Integrand size = 21, antiderivative size = 57 \[ \int \frac {1}{(2-2 x) \left (2 x-x^2\right )^{5/2}} \, dx=-\frac {1}{6 (2-x)^{3/2} x^{3/2}}-\frac {1}{2 \sqrt {2-x} \sqrt {x}}+\frac {1}{2} \text {arctanh}\left (\sqrt {2-x} \sqrt {x}\right ) \] Output:
-1/6/(2-x)^(3/2)/x^(3/2)-1/2/(2-x)^(1/2)/x^(1/2)+1/2*arctanh((2-x)^(1/2)*x ^(1/2))
Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(2-2 x) \left (2 x-x^2\right )^{5/2}} \, dx=\frac {-1-6 x+3 x^2+6 (-2+x)^{3/2} x^{3/2} \arctan \left (1+\sqrt {-2+x} \sqrt {x}-x\right )}{6 (-((-2+x) x))^{3/2}} \] Input:
Integrate[1/((2 - 2*x)*(2*x - x^2)^(5/2)),x]
Output:
(-1 - 6*x + 3*x^2 + 6*(-2 + x)^(3/2)*x^(3/2)*ArcTan[1 + Sqrt[-2 + x]*Sqrt[ x] - x])/(6*(-((-2 + x)*x))^(3/2))
Time = 0.33 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1111, 27, 1111, 1112, 220}
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.
\(\displaystyle \int \frac {1}{(2-2 x) \left (2 x-x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1111 |
\(\displaystyle \int \frac {1}{2 (1-x) \left (2 x-x^2\right )^{3/2}}dx-\frac {1}{6 \left (2 x-x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {1}{(1-x) \left (2 x-x^2\right )^{3/2}}dx-\frac {1}{6 \left (2 x-x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1111 |
\(\displaystyle \frac {1}{2} \left (\int \frac {1}{(1-x) \sqrt {2 x-x^2}}dx-\frac {1}{\sqrt {2 x-x^2}}\right )-\frac {1}{6 \left (2 x-x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1112 |
\(\displaystyle \frac {1}{2} \left (-4 \int \frac {1}{4 \left (2 x-x^2\right )-4}d\sqrt {2 x-x^2}-\frac {1}{\sqrt {2 x-x^2}}\right )-\frac {1}{6 \left (2 x-x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{2} \left (\text {arctanh}\left (\sqrt {2 x-x^2}\right )-\frac {1}{\sqrt {2 x-x^2}}\right )-\frac {1}{6 \left (2 x-x^2\right )^{3/2}}\) |
Input:
Int[1/((2 - 2*x)*(2*x - x^2)^(5/2)),x]
Output:
-1/6*1/(2*x - x^2)^(3/2) + (-(1/Sqrt[2*x - x^2]) + ArcTanh[Sqrt[2*x - x^2] ])/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[2*c*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)* (b^2 - 4*a*c))), x] - Simp[2*c*e*((m + 2*p + 3)/(e*(p + 1)*(b^2 - 4*a*c))) Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e , m}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[p, -1] && !G tQ[m, 1] && RationalQ[m] && IntegerQ[2*p]
Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symb ol] :> Simp[4*c Subst[Int[1/(b^2*e - 4*a*c*e + 4*c*e*x^2), x], x, Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Time = 0.44 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.74
method | result | size |
default | \(-\frac {1}{6 \left (1-\left (x -1\right )^{2}\right )^{\frac {3}{2}}}-\frac {1}{2 \sqrt {1-\left (x -1\right )^{2}}}+\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {1-\left (x -1\right )^{2}}}\right )}{2}\) | \(42\) |
trager | \(\frac {\left (3 x^{2}-6 x -1\right ) \sqrt {-x^{2}+2 x}}{6 \left (x -2\right )^{2} x^{2}}+\frac {\ln \left (\frac {\sqrt {-x^{2}+2 x}+1}{x -1}\right )}{2}\) | \(55\) |
pseudoelliptic | \(-\frac {x \sqrt {-x \left (x -2\right )}\, \left (x -2\right ) \ln \left (\sqrt {-x \left (x -2\right )}-1\right )-x \sqrt {-x \left (x -2\right )}\, \left (x -2\right ) \ln \left (\sqrt {-x \left (x -2\right )}+1\right )+2 x^{2}-4 x -\frac {2}{3}}{4 \sqrt {-x \left (x -2\right )}\, x \left (x -2\right )}\) | \(78\) |
Input:
int(1/(-2*x+2)/(-x^2+2*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/6/(1-(x-1)^2)^(3/2)-1/2/(1-(x-1)^2)^(1/2)+1/2*arctanh(1/(1-(x-1)^2)^(1/ 2))
Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (39) = 78\).
Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.96 \[ \int \frac {1}{(2-2 x) \left (2 x-x^2\right )^{5/2}} \, dx=\frac {3 \, {\left (x^{4} - 4 \, x^{3} + 4 \, x^{2}\right )} \log \left (\frac {x + \sqrt {-x^{2} + 2 \, x}}{x}\right ) - 3 \, {\left (x^{4} - 4 \, x^{3} + 4 \, x^{2}\right )} \log \left (-\frac {x - \sqrt {-x^{2} + 2 \, x}}{x}\right ) + {\left (3 \, x^{2} - 6 \, x - 1\right )} \sqrt {-x^{2} + 2 \, x}}{6 \, {\left (x^{4} - 4 \, x^{3} + 4 \, x^{2}\right )}} \] Input:
integrate(1/(2-2*x)/(-x^2+2*x)^(5/2),x, algorithm="fricas")
Output:
1/6*(3*(x^4 - 4*x^3 + 4*x^2)*log((x + sqrt(-x^2 + 2*x))/x) - 3*(x^4 - 4*x^ 3 + 4*x^2)*log(-(x - sqrt(-x^2 + 2*x))/x) + (3*x^2 - 6*x - 1)*sqrt(-x^2 + 2*x))/(x^4 - 4*x^3 + 4*x^2)
\[ \int \frac {1}{(2-2 x) \left (2 x-x^2\right )^{5/2}} \, dx=- \frac {\int \frac {1}{x^{5} \sqrt {- x^{2} + 2 x} - 5 x^{4} \sqrt {- x^{2} + 2 x} + 8 x^{3} \sqrt {- x^{2} + 2 x} - 4 x^{2} \sqrt {- x^{2} + 2 x}}\, dx}{2} \] Input:
integrate(1/(2-2*x)/(-x**2+2*x)**(5/2),x)
Output:
-Integral(1/(x**5*sqrt(-x**2 + 2*x) - 5*x**4*sqrt(-x**2 + 2*x) + 8*x**3*sq rt(-x**2 + 2*x) - 4*x**2*sqrt(-x**2 + 2*x)), x)/2
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(2-2 x) \left (2 x-x^2\right )^{5/2}} \, dx=-\frac {1}{2 \, \sqrt {-x^{2} + 2 \, x}} - \frac {1}{6 \, {\left (-x^{2} + 2 \, x\right )}^{\frac {3}{2}}} + \frac {1}{2} \, \log \left (\frac {2 \, \sqrt {-x^{2} + 2 \, x}}{{\left | x - 1 \right |}} + \frac {2}{{\left | x - 1 \right |}}\right ) \] Input:
integrate(1/(2-2*x)/(-x^2+2*x)^(5/2),x, algorithm="maxima")
Output:
-1/2/sqrt(-x^2 + 2*x) - 1/6/(-x^2 + 2*x)^(3/2) + 1/2*log(2*sqrt(-x^2 + 2*x )/abs(x - 1) + 2/abs(x - 1))
Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(2-2 x) \left (2 x-x^2\right )^{5/2}} \, dx=\frac {{\left (3 \, {\left (x - 2\right )} x - 1\right )} \sqrt {-x^{2} + 2 \, x}}{6 \, {\left (x^{2} - 2 \, x\right )}^{2}} - \frac {1}{2} \, \log \left (-\frac {2 \, {\left (\sqrt {-x^{2} + 2 \, x} - 1\right )}}{{\left | -2 \, x + 2 \right |}}\right ) \] Input:
integrate(1/(2-2*x)/(-x^2+2*x)^(5/2),x, algorithm="giac")
Output:
1/6*(3*(x - 2)*x - 1)*sqrt(-x^2 + 2*x)/(x^2 - 2*x)^2 - 1/2*log(-2*(sqrt(-x ^2 + 2*x) - 1)/abs(-2*x + 2))
Timed out. \[ \int \frac {1}{(2-2 x) \left (2 x-x^2\right )^{5/2}} \, dx=-\int \frac {1}{\left (2\,x-2\right )\,{\left (2\,x-x^2\right )}^{5/2}} \,d x \] Input:
int(-1/((2*x - 2)*(2*x - x^2)^(5/2)),x)
Output:
-int(1/((2*x - 2)*(2*x - x^2)^(5/2)), x)
Time = 0.21 (sec) , antiderivative size = 133, normalized size of antiderivative = 2.33 \[ \int \frac {1}{(2-2 x) \left (2 x-x^2\right )^{5/2}} \, dx=\frac {-6 \sqrt {-x +2}\, \mathit {atan} \left (\sqrt {-x +2}+\sqrt {x}\, i -1\right ) i \,x^{3}+12 \sqrt {-x +2}\, \mathit {atan} \left (\sqrt {-x +2}+\sqrt {x}\, i -1\right ) i \,x^{2}+6 \sqrt {-x +2}\, \mathit {atan} \left (\sqrt {-x +2}+\sqrt {x}\, i +1\right ) i \,x^{3}-12 \sqrt {-x +2}\, \mathit {atan} \left (\sqrt {-x +2}+\sqrt {x}\, i +1\right ) i \,x^{2}-3 \sqrt {x}\, x^{2}+6 \sqrt {x}\, x +\sqrt {x}}{6 \sqrt {-x +2}\, x^{2} \left (x -2\right )} \] Input:
int(1/(2-2*x)/(-x^2+2*x)^(5/2),x)
Output:
( - 6*sqrt( - x + 2)*atan(sqrt( - x + 2) + sqrt(x)*i - 1)*i*x**3 + 12*sqrt ( - x + 2)*atan(sqrt( - x + 2) + sqrt(x)*i - 1)*i*x**2 + 6*sqrt( - x + 2)* atan(sqrt( - x + 2) + sqrt(x)*i + 1)*i*x**3 - 12*sqrt( - x + 2)*atan(sqrt( - x + 2) + sqrt(x)*i + 1)*i*x**2 - 3*sqrt(x)*x**2 + 6*sqrt(x)*x + sqrt(x) )/(6*sqrt( - x + 2)*x**2*(x - 2))