3.2.30 \(\int \frac {-1+x^4}{x^2 \sqrt {-x+x^3}} \, dx\)

Optimal. Leaf size=18 \[ \frac {2 \left (x^3-x\right )^{3/2}}{3 x^3} \]

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Rubi [C]  time = 0.10, antiderivative size = 96, normalized size of antiderivative = 5.33, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2048, 2025, 2011, 329, 222} \begin {gather*} \frac {2 \sqrt {x^3-x}}{3}-\frac {\sqrt {2} \sqrt {x-1} \sqrt {x} \sqrt {x+1} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {x}}{\sqrt {x-1}}\right )|\frac {1}{2}\right )}{3 \sqrt {x^3-x}}-\frac {2 \sqrt {x^3-x}}{3 x^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[(-1 + x^4)/(x^2*Sqrt[-x + x^3]),x]

[Out]

(2*Sqrt[-x + x^3])/3 - (2*Sqrt[-x + x^3])/(3*x^2) - (Sqrt[2]*Sqrt[-1 + x]*Sqrt[x]*Sqrt[1 + x]*EllipticF[ArcSin
[(Sqrt[2]*Sqrt[x])/Sqrt[-1 + x]], 1/2])/(3*Sqrt[-x + x^3])

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(a*b), 2]}, Simp[(Sqrt[-a + q*x^2]*Sqrt[(a + q*x^2
)/q]*EllipticF[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]], 1/2])/(Sqrt[2]*Sqrt[-a]*Sqrt[a + b*x^4]), x] /; IntegerQ[q]]
/; FreeQ[{a, b}, x] && LtQ[a, 0] && GtQ[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 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2025

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

Rule 2048

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]},
With[{Pqq = Coeff[Pq, x, q]}, Int[(c*x)^m*ExpandToSum[Pq - Pqq*x^q - (a*Pqq*(m + q - n + 1)*x^(q - n))/(b*(m +
q + n*p + 1)), x]*(a*x^j + b*x^n)^p, x] + Simp[(Pqq*(c*x)^(m + q - n + 1)*(a*x^j + b*x^n)^(p + 1))/(b*c^(q -
n + 1)*(m + q + n*p + 1)), x]] /; GtQ[q, n - 1] && NeQ[m + q + n*p + 1, 0] && (IntegerQ[2*p] || IntegerQ[p + (
q + 1)/(2*n)])] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] &&  !IntegerQ[p] && IGtQ[j, 0] && IGtQ[n, 0] && L
tQ[j, n]

Rubi steps

\begin {align*} \int \frac {-1+x^4}{x^2 \sqrt {-x+x^3}} \, dx &=\frac {2}{3} \sqrt {-x+x^3}-\int \frac {1}{x^2 \sqrt {-x+x^3}} \, dx\\ &=\frac {2}{3} \sqrt {-x+x^3}-\frac {2 \sqrt {-x+x^3}}{3 x^2}-\frac {1}{3} \int \frac {1}{\sqrt {-x+x^3}} \, dx\\ &=\frac {2}{3} \sqrt {-x+x^3}-\frac {2 \sqrt {-x+x^3}}{3 x^2}-\frac {\left (\sqrt {x} \sqrt {-1+x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {-1+x^2}} \, dx}{3 \sqrt {-x+x^3}}\\ &=\frac {2}{3} \sqrt {-x+x^3}-\frac {2 \sqrt {-x+x^3}}{3 x^2}-\frac {\left (2 \sqrt {x} \sqrt {-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x^4}} \, dx,x,\sqrt {x}\right )}{3 \sqrt {-x+x^3}}\\ &=\frac {2}{3} \sqrt {-x+x^3}-\frac {2 \sqrt {-x+x^3}}{3 x^2}-\frac {\sqrt {2} \sqrt {-1+x} \sqrt {x} \sqrt {1+x} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {x}}{\sqrt {-1+x}}\right )|\frac {1}{2}\right )}{3 \sqrt {-x+x^3}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \begin {gather*} \frac {2 \left (x \left (x^2-1\right )\right )^{3/2}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x^4)/(x^2*Sqrt[-x + x^3]),x]

[Out]

(2*(x*(-1 + x^2))^(3/2))/(3*x^3)

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IntegrateAlgebraic [A]  time = 0.16, size = 18, normalized size = 1.00 \begin {gather*} \frac {2 \left (-x+x^3\right )^{3/2}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-x + x^3)^(3/2))/(3*x^3)

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fricas [A]  time = 0.45, size = 19, normalized size = 1.06 \begin {gather*} \frac {2 \, \sqrt {x^{3} - x} {\left (x^{2} - 1\right )}}{3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(x^3 - x)*(x^2 - 1)/x^2

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giac [A]  time = 0.54, size = 25, normalized size = 1.39 \begin {gather*} \frac {2}{3} \, \sqrt {x^{3} - x} - \frac {2}{3} \, \sqrt {\frac {1}{x} - \frac {1}{x^{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(x^3 - x) - 2/3*sqrt(1/x - 1/x^3)

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maple [A]  time = 0.11, size = 20, normalized size = 1.11

method	result	size
trager	\(\frac {2 \left (x^{2}-1\right ) \sqrt {x^{3}-x}}{3 x^{2}}\)	\(20\)
risch	\(\frac {\frac {2}{3} x^{4}-\frac {4}{3} x^{2}+\frac {2}{3}}{x \sqrt {x \left (x^{2}-1\right )}}\)	\(25\)
gosper	\(\frac {2 \left (x^{2}-1\right ) \left (-1+x \right ) \left (1+x \right )}{3 \sqrt {x^{3}-x}\, x}\)	\(26\)
default	\(\frac {2 \sqrt {x^{3}-x}}{3}-\frac {2 \sqrt {x^{3}-x}}{3 x^{2}}\)	\(27\)
elliptic	\(\frac {2 \sqrt {x^{3}-x}}{3}-\frac {2 \sqrt {x^{3}-x}}{3 x^{2}}\)	\(27\)
meijerg	\(\frac {2 \sqrt {-\mathrm {signum}\left (x^{2}-1\right )}\, \hypergeom \left (\left [\frac {1}{2}, \frac {5}{4}\right ], \left [\frac {9}{4}\right ], x^{2}\right ) x^{\frac {5}{2}}}{5 \sqrt {\mathrm {signum}\left (x^{2}-1\right )}}+\frac {2 \sqrt {-\mathrm {signum}\left (x^{2}-1\right )}\, \hypergeom \left (\left [-\frac {3}{4}, \frac {1}{2}\right ], \left [\frac {1}{4}\right ], x^{2}\right )}{3 \sqrt {\mathrm {signum}\left (x^{2}-1\right )}\, x^{\frac {3}{2}}}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(x^2-1)/x^2*(x^3-x)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} - 1}{\sqrt {x^{3} - x} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.01, size = 1, normalized size = 0.06 \begin {gather*} 0 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 - 1)/(x^2*(x^3 - x)^(1/2)),x)

[Out]

0

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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