Integrand size = 29, antiderivative size = 177 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^3+x^4}}{x^4 \left (-1+x^4\right )} \, dx=-\frac {4 \left (-5+x+4 x^2\right ) \sqrt [4]{-x^3+x^4}}{45 x^3}+\sqrt [4]{2} \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-x^3+x^4}}\right )-\sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-x^3+x^4}}\right )+\frac {1}{2} \text {RootSum}\left [2-2 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-2 \log (x)+2 \log \left (\sqrt [4]{-x^3+x^4}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-\log \left (\sqrt [4]{-x^3+x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-\text {$\#$1}^3+\text {$\#$1}^7}\&\right ] \] Output:
Unintegrable
Time = 0.12 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.10 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^3+x^4}}{x^4 \left (-1+x^4\right )} \, dx=\frac {(-1+x)^{3/4} \left (-8 \left (4 \sqrt [4]{-1+x} \left (-5+x+4 x^2\right )-45 \sqrt [4]{2} x^{9/4} \arctan \left (\frac {\sqrt [4]{2}}{\sqrt [4]{\frac {-1+x}{x}}}\right )+45 \sqrt [4]{2} x^{9/4} \text {arctanh}\left (\frac {\sqrt [4]{2}}{\sqrt [4]{\frac {-1+x}{x}}}\right )\right )+45 x^{9/4} \text {RootSum}\left [2-2 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-2 \log (x)+8 \log \left (\sqrt [4]{-1+x}-\sqrt [4]{x} \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-4 \log \left (\sqrt [4]{-1+x}-\sqrt [4]{x} \text {$\#$1}\right ) \text {$\#$1}^4}{-\text {$\#$1}^3+\text {$\#$1}^7}\&\right ]\right )}{360 \left ((-1+x) x^3\right )^{3/4}} \] Input:
Integrate[((1 + x^4)*(-x^3 + x^4)^(1/4))/(x^4*(-1 + x^4)),x]
Output:
((-1 + x)^(3/4)*(-8*(4*(-1 + x)^(1/4)*(-5 + x + 4*x^2) - 45*2^(1/4)*x^(9/4 )*ArcTan[2^(1/4)/((-1 + x)/x)^(1/4)] + 45*2^(1/4)*x^(9/4)*ArcTanh[2^(1/4)/ ((-1 + x)/x)^(1/4)]) + 45*x^(9/4)*RootSum[2 - 2*#1^4 + #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)/(-#1^3 + #1^7) & ]))/(360*((-1 + x)*x^3)^(3/4))
Result contains complex when optimal does not.
Time = 1.73 (sec) , antiderivative size = 420, normalized size of antiderivative = 2.37, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2467, 25, 2019, 2035, 25, 2461, 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.
\(\displaystyle \int \frac {\left (x^4+1\right ) \sqrt [4]{x^4-x^3}}{x^4 \left (x^4-1\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x^4-x^3} \int -\frac {\sqrt [4]{x-1} \left (x^4+1\right )}{x^{13/4} \left (1-x^4\right )}dx}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^3} \int \frac {\sqrt [4]{x-1} \left (x^4+1\right )}{x^{13/4} \left (1-x^4\right )}dx}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 2019 |
\(\displaystyle -\frac {\sqrt [4]{x^4-x^3} \int \frac {x^4+1}{(x-1)^{3/4} x^{13/4} \left (-x^3-x^2-x-1\right )}dx}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {4 \sqrt [4]{x^4-x^3} \int -\frac {x^4+1}{(x-1)^{3/4} x^{5/2} \left (x^3+x^2+x+1\right )}d\sqrt [4]{x}}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {4 \sqrt [4]{x^4-x^3} \int \frac {x^4+1}{(x-1)^{3/4} x^{5/2} \left (x^3+x^2+x+1\right )}d\sqrt [4]{x}}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 2461 |
\(\displaystyle \frac {4 \sqrt [4]{x^4-x^3} \int \left (\frac {x^4+1}{2 (x-1)^{3/4} x^{5/2} (x+1)}+\frac {(1-x) \left (x^4+1\right )}{2 (x-1)^{3/4} x^{5/2} \left (x^2+1\right )}\right )d\sqrt [4]{x}}{\sqrt [4]{x-1} x^{3/4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 \sqrt [4]{x^4-x^3} \left (\frac {1}{8} (1-i)^{5/4} \arctan \left (\frac {\sqrt [4]{1-i} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )+\frac {\arctan \left (\frac {\sqrt [4]{1-i} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )}{4 (1-i)^{3/4}}+\frac {1}{8} (1+i)^{5/4} \arctan \left (\frac {\sqrt [4]{1+i} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )+\frac {\arctan \left (\frac {\sqrt [4]{1+i} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )}{4 (1+i)^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )}{2\ 2^{3/4}}-\frac {1}{8} (1-i)^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{1-i} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )-\frac {\text {arctanh}\left (\frac {\sqrt [4]{1-i} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )}{4 (1-i)^{3/4}}-\frac {1}{8} (1+i)^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{1+i} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )-\frac {\text {arctanh}\left (\frac {\sqrt [4]{1+i} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )}{4 (1+i)^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{x-1}}\right )}{2\ 2^{3/4}}+\frac {2 (x-1)^{5/4}}{45 x^{5/4}}+\frac {(x-1)^{5/4}}{18 x^{9/4}}+\frac {\sqrt [4]{x-1}}{90 x^{5/4}}-\frac {\sqrt [4]{x-1}}{18 x^{9/4}}+\frac {2 \sqrt [4]{x-1}}{45 \sqrt [4]{x}}\right )}{\sqrt [4]{x-1} x^{3/4}}\) |
Input:
Int[((1 + x^4)*(-x^3 + x^4)^(1/4))/(x^4*(-1 + x^4)),x]
Output:
(-4*(-x^3 + x^4)^(1/4)*(-1/18*(-1 + x)^(1/4)/x^(9/4) + (-1 + x)^(5/4)/(18* x^(9/4)) + (-1 + x)^(1/4)/(90*x^(5/4)) + (2*(-1 + x)^(5/4))/(45*x^(5/4)) + (2*(-1 + x)^(1/4))/(45*x^(1/4)) + ArcTan[((1 - I)^(1/4)*x^(1/4))/(-1 + x) ^(1/4)]/(4*(1 - I)^(3/4)) + ((1 - I)^(5/4)*ArcTan[((1 - I)^(1/4)*x^(1/4))/ (-1 + x)^(1/4)])/8 + ArcTan[((1 + I)^(1/4)*x^(1/4))/(-1 + x)^(1/4)]/(4*(1 + I)^(3/4)) + ((1 + I)^(5/4)*ArcTan[((1 + I)^(1/4)*x^(1/4))/(-1 + x)^(1/4) ])/8 - ArcTan[(2^(1/4)*x^(1/4))/(-1 + x)^(1/4)]/(2*2^(3/4)) - ArcTanh[((1 - I)^(1/4)*x^(1/4))/(-1 + x)^(1/4)]/(4*(1 - I)^(3/4)) - ((1 - I)^(5/4)*Arc Tanh[((1 - I)^(1/4)*x^(1/4))/(-1 + x)^(1/4)])/8 - ArcTanh[((1 + I)^(1/4)*x ^(1/4))/(-1 + x)^(1/4)]/(4*(1 + I)^(3/4)) - ((1 + I)^(5/4)*ArcTanh[((1 + I )^(1/4)*x^(1/4))/(-1 + x)^(1/4)])/8 + ArcTanh[(2^(1/4)*x^(1/4))/(-1 + x)^( 1/4)]/(2*2^(3/4))))/((-1 + x)^(1/4)*x^(3/4))
Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px , Qx, x]^p*Qx^(p + q), x] /; FreeQ[q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 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[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px /. x -> Sqrt[x]]}, Int[ExpandIntegrand[u, (Qx /. x -> x^2)^p, x], x] /; !SumQ[NonfreeFactors[ Qx, x]]] /; PolyQ[Px, x^2] && GtQ[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
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]
Time = 73.62 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.83
method | result | size |
pseudoelliptic | \(\frac {-45 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\left (\textit {\_R}^{4}-2\right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3} \left (\textit {\_R}^{4}-1\right )}\right ) x^{3}-45 \,2^{\frac {1}{4}} x^{3} \ln \left (\frac {2^{\frac {1}{4}} x +\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{-2^{\frac {1}{4}} x +\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}\right )-90 \,2^{\frac {1}{4}} x^{3} \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{2 x}\right )-32 \left (x +\frac {5}{4}\right ) \left (-1+x \right ) \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{90 x^{3}}\) | \(147\) |
trager | \(\text {Expression too large to display}\) | \(4007\) |
risch | \(\text {Expression too large to display}\) | \(8063\) |
Input:
int((x^4+1)*(x^4-x^3)^(1/4)/x^4/(x^4-1),x,method=_RETURNVERBOSE)
Output:
1/90*(-45*sum((_R^4-2)*ln((-_R*x+(x^3*(-1+x))^(1/4))/x)/_R^3/(_R^4-1),_R=R ootOf(_Z^8-2*_Z^4+2))*x^3-45*2^(1/4)*x^3*ln((2^(1/4)*x+(x^3*(-1+x))^(1/4)) /(-2^(1/4)*x+(x^3*(-1+x))^(1/4)))-90*2^(1/4)*x^3*arctan(1/2*2^(3/4)*(x^3*( -1+x))^(1/4)/x)-32*(x+5/4)*(-1+x)*(x^3*(-1+x))^(1/4))/x^3
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.09 (sec) , antiderivative size = 412, normalized size of antiderivative = 2.33 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^3+x^4}}{x^4 \left (-1+x^4\right )} \, dx=\frac {90 \cdot 8^{\frac {3}{4}} x^{3} \arctan \left (\frac {8^{\frac {3}{4}} {\left (x^{4} - x^{3}\right )}^{\frac {3}{4}}}{4 \, {\left (x^{3} - x^{2}\right )}}\right ) - 45 \cdot 8^{\frac {3}{4}} x^{3} \log \left (\frac {8^{\frac {3}{4}} x + 4 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 45 \cdot 8^{\frac {3}{4}} x^{3} \log \left (-\frac {8^{\frac {3}{4}} x - 4 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 180 \, x^{3} \sqrt {-\sqrt {i + 1}} \log \left (\frac {x \sqrt {-\sqrt {i + 1}} + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - 180 \, x^{3} \sqrt {-\sqrt {i + 1}} \log \left (-\frac {x \sqrt {-\sqrt {i + 1}} - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 180 \, x^{3} \sqrt {-\sqrt {-i + 1}} \log \left (\frac {x \sqrt {-\sqrt {-i + 1}} + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - 180 \, x^{3} \sqrt {-\sqrt {-i + 1}} \log \left (-\frac {x \sqrt {-\sqrt {-i + 1}} - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 180 \, \left (i + 1\right )^{\frac {1}{4}} x^{3} \log \left (\frac {\left (i + 1\right )^{\frac {1}{4}} x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - 180 \, \left (i + 1\right )^{\frac {1}{4}} x^{3} \log \left (-\frac {\left (i + 1\right )^{\frac {1}{4}} x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + 180 \, \left (-i + 1\right )^{\frac {1}{4}} x^{3} \log \left (\frac {\left (-i + 1\right )^{\frac {1}{4}} x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - 180 \, \left (-i + 1\right )^{\frac {1}{4}} x^{3} \log \left (-\frac {\left (-i + 1\right )^{\frac {1}{4}} x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - 32 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} {\left (4 \, x^{2} + x - 5\right )}}{360 \, x^{3}} \] Input:
integrate((x^4+1)*(x^4-x^3)^(1/4)/x^4/(x^4-1),x, algorithm="fricas")
Output:
1/360*(90*8^(3/4)*x^3*arctan(1/4*8^(3/4)*(x^4 - x^3)^(3/4)/(x^3 - x^2)) - 45*8^(3/4)*x^3*log((8^(3/4)*x + 4*(x^4 - x^3)^(1/4))/x) + 45*8^(3/4)*x^3*l og(-(8^(3/4)*x - 4*(x^4 - x^3)^(1/4))/x) + 180*x^3*sqrt(-sqrt(I + 1))*log( (x*sqrt(-sqrt(I + 1)) + (x^4 - x^3)^(1/4))/x) - 180*x^3*sqrt(-sqrt(I + 1)) *log(-(x*sqrt(-sqrt(I + 1)) - (x^4 - x^3)^(1/4))/x) + 180*x^3*sqrt(-sqrt(- I + 1))*log((x*sqrt(-sqrt(-I + 1)) + (x^4 - x^3)^(1/4))/x) - 180*x^3*sqrt( -sqrt(-I + 1))*log(-(x*sqrt(-sqrt(-I + 1)) - (x^4 - x^3)^(1/4))/x) + 180*( I + 1)^(1/4)*x^3*log(((I + 1)^(1/4)*x + (x^4 - x^3)^(1/4))/x) - 180*(I + 1 )^(1/4)*x^3*log(-((I + 1)^(1/4)*x - (x^4 - x^3)^(1/4))/x) + 180*(-I + 1)^( 1/4)*x^3*log(((-I + 1)^(1/4)*x + (x^4 - x^3)^(1/4))/x) - 180*(-I + 1)^(1/4 )*x^3*log(-((-I + 1)^(1/4)*x - (x^4 - x^3)^(1/4))/x) - 32*(x^4 - x^3)^(1/4 )*(4*x^2 + x - 5))/x^3
Not integrable
Time = 1.43 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.18 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^3+x^4}}{x^4 \left (-1+x^4\right )} \, dx=\int \frac {\sqrt [4]{x^{3} \left (x - 1\right )} \left (x^{4} + 1\right )}{x^{4} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}\, dx \] Input:
integrate((x**4+1)*(x**4-x**3)**(1/4)/x**4/(x**4-1),x)
Output:
Integral((x**3*(x - 1))**(1/4)*(x**4 + 1)/(x**4*(x - 1)*(x + 1)*(x**2 + 1) ), x)
Not integrable
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.16 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^3+x^4}}{x^4 \left (-1+x^4\right )} \, dx=\int { \frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )}}{{\left (x^{4} - 1\right )} x^{4}} \,d x } \] Input:
integrate((x^4+1)*(x^4-x^3)^(1/4)/x^4/(x^4-1),x, algorithm="maxima")
Output:
integrate((x^4 - x^3)^(1/4)*(x^4 + 1)/((x^4 - 1)*x^4), x)
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 0.30 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.54 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^3+x^4}}{x^4 \left (-1+x^4\right )} \, dx =\text {Too large to display} \] Input:
integrate((x^4+1)*(x^4-x^3)^(1/4)/x^4/(x^4-1),x, algorithm="giac")
Output:
4/9*(1/x - 1)^2*(-1/x + 1)^(1/4) - 4/5*(-1/x + 1)^(5/4) - 2^(1/4)*arctan(1 /2*2^(3/4)*(-1/x + 1)^(1/4)) - 1/2*2^(1/4)*log(2^(1/4) + (-1/x + 1)^(1/4)) - (-1/16*I + 1/16)^(1/4)*log(I*(73786976294838206464*I - 7378697629483820 6464)^(1/4) - (65536*I - 65536)*(-1/x + 1)^(1/4)) + (-1/16*I + 1/16)^(1/4) *log(I*(73786976294838206464*I - 73786976294838206464)^(1/4) + (65536*I - 65536)*(-1/x + 1)^(1/4)) + I*(1/16*I + 1/16)^(1/4)*log(I*(-737869762948382 06464*I - 73786976294838206464)^(1/4) - (65536*I - 65536)*(-1/x + 1)^(1/4) ) - I*(1/16*I + 1/16)^(1/4)*log(I*(-73786976294838206464*I - 7378697629483 8206464)^(1/4) + (65536*I - 65536)*(-1/x + 1)^(1/4)) - (1/16*I + 1/16)^(1/ 4)*log(-I*(-73786976294838206464*I - 73786976294838206464)^(1/4) + (65536* I + 65536)*(-1/x + 1)^(1/4)) + (1/16*I + 1/16)^(1/4)*log(-I*(-737869762948 38206464*I - 73786976294838206464)^(1/4) - (65536*I + 65536)*(-1/x + 1)^(1 /4)) - 2*I*(-1/256*I + 1/256)^(1/4)*log(-I*(850705917302346158658436518579 42052864*I - 85070591730234615865843651857942052864)^(1/4) + (2147483648*I + 2147483648)*(-1/x + 1)^(1/4)) + 2*I*(-1/256*I + 1/256)^(1/4)*log(-I*(85 070591730234615865843651857942052864*I - 850705917302346158658436518579420 52864)^(1/4) - (2147483648*I + 2147483648)*(-1/x + 1)^(1/4)) + 1/2*2^(1/4) *log(abs(-2^(1/4) + (-1/x + 1)^(1/4)))
Not integrable
Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.16 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^3+x^4}}{x^4 \left (-1+x^4\right )} \, dx=\int \frac {\left (x^4+1\right )\,{\left (x^4-x^3\right )}^{1/4}}{x^4\,\left (x^4-1\right )} \,d x \] Input:
int(((x^4 + 1)*(x^4 - x^3)^(1/4))/(x^4*(x^4 - 1)),x)
Output:
int(((x^4 + 1)*(x^4 - x^3)^(1/4))/(x^4*(x^4 - 1)), x)
Not integrable
Time = 1.60 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.58 \[ \int \frac {\left (1+x^4\right ) \sqrt [4]{-x^3+x^4}}{x^4 \left (-1+x^4\right )} \, dx=\frac {\frac {344 \left (x -1\right )^{\frac {1}{4}} x^{2}}{45}-\frac {4 \left (x -1\right )^{\frac {1}{4}} x}{45}+\frac {4 \left (x -1\right )^{\frac {1}{4}}}{9}-2 x^{\frac {9}{4}} \left (\int \frac {\left (x -1\right )^{\frac {1}{4}}}{x^{\frac {21}{4}}-x^{\frac {5}{4}}}d x \right )-2 x^{\frac {9}{4}} \left (\int \frac {\left (x -1\right )^{\frac {1}{4}}}{x^{\frac {17}{4}}-x^{\frac {1}{4}}}d x \right )-2 x^{\frac {9}{4}} \left (\int \frac {x^{\frac {7}{4}} \left (x -1\right )^{\frac {1}{4}}}{x^{4}-1}d x \right )}{x^{\frac {9}{4}}} \] Input:
int((x^4+1)*(x^4-x^3)^(1/4)/x^4/(x^4-1),x)
Output:
(2*(172*(x - 1)**(1/4)*x**2 - 2*(x - 1)**(1/4)*x + 10*(x - 1)**(1/4) - 45* x**(1/4)*int((x - 1)**(1/4)/(x**(1/4)*x**5 - x**(1/4)*x),x)*x**2 - 45*x**( 1/4)*int((x - 1)**(1/4)/(x**(1/4)*x**4 - x**(1/4)),x)*x**2 - 45*x**(1/4)*i nt((x**(3/4)*(x - 1)**(1/4)*x)/(x**4 - 1),x)*x**2))/(45*x**(1/4)*x**2)