\(\int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x (1+x^3)} \, dx\) [2139]

Optimal result

Integrand size = 27, antiderivative size = 156 \[ \int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x \left (1+x^3\right )} \, dx=-\frac {4}{3} \sqrt [4]{2} \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-x^3+x^4}}\right )+\frac {4}{3} \sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-x^3+x^4}}\right )-\frac {1}{3} \text {RootSum}\left [1-\text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-2 \log (x)+2 \log \left (\sqrt [4]{-x^3+x^4}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-\log \left (\sqrt [4]{-x^3+x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-\text {$\#$1}^3+2 \text {$\#$1}^7}\&\right ] \]

[Out]

Unintegrable

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.85, number of steps used = 12, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2081, 6857, 129, 525, 524} \[ \int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x \left (1+x^3\right )} \, dx=-\frac {4 \sqrt [4]{x^4-x^3} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {5}{4},1,\frac {7}{4},x,\sqrt [3]{-1} x\right )}{9 \sqrt [4]{1-x}}-\frac {4 \sqrt [4]{x^4-x^3} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {5}{4},1,\frac {7}{4},x,-(-1)^{2/3} x\right )}{9 \sqrt [4]{1-x}}-\frac {4 \sqrt [4]{x^4-x^3} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},-x,x\right )}{9 \sqrt [4]{1-x}} \]

[In]

Int[((-1 + x)*(-x^3 + x^4)^(1/4))/(x*(1 + x^3)),x]

[Out]

(-4*(-x^3 + x^4)^(1/4)*AppellF1[3/4, -5/4, 1, 7/4, x, (-1)^(1/3)*x])/(9*(1 - x)^(1/4)) - (4*(-x^3 + x^4)^(1/4)
*AppellF1[3/4, -5/4, 1, 7/4, x, -((-1)^(2/3)*x)])/(9*(1 - x)^(1/4)) - (4*(-x^3 + x^4)^(1/4)*AppellF1[3/4, 1, -
5/4, 7/4, -x, x])/(9*(1 - x)^(1/4))

Rule 129

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + b*(x^k/e))^m*(c + d*(x^k/e))^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [4]{-x^3+x^4} \int \frac {(-1+x)^{5/4}}{\sqrt [4]{x} \left (1+x^3\right )} \, dx}{\sqrt [4]{-1+x} x^{3/4}} \\ & = \frac {\sqrt [4]{-x^3+x^4} \int \left (-\frac {(-1+x)^{5/4}}{3 (-1-x) \sqrt [4]{x}}-\frac {(-1+x)^{5/4}}{3 \sqrt [4]{x} \left (-1+\sqrt [3]{-1} x\right )}-\frac {(-1+x)^{5/4}}{3 \sqrt [4]{x} \left (-1-(-1)^{2/3} x\right )}\right ) \, dx}{\sqrt [4]{-1+x} x^{3/4}} \\ & = -\frac {\sqrt [4]{-x^3+x^4} \int \frac {(-1+x)^{5/4}}{(-1-x) \sqrt [4]{x}} \, dx}{3 \sqrt [4]{-1+x} x^{3/4}}-\frac {\sqrt [4]{-x^3+x^4} \int \frac {(-1+x)^{5/4}}{\sqrt [4]{x} \left (-1+\sqrt [3]{-1} x\right )} \, dx}{3 \sqrt [4]{-1+x} x^{3/4}}-\frac {\sqrt [4]{-x^3+x^4} \int \frac {(-1+x)^{5/4}}{\sqrt [4]{x} \left (-1-(-1)^{2/3} x\right )} \, dx}{3 \sqrt [4]{-1+x} x^{3/4}} \\ & = -\frac {\left (4 \sqrt [4]{-x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (-1+x^4\right )^{5/4}}{-1-x^4} \, dx,x,\sqrt [4]{x}\right )}{3 \sqrt [4]{-1+x} x^{3/4}}-\frac {\left (4 \sqrt [4]{-x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (-1+x^4\right )^{5/4}}{-1+\sqrt [3]{-1} x^4} \, dx,x,\sqrt [4]{x}\right )}{3 \sqrt [4]{-1+x} x^{3/4}}-\frac {\left (4 \sqrt [4]{-x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (-1+x^4\right )^{5/4}}{-1-(-1)^{2/3} x^4} \, dx,x,\sqrt [4]{x}\right )}{3 \sqrt [4]{-1+x} x^{3/4}} \\ & = \frac {\left (4 \sqrt [4]{-x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (1-x^4\right )^{5/4}}{-1-x^4} \, dx,x,\sqrt [4]{x}\right )}{3 \sqrt [4]{1-x} x^{3/4}}+\frac {\left (4 \sqrt [4]{-x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (1-x^4\right )^{5/4}}{-1+\sqrt [3]{-1} x^4} \, dx,x,\sqrt [4]{x}\right )}{3 \sqrt [4]{1-x} x^{3/4}}+\frac {\left (4 \sqrt [4]{-x^3+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (1-x^4\right )^{5/4}}{-1-(-1)^{2/3} x^4} \, dx,x,\sqrt [4]{x}\right )}{3 \sqrt [4]{1-x} x^{3/4}} \\ & = -\frac {4 \sqrt [4]{-x^3+x^4} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {5}{4},1,\frac {7}{4},x,\sqrt [3]{-1} x\right )}{9 \sqrt [4]{1-x}}-\frac {4 \sqrt [4]{-x^3+x^4} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {5}{4},1,\frac {7}{4},x,-(-1)^{2/3} x\right )}{9 \sqrt [4]{1-x}}-\frac {4 \sqrt [4]{-x^3+x^4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},-x,x\right )}{9 \sqrt [4]{1-x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03 \[ \int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x \left (1+x^3\right )} \, dx=-\frac {(-1+x)^{3/4} x^{9/4} \left (16 \sqrt [4]{2} \left (\arctan \left (\frac {\sqrt [4]{2}}{\sqrt [4]{\frac {-1+x}{x}}}\right )-\text {arctanh}\left (\frac {\sqrt [4]{2}}{\sqrt [4]{\frac {-1+x}{x}}}\right )\right )+\text {RootSum}\left [1-\text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-2 \log (x)+8 \log \left (\sqrt [4]{-1+x}-\sqrt [4]{x} \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-4 \log \left (\sqrt [4]{-1+x}-\sqrt [4]{x} \text {$\#$1}\right ) \text {$\#$1}^4}{-\text {$\#$1}^3+2 \text {$\#$1}^7}\&\right ]\right )}{12 \left ((-1+x) x^3\right )^{3/4}} \]

[In]

Integrate[((-1 + x)*(-x^3 + x^4)^(1/4))/(x*(1 + x^3)),x]

[Out]

-1/12*((-1 + x)^(3/4)*x^(9/4)*(16*2^(1/4)*(ArcTan[2^(1/4)/((-1 + x)/x)^(1/4)] - ArcTanh[2^(1/4)/((-1 + x)/x)^(
1/4)]) + RootSum[1 - #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 + 2*#1^7) & ]))/((-1 + x)*x^3)^(3/4)

Maple [N/A] (verified)

Time = 8.13 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.78

method	result	size
pseudoelliptic	\(\frac {2 \ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}\right ) 2^{\frac {1}{4}}}{3}+\frac {4 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{2 x}\right ) 2^{\frac {1}{4}}}{3}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-\textit {\_Z}^{4}+1\right )}{\sum }\frac {\left (\textit {\_R}^{4}-2\right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3} \left (2 \textit {\_R}^{4}-1\right )}\right )}{3}\)	\(122\)
trager	\(\text {Expression too large to display}\)	\(3101\)

[In]

int((-1+x)*(x^4-x^3)^(1/4)/x/(x^3+1),x,method=_RETURNVERBOSE)

[Out]

2/3*ln((-2^(1/4)*x-(x^3*(-1+x))^(1/4))/(2^(1/4)*x-(x^3*(-1+x))^(1/4)))*2^(1/4)+4/3*arctan(1/2*2^(3/4)/x*(x^3*(
-1+x))^(1/4))*2^(1/4)+1/3*sum((_R^4-2)*ln((-_R*x+(x^3*(-1+x))^(1/4))/x)/_R^3/(2*_R^4-1),_R=RootOf(_Z^8-_Z^4+1)
)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 0.29 (sec) , antiderivative size = 526, normalized size of antiderivative = 3.37 \[ \int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x \left (1+x^3\right )} \, dx=-\frac {1}{6} \, \sqrt {2} \sqrt {-\sqrt {2 i \, \sqrt {3} + 2}} \log \left (\frac {\sqrt {2} x \sqrt {-\sqrt {2 i \, \sqrt {3} + 2}} + 2 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{6} \, \sqrt {2} \sqrt {-\sqrt {2 i \, \sqrt {3} + 2}} \log \left (-\frac {\sqrt {2} x \sqrt {-\sqrt {2 i \, \sqrt {3} + 2}} - 2 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{6} \, \sqrt {2} \sqrt {-\sqrt {-2 i \, \sqrt {3} + 2}} \log \left (\frac {\sqrt {2} x \sqrt {-\sqrt {-2 i \, \sqrt {3} + 2}} + 2 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{6} \, \sqrt {2} \sqrt {-\sqrt {-2 i \, \sqrt {3} + 2}} \log \left (-\frac {\sqrt {2} x \sqrt {-\sqrt {-2 i \, \sqrt {3} + 2}} - 2 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{6} \, \sqrt {2} {\left (2 i \, \sqrt {3} + 2\right )}^{\frac {1}{4}} \log \left (\frac {\sqrt {2} x {\left (2 i \, \sqrt {3} + 2\right )}^{\frac {1}{4}} + 2 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{6} \, \sqrt {2} {\left (2 i \, \sqrt {3} + 2\right )}^{\frac {1}{4}} \log \left (-\frac {\sqrt {2} x {\left (2 i \, \sqrt {3} + 2\right )}^{\frac {1}{4}} - 2 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{6} \, \sqrt {2} {\left (-2 i \, \sqrt {3} + 2\right )}^{\frac {1}{4}} \log \left (\frac {\sqrt {2} x {\left (-2 i \, \sqrt {3} + 2\right )}^{\frac {1}{4}} + 2 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{6} \, \sqrt {2} {\left (-2 i \, \sqrt {3} + 2\right )}^{\frac {1}{4}} \log \left (-\frac {\sqrt {2} x {\left (-2 i \, \sqrt {3} + 2\right )}^{\frac {1}{4}} - 2 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {2}{3} \cdot 2^{\frac {1}{4}} \log \left (\frac {2^{\frac {1}{4}} x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {2}{3} \cdot 2^{\frac {1}{4}} \log \left (-\frac {2^{\frac {1}{4}} x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {2}{3} i \cdot 2^{\frac {1}{4}} \log \left (\frac {i \cdot 2^{\frac {1}{4}} x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {2}{3} i \cdot 2^{\frac {1}{4}} \log \left (\frac {-i \cdot 2^{\frac {1}{4}} x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

integrate((-1+x)*(x^4-x^3)^(1/4)/x/(x^3+1),x, algorithm="fricas")

[Out]

-1/6*sqrt(2)*sqrt(-sqrt(2*I*sqrt(3) + 2))*log((sqrt(2)*x*sqrt(-sqrt(2*I*sqrt(3) + 2)) + 2*(x^4 - x^3)^(1/4))/x
) + 1/6*sqrt(2)*sqrt(-sqrt(2*I*sqrt(3) + 2))*log(-(sqrt(2)*x*sqrt(-sqrt(2*I*sqrt(3) + 2)) - 2*(x^4 - x^3)^(1/4
))/x) - 1/6*sqrt(2)*sqrt(-sqrt(-2*I*sqrt(3) + 2))*log((sqrt(2)*x*sqrt(-sqrt(-2*I*sqrt(3) + 2)) + 2*(x^4 - x^3)
^(1/4))/x) + 1/6*sqrt(2)*sqrt(-sqrt(-2*I*sqrt(3) + 2))*log(-(sqrt(2)*x*sqrt(-sqrt(-2*I*sqrt(3) + 2)) - 2*(x^4
- x^3)^(1/4))/x) - 1/6*sqrt(2)*(2*I*sqrt(3) + 2)^(1/4)*log((sqrt(2)*x*(2*I*sqrt(3) + 2)^(1/4) + 2*(x^4 - x^3)^
(1/4))/x) + 1/6*sqrt(2)*(2*I*sqrt(3) + 2)^(1/4)*log(-(sqrt(2)*x*(2*I*sqrt(3) + 2)^(1/4) - 2*(x^4 - x^3)^(1/4))
/x) - 1/6*sqrt(2)*(-2*I*sqrt(3) + 2)^(1/4)*log((sqrt(2)*x*(-2*I*sqrt(3) + 2)^(1/4) + 2*(x^4 - x^3)^(1/4))/x) +
1/6*sqrt(2)*(-2*I*sqrt(3) + 2)^(1/4)*log(-(sqrt(2)*x*(-2*I*sqrt(3) + 2)^(1/4) - 2*(x^4 - x^3)^(1/4))/x) + 2/3
*2^(1/4)*log((2^(1/4)*x + (x^4 - x^3)^(1/4))/x) - 2/3*2^(1/4)*log(-(2^(1/4)*x - (x^4 - x^3)^(1/4))/x) + 2/3*I*
2^(1/4)*log((I*2^(1/4)*x + (x^4 - x^3)^(1/4))/x) - 2/3*I*2^(1/4)*log((-I*2^(1/4)*x + (x^4 - x^3)^(1/4))/x)

Sympy [N/A]

Not integrable

Time = 2.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.17 \[ \int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x \left (1+x^3\right )} \, dx=\int \frac {\sqrt [4]{x^{3} \left (x - 1\right )} \left (x - 1\right )}{x \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \]

[In]

integrate((-1+x)*(x**4-x**3)**(1/4)/x/(x**3+1),x)

[Out]

Integral((x**3*(x - 1))**(1/4)*(x - 1)/(x*(x + 1)*(x**2 - x + 1)), x)

Maxima [N/A]

Not integrable

Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.17 \[ \int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x \left (1+x^3\right )} \, dx=\int { \frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}} {\left (x - 1\right )}}{{\left (x^{3} + 1\right )} x} \,d x } \]

[In]

integrate((-1+x)*(x^4-x^3)^(1/4)/x/(x^3+1),x, algorithm="maxima")

[Out]

integrate((x^4 - x^3)^(1/4)*(x - 1)/((x^3 + 1)*x), x)

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 0.36 (sec) , antiderivative size = 392, normalized size of antiderivative = 2.51 \[ \int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x \left (1+x^3\right )} \, dx=-\frac {1}{6} \, {\left (\sqrt {6} + \sqrt {2}\right )} \arctan \left (\frac {\sqrt {6} - \sqrt {2} + 4 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}}{\sqrt {6} + \sqrt {2}}\right ) - \frac {1}{6} \, {\left (\sqrt {6} + \sqrt {2}\right )} \arctan \left (-\frac {\sqrt {6} - \sqrt {2} - 4 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}}{\sqrt {6} + \sqrt {2}}\right ) - \frac {1}{6} \, {\left (\sqrt {6} - \sqrt {2}\right )} \arctan \left (\frac {\sqrt {6} + \sqrt {2} + 4 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}}{\sqrt {6} - \sqrt {2}}\right ) - \frac {1}{6} \, {\left (\sqrt {6} - \sqrt {2}\right )} \arctan \left (-\frac {\sqrt {6} + \sqrt {2} - 4 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}}{\sqrt {6} - \sqrt {2}}\right ) - \frac {1}{12} \, {\left (\sqrt {6} + \sqrt {2}\right )} \log \left (\frac {1}{2} \, {\left (\sqrt {6} + \sqrt {2}\right )} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + \sqrt {-\frac {1}{x} + 1} + 1\right ) + \frac {1}{12} \, {\left (\sqrt {6} + \sqrt {2}\right )} \log \left (-\frac {1}{2} \, {\left (\sqrt {6} + \sqrt {2}\right )} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + \sqrt {-\frac {1}{x} + 1} + 1\right ) - \frac {1}{12} \, {\left (\sqrt {6} - \sqrt {2}\right )} \log \left (\frac {1}{2} \, {\left (\sqrt {6} - \sqrt {2}\right )} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + \sqrt {-\frac {1}{x} + 1} + 1\right ) + \frac {1}{12} \, {\left (\sqrt {6} - \sqrt {2}\right )} \log \left (-\frac {1}{2} \, {\left (\sqrt {6} - \sqrt {2}\right )} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + \sqrt {-\frac {1}{x} + 1} + 1\right ) + \frac {1}{3} \cdot 8^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) + \frac {2}{3} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {1}{4}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - \frac {2}{3} \cdot 2^{\frac {1}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} \right |}\right ) \]

[In]

integrate((-1+x)*(x^4-x^3)^(1/4)/x/(x^3+1),x, algorithm="giac")

[Out]

-1/6*(sqrt(6) + sqrt(2))*arctan((sqrt(6) - sqrt(2) + 4*(-1/x + 1)^(1/4))/(sqrt(6) + sqrt(2))) - 1/6*(sqrt(6) +
sqrt(2))*arctan(-(sqrt(6) - sqrt(2) - 4*(-1/x + 1)^(1/4))/(sqrt(6) + sqrt(2))) - 1/6*(sqrt(6) - sqrt(2))*arct
an((sqrt(6) + sqrt(2) + 4*(-1/x + 1)^(1/4))/(sqrt(6) - sqrt(2))) - 1/6*(sqrt(6) - sqrt(2))*arctan(-(sqrt(6) +
sqrt(2) - 4*(-1/x + 1)^(1/4))/(sqrt(6) - sqrt(2))) - 1/12*(sqrt(6) + sqrt(2))*log(1/2*(sqrt(6) + sqrt(2))*(-1/
x + 1)^(1/4) + sqrt(-1/x + 1) + 1) + 1/12*(sqrt(6) + sqrt(2))*log(-1/2*(sqrt(6) + sqrt(2))*(-1/x + 1)^(1/4) +
sqrt(-1/x + 1) + 1) - 1/12*(sqrt(6) - sqrt(2))*log(1/2*(sqrt(6) - sqrt(2))*(-1/x + 1)^(1/4) + sqrt(-1/x + 1) +
1) + 1/12*(sqrt(6) - sqrt(2))*log(-1/2*(sqrt(6) - sqrt(2))*(-1/x + 1)^(1/4) + sqrt(-1/x + 1) + 1) + 1/3*8^(3/
4)*arctan(1/2*2^(3/4)*(-1/x + 1)^(1/4)) + 2/3*2^(1/4)*log(2^(1/4) + (-1/x + 1)^(1/4)) - 2/3*2^(1/4)*log(abs(-2
^(1/4) + (-1/x + 1)^(1/4)))

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.17 \[ \int \frac {(-1+x) \sqrt [4]{-x^3+x^4}}{x \left (1+x^3\right )} \, dx=\int \frac {{\left (x^4-x^3\right )}^{1/4}\,\left (x-1\right )}{x\,\left (x^3+1\right )} \,d x \]

[In]

int(((x^4 - x^3)^(1/4)*(x - 1))/(x*(x^3 + 1)),x)

[Out]

int(((x^4 - x^3)^(1/4)*(x - 1))/(x*(x^3 + 1)), x)