Integrand size = 25, antiderivative size = 50 \[ \int (b x)^{-2-2 m} (1-a x)^m (1+a x)^m \, dx=-\frac {(b x)^{-1-2 m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-2 m),-m,\frac {1}{2} (1-2 m),a^2 x^2\right )}{b (1+2 m)} \] Output:
-(b*x)^(-1-2*m)*hypergeom([-m, -1/2-m],[1/2-m],a^2*x^2)/b/(1+2*m)
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.88 \[ \int (b x)^{-2-2 m} (1-a x)^m (1+a x)^m \, dx=-\frac {(b x)^{-1-2 m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}-m,-m,\frac {1}{2}-m,a^2 x^2\right )}{b+2 b m} \] Input:
Integrate[(b*x)^(-2 - 2*m)*(1 - a*x)^m*(1 + a*x)^m,x]
Output:
-(((b*x)^(-1 - 2*m)*Hypergeometric2F1[-1/2 - m, -m, 1/2 - m, a^2*x^2])/(b + 2*b*m))
Time = 0.17 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {135, 278}
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.
\(\displaystyle \int (1-a x)^m (a x+1)^m (b x)^{-2 m-2} \, dx\) |
\(\Big \downarrow \) 135 |
\(\displaystyle \int \left (1-a^2 x^2\right )^m (b x)^{-2 m-2}dx\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {(b x)^{-2 m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-2 m-1),-m,\frac {1}{2} (1-2 m),a^2 x^2\right )}{b (2 m+1)}\) |
Input:
Int[(b*x)^(-2 - 2*m)*(1 - a*x)^m*(1 + a*x)^m,x]
Output:
-(((b*x)^(-1 - 2*m)*Hypergeometric2F1[(-1 - 2*m)/2, -m, (1 - 2*m)/2, a^2*x ^2])/(b*(1 + 2*m)))
Int[((f_.)*(x_))^(p_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_] :> Int[(a*c + b*d*x^2)^m*(f*x)^p, x] /; FreeQ[{a, b, c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && GtQ[a, 0] && GtQ[c, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
\[\int \left (b x \right )^{-2 m -2} \left (-a x +1\right )^{m} \left (a x +1\right )^{m}d x\]
Input:
int((b*x)^(-2*m-2)*(-a*x+1)^m*(a*x+1)^m,x)
Output:
int((b*x)^(-2*m-2)*(-a*x+1)^m*(a*x+1)^m,x)
\[ \int (b x)^{-2-2 m} (1-a x)^m (1+a x)^m \, dx=\int { {\left (a x + 1\right )}^{m} {\left (-a x + 1\right )}^{m} \left (b x\right )^{-2 \, m - 2} \,d x } \] Input:
integrate((b*x)^(-2-2*m)*(-a*x+1)^m*(a*x+1)^m,x, algorithm="fricas")
Output:
integral((a*x + 1)^m*(-a*x + 1)^m*(b*x)^(-2*m - 2), x)
Result contains complex when optimal does not.
Time = 12.02 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.76 \[ \int (b x)^{-2-2 m} (1-a x)^m (1+a x)^m \, dx=\frac {a^{2 m + 1} b^{- 2 m - 2} {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {m}{2} + 1, \frac {m}{2} + \frac {3}{2}, 1 & m + \frac {3}{2}, 1, \frac {3}{2} \\\frac {1}{2}, 1, \frac {m}{2} + 1, \frac {3}{2}, \frac {m}{2} + \frac {3}{2} & 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{a^{2} x^{2}}} \right )} e^{i \pi m}}{4 \pi \Gamma \left (- m\right )} - \frac {a^{2 m + 1} b^{- 2 m - 2} {G_{6, 6}^{2, 6}\left (\begin {matrix} m + \frac {1}{2}, m + 1, m + \frac {3}{2}, \frac {m}{2} + \frac {1}{2}, \frac {m}{2} + 1, 1 & \\\frac {m}{2} + \frac {1}{2}, \frac {m}{2} + 1 & m + \frac {1}{2}, m + 1, \frac {1}{2}, 0 \end {matrix} \middle | {\frac {1}{a^{2} x^{2}}} \right )}}{4 \pi \Gamma \left (- m\right )} \] Input:
integrate((b*x)**(-2-2*m)*(-a*x+1)**m*(a*x+1)**m,x)
Output:
a**(2*m + 1)*b**(-2*m - 2)*meijerg(((m/2 + 1, m/2 + 3/2, 1), (m + 3/2, 1, 3/2)), ((1/2, 1, m/2 + 1, 3/2, m/2 + 3/2), (0,)), exp_polar(-2*I*pi)/(a**2 *x**2))*exp(I*pi*m)/(4*pi*gamma(-m)) - a**(2*m + 1)*b**(-2*m - 2)*meijerg( ((m + 1/2, m + 1, m + 3/2, m/2 + 1/2, m/2 + 1, 1), ()), ((m/2 + 1/2, m/2 + 1), (m + 1/2, m + 1, 1/2, 0)), 1/(a**2*x**2))/(4*pi*gamma(-m))
\[ \int (b x)^{-2-2 m} (1-a x)^m (1+a x)^m \, dx=\int { {\left (a x + 1\right )}^{m} {\left (-a x + 1\right )}^{m} \left (b x\right )^{-2 \, m - 2} \,d x } \] Input:
integrate((b*x)^(-2-2*m)*(-a*x+1)^m*(a*x+1)^m,x, algorithm="maxima")
Output:
integrate((a*x + 1)^m*(-a*x + 1)^m*(b*x)^(-2*m - 2), x)
\[ \int (b x)^{-2-2 m} (1-a x)^m (1+a x)^m \, dx=\int { {\left (a x + 1\right )}^{m} {\left (-a x + 1\right )}^{m} \left (b x\right )^{-2 \, m - 2} \,d x } \] Input:
integrate((b*x)^(-2-2*m)*(-a*x+1)^m*(a*x+1)^m,x, algorithm="giac")
Output:
integrate((a*x + 1)^m*(-a*x + 1)^m*(b*x)^(-2*m - 2), x)
Timed out. \[ \int (b x)^{-2-2 m} (1-a x)^m (1+a x)^m \, dx=\int \frac {{\left (1-a\,x\right )}^m\,{\left (a\,x+1\right )}^m}{{\left (b\,x\right )}^{2\,m+2}} \,d x \] Input:
int(((1 - a*x)^m*(a*x + 1)^m)/(b*x)^(2*m + 2),x)
Output:
int(((1 - a*x)^m*(a*x + 1)^m)/(b*x)^(2*m + 2), x)
\[ \int (b x)^{-2-2 m} (1-a x)^m (1+a x)^m \, dx=\frac {-\left (a x +1\right )^{m} \left (-a x +1\right )^{m}+2 x^{2 m} \left (\int \frac {\left (a x +1\right )^{m} \left (-a x +1\right )^{m}}{x^{2 m} a^{2} x^{4}-x^{2 m} x^{2}}d x \right ) m x}{x^{2 m} b^{2 m} b^{2} x} \] Input:
int((b*x)^(-2-2*m)*(-a*x+1)^m*(a*x+1)^m,x)
Output:
( - (a*x + 1)**m*( - a*x + 1)**m + 2*x**(2*m)*int(((a*x + 1)**m*( - a*x + 1)**m)/(x**(2*m)*a**2*x**4 - x**(2*m)*x**2),x)*m*x)/(x**(2*m)*b**(2*m)*b** 2*x)