Integrand size = 28, antiderivative size = 38 \[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx=\frac {3 \left (-1+x^5\right )^{2/3} \left (5-8 x^3-10 x^5+8 x^8+5 x^{10}\right )}{40 x^8} \]
Time = 0.80 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx=\frac {3 \left (-1+x^5\right )^{5/3} \left (-5+8 x^3+5 x^5\right )}{40 x^8} \]
Leaf count is larger than twice the leaf count of optimal. \(95\) vs. \(2(38)=76\).
Time = 0.45 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.50, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {2365, 27, 2374, 9, 27, 2374, 9, 27, 951}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\left (x^5-1\right )^{2/3} \left (x^5+x^3-1\right ) \left (2 x^5+3\right )}{x^9} \, dx\) |
| \(\Big \downarrow \) 2365 |
| \(\displaystyle \frac {3 \left (x^5-1\right )^{2/3} \left (35 x^{11}+56 x^9+280 x^6-168 x^4+60 x\right )}{280 x^9}-\frac {10}{3} \int \frac {3 \left (35 x^{10}+56 x^8+280 x^5-168 x^3+60\right )}{280 x^9 \sqrt [3]{x^5-1}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 \left (x^5-1\right )^{2/3} \left (35 x^{11}+56 x^9+280 x^6-168 x^4+60 x\right )}{280 x^9}-\frac {1}{28} \int \frac {35 x^{10}+56 x^8+280 x^5-168 x^3+60}{x^9 \sqrt [3]{x^5-1}}dx\) |
| \(\Big \downarrow \) 2374 |
| \(\displaystyle \frac {1}{28} \left (-\frac {1}{16} \int -\frac {112 \left (-5 x^9-8 x^7-45 x^4+24 x^2\right )}{x^8 \sqrt [3]{x^5-1}}dx-\frac {15 \left (x^5-1\right )^{2/3}}{2 x^8}\right )+\frac {3 \left (x^5-1\right )^{2/3} \left (35 x^{11}+56 x^9+280 x^6-168 x^4+60 x\right )}{280 x^9}\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \frac {1}{28} \left (-\frac {1}{16} \int -\frac {112 \left (-5 x^7-8 x^5-45 x^2+24\right )}{x^6 \sqrt [3]{x^5-1}}dx-\frac {15 \left (x^5-1\right )^{2/3}}{2 x^8}\right )+\frac {3 \left (x^5-1\right )^{2/3} \left (35 x^{11}+56 x^9+280 x^6-168 x^4+60 x\right )}{280 x^9}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{28} \left (7 \int \frac {-5 x^7-8 x^5-45 x^2+24}{x^6 \sqrt [3]{x^5-1}}dx-\frac {15 \left (x^5-1\right )^{2/3}}{2 x^8}\right )+\frac {3 \left (x^5-1\right )^{2/3} \left (35 x^{11}+56 x^9+280 x^6-168 x^4+60 x\right )}{280 x^9}\) |
| \(\Big \downarrow \) 2374 |
| \(\displaystyle \frac {1}{28} \left (7 \left (\frac {1}{10} \int -\frac {50 \left (x^6+9 x\right )}{x^5 \sqrt [3]{x^5-1}}dx+\frac {24 \left (x^5-1\right )^{2/3}}{5 x^5}\right )-\frac {15 \left (x^5-1\right )^{2/3}}{2 x^8}\right )+\frac {3 \left (x^5-1\right )^{2/3} \left (35 x^{11}+56 x^9+280 x^6-168 x^4+60 x\right )}{280 x^9}\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \frac {1}{28} \left (7 \left (\frac {1}{10} \int -\frac {50 \left (x^5+9\right )}{x^4 \sqrt [3]{x^5-1}}dx+\frac {24 \left (x^5-1\right )^{2/3}}{5 x^5}\right )-\frac {15 \left (x^5-1\right )^{2/3}}{2 x^8}\right )+\frac {3 \left (x^5-1\right )^{2/3} \left (35 x^{11}+56 x^9+280 x^6-168 x^4+60 x\right )}{280 x^9}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{28} \left (7 \left (\frac {24 \left (x^5-1\right )^{2/3}}{5 x^5}-5 \int \frac {x^5+9}{x^4 \sqrt [3]{x^5-1}}dx\right )-\frac {15 \left (x^5-1\right )^{2/3}}{2 x^8}\right )+\frac {3 \left (x^5-1\right )^{2/3} \left (35 x^{11}+56 x^9+280 x^6-168 x^4+60 x\right )}{280 x^9}\) |
| \(\Big \downarrow \) 951 |
| \(\displaystyle \frac {1}{28} \left (7 \left (\frac {24 \left (x^5-1\right )^{2/3}}{5 x^5}-\frac {15 \left (x^5-1\right )^{2/3}}{x^3}\right )-\frac {15 \left (x^5-1\right )^{2/3}}{2 x^8}\right )+\frac {3 \left (x^5-1\right )^{2/3} \left (35 x^{11}+56 x^9+280 x^6-168 x^4+60 x\right )}{280 x^9}\) |
(3*(-1 + x^5)^(2/3)*(60*x - 168*x^4 + 280*x^6 + 56*x^9 + 35*x^11))/(280*x^ 9) + ((-15*(-1 + x^5)^(2/3))/(2*x^8) + 7*((24*(-1 + x^5)^(2/3))/(5*x^5) - (15*(-1 + x^5)^(2/3))/x^3))/28
3.5.87.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> M odule[{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] + Simp[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]
Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Wit h[{Pq0 = Coeff[Pq, x, 0]}, Simp[Pq0*(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c *(m + 1))), x] + Simp[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]]
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}}\) | \(276\) |
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx=\frac {3 \, {\left (5 \, x^{10} + 8 \, x^{8} - 10 \, x^{5} - 8 \, x^{3} + 5\right )} {\left (x^{5} - 1\right )}^{\frac {2}{3}}}{40 \, x^{8}} \]
Result contains complex when optimal does not.
Time = 2.91 (sec) , antiderivative size = 194, normalized size of antiderivative = 5.11 \[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx=- \frac {2 x^{\frac {10}{3}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{5}}} \right )}}{5 \Gamma \left (\frac {1}{3}\right )} + \frac {2 x^{2} e^{\frac {2 i \pi }{3}} \Gamma \left (\frac {2}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {2}{5} \\ \frac {7}{5} \end {matrix}\middle | {x^{5}} \right )}}{5 \Gamma \left (\frac {7}{5}\right )} - \frac {e^{- \frac {i \pi }{3}} \Gamma \left (- \frac {3}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {3}{5} \\ \frac {2}{5} \end {matrix}\middle | {x^{5}} \right )}}{5 x^{3} \Gamma \left (\frac {2}{5}\right )} + \frac {3 e^{- \frac {i \pi }{3}} \Gamma \left (- \frac {8}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {8}{5}, - \frac {2}{3} \\ - \frac {3}{5} \end {matrix}\middle | {x^{5}} \right )}}{5 x^{8} \Gamma \left (- \frac {3}{5}\right )} - \frac {3 \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e^{2 i \pi }}{x^{5}}} \right )}}{5 x^{\frac {5}{3}} \Gamma \left (\frac {4}{3}\right )} \]
-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*pi/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*gamma(-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))
Time = 0.33 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx=\frac {3 \, {\left (5 \, x^{10} + 8 \, x^{8} - 10 \, x^{5} - 8 \, x^{3} + 5\right )} {\left (x^{4} + x^{3} + x^{2} + x + 1\right )}^{\frac {2}{3}} {\left (x - 1\right )}^{\frac {2}{3}}}{40 \, x^{8}} \]
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
\[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx=\int { \frac {{\left (2 \, x^{5} + 3\right )} {\left (x^{5} + x^{3} - 1\right )} {\left (x^{5} - 1\right )}^{\frac {2}{3}}}{x^{9}} \,d x } \]
Time = 5.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.37 \[ \int \frac {\left (-1+x^5\right )^{2/3} \left (-1+x^3+x^5\right ) \left (3+2 x^5\right )}{x^9} \, dx={\left (x^5-1\right )}^{2/3}\,\left (\frac {3\,x^2}{8}+\frac {3}{5}\right )-\frac {3\,{\left (x^5-1\right )}^{2/3}}{4\,x^3}-\frac {3\,{\left (x^5-1\right )}^{2/3}}{5\,x^5}+\frac {3\,{\left (x^5-1\right )}^{2/3}}{8\,x^8} \]