Integrand size = 23, antiderivative size = 63 \[ \int \frac {(-1-b x)^n (-1+b x)^p}{(-x)^{3/2}} \, dx=\frac {2 (-1-b x)^n (1-b x)^{-p} (-1+b x)^p (1+b x)^{-n} \operatorname {AppellF1}\left (-\frac {1}{2},-p,-n,\frac {1}{2},b x,-b x\right )}{\sqrt {-x}} \] Output:
2*(-b*x-1)^n*(b*x-1)^p*AppellF1(-1/2,-p,-n,1/2,b*x,-b*x)/(-x)^(1/2)/((-b*x +1)^p)/((b*x+1)^n)
Time = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02 \[ \int \frac {(-1-b x)^n (-1+b x)^p}{(-x)^{3/2}} \, dx=-\frac {2 x (-1-b x)^n (1-b x)^{-p} (-1+b x)^p (1+b x)^{-n} \operatorname {AppellF1}\left (-\frac {1}{2},-n,-p,\frac {1}{2},-b x,b x\right )}{(-x)^{3/2}} \] Input:
Integrate[((-1 - b*x)^n*(-1 + b*x)^p)/(-x)^(3/2),x]
Output:
(-2*x*(-1 - b*x)^n*(-1 + b*x)^p*AppellF1[-1/2, -n, -p, 1/2, -(b*x), b*x])/ ((-x)^(3/2)*(1 - b*x)^p*(1 + b*x)^n)
Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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.
\(\displaystyle \int \frac {(-b x-1)^n (b x-1)^p}{(-x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 152 |
\(\displaystyle (-b x-1)^n (b x+1)^{-n} \int \frac {(b x-1)^p (b x+1)^n}{(-x)^{3/2}}dx\) |
\(\Big \downarrow \) 152 |
\(\displaystyle (-b x-1)^n (b x+1)^{-n} (1-b x)^{-p} (b x-1)^p \int \frac {(1-b x)^p (b x+1)^n}{(-x)^{3/2}}dx\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {2 (-b x-1)^n (b x+1)^{-n} (1-b x)^{-p} (b x-1)^p \operatorname {AppellF1}\left (-\frac {1}{2},-p,-n,\frac {1}{2},b x,-b x\right )}{\sqrt {-x}}\) |
Input:
Int[((-1 - b*x)^n*(-1 + b*x)^p)/(-x)^(3/2),x]
Output:
(2*(-1 - b*x)^n*(-1 + b*x)^p*AppellF1[-1/2, -p, -n, 1/2, b*x, -(b*x)])/(Sq rt[-x]*(1 - b*x)^p*(1 + b*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]
\[\int \frac {\left (-b x -1\right )^{n} \left (b x -1\right )^{p}}{\left (-x \right )^{\frac {3}{2}}}d x\]
Input:
int((-b*x-1)^n*(b*x-1)^p/(-x)^(3/2),x)
Output:
int((-b*x-1)^n*(b*x-1)^p/(-x)^(3/2),x)
\[ \int \frac {(-1-b x)^n (-1+b x)^p}{(-x)^{3/2}} \, dx=\int { \frac {{\left (b x - 1\right )}^{p} {\left (-b x - 1\right )}^{n}}{\left (-x\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((-b*x-1)^n*(b*x-1)^p/(-x)^(3/2),x, algorithm="fricas")
Output:
integral((b*x - 1)^p*(-b*x - 1)^n*sqrt(-x)/x^2, x)
\[ \int \frac {(-1-b x)^n (-1+b x)^p}{(-x)^{3/2}} \, dx=\int \frac {\left (- b x - 1\right )^{n} \left (b x - 1\right )^{p}}{\left (- x\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((-b*x-1)**n*(b*x-1)**p/(-x)**(3/2),x)
Output:
Integral((-b*x - 1)**n*(b*x - 1)**p/(-x)**(3/2), x)
\[ \int \frac {(-1-b x)^n (-1+b x)^p}{(-x)^{3/2}} \, dx=\int { \frac {{\left (b x - 1\right )}^{p} {\left (-b x - 1\right )}^{n}}{\left (-x\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((-b*x-1)^n*(b*x-1)^p/(-x)^(3/2),x, algorithm="maxima")
Output:
integrate((b*x - 1)^p*(-b*x - 1)^n/(-x)^(3/2), x)
\[ \int \frac {(-1-b x)^n (-1+b x)^p}{(-x)^{3/2}} \, dx=\int { \frac {{\left (b x - 1\right )}^{p} {\left (-b x - 1\right )}^{n}}{\left (-x\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate((-b*x-1)^n*(b*x-1)^p/(-x)^(3/2),x, algorithm="giac")
Output:
integrate((b*x - 1)^p*(-b*x - 1)^n/(-x)^(3/2), x)
Timed out. \[ \int \frac {(-1-b x)^n (-1+b x)^p}{(-x)^{3/2}} \, dx=\int \frac {{\left (b\,x-1\right )}^p\,{\left (-b\,x-1\right )}^n}{{\left (-x\right )}^{3/2}} \,d x \] Input:
int(((b*x - 1)^p*(- b*x - 1)^n)/(-x)^(3/2),x)
Output:
int(((b*x - 1)^p*(- b*x - 1)^n)/(-x)^(3/2), x)
\[ \int \frac {(-1-b x)^n (-1+b x)^p}{(-x)^{3/2}} \, dx=\left (\int \frac {\left (b x -1\right )^{p} \left (-b x -1\right )^{n}}{\sqrt {x}\, x}d x \right ) i \] Input:
int((-b*x-1)^n*(b*x-1)^p/(-x)^(3/2),x)
Output:
int(((b*x - 1)**p*( - b*x - 1)**n)/(sqrt(x)*x),x)*i