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

Optimal. Leaf size=69 \[ \frac {1}{3} 2^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \left (x^4-x\right )^{3/4}}{x^3-1}\right )+\frac {1}{3} 2^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \left (x^4-x\right )^{3/4}}{x^3-1}\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 111, normalized size of antiderivative = 1.61, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2056, 466, 465, 377, 212, 206, 203} \begin {gather*} \frac {2^{3/4} \sqrt [4]{x} \sqrt [4]{x^3-1} \tan ^{-1}\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} \tanh ^{-1}\left (\frac {\sqrt [4]{2} x^{3/4}}{\sqrt [4]{x^3-1}}\right )}{3 \sqrt [4]{x^4-x}} \end {gather*}

Antiderivative was successfully verified.

[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 203

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

Rule 206

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

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 377

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 465

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 466

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 2056

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*} \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx &=\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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} \tan ^{-1}\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} \tanh ^{-1}\left (\frac {\sqrt [4]{2} x^{3/4}}{\sqrt [4]{-1+x^3}}\right )}{3 \sqrt [4]{-x+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 58, normalized size = 0.84 \begin {gather*} \frac {4 x \sqrt [4]{1-x^3} \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};\frac {2 x^3}{x^3+1}\right )}{3 \sqrt [4]{x \left (x^3-1\right )} \sqrt [4]{x^3+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[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))

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IntegrateAlgebraic [A]  time = 0.37, size = 69, normalized size = 1.00 \begin {gather*} \frac {1}{3} 2^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \left (-x+x^4\right )^{3/4}}{-1+x^3}\right )+\frac {1}{3} 2^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \left (-x+x^4\right )^{3/4}}{-1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.88, size = 218, normalized size = 3.16 \begin {gather*} -\frac {1}{3} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{4} - x\right )}^{\frac {1}{4}} x^{2} + 2^{\frac {3}{4}} {\left (2 \cdot 2^{\frac {3}{4}} \sqrt {x^{4} - x} x + 2^{\frac {1}{4}} {\left (3 \, x^{3} - 1\right )}\right )} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{4} - x\right )}^{\frac {3}{4}}}{2 \, {\left (x^{3} + 1\right )}}\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} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.34, size = 62, normalized size = 0.90 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maple [C]  time = 5.29, size = 230, normalized size = 3.33

method	result	size
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {\sqrt {x^{4}-x}\, \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x -2 \left (x^{4}-x \right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{3}+4 \left (x^{4}-x \right )^{\frac {3}{4}}+\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-8\right )^{2}\right )}{\left (x^{2}-x +1\right ) \left (1+x \right )}\right )}{6}+\frac {\RootOf \left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {\sqrt {x^{4}-x}\, \RootOf \left (\textit {\_Z}^{4}-8\right )^{3} x +2 \left (x^{4}-x \right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+3 \RootOf \left (\textit {\_Z}^{4}-8\right ) x^{3}+4 \left (x^{4}-x \right )^{\frac {3}{4}}-\RootOf \left (\textit {\_Z}^{4}-8\right )}{\left (x^{2}-x +1\right ) \left (1+x \right )}\right )}{6}\)	\(230\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*RootOf(_Z^2+RootOf(_Z^4-8)^2)*ln(-((x^4-x)^(1/2)*RootOf(_Z^2+RootOf(_Z^4-8)^2)*RootOf(_Z^4-8)^2*x-2*(x^4-
x)^(1/4)*RootOf(_Z^4-8)^2*x^2-3*RootOf(_Z^2+RootOf(_Z^4-8)^2)*x^3+4*(x^4-x)^(3/4)+RootOf(_Z^2+RootOf(_Z^4-8)^2
))/(x^2-x+1)/(1+x))+1/6*RootOf(_Z^4-8)*ln(-((x^4-x)^(1/2)*RootOf(_Z^4-8)^3*x+2*(x^4-x)^(1/4)*RootOf(_Z^4-8)^2*
x^2+3*RootOf(_Z^4-8)*x^3+4*(x^4-x)^(3/4)-RootOf(_Z^4-8))/(x^2-x+1)/(1+x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{4} - x\right )}^{\frac {1}{4}} {\left (x^{3} + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (x^4-x\right )}^{1/4}\,\left (x^3+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x^4 - x)^(1/4)*(x^3 + 1)),x)

[Out]

int(1/((x^4 - x)^(1/4)*(x^3 + 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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