3.8.8 \(\int \frac {\sqrt [4]{-x^3+x^4}}{x^2} \, dx\)

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

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Rubi [A]  time = 0.07, antiderivative size = 101, normalized size of antiderivative = 1.84, number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {2020, 2032, 63, 240, 212, 206, 203} \begin {gather*} -\frac {4 \sqrt [4]{x^4-x^3}}{x}+\frac {2 (x-1)^{3/4} x^{9/4} \tan ^{-1}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{\left (x^4-x^3\right )^{3/4}}+\frac {2 (x-1)^{3/4} x^{9/4} \tanh ^{-1}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{\left (x^4-x^3\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x^3 + x^4)^(1/4)/x^2,x]

[Out]

(-4*(-x^3 + x^4)^(1/4))/x + (2*(-1 + x)^(3/4)*x^(9/4)*ArcTan[(-1 + x)^(1/4)/x^(1/4)])/(-x^3 + x^4)^(3/4) + (2*
(-1 + x)^(3/4)*x^(9/4)*ArcTanh[(-1 + x)^(1/4)/x^(1/4)])/(-x^3 + x^4)^(3/4)

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 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^3+x^4}}{x^2} \, dx &=-\frac {4 \sqrt [4]{-x^3+x^4}}{x}+\int \frac {x^2}{\left (-x^3+x^4\right )^{3/4}} \, dx\\ &=-\frac {4 \sqrt [4]{-x^3+x^4}}{x}+\frac {\left ((-1+x)^{3/4} x^{9/4}\right ) \int \frac {1}{(-1+x)^{3/4} \sqrt [4]{x}} \, dx}{\left (-x^3+x^4\right )^{3/4}}\\ &=-\frac {4 \sqrt [4]{-x^3+x^4}}{x}+\frac {\left (4 (-1+x)^{3/4} x^{9/4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^4}} \, dx,x,\sqrt [4]{-1+x}\right )}{\left (-x^3+x^4\right )^{3/4}}\\ &=-\frac {4 \sqrt [4]{-x^3+x^4}}{x}+\frac {\left (4 (-1+x)^{3/4} x^{9/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\left (-x^3+x^4\right )^{3/4}}\\ &=-\frac {4 \sqrt [4]{-x^3+x^4}}{x}+\frac {\left (2 (-1+x)^{3/4} x^{9/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\left (-x^3+x^4\right )^{3/4}}+\frac {\left (2 (-1+x)^{3/4} x^{9/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\left (-x^3+x^4\right )^{3/4}}\\ &=-\frac {4 \sqrt [4]{-x^3+x^4}}{x}+\frac {2 (-1+x)^{3/4} x^{9/4} \tan ^{-1}\left (\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\left (-x^3+x^4\right )^{3/4}}+\frac {2 (-1+x)^{3/4} x^{9/4} \tanh ^{-1}\left (\frac {\sqrt [4]{-1+x}}{\sqrt [4]{x}}\right )}{\left (-x^3+x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.64 \begin {gather*} \frac {4 \left ((x-1) x^3\right )^{5/4} \, _2F_1\left (\frac {5}{4},\frac {5}{4};\frac {9}{4};1-x\right )}{5 x^{15/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^3 + x^4)^(1/4)/x^2,x]

[Out]

(4*((-1 + x)*x^3)^(5/4)*Hypergeometric2F1[5/4, 5/4, 9/4, 1 - x])/(5*x^(15/4))

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IntegrateAlgebraic [A]  time = 0.22, size = 55, normalized size = 1.00 \begin {gather*} -\frac {4 \sqrt [4]{-x^3+x^4}}{x}-2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right )+2 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-x^3+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-x^3 + x^4)^(1/4)/x^2,x]

[Out]

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

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fricas [A]  time = 0.47, size = 81, normalized size = 1.47 \begin {gather*} \frac {2 \, x \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + x \log \left (\frac {x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - x \log \left (-\frac {x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - 4 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-x^3)^(1/4)/x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 51, normalized size = 0.93 \begin {gather*} 4 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 2 \, \arctan \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) + \log \left ({\left | {\left (-\frac {1}{x} + 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^3)^(1/4)/x^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.73, size = 27, normalized size = 0.49

method	result	size
meijerg	\(-\frac {4 \mathrm {signum}\left (-1+x \right )^{\frac {1}{4}} \hypergeom \left (\left [-\frac {1}{4}, -\frac {1}{4}\right ], \left [\frac {3}{4}\right ], x\right )}{\left (-\mathrm {signum}\left (-1+x \right )\right )^{\frac {1}{4}} x^{\frac {1}{4}}}\)	\(27\)
trager	\(-\frac {4 \left (x^{4}-x^{3}\right )^{\frac {1}{4}}}{x}-\ln \left (\frac {2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}-2 \sqrt {x^{4}-x^{3}}\, x +2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}-2 x^{3}+x^{2}}{x^{2}}\right )+\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}-x^{3}}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x -2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}+2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}}{x^{2}}\right )\)	\(160\)
risch	\(-\frac {4 \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}}}{x}+\frac {\left (-\ln \left (\frac {2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {3}{4}}-2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, x +2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x^{2}-2 x^{3}+2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}-4 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x +5 x^{2}+2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}}-4 x +1}{\left (-1+x \right )^{2}}\right )+\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x -2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}-2 \sqrt {x^{4}-3 x^{3}+3 x^{2}-x}\, \RootOf \left (\textit {\_Z}^{2}+1\right )+5 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {3}{4}}+2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x^{2}-4 \RootOf \left (\textit {\_Z}^{2}+1\right ) x -4 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}} x +\RootOf \left (\textit {\_Z}^{2}+1\right )+2 \left (x^{4}-3 x^{3}+3 x^{2}-x \right )^{\frac {1}{4}}}{\left (-1+x \right )^{2}}\right )\right ) \left (x^{3} \left (-1+x \right )\right )^{\frac {1}{4}} \left (x \left (-1+x \right )^{3}\right )^{\frac {1}{4}}}{x \left (-1+x \right )}\)	\(395\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-x^3)^(1/4)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} - x^{3}\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^3)^(1/4)/x^2,x, algorithm="maxima")

[Out]

integrate((x^4 - x^3)^(1/4)/x^2, x)

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mupad [B]  time = 0.72, size = 29, normalized size = 0.53 \begin {gather*} -\frac {4\,{\left (x^4-x^3\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},-\frac {1}{4};\ \frac {3}{4};\ x\right )}{x\,{\left (1-x\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 - x^3)^(1/4)/x^2,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{x^{3} \left (x - 1\right )}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-x**3)**(1/4)/x**2,x)

[Out]

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

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