\(\int (1+x) (a+b x)^m (1-x^2)^{\frac {1}{2} (-5-m)} \, dx\) [12]

Optimal result

Integrand size = 28, antiderivative size = 244 \[ \int (1+x) (a+b x)^m \left (1-x^2\right )^{\frac {1}{2} (-5-m)} \, dx=\frac {(1-x)^{\frac {1}{2} (-3-m)} (1+x)^{\frac {1}{2} (-1-m)} (a+b x)^{1+m}}{(a+b) (3+m)}-\frac {b (a+2 b+a m) (a+b x)^{1+m} \left (1-x^2\right )^{\frac {1}{2} (-1-m)}}{(a-b) (a+b)^2 (1+m) (3+m)}+\frac {\left (2 a b-b^2+a^2 (2+m)\right ) (1-x) \left (-\frac {(a+b) (1+x)}{(a-b) (1-x)}\right )^{\frac {3+m}{2}} (a+b x)^{2+m} \left (1-x^2\right )^{\frac {1}{2} (-3-m)} \operatorname {Hypergeometric2F1}\left (2+m,\frac {3+m}{2},3+m,\frac {2 (a+b x)}{(a-b) (1-x)}\right )}{(a-b) (a+b)^3 (2+m) (3+m)} \] 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 [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.50 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.36, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {689, 25, 27, 689, 25, 679, 489}

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.

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

\(\Big \downarrow \) 689

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 689

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 679

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

\(\Big \downarrow \) 489

\(\displaystyle \frac {(a-b) \left (\frac {\frac {\left (1-x^2\right )^{\frac {1}{2} (-m-3)} \left (2 a^2-a b (m+1)^2-2 b^2 (m+2)\right ) (a+b x)^{m+3}}{(m+3) \left (a^2-b^2\right )}-\frac {(m+1) (x+1) \left (1-x^2\right )^{\frac {1}{2} (-m-5)} \left (a^2 (m+2)+2 a b-b^2\right ) \left (-\frac {(1-x) (a-b)}{(x+1) (a+b)}\right )^{\frac {m+5}{2}} (a+b x)^{m+4} \operatorname {Hypergeometric2F1}\left (m+4,\frac {m+5}{2},m+5,\frac {2 (a+b x)}{(a+b) (x+1)}\right )}{(m+4) (a-b) \left (a^2-b^2\right )}}{(m+2) \left (a^2-b^2\right )}-\frac {\left (1-x^2\right )^{\frac {1}{2} (-m-3)} (2 a+b m+b) (a+b x)^{m+2}}{(m+2) \left (a^2-b^2\right )}\right )}{(m+1) \left (a^2-b^2\right )}+\frac {\left (1-x^2\right )^{\frac {1}{2} (-m-3)} (a+b x)^{m+1}}{(m+1) (a+b)}\)

Input:

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

Output:

((a + b*x)^(1 + m)*(1 - x^2)^((-3 - m)/2))/((a + b)*(1 + m)) + ((a - b)*(- 
(((2*a + b + b*m)*(a + b*x)^(2 + m)*(1 - x^2)^((-3 - m)/2))/((a^2 - b^2)*( 
2 + m))) + (((2*a^2 - a*b*(1 + m)^2 - 2*b^2*(2 + m))*(a + b*x)^(3 + m)*(1 
- x^2)^((-3 - m)/2))/((a^2 - b^2)*(3 + m)) - ((1 + m)*(2*a*b - b^2 + a^2*( 
2 + m))*(-(((a - b)*(1 - x))/((a + b)*(1 + x))))^((5 + m)/2)*(1 + x)*(a + 
b*x)^(4 + m)*(1 - x^2)^((-5 - m)/2)*Hypergeometric2F1[4 + m, (5 + m)/2, 5 
+ m, (2*(a + b*x))/((a + b)*(1 + x))])/((a - b)*(a^2 - b^2)*(4 + m)))/((a^ 
2 - b^2)*(2 + m))))/((a^2 - b^2)*(1 + 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 489

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Rt[(-a)*b, 2]}, Simp[(q - b*x)*(c + d*x)^(n + 1)*((a + b*x^2)^p/((n + 
1)*(b*c + d*q)*((b*c + d*q)*((q + b*x)/((b*c - d*q)*(-q + b*x))))^p))*Hyper 
geometric2F1[n + 1, -p, n + 2, 2*b*q*((c + d*x)/((b*c - d*q)*(q - b*x)))], 
x]] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[n + 2*p + 2, 0]
 

rule 679

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1 
)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[(c*d*f + a*e*g)/(c*d^2 + a*e^2) 
 Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, 
 p}, x] && EqQ[Simplify[m + 2*p + 3], 0]
 

rule 689

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/( 
(m + 1)*(c*d^2 + a*e^2))), x] + Simp[1/((m + 1)*(c*d^2 + a*e^2))   Int[(d + 
 e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m 
 + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && ILtQ[Sim 
plify[m + 2*p + 3], 0] && NeQ[m, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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