Integrand size = 24, antiderivative size = 86 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{-1-2 x+x^2} \, dx=-2 \arctan \left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )+2 \text {arctanh}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )+\frac {1}{2} \text {RootSum}\left [2-4 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{-x^3+x^4}-x \text {$\#$1}\right )}{\text {$\#$1}^3}\&\right ] \] Output:
Unintegrable
Time = 0.02 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{-1-2 x+x^2} \, dx=\frac {(-1+x)^{3/4} x^{9/4} \left (-4 \arctan \left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )+4 \text {arctanh}\left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )+\text {RootSum}\left [2-4 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log \left (\sqrt [4]{x}\right )+\log \left (\sqrt [4]{-1+x}-\sqrt [4]{x} \text {$\#$1}\right )}{\text {$\#$1}^3}\&\right ]\right )}{2 \left ((-1+x) x^3\right )^{3/4}} \] Input:
Integrate[(-x^3 + x^4)^(1/4)/(-1 - 2*x + x^2),x]
Output:
((-1 + x)^(3/4)*x^(9/4)*(-4*ArcTan[((-1 + x)/x)^(-1/4)] + 4*ArcTanh[((-1 + x)/x)^(-1/4)] + RootSum[2 - 4*#1^4 + #1^8 & , (-Log[x^(1/4)] + Log[(-1 + x)^(1/4) - x^(1/4)*#1])/#1^3 & ]))/(2*((-1 + x)*x^3)^(3/4))
Leaf count is larger than twice the leaf count of optimal. \(410\) vs. \(2(86)=172\).
Time = 1.16 (sec) , antiderivative size = 410, normalized size of antiderivative = 4.77, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2467, 25, 1202, 25, 73, 770, 756, 216, 219, 2035, 7279, 2009}
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.
\(\displaystyle \int \frac {\sqrt [4]{x^4-x^3}}{x^2-2 x-1} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x^4-x^3} \int -\frac {\sqrt [4]{x-1} x^{3/4}}{-x^2+2 x+1}dx}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^3} \int \frac {\sqrt [4]{x-1} x^{3/4}}{-x^2+2 x+1}dx}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 1202 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^3} \left (-\int -\frac {x+1}{(x-1)^{3/4} \sqrt [4]{x} \left (-x^2+2 x+1\right )}dx-\int \frac {1}{(x-1)^{3/4} \sqrt [4]{x}}dx\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^3} \left (\int \frac {x+1}{(x-1)^{3/4} \sqrt [4]{x} \left (-x^2+2 x+1\right )}dx-\int \frac {1}{(x-1)^{3/4} \sqrt [4]{x}}dx\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^3} \left (\int \frac {x+1}{(x-1)^{3/4} \sqrt [4]{x} \left (-x^2+2 x+1\right )}dx-4 \int \frac {1}{\sqrt [4]{x}}d\sqrt [4]{x-1}\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^3} \left (\int \frac {x+1}{(x-1)^{3/4} \sqrt [4]{x} \left (-x^2+2 x+1\right )}dx-4 \int \frac {1}{2-x}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^3} \left (\int \frac {x+1}{(x-1)^{3/4} \sqrt [4]{x} \left (-x^2+2 x+1\right )}dx-4 \left (\frac {1}{2} \int \frac {1}{1-\sqrt {x-1}}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}+\frac {1}{2} \int \frac {1}{\sqrt {x-1}+1}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^3} \left (\int \frac {x+1}{(x-1)^{3/4} \sqrt [4]{x} \left (-x^2+2 x+1\right )}dx-4 \left (\frac {1}{2} \int \frac {1}{1-\sqrt {x-1}}d\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}+\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )\right )\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^3} \left (\int \frac {x+1}{(x-1)^{3/4} \sqrt [4]{x} \left (-x^2+2 x+1\right )}dx-4 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )\right )\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^3} \left (4 \int \frac {\sqrt {x} (x+1)}{(x-1)^{3/4} \left (-x^2+2 x+1\right )}d\sqrt [4]{x}-4 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )\right )\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^3} \left (4 \int \left (-\frac {x^{3/2}}{(x-1)^{3/4} \left (x^2-2 x-1\right )}-\frac {\sqrt {x}}{(x-1)^{3/4} \left (x^2-2 x-1\right )}\right )d\sqrt [4]{x}-4 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )\right )\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^3} \left (4 \left (-\frac {1}{8} \sqrt [4]{10+7 \sqrt {2}} \arctan \left (\frac {\sqrt [4]{2-\sqrt {2}} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )-\frac {1}{8} \sqrt [4]{2-\sqrt {2}} \arctan \left (\frac {\sqrt [4]{2-\sqrt {2}} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )-\frac {1}{8} \sqrt [4]{2+\sqrt {2}} \arctan \left (\frac {\sqrt [4]{2+\sqrt {2}} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )+\frac {1}{8} \sqrt [4]{10-7 \sqrt {2}} \arctan \left (\frac {\sqrt [4]{2+\sqrt {2}} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )+\frac {1}{8} \sqrt [4]{10+7 \sqrt {2}} \text {arctanh}\left (\frac {\sqrt [4]{2-\sqrt {2}} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )+\frac {1}{8} \sqrt [4]{2-\sqrt {2}} \text {arctanh}\left (\frac {\sqrt [4]{2-\sqrt {2}} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )+\frac {1}{8} \sqrt [4]{2+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt [4]{2+\sqrt {2}} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )-\frac {1}{8} \sqrt [4]{10-7 \sqrt {2}} \text {arctanh}\left (\frac {\sqrt [4]{2+\sqrt {2}} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )\right )-4 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )\right )\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
Input:
Int[(-x^3 + x^4)^(1/4)/(-1 - 2*x + x^2),x]
Output:
-(((-x^3 + x^4)^(1/4)*(-4*(ArcTan[(-1 + x)^(1/4)/x^(1/4)]/2 + ArcTanh[(-1 + x)^(1/4)/x^(1/4)]/2) + 4*(-1/8*((2 - Sqrt[2])^(1/4)*ArcTan[((2 - Sqrt[2] )^(1/4)*x^(1/4))/(-1 + x)^(1/4)]) - ((10 + 7*Sqrt[2])^(1/4)*ArcTan[((2 - S qrt[2])^(1/4)*x^(1/4))/(-1 + x)^(1/4)])/8 + ((10 - 7*Sqrt[2])^(1/4)*ArcTan [((2 + Sqrt[2])^(1/4)*x^(1/4))/(-1 + x)^(1/4)])/8 - ((2 + Sqrt[2])^(1/4)*A rcTan[((2 + Sqrt[2])^(1/4)*x^(1/4))/(-1 + x)^(1/4)])/8 + ((2 - Sqrt[2])^(1 /4)*ArcTanh[((2 - Sqrt[2])^(1/4)*x^(1/4))/(-1 + x)^(1/4)])/8 + ((10 + 7*Sq rt[2])^(1/4)*ArcTanh[((2 - Sqrt[2])^(1/4)*x^(1/4))/(-1 + x)^(1/4)])/8 - (( 10 - 7*Sqrt[2])^(1/4)*ArcTanh[((2 + Sqrt[2])^(1/4)*x^(1/4))/(-1 + x)^(1/4) ])/8 + ((2 + Sqrt[2])^(1/4)*ArcTanh[((2 + Sqrt[2])^(1/4)*x^(1/4))/(-1 + x) ^(1/4)])/8)))/((-1 + x)^(1/4)*x^(3/4)))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 /n]
Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_.) + (b_.)*(x _) + (c_.)*(x_)^2), x_Symbol] :> Simp[e*(g/c) Int[(d + e*x)^(m - 1)*(f + g*x)^(n - 1), x], x] + Simp[1/c Int[Simp[c*d*f - a*e*g + (c*e*f + c*d*g - b*e*g)*x, x]*(d + e*x)^(m - 1)*((f + g*x)^(n - 1)/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[m, 0] && GtQ[n, 0]
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 11.88 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08
method | result | size |
pseudoelliptic | \(2 \arctan \left (\frac {\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x}\right )+\ln \left (\frac {x +\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x}\right )+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-4 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3}}\right )}{2}-\ln \left (\frac {\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}-x}{x}\right )\) | \(93\) |
trager | \(\text {Expression too large to display}\) | \(3973\) |
Input:
int((x^4-x^3)^(1/4)/(x^2-2*x-1),x,method=_RETURNVERBOSE)
Output:
2*arctan((x^3*(-1+x))^(1/4)/x)+ln((x+(x^3*(-1+x))^(1/4))/x)+1/2*sum(ln((-_ R*x+(x^3*(-1+x))^(1/4))/x)/_R^3,_R=RootOf(_Z^8-4*_Z^4+2))-ln(((x^3*(-1+x)) ^(1/4)-x)/x)
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.11 (sec) , antiderivative size = 536, normalized size of antiderivative = 6.23 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{-1-2 x+x^2} \, dx =\text {Too large to display} \] Input:
integrate((x^4-x^3)^(1/4)/(x^2-2*x-1),x, algorithm="fricas")
Output:
1/2*sqrt(1/2)*sqrt(-sqrt(7*sqrt(2) + 10))*log((sqrt(1/2)*(sqrt(2)*x - 2*x) *sqrt(-sqrt(7*sqrt(2) + 10)) + (x^4 - x^3)^(1/4))/x) - 1/2*sqrt(1/2)*sqrt( -sqrt(7*sqrt(2) + 10))*log(-(sqrt(1/2)*(sqrt(2)*x - 2*x)*sqrt(-sqrt(7*sqrt (2) + 10)) - (x^4 - x^3)^(1/4))/x) - 1/2*sqrt(1/2)*sqrt(-sqrt(-7*sqrt(2) + 10))*log((sqrt(1/2)*(sqrt(2)*x + 2*x)*sqrt(-sqrt(-7*sqrt(2) + 10)) + (x^4 - x^3)^(1/4))/x) + 1/2*sqrt(1/2)*sqrt(-sqrt(-7*sqrt(2) + 10))*log(-(sqrt( 1/2)*(sqrt(2)*x + 2*x)*sqrt(-sqrt(-7*sqrt(2) + 10)) - (x^4 - x^3)^(1/4))/x ) + 1/2*sqrt(1/2)*(7*sqrt(2) + 10)^(1/4)*log((sqrt(1/2)*(sqrt(2)*x - 2*x)* (7*sqrt(2) + 10)^(1/4) + (x^4 - x^3)^(1/4))/x) - 1/2*sqrt(1/2)*(7*sqrt(2) + 10)^(1/4)*log(-(sqrt(1/2)*(sqrt(2)*x - 2*x)*(7*sqrt(2) + 10)^(1/4) - (x^ 4 - x^3)^(1/4))/x) - 1/2*sqrt(1/2)*(-7*sqrt(2) + 10)^(1/4)*log((sqrt(1/2)* (sqrt(2)*x + 2*x)*(-7*sqrt(2) + 10)^(1/4) + (x^4 - x^3)^(1/4))/x) + 1/2*sq rt(1/2)*(-7*sqrt(2) + 10)^(1/4)*log(-(sqrt(1/2)*(sqrt(2)*x + 2*x)*(-7*sqrt (2) + 10)^(1/4) - (x^4 - x^3)^(1/4))/x) - 2*arctan((x^4 - x^3)^(3/4)/(x^3 - x^2)) + log((x + (x^4 - x^3)^(1/4))/x) - log(-(x - (x^4 - x^3)^(1/4))/x)
Not integrable
Time = 0.49 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.22 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{-1-2 x+x^2} \, dx=\int \frac {\sqrt [4]{x^{3} \left (x - 1\right )}}{x^{2} - 2 x - 1}\, dx \] Input:
integrate((x**4-x**3)**(1/4)/(x**2-2*x-1),x)
Output:
Integral((x**3*(x - 1))**(1/4)/(x**2 - 2*x - 1), x)
Not integrable
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.28 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{-1-2 x+x^2} \, dx=\int { \frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x^{2} - 2 \, x - 1} \,d x } \] Input:
integrate((x^4-x^3)^(1/4)/(x^2-2*x-1),x, algorithm="maxima")
Output:
integrate((x^4 - x^3)^(1/4)/(x^2 - 2*x - 1), x)
Not integrable
Time = 0.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.28 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{-1-2 x+x^2} \, dx=\int { \frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x^{2} - 2 \, x - 1} \,d x } \] Input:
integrate((x^4-x^3)^(1/4)/(x^2-2*x-1),x, algorithm="giac")
Output:
integrate((x^4 - x^3)^(1/4)/(x^2 - 2*x - 1), x)
Not integrable
Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.33 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{-1-2 x+x^2} \, dx=-\int \frac {{\left (x^4-x^3\right )}^{1/4}}{-x^2+2\,x+1} \,d x \] Input:
int(-(x^4 - x^3)^(1/4)/(2*x - x^2 + 1),x)
Output:
-int((x^4 - x^3)^(1/4)/(2*x - x^2 + 1), x)
Not integrable
Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.24 \[ \int \frac {\sqrt [4]{-x^3+x^4}}{-1-2 x+x^2} \, dx=\int \frac {x^{\frac {3}{4}} \left (x -1\right )^{\frac {1}{4}}}{x^{2}-2 x -1}d x \] Input:
int((x^4-x^3)^(1/4)/(x^2-2*x-1),x)
Output:
int((x**(3/4)*(x - 1)**(1/4))/(x**2 - 2*x - 1),x)