3.2.33 \(\int \frac {-1+x^3}{x^6 \sqrt [4]{-x+x^4}} \, dx\)

Optimal. Leaf size=18 \[ \frac {4 \left (x^4-x\right )^{7/4}}{21 x^7} \]

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Rubi [B]  time = 0.12, antiderivative size = 37, normalized size of antiderivative = 2.06, number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2052, 2016, 2014} \begin {gather*} \frac {4 \left (x^4-x\right )^{3/4}}{21 x^3}-\frac {4 \left (x^4-x\right )^{3/4}}{21 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + x^3)/(x^6*(-x + x^4)^(1/4)),x]

[Out]

(-4*(-x + x^4)^(3/4))/(21*x^6) + (4*(-x + x^4)^(3/4))/(21*x^3)

Rule 2014

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*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2016

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, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
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^3}{x^6 \sqrt [4]{-x+x^4}} \, dx &=\int \left (-\frac {1}{x^6 \sqrt [4]{-x+x^4}}+\frac {1}{x^3 \sqrt [4]{-x+x^4}}\right ) \, dx\\ &=-\int \frac {1}{x^6 \sqrt [4]{-x+x^4}} \, dx+\int \frac {1}{x^3 \sqrt [4]{-x+x^4}} \, dx\\ &=-\frac {4 \left (-x+x^4\right )^{3/4}}{21 x^6}+\frac {4 \left (-x+x^4\right )^{3/4}}{9 x^3}-\frac {4}{7} \int \frac {1}{x^3 \sqrt [4]{-x+x^4}} \, dx\\ &=-\frac {4 \left (-x+x^4\right )^{3/4}}{21 x^6}+\frac {4 \left (-x+x^4\right )^{3/4}}{21 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \begin {gather*} \frac {4 \left (x \left (x^3-1\right )\right )^{7/4}}{21 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x^3)/(x^6*(-x + x^4)^(1/4)),x]

[Out]

(4*(x*(-1 + x^3))^(7/4))/(21*x^7)

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IntegrateAlgebraic [A]  time = 0.24, size = 18, normalized size = 1.00 \begin {gather*} \frac {4 \left (-x+x^4\right )^{7/4}}{21 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-1 + x^3)/(x^6*(-x + x^4)^(1/4)),x]

[Out]

(4*(-x + x^4)^(7/4))/(21*x^7)

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fricas [A]  time = 0.45, size = 19, normalized size = 1.06 \begin {gather*} \frac {4 \, {\left (x^{4} - x\right )}^{\frac {3}{4}} {\left (x^{3} - 1\right )}}{21 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)/x^6/(x^4-x)^(1/4),x, algorithm="fricas")

[Out]

4/21*(x^4 - x)^(3/4)*(x^3 - 1)/x^6

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giac [A]  time = 0.50, size = 11, normalized size = 0.61 \begin {gather*} -\frac {4}{21} \, {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {7}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)/x^6/(x^4-x)^(1/4),x, algorithm="giac")

[Out]

-4/21*(-1/x^3 + 1)^(7/4)

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maple [A]  time = 0.07, size = 20, normalized size = 1.11

method	result	size
trager	\(\frac {4 \left (x^{3}-1\right ) \left (x^{4}-x \right )^{\frac {3}{4}}}{21 x^{6}}\)	\(20\)
risch	\(\frac {\frac {4}{21} x^{6}-\frac {8}{21} x^{3}+\frac {4}{21}}{x^{5} \left (x \left (x^{3}-1\right )\right )^{\frac {1}{4}}}\)	\(25\)
gosper	\(\frac {4 \left (-1+x \right ) \left (x^{2}+x +1\right ) \left (x^{3}-1\right )}{21 x^{5} \left (x^{4}-x \right )^{\frac {1}{4}}}\)	\(29\)
meijerg	\(\frac {4 \left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {1}{4}} \left (1+\frac {4 x^{3}}{3}\right ) \left (-x^{3}+1\right )^{\frac {3}{4}}}{21 \mathrm {signum}\left (x^{3}-1\right )^{\frac {1}{4}} x^{\frac {21}{4}}}-\frac {4 \left (-\mathrm {signum}\left (x^{3}-1\right )\right )^{\frac {1}{4}} \left (-x^{3}+1\right )^{\frac {3}{4}}}{9 \mathrm {signum}\left (x^{3}-1\right )^{\frac {1}{4}} x^{\frac {9}{4}}}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3-1)/x^6/(x^4-x)^(1/4),x,method=_RETURNVERBOSE)

[Out]

4/21*(x^3-1)*(x^4-x)^(3/4)/x^6

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maxima [B]  time = 0.75, size = 58, normalized size = 3.22 \begin {gather*} \frac {4 \, {\left (x^{4} - x\right )}}{9 \, {\left (x^{2} + x + 1\right )}^{\frac {1}{4}} {\left (x - 1\right )}^{\frac {1}{4}} x^{\frac {13}{4}}} - \frac {4 \, {\left (4 \, x^{7} - x^{4} - 3 \, x\right )}}{63 \, {\left (x^{2} + x + 1\right )}^{\frac {1}{4}} {\left (x - 1\right )}^{\frac {1}{4}} x^{\frac {25}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-1)/x^6/(x^4-x)^(1/4),x, algorithm="maxima")

[Out]

4/9*(x^4 - x)/((x^2 + x + 1)^(1/4)*(x - 1)^(1/4)*x^(13/4)) - 4/63*(4*x^7 - x^4 - 3*x)/((x^2 + x + 1)^(1/4)*(x
- 1)^(1/4)*x^(25/4))

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mupad [B]  time = 0.24, size = 31, normalized size = 1.72 \begin {gather*} -\frac {4\,{\left (x^4-x\right )}^{3/4}-4\,x^3\,{\left (x^4-x\right )}^{3/4}}{21\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3 - 1)/(x^6*(x^4 - x)^(1/4)),x)

[Out]

-(4*(x^4 - x)^(3/4) - 4*x^3*(x^4 - x)^(3/4))/(21*x^6)

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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^{2} + x + 1\right )}{x^{6} \sqrt [4]{x \left (x - 1\right ) \left (x^{2} + x + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3-1)/x**6/(x**4-x)**(1/4),x)

[Out]

Integral((x - 1)*(x**2 + x + 1)/(x**6*(x*(x - 1)*(x**2 + x + 1))**(1/4)), x)

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