\(\int \frac {(-1+x^5)^{2/3} (-1+x^3+x^5) (3+2 x^5)}{x^9} \, dx\) [487]

Optimal result

Integrand size = 28, antiderivative size = 38 \[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx=\frac {3 \left (-1+x^5\right )^{2/3} \left (5-8 x^3-10 x^5+8 x^8+5 x^{10}\right )}{40 x^8} \]

[Out]

3/40*(x^5-1)^(2/3)*(5*x^10+8*x^8-10*x^5-8*x^3+5)/x^8

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(89\) vs. \(2(38)=76\).

Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.34, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1840, 1849, 1600, 1598, 460} \[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx=\frac {6 \left (x^5-1\right )^{2/3}}{5 x^5}-\frac {15 \left (x^5-1\right )^{2/3}}{56 x^8}-\frac {15 \left (x^5-1\right )^{2/3}}{4 x^3}+\frac {3 \left (x^5-1\right )^{2/3} \left (35 x^{11}+56 x^9+280 x^6-168 x^4+60 x\right )}{280 x^9} \]

[In]

Int[((-1 + x^5)^(2/3)*(-1 + x^3 + x^5)*(3 + 2*x^5))/x^9,x]

[Out]

(-15*(-1 + x^5)^(2/3))/(56*x^8) + (6*(-1 + x^5)^(2/3))/(5*x^5) - (15*(-1 + x^5)^(2/3))/(4*x^3) + (3*(-1 + x^5)
^(2/3)*(60*x - 168*x^4 + 280*x^6 + 56*x^9 + 35*x^11))/(280*x^9)

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 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 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1840

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

Rule 1849

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{Pq0 = Coeff[Pq, x, 0]}, Simp[Pq0
*(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] + Dist[1/(2*a*c*(m + 1)), Int[(c*x)^(m + 1)*ExpandToSum
[2*a*(m + 1)*((Pq - Pq0)/x) - 2*b*Pq0*(m + n*(p + 1) + 1)*x^(n - 1), x]*(a + b*x^n)^p, x], x] /; NeQ[Pq0, 0]]
/; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[m, -1] && LeQ[n - 1, Expon[Pq, x]]

Rubi steps \begin{align*} \text {integral}& = \frac {3 \left (-1+x^5\right )^{2/3} \left (60 x-168 x^4+280 x^6+56 x^9+35 x^{11}\right )}{280 x^9}-\frac {10}{3} \int \frac {\frac {9}{14}-\frac {9 x^3}{5}+3 x^5+\frac {3 x^8}{5}+\frac {3 x^{10}}{8}}{x^9 \sqrt [3]{-1+x^5}} \, dx \\ & = -\frac {15 \left (-1+x^5\right )^{2/3}}{56 x^8}+\frac {3 \left (-1+x^5\right )^{2/3} \left (60 x-168 x^4+280 x^6+56 x^9+35 x^{11}\right )}{280 x^9}-\frac {5}{24} \int \frac {-\frac {144 x^2}{5}+54 x^4+\frac {48 x^7}{5}+6 x^9}{x^8 \sqrt [3]{-1+x^5}} \, dx \\ & = -\frac {15 \left (-1+x^5\right )^{2/3}}{56 x^8}+\frac {3 \left (-1+x^5\right )^{2/3} \left (60 x-168 x^4+280 x^6+56 x^9+35 x^{11}\right )}{280 x^9}-\frac {5}{24} \int \frac {-\frac {144 x}{5}+54 x^3+\frac {48 x^6}{5}+6 x^8}{x^7 \sqrt [3]{-1+x^5}} \, dx \\ & = -\frac {15 \left (-1+x^5\right )^{2/3}}{56 x^8}+\frac {3 \left (-1+x^5\right )^{2/3} \left (60 x-168 x^4+280 x^6+56 x^9+35 x^{11}\right )}{280 x^9}-\frac {5}{24} \int \frac {-\frac {144}{5}+54 x^2+\frac {48 x^5}{5}+6 x^7}{x^6 \sqrt [3]{-1+x^5}} \, dx \\ & = -\frac {15 \left (-1+x^5\right )^{2/3}}{56 x^8}+\frac {6 \left (-1+x^5\right )^{2/3}}{5 x^5}+\frac {3 \left (-1+x^5\right )^{2/3} \left (60 x-168 x^4+280 x^6+56 x^9+35 x^{11}\right )}{280 x^9}-\frac {1}{48} \int \frac {540 x+60 x^6}{x^5 \sqrt [3]{-1+x^5}} \, dx \\ & = -\frac {15 \left (-1+x^5\right )^{2/3}}{56 x^8}+\frac {6 \left (-1+x^5\right )^{2/3}}{5 x^5}+\frac {3 \left (-1+x^5\right )^{2/3} \left (60 x-168 x^4+280 x^6+56 x^9+35 x^{11}\right )}{280 x^9}-\frac {1}{48} \int \frac {540+60 x^5}{x^4 \sqrt [3]{-1+x^5}} \, dx \\ & = -\frac {15 \left (-1+x^5\right )^{2/3}}{56 x^8}+\frac {6 \left (-1+x^5\right )^{2/3}}{5 x^5}-\frac {15 \left (-1+x^5\right )^{2/3}}{4 x^3}+\frac {3 \left (-1+x^5\right )^{2/3} \left (60 x-168 x^4+280 x^6+56 x^9+35 x^{11}\right )}{280 x^9} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx=\frac {3 \left (-1+x^5\right )^{5/3} \left (-5+8 x^3+5 x^5\right )}{40 x^8} \]

[In]

Integrate[((-1 + x^5)^(2/3)*(-1 + x^3 + x^5)*(3 + 2*x^5))/x^9,x]

[Out]

(3*(-1 + x^5)^(5/3)*(-5 + 8*x^3 + 5*x^5))/(40*x^8)

Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66

method	result	size
pseudoelliptic	\(\frac {3 \left (x^{5}-1\right )^{\frac {5}{3}} \left (5 x^{5}+8 x^{3}-5\right )}{40 x^{8}}\)	\(25\)
trager	\(\frac {3 \left (x^{5}-1\right )^{\frac {2}{3}} \left (5 x^{10}+8 x^{8}-10 x^{5}-8 x^{3}+5\right )}{40 x^{8}}\)	\(35\)
gosper	\(\frac {3 \left (x -1\right ) \left (x^{4}+x^{3}+x^{2}+x +1\right ) \left (5 x^{5}+8 x^{3}-5\right ) \left (x^{5}-1\right )^{\frac {2}{3}}}{40 x^{8}}\)	\(40\)
risch	\(\frac {-\frac {6}{5} x^{8}+\frac {3}{5} x^{3}-\frac {9}{8} x^{10}+\frac {9}{8} x^{5}-\frac {3}{8}+\frac {3}{8} x^{15}+\frac {3}{5} x^{13}}{x^{8} \left (x^{5}-1\right )^{\frac {1}{3}}}\)	\(45\)
meijerg	\(\frac {\operatorname {signum}\left (x^{5}-1\right )^{\frac {2}{3}} x^{2} \operatorname {hypergeom}\left (\left [-\frac {2}{3}, \frac {2}{5}\right ], \left [\frac {7}{5}\right ], x^{5}\right )}{{\left (-\operatorname {signum}\left (x^{5}-1\right )\right )}^{\frac {2}{3}}}-\frac {\operatorname {signum}\left (x^{5}-1\right )^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [-\frac {2}{3}, -\frac {3}{5}\right ], \left [\frac {2}{5}\right ], x^{5}\right )}{3 {\left (-\operatorname {signum}\left (x^{5}-1\right )\right )}^{\frac {2}{3}} x^{3}}-\frac {2 \sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \operatorname {signum}\left (x^{5}-1\right )^{\frac {2}{3}} \left (\frac {2 \pi \sqrt {3}\, x^{5} \operatorname {hypergeom}\left (\left [\frac {1}{3}, 1, 1\right ], \left [2, 2\right ], x^{5}\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}+5 \ln \left (x \right )+i \pi \right ) \pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right )}\right )}{15 \pi {\left (-\operatorname {signum}\left (x^{5}-1\right )\right )}^{\frac {2}{3}}}+\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \operatorname {signum}\left (x^{5}-1\right )^{\frac {2}{3}} \left (-\frac {\pi \sqrt {3}\, x^{5} \operatorname {hypergeom}\left (\left [1, 1, \frac {4}{3}\right ], \left [2, 3\right ], x^{5}\right )}{9 \Gamma \left (\frac {2}{3}\right )}-\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}-1+5 \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^{5}}\right )}{5 \pi {\left (-\operatorname {signum}\left (x^{5}-1\right )\right )}^{\frac {2}{3}}}+\frac {3 \operatorname {signum}\left (x^{5}-1\right )^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [-\frac {8}{5}, -\frac {2}{3}\right ], \left [-\frac {3}{5}\right ], x^{5}\right )}{8 {\left (-\operatorname {signum}\left (x^{5}-1\right )\right )}^{\frac {2}{3}} x^{8}}\)	\(276\)

[In]

int((x^5-1)^(2/3)*(x^5+x^3-1)*(2*x^5+3)/x^9,x,method=_RETURNVERBOSE)

[Out]

3/40*(x^5-1)^(5/3)*(5*x^5+8*x^3-5)/x^8

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx=\frac {3 \, {\left (5 \, x^{10} + 8 \, x^{8} - 10 \, x^{5} - 8 \, x^{3} + 5\right )} {\left (x^{5} - 1\right )}^{\frac {2}{3}}}{40 \, x^{8}} \]

[In]

integrate((x^5-1)^(2/3)*(x^5+x^3-1)*(2*x^5+3)/x^9,x, algorithm="fricas")

[Out]

3/40*(5*x^10 + 8*x^8 - 10*x^5 - 8*x^3 + 5)*(x^5 - 1)^(2/3)/x^8

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.91 (sec) , antiderivative size = 194, normalized size of antiderivative = 5.11 \[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx=- \frac {2 x^{\frac {10}{3}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{5}}} \right )}}{5 \Gamma \left (\frac {1}{3}\right )} + \frac {2 x^{2} e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {2}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {2}{5} \\ \frac {7}{5} \end {matrix}\middle | {x^{5}} \right )}}{5 \Gamma \left (\frac {7}{5}\right )} - \frac {e^{- \frac {i \pi }{3}} \Gamma \left (- \frac {3}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {3}{5} \\ \frac {2}{5} \end {matrix}\middle | {x^{5}} \right )}}{5 x^{3} \Gamma \left (\frac {2}{5}\right )} + \frac {3 e^{- \frac {i \pi }{3}} \Gamma \left (- \frac {8}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {8}{5}, - \frac {2}{3} \\ - \frac {3}{5} \end {matrix}\middle | {x^{5}} \right )}}{5 x^{8} \Gamma \left (- \frac {3}{5}\right )} - \frac {3 \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{5}}} \right )}}{5 x^{\frac {5}{3}} \Gamma \left (\frac {4}{3}\right )} \]

[In]

integrate((x**5-1)**(2/3)*(x**5+x**3-1)*(2*x**5+3)/x**9,x)

[Out]

-2*x**(10/3)*gamma(-2/3)*hyper((-2/3, -2/3), (1/3,), exp_polar(2*I*pi)/x**5)/(5*gamma(1/3)) + 2*x**2*exp(2*I*p
i/3)*gamma(2/5)*hyper((-2/3, 2/5), (7/5,), x**5)/(5*gamma(7/5)) - exp(-I*pi/3)*gamma(-3/5)*hyper((-2/3, -3/5),
(2/5,), x**5)/(5*x**3*gamma(2/5)) + 3*exp(-I*pi/3)*gamma(-8/5)*hyper((-8/5, -2/3), (-3/5,), x**5)/(5*x**8*gam
ma(-3/5)) - 3*gamma(1/3)*hyper((-2/3, 1/3), (4/3,), exp_polar(2*I*pi)/x**5)/(5*x**(5/3)*gamma(4/3))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx=\frac {3 \, {\left (5 \, x^{10} + 8 \, x^{8} - 10 \, x^{5} - 8 \, x^{3} + 5\right )} {\left (x^{4} + x^{3} + x^{2} + x + 1\right )}^{\frac {2}{3}} {\left (x - 1\right )}^{\frac {2}{3}}}{40 \, x^{8}} \]

[In]

integrate((x^5-1)^(2/3)*(x^5+x^3-1)*(2*x^5+3)/x^9,x, algorithm="maxima")

[Out]

3/40*(5*x^10 + 8*x^8 - 10*x^5 - 8*x^3 + 5)*(x^4 + x^3 + x^2 + x + 1)^(2/3)*(x - 1)^(2/3)/x^8

Giac [F]

\[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx=\int { \frac {{\left (2 \, x^{5} + 3\right )} {\left (x^{5} + x^{3} - 1\right )} {\left (x^{5} - 1\right )}^{\frac {2}{3}}}{x^{9}} \,d x } \]

[In]

integrate((x^5-1)^(2/3)*(x^5+x^3-1)*(2*x^5+3)/x^9,x, algorithm="giac")

[Out]

integrate((2*x^5 + 3)*(x^5 + x^3 - 1)*(x^5 - 1)^(2/3)/x^9, x)

Mupad [B] (verification not implemented)

Time = 5.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.37 \[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx={\left (x^5-1\right )}^{2/3}\,\left (\frac {3\,x^2}{8}+\frac {3}{5}\right )-\frac {3\,{\left (x^5-1\right )}^{2/3}}{4\,x^3}-\frac {3\,{\left (x^5-1\right )}^{2/3}}{5\,x^5}+\frac {3\,{\left (x^5-1\right )}^{2/3}}{8\,x^8} \]

[In]

int(((x^5 - 1)^(2/3)*(2*x^5 + 3)*(x^3 + x^5 - 1))/x^9,x)

[Out]

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