3.1.21 \(\int \frac {\sqrt [5]{1-2 x+x^2}}{1-x} \, dx\) [21]

Optimal. Leaf size=16 \[ -\frac {5}{2} \sqrt [5]{1-2 x+x^2} \]

[Out]

-5/2*(x^2-2*x+1)^(1/5)

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Rubi [A]
time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {657, 643} \begin {gather*} -\frac {5}{2} \sqrt [5]{x^2-2 x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x + x^2)^(1/5)/(1 - x),x]

[Out]

(-5*(1 - 2*x + x^2)^(1/5))/2

Rule 643

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*((a + b*x + c*x^2)^(p +
1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 657

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^(m - 1)/c^((m - 1)/2
), Int[(d + e*x)*(a + b*x + c*x^2)^(p + (m - 1)/2), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[b^2 - 4*a*c,
0] &&  !IntegerQ[p] && EqQ[2*c*d - b*e, 0] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 13, normalized size = 0.81 \begin {gather*} -\frac {5}{2} \sqrt [5]{(-1+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 2*x + x^2)^(1/5)/(1 - x),x]

[Out]

(-5*((-1 + x)^2)^(1/5))/2

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Maple [A]
time = 0.08, size = 10, normalized size = 0.62

method	result	size
risch	\(-\frac {5 \left (\left (-1+x \right )^{2}\right )^{\frac {1}{5}}}{2}\)	\(10\)
gosper	\(-\frac {5 \left (x^{2}-2 x +1\right )^{\frac {1}{5}}}{2}\)	\(13\)
trager	\(-\frac {5 \left (x^{2}-2 x +1\right )^{\frac {1}{5}}}{2}\)	\(13\)
meijerg	\(\frac {\mathrm {signum}\left (-1+x \right )^{\frac {2}{5}} x \hypergeom \left (\left [\frac {3}{5}, 1\right ], \left [2\right ], x\right )}{\left (-\mathrm {signum}\left (-1+x \right )\right )^{\frac {2}{5}}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-2*x+1)^(1/5)/(1-x),x,method=_RETURNVERBOSE)

[Out]

-5/2*((-1+x)^2)^(1/5)

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Maxima [A]
time = 3.98, size = 7, normalized size = 0.44 \begin {gather*} -\frac {5}{2} \, {\left (x - 1\right )}^{\frac {2}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-2*x+1)^(1/5)/(1-x),x, algorithm="maxima")

[Out]

-5/2*(x - 1)^(2/5)

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Fricas [A]
time = 0.42, size = 12, normalized size = 0.75 \begin {gather*} -\frac {5}{2} \, {\left (x^{2} - 2 \, x + 1\right )}^{\frac {1}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-2*x+1)^(1/5)/(1-x),x, algorithm="fricas")

[Out]

-5/2*(x^2 - 2*x + 1)^(1/5)

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Sympy [A]
time = 0.62, size = 15, normalized size = 0.94 \begin {gather*} - \frac {5 \sqrt [5]{x^{2} - 2 x + 1}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-2*x+1)**(1/5)/(1-x),x)

[Out]

-5*(x**2 - 2*x + 1)**(1/5)/2

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Giac [A]
time = 0.61, size = 12, normalized size = 0.75 \begin {gather*} -\frac {5}{2} \, {\left (x^{2} - 2 \, x + 1\right )}^{\frac {1}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-2*x+1)^(1/5)/(1-x),x, algorithm="giac")

[Out]

-5/2*(x^2 - 2*x + 1)^(1/5)

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Mupad [B]
time = 0.16, size = 9, normalized size = 0.56 \begin {gather*} -\frac {5\,{\left ({\left (x-1\right )}^2\right )}^{1/5}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2 - 2*x + 1)^(1/5)/(x - 1),x)

[Out]

-(5*((x - 1)^2)^(1/5))/2

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