Integrand size = 32, antiderivative size = 261 \[ \int \frac {\left (-1+x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-1-x+x^2} \, dx=\frac {1}{4} (11+4 x) \sqrt [4]{-x^3+x^4}-\frac {49}{8} \arctan \left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )+\sqrt {2 \left (11+5 \sqrt {5}\right )} \arctan \left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-x^3+x^4}}\right )-\sqrt {2 \left (-11+5 \sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-x^3+x^4}}\right )+\frac {49}{8} \text {arctanh}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )-\sqrt {2 \left (11+5 \sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-x^3+x^4}}\right )+\sqrt {2 \left (-11+5 \sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-x^3+x^4}}\right ) \] Output:
1/4*(11+4*x)*(x^4-x^3)^(1/4)-49/8*arctan(x/(x^4-x^3)^(1/4))+(22+10*5^(1/2) )^(1/2)*arctan(1/2*(-2+2*5^(1/2))^(1/2)*x/(x^4-x^3)^(1/4))-(-22+10*5^(1/2) )^(1/2)*arctan(1/2*(2+2*5^(1/2))^(1/2)*x/(x^4-x^3)^(1/4))+49/8*arctanh(x/( x^4-x^3)^(1/4))-(22+10*5^(1/2))^(1/2)*arctanh(1/2*(-2+2*5^(1/2))^(1/2)*x/( x^4-x^3)^(1/4))+(-22+10*5^(1/2))^(1/2)*arctanh(1/2*(2+2*5^(1/2))^(1/2)*x/( x^4-x^3)^(1/4))
Time = 1.36 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.50 \[ \int \frac {\left (-1+x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-1-x+x^2} \, dx=\frac {(-1+x)^{3/4} x^{9/4} \left (22 \sqrt [4]{-1+x} x^{3/4}+8 \sqrt [4]{-1+x} x^{7/4}-49 \arctan \left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )+12 \sqrt {2 \left (1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )}}{\sqrt [4]{\frac {-1+x}{x}}}\right )+4 \sqrt {10 \left (1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )}}{\sqrt [4]{\frac {-1+x}{x}}}\right )-12 \sqrt {2 \left (-1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )}}{\sqrt [4]{\frac {-1+x}{x}}}\right )+4 \sqrt {10 \left (-1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )}}{\sqrt [4]{\frac {-1+x}{x}}}\right )+49 \text {arctanh}\left (\frac {1}{\sqrt [4]{\frac {-1+x}{x}}}\right )-4 \sqrt {2 \left (1+\sqrt {5}\right )} \left (3+\sqrt {5}\right ) \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )}}{\sqrt [4]{\frac {-1+x}{x}}}\right )+12 \sqrt {2 \left (-1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )}}{\sqrt [4]{\frac {-1+x}{x}}}\right )-4 \sqrt {10 \left (-1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )}}{\sqrt [4]{\frac {-1+x}{x}}}\right )\right )}{8 \left ((-1+x) x^3\right )^{3/4}} \] Input:
Integrate[((-1 + x + 2*x^2)*(-x^3 + x^4)^(1/4))/(-1 - 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) - 49*ArcTan[((-1 + x)/x)^(-1/4)] + 12*Sqrt[2*(1 + Sqrt[5])]*ArcTan[Sqr t[(-1 + Sqrt[5])/2]/((-1 + x)/x)^(1/4)] + 4*Sqrt[10*(1 + Sqrt[5])]*ArcTan[ Sqrt[(-1 + Sqrt[5])/2]/((-1 + x)/x)^(1/4)] - 12*Sqrt[2*(-1 + Sqrt[5])]*Arc Tan[Sqrt[(1 + Sqrt[5])/2]/((-1 + x)/x)^(1/4)] + 4*Sqrt[10*(-1 + Sqrt[5])]* ArcTan[Sqrt[(1 + Sqrt[5])/2]/((-1 + x)/x)^(1/4)] + 49*ArcTanh[((-1 + x)/x) ^(-1/4)] - 4*Sqrt[2*(1 + Sqrt[5])]*(3 + Sqrt[5])*ArcTanh[Sqrt[(-1 + Sqrt[5 ])/2]/((-1 + x)/x)^(1/4)] + 12*Sqrt[2*(-1 + Sqrt[5])]*ArcTanh[Sqrt[(1 + Sq rt[5])/2]/((-1 + x)/x)^(1/4)] - 4*Sqrt[10*(-1 + Sqrt[5])]*ArcTanh[Sqrt[(1 + Sqrt[5])/2]/((-1 + x)/x)^(1/4)]))/(8*((-1 + x)*x^3)^(3/4))
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 1.78 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.66, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2467, 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.
\(\displaystyle \int \frac {\left (2 x^2+x-1\right ) \sqrt [4]{x^4-x^3}}{x^2-x-1} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x^4-x^3} \int \frac {\sqrt [4]{x-1} x^{3/4} \left (-2 x^2-x+1\right )}{-x^2+x+1}dx}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {4 \sqrt [4]{x^4-x^3} \int \frac {\sqrt [4]{x-1} x^{3/2} \left (-2 x^2-x+1\right )}{-x^2+x+1}d\sqrt [4]{x}}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \frac {4 \sqrt [4]{x^4-x^3} \int \left (2 \sqrt [4]{x-1} x^{3/2}+3 \sqrt [4]{x-1} \sqrt {x}+\frac {(-4 x-3) \sqrt [4]{x-1} \sqrt {x}}{-x^2+x+1}\right )d\sqrt [4]{x}}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 \sqrt [4]{x^4-x^3} \left (\frac {2 \sqrt [4]{x-1} x^{3/4} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{4},1,\frac {7}{4},x,\frac {2 x}{1-\sqrt {5}}\right )}{\sqrt {5} \left (1-\sqrt {5}\right ) \sqrt [4]{1-x}}-\frac {2 \sqrt [4]{x-1} x^{3/4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\frac {2 x}{1+\sqrt {5}},x\right )}{\sqrt {5} \left (1+\sqrt {5}\right ) \sqrt [4]{1-x}}-\frac {49}{32} \arctan \left (\frac {\sqrt [4]{x}}{\sqrt [4]{x-1}}\right )+\frac {2^{3/4} \sqrt [4]{3+\sqrt {5}} \arctan \left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )}{\sqrt {5}}+\frac {2^{3/4} \sqrt [4]{3-\sqrt {5}} \arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )}{\sqrt {5}}+\frac {49}{32} \text {arctanh}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x-1}}\right )-\frac {2^{3/4} \sqrt [4]{3+\sqrt {5}} \text {arctanh}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )}{\sqrt {5}}-\frac {2^{3/4} \sqrt [4]{3-\sqrt {5}} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )}{\sqrt {5}}+\frac {1}{4} \sqrt [4]{x-1} x^{7/4}+\frac {11}{16} \sqrt [4]{x-1} x^{3/4}\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
Input:
Int[((-1 + x + 2*x^2)*(-x^3 + x^4)^(1/4))/(-1 - 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 + (2*(-1 + x)^(1/4)*x^(3/4)*AppellF1[3/4, -1/4, 1, 7/4, x, (2*x)/ (1 - Sqrt[5])])/(Sqrt[5]*(1 - Sqrt[5])*(1 - x)^(1/4)) - (2*(-1 + x)^(1/4)* x^(3/4)*AppellF1[3/4, 1, -1/4, 7/4, (2*x)/(1 + Sqrt[5]), x])/(Sqrt[5]*(1 + Sqrt[5])*(1 - x)^(1/4)) - (49*ArcTan[x^(1/4)/(-1 + x)^(1/4)])/32 + (2^(3/ 4)*(3 + Sqrt[5])^(1/4)*ArcTan[((2/(3 + Sqrt[5]))^(1/4)*x^(1/4))/(-1 + x)^( 1/4)])/Sqrt[5] + (2^(3/4)*(3 - Sqrt[5])^(1/4)*ArcTan[(((3 + Sqrt[5])/2)^(1 /4)*x^(1/4))/(-1 + x)^(1/4)])/Sqrt[5] + (49*ArcTanh[x^(1/4)/(-1 + x)^(1/4) ])/32 - (2^(3/4)*(3 + Sqrt[5])^(1/4)*ArcTanh[((2/(3 + Sqrt[5]))^(1/4)*x^(1 /4))/(-1 + x)^(1/4)])/Sqrt[5] - (2^(3/4)*(3 - Sqrt[5])^(1/4)*ArcTanh[(((3 + Sqrt[5])/2)^(1/4)*x^(1/4))/(-1 + x)^(1/4)])/Sqrt[5]))/((-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 = 13.60 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.06
method | result | size |
pseudoelliptic | \(\frac {x^{6} \left (-\frac {\sqrt {2+2 \sqrt {5}}\, \left (3+\sqrt {5}\right ) \operatorname {arctanh}\left (\frac {2 \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x \sqrt {-2+2 \sqrt {5}}}\right )}{2}-\frac {\sqrt {-2+2 \sqrt {5}}\, \left (\sqrt {5}-3\right ) \operatorname {arctanh}\left (\frac {2 \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x \sqrt {2+2 \sqrt {5}}}\right )}{2}-\frac {\sqrt {2+2 \sqrt {5}}\, \left (3+\sqrt {5}\right ) \arctan \left (\frac {2 \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x \sqrt {-2+2 \sqrt {5}}}\right )}{2}-\frac {\sqrt {-2+2 \sqrt {5}}\, \left (\sqrt {5}-3\right ) \arctan \left (\frac {2 \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x \sqrt {2+2 \sqrt {5}}}\right )}{2}-\frac {49 \ln \left (\frac {\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}-x}{x}\right )}{16}+\frac {49 \ln \left (\frac {\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}+x}{x}\right )}{16}+\frac {49 \arctan \left (\frac {\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x}\right )}{8}+\left (\frac {11}{4}+x \right ) \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}\right )}{{\left (-\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}+x \right )}^{2} \left (x^{2}+\sqrt {x^{3} \left (-1+x \right )}\right )^{2} {\left (\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}+x \right )}^{2}}\) | \(276\) |
trager | \(\text {Expression too large to display}\) | \(1822\) |
risch | \(\text {Expression too large to display}\) | \(4137\) |
Input:
int((2*x^2+x-1)*(x^4-x^3)^(1/4)/(x^2-x-1),x,method=_RETURNVERBOSE)
Output:
x^6*(-1/2*(2+2*5^(1/2))^(1/2)*(3+5^(1/2))*arctanh(2*(x^3*(-1+x))^(1/4)/x/( -2+2*5^(1/2))^(1/2))-1/2*(-2+2*5^(1/2))^(1/2)*(5^(1/2)-3)*arctanh(2*(x^3*( -1+x))^(1/4)/x/(2+2*5^(1/2))^(1/2))-1/2*(2+2*5^(1/2))^(1/2)*(3+5^(1/2))*ar ctan(2*(x^3*(-1+x))^(1/4)/x/(-2+2*5^(1/2))^(1/2))-1/2*(-2+2*5^(1/2))^(1/2) *(5^(1/2)-3)*arctan(2*(x^3*(-1+x))^(1/4)/x/(2+2*5^(1/2))^(1/2))-49/16*ln(( (x^3*(-1+x))^(1/4)-x)/x)+49/16*ln(((x^3*(-1+x))^(1/4)+x)/x)+49/8*arctan((x ^3*(-1+x))^(1/4)/x)+(11/4+x)*(x^3*(-1+x))^(1/4))/(-(x^3*(-1+x))^(1/4)+x)^2 /(x^2+(x^3*(-1+x))^(1/2))^2/((x^3*(-1+x))^(1/4)+x)^2
Time = 0.11 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.42 \[ \int \frac {\left (-1+x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-1-x+x^2} \, dx=-2 \, \sqrt {\frac {5}{2} \, \sqrt {5} - \frac {11}{2}} \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {3}{4}} \sqrt {\frac {5}{2} \, \sqrt {5} - \frac {11}{2}} {\left (\sqrt {5} + 2\right )}}{x^{3} - x^{2}}\right ) + 2 \, \sqrt {\frac {5}{2} \, \sqrt {5} + \frac {11}{2}} \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} \sqrt {\frac {5}{2} \, \sqrt {5} + \frac {11}{2}} {\left (\sqrt {5} - 3\right )}}{2 \, x}\right ) - \sqrt {\frac {5}{2} \, \sqrt {5} + \frac {11}{2}} \log \left (\frac {{\left (\sqrt {5} x - 2 \, x\right )} \sqrt {\frac {5}{2} \, \sqrt {5} + \frac {11}{2}} + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \sqrt {\frac {5}{2} \, \sqrt {5} + \frac {11}{2}} \log \left (-\frac {{\left (\sqrt {5} x - 2 \, x\right )} \sqrt {\frac {5}{2} \, \sqrt {5} + \frac {11}{2}} - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \sqrt {\frac {5}{2} \, \sqrt {5} - \frac {11}{2}} \log \left (\frac {{\left (\sqrt {5} x + 2 \, x\right )} \sqrt {\frac {5}{2} \, \sqrt {5} - \frac {11}{2}} + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \sqrt {\frac {5}{2} \, \sqrt {5} - \frac {11}{2}} \log \left (-\frac {{\left (\sqrt {5} x + 2 \, x\right )} \sqrt {\frac {5}{2} \, \sqrt {5} - \frac {11}{2}} - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} {\left (4 \, x + 11\right )} - \frac {49}{8} \, \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {3}{4}}}{x^{3} - x^{2}}\right ) + \frac {49}{16} \, \log \left (\frac {x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {49}{16} \, \log \left (-\frac {x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \] Input:
integrate((2*x^2+x-1)*(x^4-x^3)^(1/4)/(x^2-x-1),x, algorithm="fricas")
Output:
-2*sqrt(5/2*sqrt(5) - 11/2)*arctan((x^4 - x^3)^(3/4)*sqrt(5/2*sqrt(5) - 11 /2)*(sqrt(5) + 2)/(x^3 - x^2)) + 2*sqrt(5/2*sqrt(5) + 11/2)*arctan(1/2*(x^ 4 - x^3)^(1/4)*sqrt(5/2*sqrt(5) + 11/2)*(sqrt(5) - 3)/x) - sqrt(5/2*sqrt(5 ) + 11/2)*log(((sqrt(5)*x - 2*x)*sqrt(5/2*sqrt(5) + 11/2) + (x^4 - x^3)^(1 /4))/x) + sqrt(5/2*sqrt(5) + 11/2)*log(-((sqrt(5)*x - 2*x)*sqrt(5/2*sqrt(5 ) + 11/2) - (x^4 - x^3)^(1/4))/x) + sqrt(5/2*sqrt(5) - 11/2)*log(((sqrt(5) *x + 2*x)*sqrt(5/2*sqrt(5) - 11/2) + (x^4 - x^3)^(1/4))/x) - sqrt(5/2*sqrt (5) - 11/2)*log(-((sqrt(5)*x + 2*x)*sqrt(5/2*sqrt(5) - 11/2) - (x^4 - x^3) ^(1/4))/x) + 1/4*(x^4 - x^3)^(1/4)*(4*x + 11) - 49/8*arctan((x^4 - x^3)^(3 /4)/(x^3 - x^2)) + 49/16*log((x + (x^4 - x^3)^(1/4))/x) - 49/16*log(-(x - (x^4 - x^3)^(1/4))/x)
\[ \int \frac {\left (-1+x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-1-x+x^2} \, dx=\int \frac {\sqrt [4]{x^{3} \left (x - 1\right )} \left (x + 1\right ) \left (2 x - 1\right )}{x^{2} - x - 1}\, dx \] Input:
integrate((2*x**2+x-1)*(x**4-x**3)**(1/4)/(x**2-x-1),x)
Output:
Integral((x**3*(x - 1))**(1/4)*(x + 1)*(2*x - 1)/(x**2 - x - 1), x)
\[ \int \frac {\left (-1+x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-1-x+x^2} \, dx=\int { \frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} {\left (2 \, x^{2} + x - 1\right )}}{x^{2} - x - 1} \,d x } \] Input:
integrate((2*x^2+x-1)*(x^4-x^3)^(1/4)/(x^2-x-1),x, algorithm="maxima")
Output:
integrate((x^4 - x^3)^(1/4)*(2*x^2 + x - 1)/(x^2 - x - 1), x)
Time = 0.37 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-1+x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-1-x+x^2} \, dx=-\frac {1}{4} \, {\left (11 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {5}{4}} - 15 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right )} x^{2} + \sqrt {10 \, \sqrt {5} - 22} \arctan \left (\frac {{\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}}{\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}}}\right ) - \sqrt {10 \, \sqrt {5} + 22} \arctan \left (\frac {{\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}}{\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}}}\right ) + \frac {1}{2} \, \sqrt {10 \, \sqrt {5} - 22} \log \left (\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \sqrt {10 \, \sqrt {5} + 22} \log \left (\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \sqrt {10 \, \sqrt {5} - 22} \log \left ({\left | -\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} \right |}\right ) + \frac {1}{2} \, \sqrt {10 \, \sqrt {5} + 22} \log \left ({\left | -\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} \right |}\right ) + \frac {49}{8} \, \arctan \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + \frac {49}{16} \, \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {49}{16} \, \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \] Input:
integrate((2*x^2+x-1)*(x^4-x^3)^(1/4)/(x^2-x-1),x, algorithm="giac")
Output:
-1/4*(11*(-1/x + 1)^(5/4) - 15*(-1/x + 1)^(1/4))*x^2 + sqrt(10*sqrt(5) - 2 2)*arctan((-1/x + 1)^(1/4)/sqrt(1/2*sqrt(5) + 1/2)) - sqrt(10*sqrt(5) + 22 )*arctan((-1/x + 1)^(1/4)/sqrt(1/2*sqrt(5) - 1/2)) + 1/2*sqrt(10*sqrt(5) - 22)*log(sqrt(1/2*sqrt(5) + 1/2) + (-1/x + 1)^(1/4)) - 1/2*sqrt(10*sqrt(5) + 22)*log(sqrt(1/2*sqrt(5) - 1/2) + (-1/x + 1)^(1/4)) - 1/2*sqrt(10*sqrt( 5) - 22)*log(abs(-sqrt(1/2*sqrt(5) + 1/2) + (-1/x + 1)^(1/4))) + 1/2*sqrt( 10*sqrt(5) + 22)*log(abs(-sqrt(1/2*sqrt(5) - 1/2) + (-1/x + 1)^(1/4))) + 4 9/8*arctan((-1/x + 1)^(1/4)) + 49/16*log((-1/x + 1)^(1/4) + 1) - 49/16*log (abs((-1/x + 1)^(1/4) - 1))
Timed out. \[ \int \frac {\left (-1+x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-1-x+x^2} \, dx=\int -\frac {{\left (x^4-x^3\right )}^{1/4}\,\left (2\,x^2+x-1\right )}{-x^2+x+1} \,d x \] Input:
int(-((x^4 - x^3)^(1/4)*(x + 2*x^2 - 1))/(x - x^2 + 1),x)
Output:
int(-((x^4 - x^3)^(1/4)*(x + 2*x^2 - 1))/(x - x^2 + 1), x)
\[ \int \frac {\left (-1+x+2 x^2\right ) \sqrt [4]{-x^3+x^4}}{-1-x+x^2} \, dx=x^{\frac {7}{4}} \left (x -1\right )^{\frac {1}{4}}+\frac {11 x^{\frac {3}{4}} \left (x -1\right )^{\frac {1}{4}}}{4}-\frac {33 \left (\int \frac {\left (x -1\right )^{\frac {1}{4}}}{x^{\frac {13}{4}}-2 x^{\frac {9}{4}}+x^{\frac {1}{4}}}d x \right )}{16}+\frac {49 \left (\int \frac {\left (x -1\right )^{\frac {1}{4}} x^{2}}{x^{\frac {13}{4}}-2 x^{\frac {9}{4}}+x^{\frac {1}{4}}}d x \right )}{16}-\frac {\left (\int \frac {\left (x -1\right )^{\frac {1}{4}} x}{x^{\frac {13}{4}}-2 x^{\frac {9}{4}}+x^{\frac {1}{4}}}d x \right )}{16} \] Input:
int((2*x^2+x-1)*(x^4-x^3)^(1/4)/(x^2-x-1),x)
Output:
(16*x**(3/4)*(x - 1)**(1/4)*x + 44*x**(3/4)*(x - 1)**(1/4) - 33*int((x - 1 )**(1/4)/(x**(1/4)*x**3 - 2*x**(1/4)*x**2 + x**(1/4)),x) + 49*int(((x - 1) **(1/4)*x**2)/(x**(1/4)*x**3 - 2*x**(1/4)*x**2 + x**(1/4)),x) - int(((x - 1)**(1/4)*x)/(x**(1/4)*x**3 - 2*x**(1/4)*x**2 + x**(1/4)),x))/16