Integrand size = 28, antiderivative size = 28 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=-\frac {4 \left (1+x^3\right )^{3/4} \left (-3-3 x^3+7 x^4\right )}{21 x^7} \]
Time = 0.66 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=-\frac {4 \left (1+x^3\right )^{3/4} \left (-3-3 x^3+7 x^4\right )}{21 x^7} \]
Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.86, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2374, 9, 27, 2374, 27, 948, 83}
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.
\(\displaystyle \int \frac {\left (x^3+4\right ) \left (x^4-x^3-1\right )}{x^8 \sqrt [4]{x^3+1}} \, dx\) |
\(\Big \downarrow \) 2374 |
\(\displaystyle \frac {4 \left (x^3+1\right )^{3/4}}{7 x^7}-\frac {1}{14} \int \frac {2 \left (-7 x^6+7 x^5-28 x^3+16 x^2\right )}{x^7 \sqrt [4]{x^3+1}}dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \frac {4 \left (x^3+1\right )^{3/4}}{7 x^7}-\frac {1}{14} \int \frac {2 \left (-7 x^4+7 x^3-28 x+16\right )}{x^5 \sqrt [4]{x^3+1}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 \left (x^3+1\right )^{3/4}}{7 x^7}-\frac {1}{7} \int \frac {-7 x^4+7 x^3-28 x+16}{x^5 \sqrt [4]{x^3+1}}dx\) |
\(\Big \downarrow \) 2374 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{8} \int \frac {56 \left (x^3+4\right )}{x^4 \sqrt [4]{x^3+1}}dx+\frac {4 \left (x^3+1\right )^{3/4}}{x^4}\right )+\frac {4 \left (x^3+1\right )^{3/4}}{7 x^7}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (7 \int \frac {x^3+4}{x^4 \sqrt [4]{x^3+1}}dx+\frac {4 \left (x^3+1\right )^{3/4}}{x^4}\right )+\frac {4 \left (x^3+1\right )^{3/4}}{7 x^7}\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{7} \left (\frac {7}{3} \int \frac {x^3+4}{x^6 \sqrt [4]{x^3+1}}dx^3+\frac {4 \left (x^3+1\right )^{3/4}}{x^4}\right )+\frac {4 \left (x^3+1\right )^{3/4}}{7 x^7}\) |
\(\Big \downarrow \) 83 |
\(\displaystyle \frac {4 \left (x^3+1\right )^{3/4}}{7 x^7}+\frac {1}{7} \left (\frac {4 \left (x^3+1\right )^{3/4}}{x^4}-\frac {28 \left (x^3+1\right )^{3/4}}{3 x^3}\right )\) |
3.4.25.3.1 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[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> 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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
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 = 0.96 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89
method | result | size |
trager | \(-\frac {4 \left (x^{3}+1\right )^{\frac {3}{4}} \left (7 x^{4}-3 x^{3}-3\right )}{21 x^{7}}\) | \(25\) |
risch | \(-\frac {4 \left (7 x^{7}-3 x^{6}+7 x^{4}-6 x^{3}-3\right )}{21 x^{7} \left (x^{3}+1\right )^{\frac {1}{4}}}\) | \(35\) |
gosper | \(-\frac {4 \left (1+x \right ) \left (x^{2}-x +1\right ) \left (7 x^{4}-3 x^{3}-3\right )}{21 x^{7} \left (x^{3}+1\right )^{\frac {1}{4}}}\) | \(36\) |
meijerg | \(\frac {4 \operatorname {hypergeom}\left (\left [-\frac {7}{3}, \frac {1}{4}\right ], \left [-\frac {4}{3}\right ], -x^{3}\right )}{7 x^{7}}+\frac {2 \sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (\frac {5 \pi \sqrt {2}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {9}{4}\right ], \left [2, 3\right ], -x^{3}\right )}{32 \Gamma \left (\frac {3}{4}\right )}-\frac {\left (3-3 \ln \left (2\right )-\frac {\pi }{2}+3 \ln \left (x \right )\right ) \pi \sqrt {2}}{4 \Gamma \left (\frac {3}{4}\right )}-\frac {\pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right ) x^{3}}\right )}{3 \pi }+\frac {5 \operatorname {hypergeom}\left (\left [-\frac {4}{3}, \frac {1}{4}\right ], \left [-\frac {1}{3}\right ], -x^{3}\right )}{4 x^{4}}+\frac {\sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (-\frac {\pi \sqrt {2}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {5}{4}\right ], \left [2, 2\right ], -x^{3}\right )}{4 \Gamma \left (\frac {3}{4}\right )}+\frac {\left (-3 \ln \left (2\right )-\frac {\pi }{2}+3 \ln \left (x \right )\right ) \pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right )}\right )}{6 \pi }+\frac {\operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {1}{4}\right ], \left [\frac {2}{3}\right ], -x^{3}\right )}{x}\) | \(180\) |
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=-\frac {4 \, {\left (7 \, x^{4} - 3 \, x^{3} - 3\right )} {\left (x^{3} + 1\right )}^{\frac {3}{4}}}{21 \, x^{7}} \]
Result contains complex when optimal does not.
Time = 2.51 (sec) , antiderivative size = 177, normalized size of antiderivative = 6.32 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=- \frac {\Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{4} \\ \frac {2}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} - \frac {5 \Gamma \left (- \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {1}{4} \\ - \frac {1}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{4} \Gamma \left (- \frac {1}{3}\right )} - \frac {4 \Gamma \left (- \frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{3}, \frac {1}{4} \\ - \frac {4}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x^{7} \Gamma \left (- \frac {4}{3}\right )} - \frac {\Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{\frac {3}{4}} \Gamma \left (\frac {5}{4}\right )} - \frac {4 \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{\frac {15}{4}} \Gamma \left (\frac {9}{4}\right )} \]
-gamma(-1/3)*hyper((-1/3, 1/4), (2/3,), x**3*exp_polar(I*pi))/(3*x*gamma(2 /3)) - 5*gamma(-4/3)*hyper((-4/3, 1/4), (-1/3,), x**3*exp_polar(I*pi))/(3* x**4*gamma(-1/3)) - 4*gamma(-7/3)*hyper((-7/3, 1/4), (-4/3,), x**3*exp_pol ar(I*pi))/(3*x**7*gamma(-4/3)) - gamma(1/4)*hyper((1/4, 1/4), (5/4,), exp_ polar(I*pi)/x**3)/(3*x**(3/4)*gamma(5/4)) - 4*gamma(5/4)*hyper((1/4, 5/4), (9/4,), exp_polar(I*pi)/x**3)/(3*x**(15/4)*gamma(9/4))
Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=-\frac {4 \, {\left (7 \, x^{7} - 3 \, x^{6} + 7 \, x^{4} - 6 \, x^{3} - 3\right )}}{21 \, {\left (x^{2} - x + 1\right )}^{\frac {1}{4}} {\left (x + 1\right )}^{\frac {1}{4}} x^{7}} \]
\[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=\int { \frac {{\left (x^{4} - x^{3} - 1\right )} {\left (x^{3} + 4\right )}}{{\left (x^{3} + 1\right )}^{\frac {1}{4}} x^{8}} \,d x } \]
Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {\left (4+x^3\right ) \left (-1-x^3+x^4\right )}{x^8 \sqrt [4]{1+x^3}} \, dx=\frac {12\,{\left (x^3+1\right )}^{3/4}+12\,x^3\,{\left (x^3+1\right )}^{3/4}-28\,x^4\,{\left (x^3+1\right )}^{3/4}}{21\,x^7} \]