3.2.81 \(\int \frac {-1+x^4}{x^2 \sqrt [4]{-x^3+x^5}} \, dx\) [181]

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

[Out]

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

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(41\) vs. \(2(20)=40\).
time = 0.10, 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 = {2077, 2050, 2036, 372, 371, 2049} \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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 2036

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 2049

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 2050

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 2077

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 = 0.19, size = 20, normalized size = 1.00 \begin {gather*} \frac {4 \left (x^3 \left (-1+x^2\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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Maple [A]
time = 0.29, 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	\(\frac {4 \left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {1}{4}} \hypergeom \left (\left [-\frac {7}{8}, \frac {1}{4}\right ], \left [\frac {1}{8}\right ], x^{2}\right )}{7 \mathrm {signum}\left (x^{2}-1\right )^{\frac {1}{4}} x^{\frac {7}{4}}}+\frac {4 \left (-\mathrm {signum}\left (x^{2}-1\right )\right )^{\frac {1}{4}} x^{\frac {9}{4}} \hypergeom \left (\left [\frac {1}{4}, \frac {9}{8}\right ], \left [\frac {17}{8}\right ], x^{2}\right )}{9 \mathrm {signum}\left (x^{2}-1\right )^{\frac {1}{4}}}\)	\(66\)

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*} \text {Failed to integrate} \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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Fricas [A]
time = 0.34, 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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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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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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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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