3.8.5 \(\int \frac {\sqrt [4]{-x+x^4}}{x^2} \, dx\)

Optimal. Leaf size=55 \[ -\frac {4 \sqrt [4]{x^4-x}}{3 x}-\frac {2}{3} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x}}\right )+\frac {2}{3} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x}}\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 109, normalized size of antiderivative = 1.98, number of steps used = 8, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {2020, 2032, 329, 275, 331, 298, 203, 206} \begin {gather*} -\frac {4 \sqrt [4]{x^4-x}}{3 x}-\frac {2 x^{3/4} \left (x^3-1\right )^{3/4} \tan ^{-1}\left (\frac {x^{3/4}}{\sqrt [4]{x^3-1}}\right )}{3 \left (x^4-x\right )^{3/4}}+\frac {2 x^{3/4} \left (x^3-1\right )^{3/4} \tanh ^{-1}\left (\frac {x^{3/4}}{\sqrt [4]{x^3-1}}\right )}{3 \left (x^4-x\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x + x^4)^(1/4)/x^2,x]

[Out]

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

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 275

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

Rule 298

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 2020

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b*
x^n)^p)/(c*(m + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{-x+x^4}}{x^2} \, dx &=-\frac {4 \sqrt [4]{-x+x^4}}{3 x}+\int \frac {x^2}{\left (-x+x^4\right )^{3/4}} \, dx\\ &=-\frac {4 \sqrt [4]{-x+x^4}}{3 x}+\frac {\left (x^{3/4} \left (-1+x^3\right )^{3/4}\right ) \int \frac {x^{5/4}}{\left (-1+x^3\right )^{3/4}} \, dx}{\left (-x+x^4\right )^{3/4}}\\ &=-\frac {4 \sqrt [4]{-x+x^4}}{3 x}+\frac {\left (4 x^{3/4} \left (-1+x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^8}{\left (-1+x^{12}\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{\left (-x+x^4\right )^{3/4}}\\ &=-\frac {4 \sqrt [4]{-x+x^4}}{3 x}+\frac {\left (4 x^{3/4} \left (-1+x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-1+x^4\right )^{3/4}} \, dx,x,x^{3/4}\right )}{3 \left (-x+x^4\right )^{3/4}}\\ &=-\frac {4 \sqrt [4]{-x+x^4}}{3 x}+\frac {\left (4 x^{3/4} \left (-1+x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-x^4} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{-1+x^3}}\right )}{3 \left (-x+x^4\right )^{3/4}}\\ &=-\frac {4 \sqrt [4]{-x+x^4}}{3 x}+\frac {\left (2 x^{3/4} \left (-1+x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{-1+x^3}}\right )}{3 \left (-x+x^4\right )^{3/4}}-\frac {\left (2 x^{3/4} \left (-1+x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{-1+x^3}}\right )}{3 \left (-x+x^4\right )^{3/4}}\\ &=-\frac {4 \sqrt [4]{-x+x^4}}{3 x}-\frac {2 x^{3/4} \left (-1+x^3\right )^{3/4} \tan ^{-1}\left (\frac {x^{3/4}}{\sqrt [4]{-1+x^3}}\right )}{3 \left (-x+x^4\right )^{3/4}}+\frac {2 x^{3/4} \left (-1+x^3\right )^{3/4} \tanh ^{-1}\left (\frac {x^{3/4}}{\sqrt [4]{-1+x^3}}\right )}{3 \left (-x+x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 42, normalized size = 0.76 \begin {gather*} -\frac {4 \sqrt [4]{x \left (x^3-1\right )} \, _2F_1\left (-\frac {1}{4},-\frac {1}{4};\frac {3}{4};x^3\right )}{3 x \sqrt [4]{1-x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x + x^4)^(1/4)/x^2,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.21, size = 55, normalized size = 1.00 \begin {gather*} -\frac {4 \sqrt [4]{-x+x^4}}{3 x}-\frac {2}{3} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-x+x^4}}\right )+\frac {2}{3} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-x+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-x + x^4)^(1/4)/x^2,x]

[Out]

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

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fricas [B]  time = 1.33, size = 93, normalized size = 1.69 \begin {gather*} \frac {x \arctan \left (2 \, {\left (x^{4} - x\right )}^{\frac {1}{4}} x^{2} + 2 \, {\left (x^{4} - x\right )}^{\frac {3}{4}}\right ) + x \log \left (-2 \, x^{3} - 2 \, {\left (x^{4} - x\right )}^{\frac {1}{4}} x^{2} - 2 \, \sqrt {x^{4} - x} x - 2 \, {\left (x^{4} - x\right )}^{\frac {3}{4}} + 1\right ) - 4 \, {\left (x^{4} - x\right )}^{\frac {1}{4}}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-x)^(1/4)/x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.24, size = 53, normalized size = 0.96 \begin {gather*} \frac {4}{3} \, {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}} - \frac {2}{3} \, \arctan \left ({\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{3} \, \log \left ({\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{3} \, \log \left ({\left | {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-x)^(1/4)/x^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 5.47, size = 33, normalized size = 0.60

method	result	size
meijerg	\(-\frac {4 \mathrm {signum}\left (x^{3}-1\right )^{\frac {1}{4}} \hypergeom \left (\left [-\frac {1}{4}, -\frac {1}{4}\right ], \left [\frac {3}{4}\right ], x^{3}\right )}{3 \left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {1}{4}} x^{\frac {3}{4}}}\)	\(33\)
trager	\(-\frac {4 \left (x^{4}-x \right )^{\frac {1}{4}}}{3 x}-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (2 \sqrt {x^{4}-x}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x -2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}-2 \left (x^{4}-x \right )^{\frac {3}{4}}+2 x^{2} \left (x^{4}-x \right )^{\frac {1}{4}}+\RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{3}-\frac {\ln \left (-2 \left (x^{4}-x \right )^{\frac {3}{4}}+2 x \sqrt {x^{4}-x}-2 x^{2} \left (x^{4}-x \right )^{\frac {1}{4}}+2 x^{3}-1\right )}{3}\)	\(133\)
risch	\(-\frac {4 \left (x \left (x^{3}-1\right )\right )^{\frac {1}{4}}}{3 x}+\frac {\left (-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 x^{9}-2 \left (x^{12}-3 x^{9}+3 x^{6}-x^{3}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{6}-5 x^{6}-2 \sqrt {x^{12}-3 x^{9}+3 x^{6}-x^{3}}\, x^{3}+4 \left (x^{12}-3 x^{9}+3 x^{6}-x^{3}\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{12}-3 x^{9}+3 x^{6}-x^{3}\right )^{\frac {3}{4}}+4 x^{3}+2 \sqrt {x^{12}-3 x^{9}+3 x^{6}-x^{3}}-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{12}-3 x^{9}+3 x^{6}-x^{3}\right )^{\frac {1}{4}}-1}{\left (-1+x \right )^{2} \left (x^{2}+x +1\right )^{2}}\right )}{3}-\frac {\ln \left (-\frac {-2 x^{9}+2 \left (x^{12}-3 x^{9}+3 x^{6}-x^{3}\right )^{\frac {1}{4}} x^{6}+5 x^{6}-2 \sqrt {x^{12}-3 x^{9}+3 x^{6}-x^{3}}\, x^{3}-4 \left (x^{12}-3 x^{9}+3 x^{6}-x^{3}\right )^{\frac {1}{4}} x^{3}+2 \left (x^{12}-3 x^{9}+3 x^{6}-x^{3}\right )^{\frac {3}{4}}-4 x^{3}+2 \sqrt {x^{12}-3 x^{9}+3 x^{6}-x^{3}}+2 \left (x^{12}-3 x^{9}+3 x^{6}-x^{3}\right )^{\frac {1}{4}}+1}{\left (-1+x \right )^{2} \left (x^{2}+x +1\right )^{2}}\right )}{3}\right ) \left (x \left (x^{3}-1\right )\right )^{\frac {1}{4}} \left (x^{3} \left (x^{3}-1\right )^{3}\right )^{\frac {1}{4}}}{x \left (x^{3}-1\right )}\)	\(444\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-x)^(1/4)/x^2,x,method=_RETURNVERBOSE)

[Out]

-4/3*signum(x^3-1)^(1/4)/(-signum(x^3-1))^(1/4)/x^(3/4)*hypergeom([-1/4,-1/4],[3/4],x^3)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} - x\right )}^{\frac {1}{4}}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-x)^(1/4)/x^2,x, algorithm="maxima")

[Out]

integrate((x^4 - x)^(1/4)/x^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (x^4-x\right )}^{1/4}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 - x)^(1/4)/x^2,x)

[Out]

int((x^4 - x)^(1/4)/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{x \left (x - 1\right ) \left (x^{2} + x + 1\right )}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-x)**(1/4)/x**2,x)

[Out]

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

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