3.2.81 \(\int \frac {-1+x^4}{x^2 \sqrt [4]{-x^3+x^5}} \, dx\)

Optimal. Leaf size=20 \[ \frac {4 \left (x^5-x^3\right )^{7/4}}{7 x^7} \]

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Rubi [B]  time = 0.15, antiderivative size = 41, normalized size of antiderivative = 2.05, number of steps used = 10, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2052, 2025, 2011, 365, 364, 2024} \begin {gather*} \frac {4 \left (x^5-x^3\right )^{3/4}}{7 x^2}-\frac {4 \left (x^5-x^3\right )^{3/4}}{7 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + x^4)/(x^2*(-x^3 + x^5)^(1/4)),x]

[Out]

(-4*(-x^3 + x^5)^(3/4))/(7*x^4) + (4*(-x^3 + x^5)^(3/4))/(7*x^2)

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
 (ILtQ[p, 0] || GtQ[a, 0])

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

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 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
 1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*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]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

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 2052

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(c*x)
^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !In
tegerQ[p] && NeQ[n, j]

Rubi steps

\begin {align*} \int \frac {-1+x^4}{x^2 \sqrt [4]{-x^3+x^5}} \, dx &=\int \left (-\frac {1}{x^2 \sqrt [4]{-x^3+x^5}}+\frac {x^2}{\sqrt [4]{-x^3+x^5}}\right ) \, dx\\ &=-\int \frac {1}{x^2 \sqrt [4]{-x^3+x^5}} \, dx+\int \frac {x^2}{\sqrt [4]{-x^3+x^5}} \, dx\\ &=-\frac {4 \left (-x^3+x^5\right )^{3/4}}{7 x^4}+\frac {4 \left (-x^3+x^5\right )^{3/4}}{7 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 1.00 \begin {gather*} \frac {4 \left (x^3 \left (x^2-1\right )\right )^{7/4}}{7 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x^4)/(x^2*(-x^3 + x^5)^(1/4)),x]

[Out]

(4*(x^3*(-1 + x^2))^(7/4))/(7*x^7)

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IntegrateAlgebraic [A]  time = 0.14, size = 20, normalized size = 1.00 \begin {gather*} \frac {4 \left (-x^3+x^5\right )^{7/4}}{7 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-1 + x^4)/(x^2*(-x^3 + x^5)^(1/4)),x]

[Out]

(4*(-x^3 + x^5)^(7/4))/(7*x^7)

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fricas [A]  time = 0.47, size = 21, normalized size = 1.05 \begin {gather*} \frac {4 \, {\left (x^{5} - x^{3}\right )}^{\frac {3}{4}} {\left (x^{2} - 1\right )}}{7 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)/x^2/(x^5-x^3)^(1/4),x, algorithm="fricas")

[Out]

4/7*(x^5 - x^3)^(3/4)*(x^2 - 1)/x^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} - 1}{{\left (x^{5} - 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^4-1)/x^2/(x^5-x^3)^(1/4),x, algorithm="giac")

[Out]

integrate((x^4 - 1)/((x^5 - x^3)^(1/4)*x^2), x)

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maple [A]  time = 0.11, size = 22, normalized size = 1.10

method result size
trager \(\frac {4 \left (x^{2}-1\right ) \left (x^{5}-x^{3}\right )^{\frac {3}{4}}}{7 x^{4}}\) \(22\)
risch \(\frac {\frac {4}{7} x^{4}-\frac {8}{7} x^{2}+\frac {4}{7}}{x \left (x^{3} \left (x^{2}-1\right )\right )^{\frac {1}{4}}}\) \(27\)
gosper \(\frac {4 \left (x^{2}-1\right ) \left (-1+x \right ) \left (1+x \right )}{7 \left (x^{5}-x^{3}\right )^{\frac {1}{4}} x}\) \(28\)
meijerg \(\frac {4 \left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {1}{4}} \hypergeom \left (\left [-\frac {7}{8}, \frac {1}{4}\right ], \left [\frac {1}{8}\right ], x^{2}\right )}{7 \mathrm {signum}\left (x^{2}-1\right )^{\frac {1}{4}} x^{\frac {7}{4}}}+\frac {4 \left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {1}{4}} \hypergeom \left (\left [\frac {1}{4}, \frac {9}{8}\right ], \left [\frac {17}{8}\right ], x^{2}\right ) x^{\frac {9}{4}}}{9 \mathrm {signum}\left (x^{2}-1\right )^{\frac {1}{4}}}\) \(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-1)/x^2/(x^5-x^3)^(1/4),x,method=_RETURNVERBOSE)

[Out]

4/7*(x^2-1)/x^4*(x^5-x^3)^(3/4)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} - 1}{{\left (x^{5} - 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^4-1)/x^2/(x^5-x^3)^(1/4),x, algorithm="maxima")

[Out]

integrate((x^4 - 1)/((x^5 - x^3)^(1/4)*x^2), x)

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mupad [B]  time = 0.28, size = 35, normalized size = 1.75 \begin {gather*} \frac {4\,x^2\,{\left (x^5-x^3\right )}^{3/4}-4\,{\left (x^5-x^3\right )}^{3/4}}{7\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 - 1)/(x^2*(x^5 - x^3)^(1/4)),x)

[Out]

(4*x^2*(x^5 - x^3)^(3/4) - 4*(x^5 - x^3)^(3/4))/(7*x^4)

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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 [4]{x^{3} \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**5-x**3)**(1/4),x)

[Out]

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

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