3.5.84 \(\int \frac {(-1+x^4)^{2/3} (3+x^4) (-2-x^3+2 x^4)}{x^9} \, dx\) [484]

Optimal. Leaf size=38 \[ \frac {3 \left (-1+x^4\right )^{2/3} \left (5+4 x^3-10 x^4-4 x^7+5 x^8\right )}{20 x^8} \]

[Out]

3/20*(x^4-1)^(2/3)*(5*x^8-4*x^7-10*x^4+4*x^3+5)/x^8

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Rubi [A]
time = 0.06, antiderivative size = 33, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1847, 1598, 460, 1488, 861, 75} \begin {gather*} \frac {3 \left (x^4-1\right )^{8/3}}{4 x^8}-\frac {3 \left (x^4-1\right )^{5/3}}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1 + x^4)^(2/3)*(3 + x^4)*(-2 - x^3 + 2*x^4))/x^9,x]

[Out]

(-3*(-1 + x^4)^(5/3))/(5*x^5) + (3*(-1 + x^4)^(8/3))/(4*x^8)

Rule 75

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

Rule 460

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

Rule 861

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

Rule 1488

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a,
b, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1847

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[
Sum[((c*x)^(m + j)/c^j)*Sum[Coeff[Pq, x, j + k*(n/2)]*x^(k*(n/2)), {k, 0, 2*((q - j)/n) + 1}]*(a + b*x^n)^p, {
j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rubi steps

\begin {align*} \int \frac {\left (-1+x^4\right )^{2/3} \left (3+x^4\right ) \left (-2-x^3+2 x^4\right )}{x^9} \, dx &=\int \left (\frac {\left (-1+x^4\right )^{2/3} \left (-3 x^2-x^6\right )}{x^8}+\frac {\left (-1+x^4\right )^{2/3} \left (-6+4 x^4+2 x^8\right )}{x^9}\right ) \, dx\\ &=\int \frac {\left (-1+x^4\right )^{2/3} \left (-3 x^2-x^6\right )}{x^8} \, dx+\int \frac {\left (-1+x^4\right )^{2/3} \left (-6+4 x^4+2 x^8\right )}{x^9} \, dx\\ &=\frac {1}{4} \text {Subst}\left (\int \frac {(-1+x)^{2/3} \left (-6+4 x+2 x^2\right )}{x^3} \, dx,x,x^4\right )+\int \frac {\left (-3-x^4\right ) \left (-1+x^4\right )^{2/3}}{x^6} \, dx\\ &=-\frac {3 \left (-1+x^4\right )^{5/3}}{5 x^5}+\frac {1}{4} \text {Subst}\left (\int \frac {(-1+x)^{5/3} (6+2 x)}{x^3} \, dx,x,x^4\right )\\ &=-\frac {3 \left (-1+x^4\right )^{5/3}}{5 x^5}+\frac {3 \left (-1+x^4\right )^{8/3}}{4 x^8}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 28, normalized size = 0.74 \begin {gather*} \frac {3 \left (-1+x^4\right )^{5/3} \left (-5-4 x^3+5 x^4\right )}{20 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1 + x^4)^(2/3)*(3 + x^4)*(-2 - x^3 + 2*x^4))/x^9,x]

[Out]

(3*(-1 + x^4)^(5/3)*(-5 - 4*x^3 + 5*x^4))/(20*x^8)

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Maple [A]
time = 0.53, size = 35, normalized size = 0.92

method	result	size
trager	\(\frac {3 \left (x^{4}-1\right )^{\frac {2}{3}} \left (5 x^{8}-4 x^{7}-10 x^{4}+4 x^{3}+5\right )}{20 x^{8}}\)	\(35\)
gosper	\(\frac {3 \left (x^{2}+1\right ) \left (1+x \right ) \left (-1+x \right ) \left (5 x^{4}-4 x^{3}-5\right ) \left (x^{4}-1\right )^{\frac {2}{3}}}{20 x^{8}}\)	\(36\)
risch	\(\frac {-\frac {3}{5} x^{11}+\frac {6}{5} x^{7}-\frac {3}{5} x^{3}-\frac {9}{4} x^{8}+\frac {9}{4} x^{4}-\frac {3}{4}+\frac {3}{4} x^{12}}{x^{8} \left (x^{4}-1\right )^{\frac {1}{3}}}\)	\(45\)
meijerg	\(-\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \mathrm {signum}\left (x^{4}-1\right )^{\frac {2}{3}} \left (\frac {2 \pi \sqrt {3}\, x^{4} \hypergeom \left (\left [\frac {1}{3}, 1, 1\right ], \left [2, 2\right ], x^{4}\right )}{3 \Gamma \left (\frac {2}{3}\right )}-\frac {\left (\frac {3}{2}-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+4 \ln \left (x \right )+i \pi \right ) \pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right )}\right )}{6 \pi \left (-\mathrm {signum}\left (x^{4}-1\right )\right )^{\frac {2}{3}}}+\frac {\mathrm {signum}\left (x^{4}-1\right )^{\frac {2}{3}} \hypergeom \left (\left [-\frac {2}{3}, -\frac {1}{4}\right ], \left [\frac {3}{4}\right ], x^{4}\right )}{\left (-\mathrm {signum}\left (x^{4}-1\right )\right )^{\frac {2}{3}} x}+\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \mathrm {signum}\left (x^{4}-1\right )^{\frac {2}{3}} \left (-\frac {\pi \sqrt {3}\, x^{4} \hypergeom \left (\left [1, 1, \frac {4}{3}\right ], \left [2, 3\right ], x^{4}\right )}{9 \Gamma \left (\frac {2}{3}\right )}-\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}-1+4 \ln \left (x \right )+i \pi \right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}-\frac {\pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right ) x^{4}}\right )}{3 \pi \left (-\mathrm {signum}\left (x^{4}-1\right )\right )^{\frac {2}{3}}}+\frac {3 \mathrm {signum}\left (x^{4}-1\right )^{\frac {2}{3}} \hypergeom \left (\left [-\frac {5}{4}, -\frac {2}{3}\right ], \left [-\frac {1}{4}\right ], x^{4}\right )}{5 \left (-\mathrm {signum}\left (x^{4}-1\right )\right )^{\frac {2}{3}} x^{5}}+\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \mathrm {signum}\left (x^{4}-1\right )^{\frac {2}{3}} \left (\frac {4 \pi \sqrt {3}\, x^{4} \hypergeom \left (\left [1, 1, \frac {7}{3}\right ], \left [2, 4\right ], x^{4}\right )}{81 \Gamma \left (\frac {2}{3}\right )}+\frac {\left (\frac {3}{2}-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+4 \ln \left (x \right )+i \pi \right ) \pi \sqrt {3}}{9 \Gamma \left (\frac {2}{3}\right )}+\frac {\pi \sqrt {3}}{2 \Gamma \left (\frac {2}{3}\right ) x^{8}}-\frac {2 \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right ) x^{4}}\right )}{2 \pi \left (-\mathrm {signum}\left (x^{4}-1\right )\right )^{\frac {2}{3}}}\)	\(353\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-1)^(2/3)*(x^4+3)*(2*x^4-x^3-2)/x^9,x,method=_RETURNVERBOSE)

[Out]

3/20*(x^4-1)^(2/3)*(5*x^8-4*x^7-10*x^4+4*x^3+5)/x^8

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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)^(2/3)*(x^4+3)*(2*x^4-x^3-2)/x^9,x, algorithm="maxima")

[Out]

-1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^4 - 1)^(1/3) - 1)) - 1/4*(2*(x^4 - 1)^(5/3) - (x^4 - 1)^(2/3))/(2*x^4 +
(x^4 - 1)^2 - 1) + integrate((2*x^5 - x^4 + 4*x - 3)*(x^2 + 1)^(2/3)*(x + 1)^(2/3)*(x - 1)^(2/3)/x^6, x) - 1/1
2*log((x^4 - 1)^(2/3) - (x^4 - 1)^(1/3) + 1) + 1/6*log((x^4 - 1)^(1/3) + 1)

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Fricas [A]
time = 0.36, size = 34, normalized size = 0.89 \begin {gather*} \frac {3 \, {\left (5 \, x^{8} - 4 \, x^{7} - 10 \, x^{4} + 4 \, x^{3} + 5\right )} {\left (x^{4} - 1\right )}^{\frac {2}{3}}}{20 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-1)^(2/3)*(x^4+3)*(2*x^4-x^3-2)/x^9,x, algorithm="fricas")

[Out]

3/20*(5*x^8 - 4*x^7 - 10*x^4 + 4*x^3 + 5)*(x^4 - 1)^(2/3)/x^8

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Sympy [C] Result contains complex when optimal does not.
time = 4.26, size = 187, normalized size = 4.92 \begin {gather*} - \frac {x^{\frac {8}{3}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{4}}} \right )}}{2 \Gamma \left (\frac {1}{3}\right )} + \frac {e^{- \frac {i \pi }{3}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} + \frac {3 e^{- \frac {i \pi }{3}} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {2}{3} \\ - \frac {1}{4} \end {matrix}\middle | {x^{4}} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} - \frac {\Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{4}}} \right )}}{x^{\frac {4}{3}} \Gamma \left (\frac {4}{3}\right )} + \frac {3 \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{4}}} \right )}}{2 x^{\frac {16}{3}} \Gamma \left (\frac {7}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-1)**(2/3)*(x**4+3)*(2*x**4-x**3-2)/x**9,x)

[Out]

-x**(8/3)*gamma(-2/3)*hyper((-2/3, -2/3), (1/3,), exp_polar(2*I*pi)/x**4)/(2*gamma(1/3)) + exp(-I*pi/3)*gamma(
-1/4)*hyper((-2/3, -1/4), (3/4,), x**4)/(4*x*gamma(3/4)) + 3*exp(-I*pi/3)*gamma(-5/4)*hyper((-5/4, -2/3), (-1/
4,), x**4)/(4*x**5*gamma(-1/4)) - gamma(1/3)*hyper((-2/3, 1/3), (4/3,), exp_polar(2*I*pi)/x**4)/(x**(4/3)*gamm
a(4/3)) + 3*gamma(4/3)*hyper((-2/3, 4/3), (7/3,), exp_polar(2*I*pi)/x**4)/(2*x**(16/3)*gamma(7/3))

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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)^(2/3)*(x^4+3)*(2*x^4-x^3-2)/x^9,x, algorithm="giac")

[Out]

integrate((2*x^4 - x^3 - 2)*(x^4 + 3)*(x^4 - 1)^(2/3)/x^9, x)

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Mupad [B]
time = 0.54, size = 58, normalized size = 1.53 \begin {gather*} \frac {3\,{\left (x^4-1\right )}^{2/3}}{4}-\frac {3\,{\left (x^4-1\right )}^{2/3}}{5\,x}-\frac {3\,{\left (x^4-1\right )}^{2/3}}{2\,x^4}+\frac {3\,{\left (x^4-1\right )}^{2/3}}{5\,x^5}+\frac {3\,{\left (x^4-1\right )}^{2/3}}{4\,x^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((x^4 - 1)^(2/3)*(x^4 + 3)*(x^3 - 2*x^4 + 2))/x^9,x)

[Out]

(3*(x^4 - 1)^(2/3))/4 - (3*(x^4 - 1)^(2/3))/(5*x) - (3*(x^4 - 1)^(2/3))/(2*x^4) + (3*(x^4 - 1)^(2/3))/(5*x^5)
+ (3*(x^4 - 1)^(2/3))/(4*x^8)

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