Integrand size = 20, antiderivative size = 54 \[ \int \frac {(-1-x)^n (-1+x)^p}{(-x)^{3/2}} \, dx=\frac {2 (-1-x)^n (1-x)^{-p} (-1+x)^p (1+x)^{-n} \operatorname {AppellF1}\left (-\frac {1}{2},-p,-n,\frac {1}{2},x,-x\right )}{\sqrt {-x}} \] Output:
2*(-1-x)^n*(-1+x)^p*AppellF1(-1/2,-p,-n,1/2,x,-x)/((1-x)^p)/(-x)^(1/2)/((1 +x)^n)
Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02 \[ \int \frac {(-1-x)^n (-1+x)^p}{(-x)^{3/2}} \, dx=-\frac {2 (-1-x)^n (1-x)^{-p} (-1+x)^p x (1+x)^{-n} \operatorname {AppellF1}\left (-\frac {1}{2},-n,-p,\frac {1}{2},-x,x\right )}{(-x)^{3/2}} \] Input:
Integrate[((-1 - x)^n*(-1 + x)^p)/(-x)^(3/2),x]
Output:
(-2*(-1 - x)^n*(-1 + x)^p*x*AppellF1[-1/2, -n, -p, 1/2, -x, x])/((1 - x)^p *(-x)^(3/2)*(1 + x)^n)
Time = 0.17 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {152, 152, 150}
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.
\(\displaystyle \int \frac {(-x-1)^n (x-1)^p}{(-x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 152 |
\(\displaystyle (-x-1)^n (x+1)^{-n} \int \frac {(x-1)^p (x+1)^n}{(-x)^{3/2}}dx\) |
\(\Big \downarrow \) 152 |
\(\displaystyle (-x-1)^n (x+1)^{-n} (1-x)^{-p} (x-1)^p \int \frac {(1-x)^p (x+1)^n}{(-x)^{3/2}}dx\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {2 (-x-1)^n (x+1)^{-n} (1-x)^{-p} (x-1)^p \operatorname {AppellF1}\left (-\frac {1}{2},-p,-n,\frac {1}{2},x,-x\right )}{\sqrt {-x}}\) |
Input:
Int[((-1 - x)^n*(-1 + x)^p)/(-x)^(3/2),x]
Output:
(2*(-1 - x)^n*(-1 + x)^p*AppellF1[-1/2, -p, -n, 1/2, x, -x])/((1 - x)^p*Sq rt[-x]*(1 + x)^n)
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 0]
\[\int \frac {\left (-1-x \right )^{n} \left (-1+x \right )^{p}}{\left (-x \right )^{\frac {3}{2}}}d x\]
Input:
int((-1-x)^n*(-1+x)^p/(-x)^(3/2),x)
Output:
int((-1-x)^n*(-1+x)^p/(-x)^(3/2),x)
\[ \int \frac {(-1-x)^n (-1+x)^p}{(-x)^{3/2}} \, dx=\int { \frac {{\left (x - 1\right )}^{p} {\left (-x - 1\right )}^{n}}{\left (-x\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((-1-x)^n*(-1+x)^p/(-x)^(3/2),x, algorithm="fricas")
Output:
integral((x - 1)^p*sqrt(-x)*(-x - 1)^n/x^2, x)
\[ \int \frac {(-1-x)^n (-1+x)^p}{(-x)^{3/2}} \, dx=\int \frac {\left (- x - 1\right )^{n} \left (x - 1\right )^{p}}{\left (- x\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((-1-x)**n*(-1+x)**p/(-x)**(3/2),x)
Output:
Integral((-x - 1)**n*(x - 1)**p/(-x)**(3/2), x)
\[ \int \frac {(-1-x)^n (-1+x)^p}{(-x)^{3/2}} \, dx=\int { \frac {{\left (x - 1\right )}^{p} {\left (-x - 1\right )}^{n}}{\left (-x\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((-1-x)^n*(-1+x)^p/(-x)^(3/2),x, algorithm="maxima")
Output:
integrate((x - 1)^p*(-x - 1)^n/(-x)^(3/2), x)
\[ \int \frac {(-1-x)^n (-1+x)^p}{(-x)^{3/2}} \, dx=\int { \frac {{\left (x - 1\right )}^{p} {\left (-x - 1\right )}^{n}}{\left (-x\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((-1-x)^n*(-1+x)^p/(-x)^(3/2),x, algorithm="giac")
Output:
integrate((x - 1)^p*(-x - 1)^n/(-x)^(3/2), x)
Timed out. \[ \int \frac {(-1-x)^n (-1+x)^p}{(-x)^{3/2}} \, dx=\int \frac {{\left (x-1\right )}^p\,{\left (-x-1\right )}^n}{{\left (-x\right )}^{3/2}} \,d x \] Input:
int(((x - 1)^p*(- x - 1)^n)/(-x)^(3/2),x)
Output:
int(((x - 1)^p*(- x - 1)^n)/(-x)^(3/2), x)
\[ \int \frac {(-1-x)^n (-1+x)^p}{(-x)^{3/2}} \, dx=\left (\int \frac {\left (x -1\right )^{p} \left (-x -1\right )^{n}}{\sqrt {x}\, x}d x \right ) i \] Input:
int((-1-x)^n*(-1+x)^p/(-x)^(3/2),x)
Output:
int(((x - 1)**p*( - x - 1)**n)/(sqrt(x)*x),x)*i