Integrand size = 28, antiderivative size = 28 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=-\frac {4 \sqrt [4]{1+x^3} \left (-1-x^3+5 x^4\right )}{5 x^5} \] Output:
-4/5*(x^3+1)^(1/4)*(5*x^4-x^3-1)/x^5
Time = 3.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=-\frac {4 \sqrt [4]{1+x^3} \left (-1-x^3+5 x^4\right )}{5 x^5} \] Input:
Integrate[((4 + x^3)*(-1 - x^3 + x^4))/(x^6*(1 + x^3)^(3/4)),x]
Output:
(-4*(1 + x^3)^(1/4)*(-1 - x^3 + 5*x^4))/(5*x^5)
Time = 0.32 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.79, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, 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 |
\(\displaystyle \frac {4 \sqrt [4]{x^3+1}}{5 x^5}-\frac {1}{10} \int \frac {2 \left (-5 x^6+5 x^5-20 x^3+8 x^2\right )}{x^5 \left (x^3+1\right )^{3/4}}dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \frac {4 \sqrt [4]{x^3+1}}{5 x^5}-\frac {1}{10} \int \frac {2 \left (-5 x^4+5 x^3-20 x+8\right )}{x^3 \left (x^3+1\right )^{3/4}}dx\) |
\(\Big \downarrow \) 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 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \frac {20 \left (x^3+4\right )}{x^2 \left (x^3+1\right )^{3/4}}dx+\frac {4 \sqrt [4]{x^3+1}}{x^2}\right )+\frac {4 \sqrt [4]{x^3+1}}{5 x^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (5 \int \frac {x^3+4}{x^2 \left (x^3+1\right )^{3/4}}dx+\frac {4 \sqrt [4]{x^3+1}}{x^2}\right )+\frac {4 \sqrt [4]{x^3+1}}{5 x^5}\) |
\(\Big \downarrow \) 951 |
\(\displaystyle \frac {4 \sqrt [4]{x^3+1}}{5 x^5}+\frac {1}{5} \left (\frac {4 \sqrt [4]{x^3+1}}{x^2}-\frac {20 \sqrt [4]{x^3+1}}{x}\right )\) |
Input:
Int[((4 + x^3)*(-1 - x^3 + x^4))/(x^6*(1 + x^3)^(3/4)),x]
Output:
(4*(1 + x^3)^(1/4))/(5*x^5) + ((4*(1 + x^3)^(1/4))/x^2 - (20*(1 + x^3)^(1/ 4))/x)/5
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 = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89
method | result | size |
trager | \(-\frac {4 \left (x^{3}+1\right )^{\frac {1}{4}} \left (5 x^{4}-x^{3}-1\right )}{5 x^{5}}\) | \(25\) |
risch | \(-\frac {4 \left (5 x^{7}-x^{6}+5 x^{4}-2 x^{3}-1\right )}{5 \left (x^{3}+1\right )^{\frac {3}{4}} x^{5}}\) | \(35\) |
gosper | \(-\frac {4 \left (1+x \right ) \left (x^{2}-x +1\right ) \left (5 x^{4}-x^{3}-1\right )}{5 x^{5} \left (x^{3}+1\right )^{\frac {3}{4}}}\) | \(36\) |
orering | \(-\frac {4 \left (1+x \right ) \left (x^{2}-x +1\right ) \left (5 x^{4}-x^{3}-1\right )}{5 x^{5} \left (x^{3}+1\right )^{\frac {3}{4}}}\) | \(36\) |
meijerg | \(\frac {x^{2} \operatorname {hypergeom}\left (\left [\frac {2}{3}, \frac {3}{4}\right ], \left [\frac {5}{3}\right ], -x^{3}\right )}{2}-x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {3}{4}\right ], \left [\frac {4}{3}\right ], -x^{3}\right )+\frac {5 \operatorname {hypergeom}\left (\left [-\frac {2}{3}, \frac {3}{4}\right ], \left [\frac {1}{3}\right ], -x^{3}\right )}{2 x^{2}}+\frac {4 \operatorname {hypergeom}\left (\left [-\frac {5}{3}, \frac {3}{4}\right ], \left [-\frac {2}{3}\right ], -x^{3}\right )}{5 x^{5}}-\frac {4 \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {3}{4}\right ], \left [\frac {2}{3}\right ], -x^{3}\right )}{x}\) | \(80\) |
Input:
int((x^3+4)*(x^4-x^3-1)/x^6/(x^3+1)^(3/4),x,method=_RETURNVERBOSE)
Output:
-4/5*(x^3+1)^(1/4)*(5*x^4-x^3-1)/x^5
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=-\frac {4 \, {\left (5 \, x^{4} - x^{3} - 1\right )} {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{5 \, x^{5}} \] Input:
integrate((x^3+4)*(x^4-x^3-1)/x^6/(x^3+1)^(3/4),x, algorithm="fricas")
Output:
-4/5*(5*x^4 - x^3 - 1)*(x^3 + 1)^(1/4)/x^5
Result contains complex when optimal does not.
Time = 2.38 (sec) , antiderivative size = 167, normalized size of antiderivative = 5.96 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=\frac {x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {3}{4} \\ \frac {5}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} - \frac {x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {3}{4} \\ \frac {4}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {4 \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{4} \\ \frac {2}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} - \frac {5 \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {3}{4} \\ \frac {1}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{2} \Gamma \left (\frac {1}{3}\right )} - \frac {4 \Gamma \left (- \frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{3}, \frac {3}{4} \\ - \frac {2}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{5} \Gamma \left (- \frac {2}{3}\right )} \] Input:
integrate((x**3+4)*(x**4-x**3-1)/x**6/(x**3+1)**(3/4),x)
Output:
x**2*gamma(2/3)*hyper((2/3, 3/4), (5/3,), x**3*exp_polar(I*pi))/(3*gamma(5 /3)) - x*gamma(1/3)*hyper((1/3, 3/4), (4/3,), x**3*exp_polar(I*pi))/(3*gam ma(4/3)) + 4*gamma(-1/3)*hyper((-1/3, 3/4), (2/3,), x**3*exp_polar(I*pi))/ (3*x*gamma(2/3)) - 5*gamma(-2/3)*hyper((-2/3, 3/4), (1/3,), x**3*exp_polar (I*pi))/(3*x**2*gamma(1/3)) - 4*gamma(-5/3)*hyper((-5/3, 3/4), (-2/3,), x* *3*exp_polar(I*pi))/(3*x**5*gamma(-2/3))
Time = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=-\frac {4 \, {\left (5 \, x^{7} - x^{6} + 5 \, x^{4} - 2 \, x^{3} - 1\right )}}{5 \, {\left (x^{2} - x + 1\right )}^{\frac {3}{4}} {\left (x + 1\right )}^{\frac {3}{4}} x^{5}} \] Input:
integrate((x^3+4)*(x^4-x^3-1)/x^6/(x^3+1)^(3/4),x, algorithm="maxima")
Output:
-4/5*(5*x^7 - x^6 + 5*x^4 - 2*x^3 - 1)/((x^2 - x + 1)^(3/4)*(x + 1)^(3/4)* x^5)
\[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=\int { \frac {{\left (x^{4} - x^{3} - 1\right )} {\left (x^{3} + 4\right )}}{{\left (x^{3} + 1\right )}^{\frac {3}{4}} x^{6}} \,d x } \] Input:
integrate((x^3+4)*(x^4-x^3-1)/x^6/(x^3+1)^(3/4),x, algorithm="giac")
Output:
integrate((x^4 - x^3 - 1)*(x^3 + 4)/((x^3 + 1)^(3/4)*x^6), x)
Time = 6.64 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=\frac {4\,{\left (x^3+1\right )}^{1/4}+4\,x^3\,{\left (x^3+1\right )}^{1/4}-20\,x^4\,{\left (x^3+1\right )}^{1/4}}{5\,x^5} \] Input:
int(-((x^3 + 4)*(x^3 - x^4 + 1))/(x^6*(x^3 + 1)^(3/4)),x)
Output:
(4*(x^3 + 1)^(1/4) + 4*x^3*(x^3 + 1)^(1/4) - 20*x^4*(x^3 + 1)^(1/4))/(5*x^ 5)
\[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^6 \left (1+x^3\right )^{3/4}} \, dx=\int \frac {x}{\left (x^{3}+1\right )^{\frac {3}{4}}}d x -\left (\int \frac {1}{\left (x^{3}+1\right )^{\frac {3}{4}}}d x \right )-4 \left (\int \frac {1}{\left (x^{3}+1\right )^{\frac {3}{4}} x^{6}}d x \right )-5 \left (\int \frac {1}{\left (x^{3}+1\right )^{\frac {3}{4}} x^{3}}d x \right )+4 \left (\int \frac {1}{\left (x^{3}+1\right )^{\frac {3}{4}} x^{2}}d x \right ) \] Input:
int((x^3+4)*(x^4-x^3-1)/x^6/(x^3+1)^(3/4),x)
Output:
int(x/(x**3 + 1)**(3/4),x) - int(1/(x**3 + 1)**(3/4),x) - 4*int(1/((x**3 + 1)**(3/4)*x**6),x) - 5*int(1/((x**3 + 1)**(3/4)*x**3),x) + 4*int(1/((x**3 + 1)**(3/4)*x**2),x)