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

Optimal. Leaf size=37 \[ 2 \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x^3}}\right )-2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x^3}}\right ) \]

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Rubi [B]  time = 0.11, antiderivative size = 83, normalized size of antiderivative = 2.24, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2056, 63, 240, 212, 206, 203} \begin {gather*} \frac {2 \sqrt [4]{x^4-x^3} \tan ^{-1}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{\sqrt [4]{x-1} x^{3/4}}+\frac {2 \sqrt [4]{x^4-x^3} \tanh ^{-1}\left (\frac {\sqrt [4]{x-1}}{\sqrt [4]{x}}\right )}{\sqrt [4]{x-1} x^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.01, size = 33, normalized size = 0.89 \begin {gather*} \frac {4 \sqrt [4]{(x-1) x^3} \, _2F_1\left (\frac {1}{4},\frac {1}{4};\frac {5}{4};1-x\right )}{x^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.21, size = 37, normalized size = 1.00 \begin {gather*} -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)/((-1 + x)*x),x]

[Out]

-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 = 0.48, size = 60, normalized size = 1.62 \begin {gather*} 2 \, \arctan \left (\frac {{\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \log \left (\frac {x + {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \log \left (-\frac {x - {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.31, size = 40, normalized size = 1.08 \begin {gather*} -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)/(-1+x)/x,x, algorithm="giac")

[Out]

-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 = 0.38, size = 27, normalized size = 0.73

method	result	size
meijerg	\(-\frac {4 \mathrm {signum}\left (-1+x \right )^{\frac {1}{4}} \hypergeom \left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], x\right ) x^{\frac {3}{4}}}{3 \left (-\mathrm {signum}\left (-1+x \right )\right )^{\frac {1}{4}}}\)	\(27\)
trager	\(\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}-x^{3}}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x -2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}+\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}}{x^{2}}\right )+\ln \left (\frac {2 \left (x^{4}-x^{3}\right )^{\frac {3}{4}}+2 \sqrt {x^{4}-x^{3}}\, x +2 x^{2} \left (x^{4}-x^{3}\right )^{\frac {1}{4}}+2 x^{3}-x^{2}}{x^{2}}\right )\)	\(144\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3/(-signum(-1+x))^(1/4)*signum(-1+x)^(1/4)*hypergeom([3/4,3/4],[7/4],x)*x^(3/4)

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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}}}{{\left (x - 1\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\left (x^4-x^3\right )}^{1/4}}{x\,\left (x-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 - x^3)^(1/4)/(x*(x - 1)),x)

[Out]

int((x^4 - x^3)^(1/4)/(x*(x - 1)), x)

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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 \left (x - 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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