Integrand size = 34, antiderivative size = 158 \[ \int \frac {\left (2-x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-2-2 x+x^2} \, dx=\frac {1}{4} (11+4 x) \sqrt [4]{-x^3+x^4}-\frac {177}{8} \arctan \left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )+\frac {177}{8} \text {arctanh}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )-9 \text {RootSum}\left [3-6 \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-2 \log (x)+2 \log \left (\sqrt [4]{-x^3+x^4}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-\log \left (\sqrt [4]{-x^3+x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-3 \text {$\#$1}^3+2 \text {$\#$1}^7}\&\right ] \] Output:
Unintegrable
Time = 0.04 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.11 \[ \int \frac {\left (2-x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-2-2 x+x^2} \, dx=\frac {(-1+x)^{3/4} x^{9/4} \left (22 \sqrt [4]{-1+x} x^{3/4}+8 \sqrt [4]{-1+x} x^{7/4}-177 \arctan \left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )+177 \text {arctanh}\left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )-18 \text {RootSum}\left [3-6 \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-2 \log (x)+8 \log \left (\sqrt [4]{-1+x}-\sqrt [4]{x} \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-4 \log \left (\sqrt [4]{-1+x}-\sqrt [4]{x} \text {$\#$1}\right ) \text {$\#$1}^4}{-3 \text {$\#$1}^3+2 \text {$\#$1}^7}\&\right ]\right )}{8 \left ((-1+x) x^3\right )^{3/4}} \] Input:
Integrate[((2 - x + 2*x^2)*(-x^3 + x^4)^(1/4))/(-2 - 2*x + x^2),x]
Output:
((-1 + x)^(3/4)*x^(9/4)*(22*(-1 + x)^(1/4)*x^(3/4) + 8*(-1 + x)^(1/4)*x^(7 /4) - 177*ArcTan[((-1 + x)/x)^(-1/4)] + 177*ArcTanh[((-1 + x)/x)^(-1/4)] - 18*RootSum[3 - 6*#1^4 + 2*#1^8 & , (-2*Log[x] + 8*Log[(-1 + x)^(1/4) - x^ (1/4)*#1] + Log[x]*#1^4 - 4*Log[(-1 + x)^(1/4) - x^(1/4)*#1]*#1^4)/(-3*#1^ 3 + 2*#1^7) & ]))/(8*((-1 + x)*x^3)^(3/4))
Leaf count is larger than twice the leaf count of optimal. \(586\) vs. \(2(158)=316\).
Time = 1.90 (sec) , antiderivative size = 586, normalized size of antiderivative = 3.71, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {2467, 25, 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.
\(\displaystyle \int \frac {\left (2 x^2-x+2\right ) \sqrt [4]{x^4-x^3}}{x^2-2 x-2} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x^4-x^3} \int -\frac {\sqrt [4]{x-1} x^{3/4} \left (2 x^2-x+2\right )}{-x^2+2 x+2}dx}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^3} \int \frac {\sqrt [4]{x-1} x^{3/4} \left (2 x^2-x+2\right )}{-x^2+2 x+2}dx}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {4 \sqrt [4]{x^4-x^3} \int \frac {\sqrt [4]{x-1} x^{3/2} \left (2 x^2-x+2\right )}{-x^2+2 x+2}d\sqrt [4]{x}}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle -\frac {4 \sqrt [4]{x^4-x^3} \int \left (-2 \sqrt [4]{x-1} x^{3/2}-3 \sqrt [4]{x-1} \sqrt {x}+\frac {6 \sqrt [4]{x-1} (2 x+1) \sqrt {x}}{-x^2+2 x+2}\right )d\sqrt [4]{x}}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 \sqrt [4]{x^4-x^3} \left (-\frac {\sqrt [4]{x-1} x^{3/4} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{4},1,\frac {7}{4},x,\frac {x}{1-\sqrt {3}}\right )}{\sqrt {3} \left (1-\sqrt {3}\right ) \sqrt [4]{1-x}}+\frac {\sqrt [4]{x-1} x^{3/4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\frac {x}{1+\sqrt {3}},x\right )}{\sqrt {3} \left (1+\sqrt {3}\right ) \sqrt [4]{1-x}}+\frac {177}{32} \arctan \left (\frac {\sqrt [4]{x}}{\sqrt [4]{x-1}}\right )-\sqrt [4]{2 \left (9+5 \sqrt {3}\right )} \arctan \left (\frac {\sqrt [4]{\frac {3}{3+\sqrt {3}}} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )-2 \sqrt [4]{\frac {3}{3+\sqrt {3}}} \arctan \left (\frac {\sqrt [4]{\frac {3}{3+\sqrt {3}}} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )-2^{3/4} \sqrt [4]{3+\sqrt {3}} \arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {3}\right )} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )+\sqrt [4]{2 \left (9-5 \sqrt {3}\right )} \arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {3}\right )} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )-\frac {177}{32} \text {arctanh}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x-1}}\right )+\sqrt [4]{2 \left (9+5 \sqrt {3}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {3}{3+\sqrt {3}}} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )+2 \sqrt [4]{\frac {3}{3+\sqrt {3}}} \text {arctanh}\left (\frac {\sqrt [4]{\frac {3}{3+\sqrt {3}}} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )+2^{3/4} \sqrt [4]{3+\sqrt {3}} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {3}\right )} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )-\sqrt [4]{2 \left (9-5 \sqrt {3}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {3}\right )} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )-\frac {1}{4} \sqrt [4]{x-1} x^{7/4}-\frac {11}{16} \sqrt [4]{x-1} x^{3/4}\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
Input:
Int[((2 - x + 2*x^2)*(-x^3 + x^4)^(1/4))/(-2 - 2*x + x^2),x]
Output:
(-4*(-x^3 + x^4)^(1/4)*((-11*(-1 + x)^(1/4)*x^(3/4))/16 - ((-1 + x)^(1/4)* x^(7/4))/4 - ((-1 + x)^(1/4)*x^(3/4)*AppellF1[3/4, -1/4, 1, 7/4, x, x/(1 - Sqrt[3])])/(Sqrt[3]*(1 - Sqrt[3])*(1 - x)^(1/4)) + ((-1 + x)^(1/4)*x^(3/4 )*AppellF1[3/4, 1, -1/4, 7/4, x/(1 + Sqrt[3]), x])/(Sqrt[3]*(1 + Sqrt[3])* (1 - x)^(1/4)) + (177*ArcTan[x^(1/4)/(-1 + x)^(1/4)])/32 - 2*(3/(3 + Sqrt[ 3]))^(1/4)*ArcTan[((3/(3 + Sqrt[3]))^(1/4)*x^(1/4))/(-1 + x)^(1/4)] - (2*( 9 + 5*Sqrt[3]))^(1/4)*ArcTan[((3/(3 + Sqrt[3]))^(1/4)*x^(1/4))/(-1 + x)^(1 /4)] + (2*(9 - 5*Sqrt[3]))^(1/4)*ArcTan[(((3 + Sqrt[3])/2)^(1/4)*x^(1/4))/ (-1 + x)^(1/4)] - 2^(3/4)*(3 + Sqrt[3])^(1/4)*ArcTan[(((3 + Sqrt[3])/2)^(1 /4)*x^(1/4))/(-1 + x)^(1/4)] - (177*ArcTanh[x^(1/4)/(-1 + x)^(1/4)])/32 + 2*(3/(3 + Sqrt[3]))^(1/4)*ArcTanh[((3/(3 + Sqrt[3]))^(1/4)*x^(1/4))/(-1 + x)^(1/4)] + (2*(9 + 5*Sqrt[3]))^(1/4)*ArcTanh[((3/(3 + Sqrt[3]))^(1/4)*x^( 1/4))/(-1 + x)^(1/4)] - (2*(9 - 5*Sqrt[3]))^(1/4)*ArcTanh[(((3 + Sqrt[3])/ 2)^(1/4)*x^(1/4))/(-1 + x)^(1/4)] + 2^(3/4)*(3 + Sqrt[3])^(1/4)*ArcTanh[(( (3 + Sqrt[3])/2)^(1/4)*x^(1/4))/(-1 + x)^(1/4)]))/((-1 + x)^(1/4)*x^(3/4))
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 = 23.23 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.15
method | result | size |
pseudoelliptic | \(\frac {x^{6} \left (16 \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}} x +354 \arctan \left (\frac {\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x}\right )+144 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{8}-6 \textit {\_Z}^{4}+3\right )}{\sum }\frac {\left (\textit {\_R}^{4}-2\right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3} \left (2 \textit {\_R}^{4}-3\right )}\right )+177 \ln \left (\frac {x +\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x}\right )-177 \ln \left (\frac {\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}-x}{x}\right )+44 \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}\right )}{16 \left (x^{2}+\sqrt {x^{3} \left (-1+x \right )}\right )^{2} {\left (x +\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}\right )}^{2} {\left (\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}-x \right )}^{2}}\) | \(182\) |
trager | \(\text {Expression too large to display}\) | \(3309\) |
risch | \(\text {Expression too large to display}\) | \(5735\) |
Input:
int((2*x^2-x+2)*(x^4-x^3)^(1/4)/(x^2-2*x-2),x,method=_RETURNVERBOSE)
Output:
1/16*x^6*(16*(x^3*(-1+x))^(1/4)*x+354*arctan((x^3*(-1+x))^(1/4)/x)+144*sum ((_R^4-2)*ln((-_R*x+(x^3*(-1+x))^(1/4))/x)/_R^3/(2*_R^4-3),_R=RootOf(2*_Z^ 8-6*_Z^4+3))+177*ln((x+(x^3*(-1+x))^(1/4))/x)-177*ln(((x^3*(-1+x))^(1/4)-x )/x)+44*(x^3*(-1+x))^(1/4))/(x^2+(x^3*(-1+x))^(1/2))^2/(x+(x^3*(-1+x))^(1/ 4))^2/((x^3*(-1+x))^(1/4)-x)^2
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.09 (sec) , antiderivative size = 602, normalized size of antiderivative = 3.81 \[ \int \frac {\left (2-x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-2-2 x+x^2} \, dx =\text {Too large to display} \] Input:
integrate((2*x^2-x+2)*(x^4-x^3)^(1/4)/(x^2-2*x-2),x, algorithm="fricas")
Output:
3/2*sqrt(2)*sqrt(sqrt(2)*sqrt(71*sqrt(3) + 123))*log(3*(sqrt(2)*(sqrt(3)*x - 2*x)*sqrt(sqrt(2)*sqrt(71*sqrt(3) + 123)) + 2*(x^4 - x^3)^(1/4))/x) - 3 /2*sqrt(2)*sqrt(sqrt(2)*sqrt(71*sqrt(3) + 123))*log(-3*(sqrt(2)*(sqrt(3)*x - 2*x)*sqrt(sqrt(2)*sqrt(71*sqrt(3) + 123)) - 2*(x^4 - x^3)^(1/4))/x) + 3 /2*sqrt(2)*sqrt(-sqrt(2)*sqrt(71*sqrt(3) + 123))*log(3*(sqrt(2)*(sqrt(3)*x - 2*x)*sqrt(-sqrt(2)*sqrt(71*sqrt(3) + 123)) + 2*(x^4 - x^3)^(1/4))/x) - 3/2*sqrt(2)*sqrt(-sqrt(2)*sqrt(71*sqrt(3) + 123))*log(-3*(sqrt(2)*(sqrt(3) *x - 2*x)*sqrt(-sqrt(2)*sqrt(71*sqrt(3) + 123)) - 2*(x^4 - x^3)^(1/4))/x) - 1/2*sqrt(2)*sqrt(-sqrt(-11502*sqrt(3) + 19926))*log((sqrt(2)*(sqrt(3)*x + 2*x)*sqrt(-sqrt(-11502*sqrt(3) + 19926)) + 6*(x^4 - x^3)^(1/4))/x) + 1/2 *sqrt(2)*sqrt(-sqrt(-11502*sqrt(3) + 19926))*log(-(sqrt(2)*(sqrt(3)*x + 2* x)*sqrt(-sqrt(-11502*sqrt(3) + 19926)) - 6*(x^4 - x^3)^(1/4))/x) - 1/2*sqr t(2)*(-11502*sqrt(3) + 19926)^(1/4)*log((sqrt(2)*(sqrt(3)*x + 2*x)*(-11502 *sqrt(3) + 19926)^(1/4) + 6*(x^4 - x^3)^(1/4))/x) + 1/2*sqrt(2)*(-11502*sq rt(3) + 19926)^(1/4)*log(-(sqrt(2)*(sqrt(3)*x + 2*x)*(-11502*sqrt(3) + 199 26)^(1/4) - 6*(x^4 - x^3)^(1/4))/x) + 1/4*(x^4 - x^3)^(1/4)*(4*x + 11) - 1 77/8*arctan((x^4 - x^3)^(3/4)/(x^3 - x^2)) + 177/16*log((x + (x^4 - x^3)^( 1/4))/x) - 177/16*log(-(x - (x^4 - x^3)^(1/4))/x)
Not integrable
Time = 1.89 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.17 \[ \int \frac {\left (2-x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-2-2 x+x^2} \, dx=\int \frac {\sqrt [4]{x^{3} \left (x - 1\right )} \left (2 x^{2} - x + 2\right )}{x^{2} - 2 x - 2}\, dx \] Input:
integrate((2*x**2-x+2)*(x**4-x**3)**(1/4)/(x**2-2*x-2),x)
Output:
Integral((x**3*(x - 1))**(1/4)*(2*x**2 - x + 2)/(x**2 - 2*x - 2), x)
Not integrable
Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.22 \[ \int \frac {\left (2-x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-2-2 x+x^2} \, dx=\int { \frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} {\left (2 \, x^{2} - x + 2\right )}}{x^{2} - 2 \, x - 2} \,d x } \] Input:
integrate((2*x^2-x+2)*(x^4-x^3)^(1/4)/(x^2-2*x-2),x, algorithm="maxima")
Output:
integrate((x^4 - x^3)^(1/4)*(2*x^2 - x + 2)/(x^2 - 2*x - 2), x)
Not integrable
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.22 \[ \int \frac {\left (2-x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-2-2 x+x^2} \, dx=\int { \frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} {\left (2 \, x^{2} - x + 2\right )}}{x^{2} - 2 \, x - 2} \,d x } \] Input:
integrate((2*x^2-x+2)*(x^4-x^3)^(1/4)/(x^2-2*x-2),x, algorithm="giac")
Output:
integrate((x^4 - x^3)^(1/4)*(2*x^2 - x + 2)/(x^2 - 2*x - 2), x)
Not integrable
Time = 0.00 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.23 \[ \int \frac {\left (2-x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-2-2 x+x^2} \, dx=\int -\frac {{\left (x^4-x^3\right )}^{1/4}\,\left (2\,x^2-x+2\right )}{-x^2+2\,x+2} \,d x \] Input:
int(-((x^4 - x^3)^(1/4)*(2*x^2 - x + 2))/(2*x - x^2 + 2),x)
Output:
int(-((x^4 - x^3)^(1/4)*(2*x^2 - x + 2))/(2*x - x^2 + 2), x)
Not integrable
Time = 0.72 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.65 \[ \int \frac {\left (2-x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-2-2 x+x^2} \, dx=x^{\frac {7}{4}} \left (x -1\right )^{\frac {1}{4}}+\frac {11 x^{\frac {3}{4}} \left (x -1\right )^{\frac {1}{4}}}{4}-\frac {33 \left (\int \frac {\left (x -1\right )^{\frac {1}{4}}}{x^{\frac {13}{4}}-3 x^{\frac {9}{4}}+2 x^{\frac {1}{4}}}d x \right )}{8}+\frac {177 \left (\int \frac {\left (x -1\right )^{\frac {1}{4}} x^{2}}{x^{\frac {13}{4}}-3 x^{\frac {9}{4}}+2 x^{\frac {1}{4}}}d x \right )}{16}-\frac {33 \left (\int \frac {\left (x -1\right )^{\frac {1}{4}} x}{x^{\frac {13}{4}}-3 x^{\frac {9}{4}}+2 x^{\frac {1}{4}}}d x \right )}{8} \] Input:
int((2*x^2-x+2)*(x^4-x^3)^(1/4)/(x^2-2*x-2),x)
Output:
(16*x**(3/4)*(x - 1)**(1/4)*x + 44*x**(3/4)*(x - 1)**(1/4) - 66*int((x - 1 )**(1/4)/(x**(1/4)*x**3 - 3*x**(1/4)*x**2 + 2*x**(1/4)),x) + 177*int(((x - 1)**(1/4)*x**2)/(x**(1/4)*x**3 - 3*x**(1/4)*x**2 + 2*x**(1/4)),x) - 66*in t(((x - 1)**(1/4)*x)/(x**(1/4)*x**3 - 3*x**(1/4)*x**2 + 2*x**(1/4)),x))/16