\(\int \frac {1}{(1+x^3) \sqrt [4]{-x+x^4}} \, dx\) [908]

Optimal result

Integrand size = 19, antiderivative size = 69 \[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\frac {1}{3} 2^{3/4} \arctan \left (\frac {\sqrt [4]{2} \left (-x+x^4\right )^{3/4}}{-1+x^3}\right )+\frac {1}{3} 2^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2} \left (-x+x^4\right )^{3/4}}{-1+x^3}\right ) \]

[Out]

1/3*2^(3/4)*arctan(2^(1/4)*(x^4-x)^(3/4)/(x^3-1))+1/3*2^(3/4)*arctanh(2^(1/4)*(x^4-x)^(3/4)/(x^3-1))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.61, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2081, 477, 476, 385, 218, 212, 209} \[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\frac {2^{3/4} \sqrt [4]{x} \sqrt [4]{x^3-1} \arctan \left (\frac {\sqrt [4]{2} x^{3/4}}{\sqrt [4]{x^3-1}}\right )}{3 \sqrt [4]{x^4-x}}+\frac {2^{3/4} \sqrt [4]{x} \sqrt [4]{x^3-1} \text {arctanh}\left (\frac {\sqrt [4]{2} x^{3/4}}{\sqrt [4]{x^3-1}}\right )}{3 \sqrt [4]{x^4-x}} \]

[In]

Int[1/((1 + x^3)*(-x + x^4)^(1/4)),x]

[Out]

(2^(3/4)*x^(1/4)*(-1 + x^3)^(1/4)*ArcTan[(2^(1/4)*x^(3/4))/(-1 + x^3)^(1/4)])/(3*(-x + x^4)^(1/4)) + (2^(3/4)*
x^(1/4)*(-1 + x^3)^(1/4)*ArcTanh[(2^(1/4)*x^(3/4))/(-1 + x^3)^(1/4)])/(3*(-x + x^4)^(1/4))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 476

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1,
n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /;
FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [4]{x} \sqrt [4]{-1+x^3}\right ) \int \frac {1}{\sqrt [4]{x} \sqrt [4]{-1+x^3} \left (1+x^3\right )} \, dx}{\sqrt [4]{-x+x^4}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{-1+x^3}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt [4]{-1+x^{12}} \left (1+x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{-x+x^4}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{-1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{-1+x^4} \left (1+x^4\right )} \, dx,x,x^{3/4}\right )}{3 \sqrt [4]{-x+x^4}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{-1+x^3}\right ) \text {Subst}\left (\int \frac {1}{1-2 x^4} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{-1+x^3}}\right )}{3 \sqrt [4]{-x+x^4}} \\ & = \frac {\left (2 \sqrt [4]{x} \sqrt [4]{-1+x^3}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{-1+x^3}}\right )}{3 \sqrt [4]{-x+x^4}}+\frac {\left (2 \sqrt [4]{x} \sqrt [4]{-1+x^3}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{-1+x^3}}\right )}{3 \sqrt [4]{-x+x^4}} \\ & = \frac {2^{3/4} \sqrt [4]{x} \sqrt [4]{-1+x^3} \arctan \left (\frac {\sqrt [4]{2} x^{3/4}}{\sqrt [4]{-1+x^3}}\right )}{3 \sqrt [4]{-x+x^4}}+\frac {2^{3/4} \sqrt [4]{x} \sqrt [4]{-1+x^3} \text {arctanh}\left (\frac {\sqrt [4]{2} x^{3/4}}{\sqrt [4]{-1+x^3}}\right )}{3 \sqrt [4]{-x+x^4}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\frac {4 x \sqrt [4]{1-x^3} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4},\frac {5}{4},\frac {2 x^3}{1+x^3}\right )}{3 \sqrt [4]{x \left (-1+x^3\right )} \sqrt [4]{1+x^3}} \]

[In]

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

[Out]

(4*x*(1 - x^3)^(1/4)*Hypergeometric2F1[1/4, 1/4, 5/4, (2*x^3)/(1 + x^3)])/(3*(x*(-1 + x^3))^(1/4)*(1 + x^3)^(1
/4))

Maple [A] (verified)

Time = 7.30 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99

method	result	size
pseudoelliptic	\(-\frac {2^{\frac {3}{4}} \left (2 \arctan \left (\frac {\left (x^{4}-x \right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x}\right )-\ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{4}-x \right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{4}-x \right )^{\frac {1}{4}}}\right )\right )}{6}\)	\(68\)
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {-\sqrt {x^{4}-x}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} x +2 \left (x^{4}-x \right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) x^{3}+4 \left (x^{4}-x \right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )}{\left (1+x \right ) \left (x^{2}-x +1\right )}\right )}{6}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {-\sqrt {x^{4}-x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x -2 \left (x^{4}-x \right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{3}+4 \left (x^{4}-x \right )^{\frac {3}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right )}{\left (1+x \right ) \left (x^{2}-x +1\right )}\right )}{6}\)	\(232\)

[In]

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

[Out]

-1/6*2^(3/4)*(2*arctan(1/2*(x^4-x)^(1/4)/x*2^(3/4))-ln((-2^(1/4)*x-(x^4-x)^(1/4))/(2^(1/4)*x-(x^4-x)^(1/4))))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.06 (sec) , antiderivative size = 282, normalized size of antiderivative = 4.09 \[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\frac {1}{12} \cdot 2^{\frac {3}{4}} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} - x\right )}^{\frac {1}{4}} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{3} - 1\right )} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} - x} x + 4 \, {\left (x^{4} - x\right )}^{\frac {3}{4}}}{x^{3} + 1}\right ) - \frac {1}{12} \cdot 2^{\frac {3}{4}} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} - x\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{3} - 1\right )} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} - x} x + 4 \, {\left (x^{4} - x\right )}^{\frac {3}{4}}}{x^{3} + 1}\right ) + \frac {1}{12} i \cdot 2^{\frac {3}{4}} \log \left (-\frac {4 \, \sqrt {2} {\left (x^{4} - x\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (3 i \, x^{3} - i\right )} + 4 i \cdot 2^{\frac {1}{4}} \sqrt {x^{4} - x} x - 4 \, {\left (x^{4} - x\right )}^{\frac {3}{4}}}{x^{3} + 1}\right ) - \frac {1}{12} i \cdot 2^{\frac {3}{4}} \log \left (-\frac {4 \, \sqrt {2} {\left (x^{4} - x\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (-3 i \, x^{3} + i\right )} - 4 i \cdot 2^{\frac {1}{4}} \sqrt {x^{4} - x} x - 4 \, {\left (x^{4} - x\right )}^{\frac {3}{4}}}{x^{3} + 1}\right ) \]

[In]

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

[Out]

1/12*2^(3/4)*log((4*sqrt(2)*(x^4 - x)^(1/4)*x^2 + 2^(3/4)*(3*x^3 - 1) + 4*2^(1/4)*sqrt(x^4 - x)*x + 4*(x^4 - x
)^(3/4))/(x^3 + 1)) - 1/12*2^(3/4)*log((4*sqrt(2)*(x^4 - x)^(1/4)*x^2 - 2^(3/4)*(3*x^3 - 1) - 4*2^(1/4)*sqrt(x
^4 - x)*x + 4*(x^4 - x)^(3/4))/(x^3 + 1)) + 1/12*I*2^(3/4)*log(-(4*sqrt(2)*(x^4 - x)^(1/4)*x^2 - 2^(3/4)*(3*I*
x^3 - I) + 4*I*2^(1/4)*sqrt(x^4 - x)*x - 4*(x^4 - x)^(3/4))/(x^3 + 1)) - 1/12*I*2^(3/4)*log(-(4*sqrt(2)*(x^4 -
x)^(1/4)*x^2 - 2^(3/4)*(-3*I*x^3 + I) - 4*I*2^(1/4)*sqrt(x^4 - x)*x - 4*(x^4 - x)^(3/4))/(x^3 + 1))

Sympy [F]

\[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\int \frac {1}{\sqrt [4]{x \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\int { \frac {1}{{\left (x^{4} - x\right )}^{\frac {1}{4}} {\left (x^{3} + 1\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=-\frac {1}{3} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{6} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} + {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{6} \cdot 2^{\frac {3}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}} \right |}\right ) \]

[In]

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

[Out]

-1/3*2^(3/4)*arctan(1/2*2^(3/4)*(-1/x^3 + 1)^(1/4)) + 1/6*2^(3/4)*log(2^(1/4) + (-1/x^3 + 1)^(1/4)) - 1/6*2^(3
/4)*log(abs(-2^(1/4) + (-1/x^3 + 1)^(1/4)))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\int \frac {1}{{\left (x^4-x\right )}^{1/4}\,\left (x^3+1\right )} \,d x \]

[In]

int(1/((x^4 - x)^(1/4)*(x^3 + 1)),x)

[Out]

int(1/((x^4 - x)^(1/4)*(x^3 + 1)), x)