Integrand size = 26, antiderivative size = 26 \[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=-\frac {4 \sqrt [4]{-1+x^3} \left (-1+x^3+5 x^4\right )}{5 x^5} \]
Time = 3.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=-\frac {4 \sqrt [4]{-1+x^3} \left (-1+x^3+5 x^4\right )}{5 x^5} \]
Time = 0.33 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.92, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2374, 9, 27, 2374, 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.
\(\displaystyle \int \frac {\left (x^3-4\right ) \left (x^4+x^3-1\right )}{x^6 \left (x^3-1\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 2374 |
\(\displaystyle \frac {1}{10} \int -\frac {2 \left (-5 x^6-5 x^5+20 x^3+8 x^2\right )}{x^5 \left (x^3-1\right )^{3/4}}dx+\frac {4 \sqrt [4]{x^3-1}}{5 x^5}\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \frac {1}{10} \int -\frac {2 \left (-5 x^4-5 x^3+20 x+8\right )}{x^3 \left (x^3-1\right )^{3/4}}dx+\frac {4 \sqrt [4]{x^3-1}}{5 x^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 \sqrt [4]{x^3-1}}{5 x^5}-\frac {1}{5} \int \frac {-5 x^4-5 x^3+20 x+8}{x^3 \left (x^3-1\right )^{3/4}}dx\) |
\(\Big \downarrow \) 2374 |
\(\displaystyle \frac {1}{5} \left (-\frac {1}{4} \int \frac {20 \left (4-x^3\right )}{x^2 \left (x^3-1\right )^{3/4}}dx-\frac {4 \sqrt [4]{x^3-1}}{x^2}\right )+\frac {4 \sqrt [4]{x^3-1}}{5 x^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (-5 \int \frac {4-x^3}{x^2 \left (x^3-1\right )^{3/4}}dx-\frac {4 \sqrt [4]{x^3-1}}{x^2}\right )+\frac {4 \sqrt [4]{x^3-1}}{5 x^5}\) |
\(\Big \downarrow \) 951 |
\(\displaystyle \frac {4 \sqrt [4]{x^3-1}}{5 x^5}+\frac {1}{5} \left (-\frac {20 \sqrt [4]{x^3-1}}{x}-\frac {4 \sqrt [4]{x^3-1}}{x^2}\right )\) |
3.3.82.3.1 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] :> 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 = 0.84 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88
method | result | size |
trager | \(-\frac {4 \left (x^{3}-1\right )^{\frac {1}{4}} \left (5 x^{4}+x^{3}-1\right )}{5 x^{5}}\) | \(23\) |
gosper | \(-\frac {4 \left (x -1\right ) \left (x^{2}+x +1\right ) \left (5 x^{4}+x^{3}-1\right )}{5 x^{5} \left (x^{3}-1\right )^{\frac {3}{4}}}\) | \(32\) |
risch | \(-\frac {4 \left (5 x^{7}+x^{6}-5 x^{4}-2 x^{3}+1\right )}{5 \left (x^{3}-1\right )^{\frac {3}{4}} x^{5}}\) | \(33\) |
meijerg | \(\frac {{\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {3}{4}} x^{2} \operatorname {hypergeom}\left (\left [\frac {2}{3}, \frac {3}{4}\right ], \left [\frac {5}{3}\right ], x^{3}\right )}{2 \operatorname {signum}\left (x^{3}-1\right )^{\frac {3}{4}}}+\frac {{\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {3}{4}} x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {3}{4}\right ], \left [\frac {4}{3}\right ], x^{3}\right )}{\operatorname {signum}\left (x^{3}-1\right )^{\frac {3}{4}}}+\frac {5 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [-\frac {2}{3}, \frac {3}{4}\right ], \left [\frac {1}{3}\right ], x^{3}\right )}{2 \operatorname {signum}\left (x^{3}-1\right )^{\frac {3}{4}} x^{2}}-\frac {4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [-\frac {5}{3}, \frac {3}{4}\right ], \left [-\frac {2}{3}\right ], x^{3}\right )}{5 \operatorname {signum}\left (x^{3}-1\right )^{\frac {3}{4}} x^{5}}+\frac {4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {3}{4}\right ], \left [\frac {2}{3}\right ], x^{3}\right )}{\operatorname {signum}\left (x^{3}-1\right )^{\frac {3}{4}} x}\) | \(159\) |
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=-\frac {4 \, {\left (5 \, x^{4} + x^{3} - 1\right )} {\left (x^{3} - 1\right )}^{\frac {1}{4}}}{5 \, x^{5}} \]
Result contains complex when optimal does not.
Time = 2.03 (sec) , antiderivative size = 178, normalized size of antiderivative = 6.85 \[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=\frac {x^{2} e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {3}{4} \\ \frac {5}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} + \frac {x e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {3}{4} \\ \frac {4}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {4 e^{\frac {i \pi }{4}} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{4} \\ \frac {2}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} + \frac {5 e^{\frac {i \pi }{4}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {3}{4} \\ \frac {1}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 x^{2} \Gamma \left (\frac {1}{3}\right )} - \frac {4 e^{\frac {i \pi }{4}} \Gamma \left (- \frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{3}, \frac {3}{4} \\ - \frac {2}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 x^{5} \Gamma \left (- \frac {2}{3}\right )} \]
x**2*exp(-3*I*pi/4)*gamma(2/3)*hyper((2/3, 3/4), (5/3,), x**3)/(3*gamma(5/ 3)) + x*exp(-3*I*pi/4)*gamma(1/3)*hyper((1/3, 3/4), (4/3,), x**3)/(3*gamma (4/3)) + 4*exp(I*pi/4)*gamma(-1/3)*hyper((-1/3, 3/4), (2/3,), x**3)/(3*x*g amma(2/3)) + 5*exp(I*pi/4)*gamma(-2/3)*hyper((-2/3, 3/4), (1/3,), x**3)/(3 *x**2*gamma(1/3)) - 4*exp(I*pi/4)*gamma(-5/3)*hyper((-5/3, 3/4), (-2/3,), x**3)/(3*x**5*gamma(-2/3))
Time = 0.30 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=-\frac {4 \, {\left (5 \, x^{7} + x^{6} - 5 \, x^{4} - 2 \, x^{3} + 1\right )}}{5 \, {\left (x^{2} + x + 1\right )}^{\frac {3}{4}} {\left (x - 1\right )}^{\frac {3}{4}} x^{5}} \]
\[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=\int { \frac {{\left (x^{4} + x^{3} - 1\right )} {\left (x^{3} - 4\right )}}{{\left (x^{3} - 1\right )}^{\frac {3}{4}} x^{6}} \,d x } \]
Time = 5.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {\left (-4+x^3\right ) \left (-1+x^3+x^4\right )}{x^6 \left (-1+x^3\right )^{3/4}} \, dx=-\frac {4\,x^3\,{\left (x^3-1\right )}^{1/4}-4\,{\left (x^3-1\right )}^{1/4}+20\,x^4\,{\left (x^3-1\right )}^{1/4}}{5\,x^5} \]