Integrand size = 27, antiderivative size = 142 \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (-1+x^2\right )} \, dx=\frac {3 \left (1+10 x^2\right ) \sqrt [3]{x+2 x^3}}{8 x^3}+3^{5/6} \arctan \left (\frac {3^{5/6} x}{\sqrt [3]{3} x+2 \sqrt [3]{x+2 x^3}}\right )+\sqrt [3]{3} \log \left (-3 x+3^{2/3} \sqrt [3]{x+2 x^3}\right )-\frac {1}{2} \sqrt [3]{3} \log \left (3 x^2+3^{2/3} x \sqrt [3]{x+2 x^3}+\sqrt [3]{3} \left (x+2 x^3\right )^{2/3}\right ) \]
3/8*(10*x^2+1)*(2*x^3+x)^(1/3)/x^3+3^(5/6)*arctan(3^(5/6)*x/(3^(1/3)*x+2*( 2*x^3+x)^(1/3)))+3^(1/3)*ln(-3*x+3^(2/3)*(2*x^3+x)^(1/3))-1/2*3^(1/3)*ln(3 *x^2+3^(2/3)*x*(2*x^3+x)^(1/3)+3^(1/3)*(2*x^3+x)^(2/3))
Time = 1.62 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.46 \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (-1+x^2\right )} \, dx=\frac {\sqrt [3]{x+2 x^3} \left (3 \sqrt [3]{1+2 x^2}+30 x^2 \sqrt [3]{1+2 x^2}+8\ 3^{5/6} x^{8/3} \arctan \left (\frac {3^{5/6} x^{2/3}}{\sqrt [3]{3} x^{2/3}+2 \sqrt [3]{1+2 x^2}}\right )+8 \sqrt [3]{3} x^{8/3} \log \left (-3 x^{2/3}+3^{2/3} \sqrt [3]{1+2 x^2}\right )-4 \sqrt [3]{3} x^{8/3} \log \left (3 x^{4/3}+3^{2/3} x^{2/3} \sqrt [3]{1+2 x^2}+\sqrt [3]{3} \left (1+2 x^2\right )^{2/3}\right )\right )}{8 x^3 \sqrt [3]{1+2 x^2}} \]
((x + 2*x^3)^(1/3)*(3*(1 + 2*x^2)^(1/3) + 30*x^2*(1 + 2*x^2)^(1/3) + 8*3^( 5/6)*x^(8/3)*ArcTan[(3^(5/6)*x^(2/3))/(3^(1/3)*x^(2/3) + 2*(1 + 2*x^2)^(1/ 3))] + 8*3^(1/3)*x^(8/3)*Log[-3*x^(2/3) + 3^(2/3)*(1 + 2*x^2)^(1/3)] - 4*3 ^(1/3)*x^(8/3)*Log[3*x^(4/3) + 3^(2/3)*x^(2/3)*(1 + 2*x^2)^(1/3) + 3^(1/3) *(1 + 2*x^2)^(2/3)]))/(8*x^3*(1 + 2*x^2)^(1/3))
Time = 0.45 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.16, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2467, 25, 442, 27, 445, 27, 368, 965, 992}
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.
| \(\displaystyle \int \frac {\left (x^2+1\right ) \sqrt [3]{2 x^3+x}}{x^4 \left (x^2-1\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{2 x^3+x} \int -\frac {\left (x^2+1\right ) \sqrt [3]{2 x^2+1}}{x^{11/3} \left (1-x^2\right )}dx}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [3]{2 x^3+x} \int \frac {\left (x^2+1\right ) \sqrt [3]{2 x^2+1}}{x^{11/3} \left (1-x^2\right )}dx}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
| \(\Big \downarrow \) 442 |
| \(\displaystyle -\frac {\sqrt [3]{2 x^3+x} \left (\frac {3}{8} \int \frac {4 \left (7 x^2+5\right )}{3 x^{5/3} \left (1-x^2\right ) \left (2 x^2+1\right )^{2/3}}dx-\frac {3 \sqrt [3]{2 x^2+1}}{8 x^{8/3}}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\sqrt [3]{2 x^3+x} \left (\frac {1}{2} \int \frac {7 x^2+5}{x^{5/3} \left (1-x^2\right ) \left (2 x^2+1\right )^{2/3}}dx-\frac {3 \sqrt [3]{2 x^2+1}}{8 x^{8/3}}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
| \(\Big \downarrow \) 445 |
| \(\displaystyle -\frac {\sqrt [3]{2 x^3+x} \left (\frac {1}{2} \left (-\frac {3}{2} \int -\frac {8 \sqrt [3]{x}}{\left (1-x^2\right ) \left (2 x^2+1\right )^{2/3}}dx-\frac {15 \sqrt [3]{2 x^2+1}}{2 x^{2/3}}\right )-\frac {3 \sqrt [3]{2 x^2+1}}{8 x^{8/3}}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\sqrt [3]{2 x^3+x} \left (\frac {1}{2} \left (12 \int \frac {\sqrt [3]{x}}{\left (1-x^2\right ) \left (2 x^2+1\right )^{2/3}}dx-\frac {15 \sqrt [3]{2 x^2+1}}{2 x^{2/3}}\right )-\frac {3 \sqrt [3]{2 x^2+1}}{8 x^{8/3}}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
| \(\Big \downarrow \) 368 |
| \(\displaystyle -\frac {\sqrt [3]{2 x^3+x} \left (\frac {1}{2} \left (36 \int \frac {x}{\left (1-x^2\right ) \left (2 x^2+1\right )^{2/3}}d\sqrt [3]{x}-\frac {15 \sqrt [3]{2 x^2+1}}{2 x^{2/3}}\right )-\frac {3 \sqrt [3]{2 x^2+1}}{8 x^{8/3}}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
| \(\Big \downarrow \) 965 |
| \(\displaystyle -\frac {\sqrt [3]{2 x^3+x} \left (\frac {1}{2} \left (18 \int \frac {x^{2/3}}{(1-x) (2 x+1)^{2/3}}dx^{2/3}-\frac {15 \sqrt [3]{2 x^2+1}}{2 x^{2/3}}\right )-\frac {3 \sqrt [3]{2 x^2+1}}{8 x^{8/3}}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
| \(\Big \downarrow \) 992 |
| \(\displaystyle -\frac {\sqrt [3]{2 x^3+x} \left (\frac {1}{2} \left (18 \left (-\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{3} x^{2/3}}{\sqrt [3]{2 x+1}}+1}{\sqrt {3}}\right )}{3 \sqrt [6]{3}}-\frac {\log \left (\sqrt [3]{3} x^{2/3}-\sqrt [3]{2 x+1}\right )}{2\ 3^{2/3}}+\frac {\log (1-x)}{6\ 3^{2/3}}\right )-\frac {15 \sqrt [3]{2 x^2+1}}{2 x^{2/3}}\right )-\frac {3 \sqrt [3]{2 x^2+1}}{8 x^{8/3}}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}}\) |
-(((x + 2*x^3)^(1/3)*((-3*(1 + 2*x^2)^(1/3))/(8*x^(8/3)) + ((-15*(1 + 2*x^ 2)^(1/3))/(2*x^(2/3)) + 18*(-1/3*ArcTan[(1 + (2*3^(1/3)*x^(2/3))/(1 + 2*x) ^(1/3))/Sqrt[3]]/3^(1/6) + Log[1 - x]/(6*3^(2/3)) - Log[3^(1/3)*x^(2/3) - (1 + 2*x)^(1/3)]/(2*3^(2/3))))/2))/(x^(1/3)*(1 + 2*x^2)^(1/3)))
3.21.7.3.1 Defintions of rubi rules used
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_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) , x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[m ] && IntegerQ[p]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(a*g*(m + 1))), x] - Simp[1/(a*g^2*(m + 1)) Int[(g*x) ^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*2 *(b*c*(p + 1) + a*d*q) + d*((b*e - a*f)*(m + 1) + b*e*2*(p + q + 1))*x^2, x ], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && GtQ[q, 0] && LtQ[m, -1] && !(EqQ[q, 1] && SimplerQ[e + f*x^2, c + d*x^2])
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /; Free Q[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]
Int[(x_)/(((a_) + (b_.)*(x_)^3)^(2/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[-ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3 ))/Sqrt[3]]/(Sqrt[3]*c*q^2), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c* q^2), x] + Simp[Log[c + d*x^3]/(6*c*q^2), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Time = 15.70 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.99
| method | result | size |
| pseudoelliptic | \(\frac {-4 \,3^{\frac {1}{3}} \ln \left (\frac {3^{\frac {2}{3}} x^{2}+3^{\frac {1}{3}} \left (2 x^{3}+x \right )^{\frac {1}{3}} x +\left (2 x^{3}+x \right )^{\frac {2}{3}}}{x^{2}}\right ) x^{3}+8 \,3^{\frac {1}{3}} \ln \left (\frac {-3^{\frac {1}{3}} x +\left (2 x^{3}+x \right )^{\frac {1}{3}}}{x}\right ) x^{3}-8 \,3^{\frac {5}{6}} \arctan \left (\frac {\sqrt {3}\, \left (2 \,3^{\frac {2}{3}} \left (2 x^{3}+x \right )^{\frac {1}{3}}+3 x \right )}{9 x}\right ) x^{3}+30 \left (2 x^{3}+x \right )^{\frac {1}{3}} x^{2}+3 \left (2 x^{3}+x \right )^{\frac {1}{3}}}{8 x^{3}}\) | \(140\) |
| trager | \(\text {Expression too large to display}\) | \(1136\) |
| risch | \(\text {Expression too large to display}\) | \(1882\) |
1/8/x^3*(-4*3^(1/3)*ln((3^(2/3)*x^2+3^(1/3)*(2*x^3+x)^(1/3)*x+(2*x^3+x)^(2 /3))/x^2)*x^3+8*3^(1/3)*ln((-3^(1/3)*x+(2*x^3+x)^(1/3))/x)*x^3-8*3^(5/6)*a rctan(1/9*3^(1/2)*(2*3^(2/3)*(2*x^3+x)^(1/3)+3*x)/x)*x^3+30*(2*x^3+x)^(1/3 )*x^2+3*(2*x^3+x)^(1/3))
Leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (112) = 224\).
Time = 1.22 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.85 \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (-1+x^2\right )} \, dx=\frac {8 \cdot 3^{\frac {5}{6}} x^{3} \arctan \left (\frac {6 \cdot 3^{\frac {5}{6}} {\left (8 \, x^{4} - 7 \, x^{2} - 1\right )} {\left (2 \, x^{3} + x\right )}^{\frac {2}{3}} - \sqrt {3} {\left (377 \, x^{6} + 300 \, x^{4} + 51 \, x^{2} + 1\right )} - 18 \cdot 3^{\frac {1}{6}} {\left (55 \, x^{5} + 25 \, x^{3} + x\right )} {\left (2 \, x^{3} + x\right )}^{\frac {1}{3}}}{3 \, {\left (487 \, x^{6} + 240 \, x^{4} + 3 \, x^{2} - 1\right )}}\right ) - 4 \cdot 3^{\frac {1}{3}} x^{3} \log \left (\frac {3 \cdot 3^{\frac {2}{3}} {\left (2 \, x^{3} + x\right )}^{\frac {2}{3}} {\left (8 \, x^{2} + 1\right )} + 3^{\frac {1}{3}} {\left (55 \, x^{4} + 25 \, x^{2} + 1\right )} + 9 \, {\left (7 \, x^{3} + 2 \, x\right )} {\left (2 \, x^{3} + x\right )}^{\frac {1}{3}}}{x^{4} - 2 \, x^{2} + 1}\right ) + 8 \cdot 3^{\frac {1}{3}} x^{3} \log \left (-\frac {3^{\frac {2}{3}} {\left (x^{2} - 1\right )} - 9 \cdot 3^{\frac {1}{3}} {\left (2 \, x^{3} + x\right )}^{\frac {1}{3}} x + 9 \, {\left (2 \, x^{3} + x\right )}^{\frac {2}{3}}}{x^{2} - 1}\right ) + 9 \, {\left (2 \, x^{3} + x\right )}^{\frac {1}{3}} {\left (10 \, x^{2} + 1\right )}}{24 \, x^{3}} \]
1/24*(8*3^(5/6)*x^3*arctan(1/3*(6*3^(5/6)*(8*x^4 - 7*x^2 - 1)*(2*x^3 + x)^ (2/3) - sqrt(3)*(377*x^6 + 300*x^4 + 51*x^2 + 1) - 18*3^(1/6)*(55*x^5 + 25 *x^3 + x)*(2*x^3 + x)^(1/3))/(487*x^6 + 240*x^4 + 3*x^2 - 1)) - 4*3^(1/3)* x^3*log((3*3^(2/3)*(2*x^3 + x)^(2/3)*(8*x^2 + 1) + 3^(1/3)*(55*x^4 + 25*x^ 2 + 1) + 9*(7*x^3 + 2*x)*(2*x^3 + x)^(1/3))/(x^4 - 2*x^2 + 1)) + 8*3^(1/3) *x^3*log(-(3^(2/3)*(x^2 - 1) - 9*3^(1/3)*(2*x^3 + x)^(1/3)*x + 9*(2*x^3 + x)^(2/3))/(x^2 - 1)) + 9*(2*x^3 + x)^(1/3)*(10*x^2 + 1))/x^3
\[ \int \frac {\left (1+x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (-1+x^2\right )} \, dx=\int \frac {\sqrt [3]{x \left (2 x^{2} + 1\right )} \left (x^{2} + 1\right )}{x^{4} \left (x - 1\right ) \left (x + 1\right )}\, dx \]
\[ \int \frac {\left (1+x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (-1+x^2\right )} \, dx=\int { \frac {{\left (2 \, x^{3} + x\right )}^{\frac {1}{3}} {\left (x^{2} + 1\right )}}{{\left (x^{2} - 1\right )} x^{4}} \,d x } \]
Time = 0.40 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.63 \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (-1+x^2\right )} \, dx=-3^{\frac {5}{6}} \arctan \left (\frac {1}{3} \cdot 3^{\frac {1}{6}} {\left (3^{\frac {1}{3}} + 2 \, {\left (\frac {1}{x^{2}} + 2\right )}^{\frac {1}{3}}\right )}\right ) + \frac {3}{8} \, {\left (\frac {1}{x^{2}} + 2\right )}^{\frac {4}{3}} - \frac {1}{2} \cdot 3^{\frac {1}{3}} \log \left (3^{\frac {2}{3}} + 3^{\frac {1}{3}} {\left (\frac {1}{x^{2}} + 2\right )}^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 2\right )}^{\frac {2}{3}}\right ) + 3^{\frac {1}{3}} \log \left ({\left | -3^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 2\right )}^{\frac {1}{3}} \right |}\right ) + 3 \, {\left (\frac {1}{x^{2}} + 2\right )}^{\frac {1}{3}} \]
-3^(5/6)*arctan(1/3*3^(1/6)*(3^(1/3) + 2*(1/x^2 + 2)^(1/3))) + 3/8*(1/x^2 + 2)^(4/3) - 1/2*3^(1/3)*log(3^(2/3) + 3^(1/3)*(1/x^2 + 2)^(1/3) + (1/x^2 + 2)^(2/3)) + 3^(1/3)*log(abs(-3^(1/3) + (1/x^2 + 2)^(1/3))) + 3*(1/x^2 + 2)^(1/3)
Timed out. \[ \int \frac {\left (1+x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (-1+x^2\right )} \, dx=\int \frac {{\left (2\,x^3+x\right )}^{1/3}\,\left (x^2+1\right )}{x^4\,\left (x^2-1\right )} \,d x \]