\(\int \frac {(c+d x)^{-3-2 p} (1-x^2)^{1+p}}{1-x} \, dx\) [1806]

Optimal result

Integrand size = 30, antiderivative size = 92 \[ \int \frac {(c+d x)^{-3-2 p} \left (1-x^2\right )^{1+p}}{1-x} \, dx=-\frac {2^{1+p} \left (\frac {(c+d) (1+x)}{c+d x}\right )^{-1-p} (c+d x)^{-2 (1+p)} \left (1-x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-1-p,1+p,2+p,\frac {(c-d) (1-x)}{2 (c+d x)}\right )}{(c+d) (1+p)} \] Output:

-2^(p+1)*((c+d)*(1+x)/(d*x+c))^(-1-p)*(-x^2+1)^(p+1)*hypergeom([p+1, -1-p] 
,[2+p],(c-d)*(1-x)/(2*d*x+2*c))/(c+d)/(p+1)/((d*x+c)^(2*p+2))
 

Mathematica [F]

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

Integrate[((c + d*x)^(-3 - 2*p)*(1 - x^2)^(1 + p))/(1 - x),x]
 

Output:

Integrate[((c + d*x)^(-3 - 2*p)*(1 - x^2)^(1 + p))/(1 - x), x]
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {717, 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.

\(\displaystyle \int \frac {\left (1-x^2\right )^{p+1} (c+d x)^{-2 p-3}}{1-x} \, dx\)

\(\Big \downarrow \) 717

\(\displaystyle \int (1-x)^p (x+1)^{p+1} (c+d x)^{-2 p-3}dx\)

\(\Big \downarrow \) 142

\(\displaystyle -\frac {2^{p+1} (1-x)^{p+1} (x+1)^{p+1} \left (\frac {(x+1) (c+d)}{c+d x}\right )^{-p-1} (c+d x)^{-2 (p+1)} \operatorname {Hypergeometric2F1}\left (-p-1,p+1,p+2,\frac {(c-d) (1-x)}{2 (c+d x)}\right )}{(p+1) (c+d)}\)

Input:

Int[((c + d*x)^(-3 - 2*p)*(1 - x^2)^(1 + p))/(1 - x),x]
 

Output:

-((2^(1 + p)*(1 - x)^(1 + p)*(1 + x)^(1 + p)*(((c + d)*(1 + x))/(c + d*x)) 
^(-1 - p)*Hypergeometric2F1[-1 - p, 1 + p, 2 + p, ((c - d)*(1 - x))/(2*(c 
+ d*x))])/((c + d)*(1 + p)*(c + d*x)^(2*(1 + p))))
 

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]
 

rule 717

Int[((d_) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_) 
^2)^(p_), x_Symbol] :> Int[(d + e*x)^(m + p)*(f + g*x)^n*(a/d + (c/e)*x)^p, 
 x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a 
, 0] && GtQ[d, 0]
 

Maple [F]

\[\int \frac {\left (x d +c \right )^{-3-2 p} \left (-x^{2}+1\right )^{p +1}}{1-x}d x\]

Input:

int((d*x+c)^(-3-2*p)*(-x^2+1)^(p+1)/(1-x),x)
 

Output:

int((d*x+c)^(-3-2*p)*(-x^2+1)^(p+1)/(1-x),x)
 

Fricas [F]

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

integrate((d*x+c)^(-3-2*p)*(-x^2+1)^(p+1)/(1-x),x, algorithm="fricas")
 

Output:

integral(-(d*x + c)^(-2*p - 3)*(-x^2 + 1)^(p + 1)/(x - 1), x)
 

Sympy [F]

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

integrate((d*x+c)**(-3-2*p)*(-x**2+1)**(p+1)/(1-x),x)
 

Output:

-Integral((1 - x**2)**(p + 1)*(c + d*x)**(-2*p - 3)/(x - 1), x)
 

Maxima [F]

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

integrate((d*x+c)^(-3-2*p)*(-x^2+1)^(p+1)/(1-x),x, algorithm="maxima")
 

Output:

-integrate((d*x + c)^(-2*p - 3)*(-x^2 + 1)^(p + 1)/(x - 1), x)
 

Giac [F]

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

integrate((d*x+c)^(-3-2*p)*(-x^2+1)^(p+1)/(1-x),x, algorithm="giac")
 

Output:

integrate(-(d*x + c)^(-2*p - 3)*(-x^2 + 1)^(p + 1)/(x - 1), x)
 

Mupad [F(-1)]

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

int(-(1 - x^2)^(p + 1)/((x - 1)*(c + d*x)^(2*p + 3)),x)
 

Output:

-int((1 - x^2)^(p + 1)/((x - 1)*(c + d*x)^(2*p + 3)), x)
 

Reduce [F]

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

int((d*x+c)^(-3-2*p)*(-x^2+1)^(p+1)/(1-x),x)
 

Output:

int(( - x**2 + 1)**p/((c + d*x)**(2*p)*c**3 + 3*(c + d*x)**(2*p)*c**2*d*x 
+ 3*(c + d*x)**(2*p)*c*d**2*x**2 + (c + d*x)**(2*p)*d**3*x**3),x) + int((( 
 - x**2 + 1)**p*x)/((c + d*x)**(2*p)*c**3 + 3*(c + d*x)**(2*p)*c**2*d*x + 
3*(c + d*x)**(2*p)*c*d**2*x**2 + (c + d*x)**(2*p)*d**3*x**3),x)