Integrand size = 37, antiderivative size = 46 \[ \int \frac {-x+x^2}{\sqrt {-2 x+4 x^2-2 x^3+x^4-2 x^5+x^6}} \, dx=\frac {2}{3} \text {arctanh}\left (\frac {(-1+x) \left (-2+x^3\right )}{x \sqrt {-2 x+4 x^2-2 x^3+x^4-2 x^5+x^6}}\right ) \] Output:
2/3*arctanh((-1+x)*(x^3-2)/x/(x^6-2*x^5+x^4-2*x^3+4*x^2-2*x)^(1/2))
Time = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.15 \[ \int \frac {-x+x^2}{\sqrt {-2 x+4 x^2-2 x^3+x^4-2 x^5+x^6}} \, dx=\frac {2 (-1+x) \sqrt {x} \sqrt {-2+x^3} \log \left (x^{3/2}+\sqrt {-2+x^3}\right )}{3 \sqrt {(-1+x)^2 x \left (-2+x^3\right )}} \] Input:
Integrate[(-x + x^2)/Sqrt[-2*x + 4*x^2 - 2*x^3 + x^4 - 2*x^5 + x^6],x]
Output:
(2*(-1 + x)*Sqrt[x]*Sqrt[-2 + x^3]*Log[x^(3/2) + Sqrt[-2 + x^3]])/(3*Sqrt[ (-1 + x)^2*x*(-2 + x^3)])
Time = 0.53 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.26, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {2027, 7239, 7269, 851, 807, 224, 219}
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.
\(\displaystyle \int \frac {x^2-x}{\sqrt {x^6-2 x^5+x^4-2 x^3+4 x^2-2 x}} \, dx\) |
\(\Big \downarrow \) 2027 |
\(\displaystyle \int \frac {(x-1) x}{\sqrt {x^6-2 x^5+x^4-2 x^3+4 x^2-2 x}}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {(x-1) x}{\sqrt {(x-1)^2 x \left (x^3-2\right )}}dx\) |
\(\Big \downarrow \) 7269 |
\(\displaystyle -\frac {(1-x) \sqrt {x} \sqrt {x^3-2} \int \frac {\sqrt {x}}{\sqrt {x^3-2}}dx}{\sqrt {-(1-x)^2 x \left (2-x^3\right )}}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle -\frac {2 (1-x) \sqrt {x} \sqrt {x^3-2} \int \frac {x}{\sqrt {x^3-2}}d\sqrt {x}}{\sqrt {-(1-x)^2 x \left (2-x^3\right )}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {2 (1-x) \sqrt {x} \sqrt {x^3-2} \int \frac {1}{\sqrt {x-2}}dx^{3/2}}{3 \sqrt {-(1-x)^2 x \left (2-x^3\right )}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {2 (1-x) \sqrt {x} \sqrt {x^3-2} \int \frac {1}{1-x}d\frac {x^{3/2}}{\sqrt {x-2}}}{3 \sqrt {-(1-x)^2 x \left (2-x^3\right )}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2 (1-x) \sqrt {x} \sqrt {x^3-2} \text {arctanh}\left (\frac {x^{3/2}}{\sqrt {x-2}}\right )}{3 \sqrt {-(1-x)^2 x \left (2-x^3\right )}}\) |
Input:
Int[(-x + x^2)/Sqrt[-2*x + 4*x^2 - 2*x^3 + x^4 - 2*x^5 + x^6],x]
Output:
(-2*(1 - x)*Sqrt[x]*Sqrt[-2 + x^3]*ArcTanh[x^(3/2)/Sqrt[-2 + x]])/(3*Sqrt[ -((1 - x)^2*x*(2 - x^3))])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^ n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[p, 1] && EqQ[u, 1])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.)*(z_)^(q_.))^(p_), x_Symbol] :> Simp[ a^IntPart[p]*((a*v^m*w^n*z^q)^FracPart[p]/(v^(m*FracPart[p])*w^(n*FracPart[ p])*z^(q*FracPart[p]))) Int[u*v^(m*p)*w^(n*p)*z^(p*q), x], x] /; FreeQ[{a , m, n, p, q}, x] && !IntegerQ[p] && !FreeQ[v, x] && !FreeQ[w, x] && !F reeQ[z, x]
Time = 0.76 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.15
method | result | size |
trager | \(\frac {\ln \left (-\frac {x^{4}-x^{3}+x \sqrt {x^{6}-2 x^{5}+x^{4}-2 x^{3}+4 x^{2}-2 x}-x +1}{-1+x}\right )}{3}\) | \(53\) |
default | \(-\frac {\left (-1+x \right ) \sqrt {x \left (x^{3}-2\right )}\, \ln \left (x^{3}-\sqrt {x^{4}-2 x}\, x -1\right )}{3 \sqrt {x^{6}-2 x^{5}+x^{4}-2 x^{3}+4 x^{2}-2 x}}\) | \(60\) |
Input:
int((x^2-x)/(x^6-2*x^5+x^4-2*x^3+4*x^2-2*x)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/3*ln(-(x^4-x^3+x*(x^6-2*x^5+x^4-2*x^3+4*x^2-2*x)^(1/2)-x+1)/(-1+x))
Time = 0.10 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.13 \[ \int \frac {-x+x^2}{\sqrt {-2 x+4 x^2-2 x^3+x^4-2 x^5+x^6}} \, dx=\frac {1}{3} \, \log \left (-\frac {x^{4} - x^{3} + \sqrt {x^{6} - 2 \, x^{5} + x^{4} - 2 \, x^{3} + 4 \, x^{2} - 2 \, x} x - x + 1}{x - 1}\right ) \] Input:
integrate((x^2-x)/(x^6-2*x^5+x^4-2*x^3+4*x^2-2*x)^(1/2),x, algorithm="fric as")
Output:
1/3*log(-(x^4 - x^3 + sqrt(x^6 - 2*x^5 + x^4 - 2*x^3 + 4*x^2 - 2*x)*x - x + 1)/(x - 1))
\[ \int \frac {-x+x^2}{\sqrt {-2 x+4 x^2-2 x^3+x^4-2 x^5+x^6}} \, dx=\int \frac {x \left (x - 1\right )}{\sqrt {x \left (x - 1\right )^{2} \left (x^{3} - 2\right )}}\, dx \] Input:
integrate((x**2-x)/(x**6-2*x**5+x**4-2*x**3+4*x**2-2*x)**(1/2),x)
Output:
Integral(x*(x - 1)/sqrt(x*(x - 1)**2*(x**3 - 2)), x)
\[ \int \frac {-x+x^2}{\sqrt {-2 x+4 x^2-2 x^3+x^4-2 x^5+x^6}} \, dx=\int { \frac {x^{2} - x}{\sqrt {x^{6} - 2 \, x^{5} + x^{4} - 2 \, x^{3} + 4 \, x^{2} - 2 \, x}} \,d x } \] Input:
integrate((x^2-x)/(x^6-2*x^5+x^4-2*x^3+4*x^2-2*x)^(1/2),x, algorithm="maxi ma")
Output:
integrate((x^2 - x)/sqrt(x^6 - 2*x^5 + x^4 - 2*x^3 + 4*x^2 - 2*x), x)
Result contains complex when optimal does not.
Time = 0.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.09 \[ \int \frac {-x+x^2}{\sqrt {-2 x+4 x^2-2 x^3+x^4-2 x^5+x^6}} \, dx=\frac {1}{6} \, {\left (\log \left (2\right ) - 2 \, \log \left (i + 1\right )\right )} \mathrm {sgn}\left (x - 1\right ) + \frac {\log \left (\sqrt {-\frac {2}{x^{3}} + 1} + 1\right ) - \log \left ({\left | \sqrt {-\frac {2}{x^{3}} + 1} - 1 \right |}\right )}{3 \, \mathrm {sgn}\left (x - 1\right )} \] Input:
integrate((x^2-x)/(x^6-2*x^5+x^4-2*x^3+4*x^2-2*x)^(1/2),x, algorithm="giac ")
Output:
1/6*(log(2) - 2*log(I + 1))*sgn(x - 1) + 1/3*(log(sqrt(-2/x^3 + 1) + 1) - log(abs(sqrt(-2/x^3 + 1) - 1)))/sgn(x - 1)
Timed out. \[ \int \frac {-x+x^2}{\sqrt {-2 x+4 x^2-2 x^3+x^4-2 x^5+x^6}} \, dx=\int -\frac {x-x^2}{\sqrt {x^6-2\,x^5+x^4-2\,x^3+4\,x^2-2\,x}} \,d x \] Input:
int(-(x - x^2)/(4*x^2 - 2*x - 2*x^3 + x^4 - 2*x^5 + x^6)^(1/2),x)
Output:
int(-(x - x^2)/(4*x^2 - 2*x - 2*x^3 + x^4 - 2*x^5 + x^6)^(1/2), x)
Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.65 \[ \int \frac {-x+x^2}{\sqrt {-2 x+4 x^2-2 x^3+x^4-2 x^5+x^6}} \, dx=-\frac {\mathrm {log}\left (\sqrt {x^{3}-2}-\sqrt {x}\, x \right )}{3}+\frac {\mathrm {log}\left (\sqrt {x^{3}-2}+\sqrt {x}\, x \right )}{3} \] Input:
int((x^2-x)/(x^6-2*x^5+x^4-2*x^3+4*x^2-2*x)^(1/2),x)
Output:
( - log(sqrt(x**3 - 2) - sqrt(x)*x) + log(sqrt(x**3 - 2) + sqrt(x)*x))/3