dx=\frac {(a+b x)^{1+m} \left (1-x^2\right )^{-\frac {3}{2}-\frac {m}{2}} \left (4 (1+x)-\frac {4 (3 b+a (2+m)) (-1+x) (1+x)}{(a+b) (1+m)}+\frac {2^{\frac {3}{2}-\frac {m}{2}} \left (2 a b-b^2+a^2 (2+m)\right ) \left (\frac {(a+b) (1+x)}{a+b x}\right )^{\frac {1}{2} (-1+m)} \left (-1+x^2\right )^2 \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {3+m}{2},\frac {3-m}{2},-\frac {(a-b) (-1+x)}{2 (a+b x)}\right )}{(-1+m) (a+b x)^2}\right )}{4 (a+b) (3+m)} \] Input:

Integrate[(1 - x)^((-5 - m)/2)*(1 + x)^((-3 - m)/2)*(a + b*x)^m,x]
 

Output:

((a + b*x)^(1 + m)*(1 - x^2)^(-3/2 - m/2)*(4*(1 + x) - (4*(3*b + a*(2 + m) 
)*(-1 + x)*(1 + x))/((a + b)*(1 + m)) + (2^(3/2 - m/2)*(2*a*b - b^2 + a^2* 
(2 + m))*(((a + b)*(1 + x))/(a + b*x))^((-1 + m)/2)*(-1 + x^2)^2*Hypergeom 
etric2F1[(1 - m)/2, (3 + m)/2, (3 - m)/2, -1/2*((a - b)*(-1 + x))/(a + b*x 
)])/((-1 + m)*(a + b*x)^2)))/(4*(a + b)*(3 + m))
 

Rubi [A] (warning: unable to verify)

Time = 0.36 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {144, 25, 172, 25, 27, 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.

\(\displaystyle \int (1-x)^{\frac {1}{2} (-m-5)} (x+1)^{\frac {1}{2} (-m-3)} (a+b x)^m \, dx\)

\(\Big \downarrow \) 144

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 172

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

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int (m+1) \left ((m+2) a^2+2 b a-b^2\right ) (1-x)^{\frac {1}{2} (-m-1)} (x+1)^{\frac {1}{2} (-m-3)} (a+b x)^mdx}{(m+1) (a+b)}+\frac {(1-x)^{\frac {1}{2} (-m-1)} (x+1)^{\frac {1}{2} (-m-1)} (a (m+2)+3 b) (a+b x)^{m+1}}{(m+1) (a+b)}}{(m+3) (a+b)}+\frac {(x+1)^{\frac {1}{2} (-m-1)} (1-x)^{\frac {1}{2} (-m-3)} (a+b x)^{m+1}}{(m+3) (a+b)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\left (a^2 (m+2)+2 a b-b^2\right ) \int (1-x)^{\frac {1}{2} (-m-1)} (x+1)^{\frac {1}{2} (-m-3)} (a+b x)^mdx}{a+b}+\frac {(1-x)^{\frac {1}{2} (-m-1)} (x+1)^{\frac {1}{2} (-m-1)} (a (m+2)+3 b) (a+b x)^{m+1}}{(m+1) (a+b)}}{(m+3) (a+b)}+\frac {(x+1)^{\frac {1}{2} (-m-1)} (1-x)^{\frac {1}{2} (-m-3)} (a+b x)^{m+1}}{(m+3) (a+b)}\)

\(\Big \downarrow \) 142

\(\displaystyle \frac {\frac {(1-x)^{\frac {1}{2} (-m-1)} (x+1)^{\frac {1}{2} (-m-1)} (a (m+2)+3 b) (a+b x)^{m+1}}{(m+1) (a+b)}-\frac {2^{-\frac {m}{2}-\frac {1}{2}} (1-x)^{\frac {1-m}{2}} (x+1)^{\frac {1}{2} (-m-3)} \left (a^2 (m+2)+2 a b-b^2\right ) \left (\frac {(x+1) (a+b)}{a+b x}\right )^{\frac {m+3}{2}} (a+b x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {m+3}{2},\frac {3-m}{2},\frac {(a-b) (1-x)}{2 (a+b x)}\right )}{(1-m) (a+b)^2}}{(m+3) (a+b)}+\frac {(x+1)^{\frac {1}{2} (-m-1)} (1-x)^{\frac {1}{2} (-m-3)} (a+b x)^{m+1}}{(m+3) (a+b)}\)

Input:

Int[(1 - x)^((-5 - m)/2)*(1 + x)^((-3 - m)/2)*(a + b*x)^m,x]
 

Output:

((1 - x)^((-3 - m)/2)*(1 + x)^((-1 - m)/2)*(a + b*x)^(1 + m))/((a + b)*(3 
+ m)) + (((3*b + a*(2 + m))*(1 - x)^((-1 - m)/2)*(1 + x)^((-1 - m)/2)*(a + 
 b*x)^(1 + m))/((a + b)*(1 + m)) - (2^(-1/2 - m/2)*(2*a*b - b^2 + a^2*(2 + 
 m))*(1 - x)^((1 - m)/2)*(1 + x)^((-3 - m)/2)*(((a + b)*(1 + x))/(a + b*x) 
)^((3 + m)/2)*(a + b*x)^(1 + m)*Hypergeometric2F1[(1 - m)/2, (3 + m)/2, (3 
 - m)/2, ((a - b)*(1 - x))/(2*(a + b*x))])/((a + b)^2*(1 - m)))/((a + b)*( 
3 + m))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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]
 

rule 144

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> With[{mnp = Simplify[m + n + p]}, Simp[b*(a + b*x)^(m + 1)*( 
c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] 
+ Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x 
)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) 
- b*d*f*(mnp + 3)*x, x], x], x] /; ILtQ[mnp + 2, 0] && (SumSimplerQ[m, 1] | 
| ( !SumSimplerQ[n, 1] &&  !SumSimplerQ[p, 1]))] /; FreeQ[{a, b, c, d, e, f 
, m, n, p}, x] && NeQ[m, -1]
 

rule 172

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> With[{mnp = Simplify[m + n + p]}, Simp[ 
(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1) 
*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) 
 Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f 
)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g 
 - a*h)*(mnp + 3)*x, x], x], x] /; ILtQ[mnp + 2, 0] && (SumSimplerQ[m, 1] | 
| ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 1 
])))] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && NeQ[m, -1]
 

Maple [F]

\[\int \left (1-x \right )^{-\frac {5}{2}-\frac {m}{2}} \left (1+x \right )^{-\frac {3}{2}-\frac {m}{2}} \left (b x +a \right )^{m}d x\]

Input:

int((1-x)^(-5/2-1/2*m)*(1+x)^(-3/2-1/2*m)*(b*x+a)^m,x)
 

Output:

int((1-x)^(-5/2-1/2*m)*(1+x)^(-3/2-1/2*m)*(b*x+a)^m,x)
 

Fricas [F]

\[ \int (1-x)^{\frac {1}{2} (-5-m)} (1+x)^{\frac {1}{2} (-3-m)} (a+b x)^m \, dx=\int { {\left (b x + a\right )}^{m} {\left (x + 1\right )}^{-\frac {1}{2} \, m - \frac {3}{2}} {\left (-x + 1\right )}^{-\frac {1}{2} \, m - \frac {5}{2}} \,d x } \] Input:

integrate((1-x)^(-5/2-1/2*m)*(1+x)^(-3/2-1/2*m)*(b*x+a)^m,x, algorithm="fr 
icas")
 

Output:

integral((b*x + a)^m*(x + 1)^(-1/2*m - 3/2)*(-x + 1)^(-1/2*m - 5/2), x)
 

Sympy [F]

\[ \int (1-x)^{\frac {1}{2} (-5-m)} (1+x)^{\frac {1}{2} (-3-m)} (a+b x)^m \, dx=\int \left (1 - x\right )^{- \frac {m}{2} - \frac {5}{2}} \left (a + b x\right )^{m} \left (x + 1\right )^{- \frac {m}{2} - \frac {3}{2}}\, dx \] Input:

integrate((1-x)**(-5/2-1/2*m)*(1+x)**(-3/2-1/2*m)*(b*x+a)**m,x)
 

Output:

Integral((1 - x)**(-m/2 - 5/2)*(a + b*x)**m*(x + 1)**(-m/2 - 3/2), x)
 

Maxima [F]

\[ \int (1-x)^{\frac {1}{2} (-5-m)} (1+x)^{\frac {1}{2} (-3-m)} (a+b x)^m \, dx=\int { {\left (b x + a\right )}^{m} {\left (x + 1\right )}^{-\frac {1}{2} \, m - \frac {3}{2}} {\left (-x + 1\right )}^{-\frac {1}{2} \, m - \frac {5}{2}} \,d x } \] Input:

integrate((1-x)^(-5/2-1/2*m)*(1+x)^(-3/2-1/2*m)*(b*x+a)^m,x, algorithm="ma 
xima")
 

Output:

integrate((b*x + a)^m*(x + 1)^(-1/2*m - 3/2)*(-x + 1)^(-1/2*m - 5/2), x)
 

Giac [F]

\[ \int (1-x)^{\frac {1}{2} (-5-m)} (1+x)^{\frac {1}{2} (-3-m)} (a+b x)^m \, dx=\int { {\left (b x + a\right )}^{m} {\left (x + 1\right )}^{-\frac {1}{2} \, m - \frac {3}{2}} {\left (-x + 1\right )}^{-\frac {1}{2} \, m - \frac {5}{2}} \,d x } \] Input:

integrate((1-x)^(-5/2-1/2*m)*(1+x)^(-3/2-1/2*m)*(b*x+a)^m,x, algorithm="gi 
ac")
 

Output:

integrate((b*x + a)^m*(x + 1)^(-1/2*m - 3/2)*(-x + 1)^(-1/2*m - 5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (1-x)^{\frac {1}{2} (-5-m)} (1+x)^{\frac {1}{2} (-3-m)} (a+b x)^m \, dx=\int \frac {{\left (a+b\,x\right )}^m}{{\left (1-x\right )}^{\frac {m}{2}+\frac {5}{2}}\,{\left (x+1\right )}^{\frac {m}{2}+\frac {3}{2}}} \,d x \] Input:

int((a + b*x)^m/((1 - x)^(m/2 + 5/2)*(x + 1)^(m/2 + 3/2)),x)
 

Output:

int((a + b*x)^m/((1 - x)^(m/2 + 5/2)*(x + 1)^(m/2 + 3/2)), x)
 

Reduce [F]

\[ \int (1-x)^{\frac {1}{2} (-5-m)} (1+x)^{\frac {1}{2} (-3-m)} (a+b x)^m \, dx=\int \frac {\left (b x +a \right )^{m}}{\left (x +1\right )^{\frac {m}{2}+\frac {1}{2}} \left (1-x \right )^{\frac {m}{2}+\frac {1}{2}} x^{3}-\left (x +1\right )^{\frac {m}{2}+\frac {1}{2}} \left (1-x \right )^{\frac {m}{2}+\frac {1}{2}} x^{2}-\left (x +1\right )^{\frac {m}{2}+\frac {1}{2}} \left (1-x \right )^{\frac {m}{2}+\frac {1}{2}} x +\left (x +1\right )^{\frac {m}{2}+\frac {1}{2}} \left (1-x \right )^{\frac {m}{2}+\frac {1}{2}}}d x \] Input:

int((1-x)^(-5/2-1/2*m)*(1+x)^(-3/2-1/2*m)*(b*x+a)^m,x)
 

Output:

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