Integrand size = 24, antiderivative size = 167 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{-x+2^{2/3} \sqrt [3]{x^2+x^4}}\right )}{2 \sqrt [3]{2}}-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{x^2+x^4}}\right )}{2 \sqrt [3]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [3]{2} x}{\sqrt [3]{x^2+x^4}}\right )}{\sqrt [3]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [3]{2} x^2+\frac {\left (x^2+x^4\right )^{2/3}}{\sqrt [3]{2}}}{x \sqrt [3]{x^2+x^4}}\right )}{2 \sqrt [3]{2}} \]
-1/4*3^(1/2)*arctan(3^(1/2)*x/(-x+2^(2/3)*(x^4+x^2)^(1/3)))*2^(2/3)-1/4*3^ (1/2)*arctan(3^(1/2)*x/(x+2^(2/3)*(x^4+x^2)^(1/3)))*2^(2/3)-1/2*arctanh(2^ (1/3)*x/(x^4+x^2)^(1/3))*2^(2/3)-1/4*arctanh((2^(1/3)*x^2+1/2*(x^4+x^2)^(2 /3)*2^(2/3))/x/(x^4+x^2)^(1/3))*2^(2/3)
Time = 0.51 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.10 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=-\frac {x^{2/3} \sqrt [3]{1+x^2} \left (\sqrt {3} \left (\arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{-\sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}}\right )+\arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}}\right )\right )+2 \text {arctanh}\left (\frac {\sqrt [3]{2} \sqrt [3]{x}}{\sqrt [3]{1+x^2}}\right )+\text {arctanh}\left (\frac {2^{2/3} \sqrt [3]{x} \sqrt [3]{1+x^2}}{2 x^{2/3}+\sqrt [3]{2} \left (1+x^2\right )^{2/3}}\right )\right )}{2 \sqrt [3]{2} \sqrt [3]{x^2+x^4}} \]
-1/2*(x^(2/3)*(1 + x^2)^(1/3)*(Sqrt[3]*(ArcTan[(Sqrt[3]*x^(1/3))/(-x^(1/3) + 2^(2/3)*(1 + x^2)^(1/3))] + ArcTan[(Sqrt[3]*x^(1/3))/(x^(1/3) + 2^(2/3) *(1 + x^2)^(1/3))]) + 2*ArcTanh[(2^(1/3)*x^(1/3))/(1 + x^2)^(1/3)] + ArcTa nh[(2^(2/3)*x^(1/3)*(1 + x^2)^(1/3))/(2*x^(2/3) + 2^(1/3)*(1 + x^2)^(2/3)) ]))/(2^(1/3)*(x^2 + x^4)^(1/3))
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.25, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2467, 25, 368, 936}
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.
\(\displaystyle \int \frac {x^2+1}{\left (x^2-1\right ) \sqrt [3]{x^4+x^2}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {x^{2/3} \sqrt [3]{x^2+1} \int -\frac {\left (x^2+1\right )^{2/3}}{x^{2/3} \left (1-x^2\right )}dx}{\sqrt [3]{x^4+x^2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {x^{2/3} \sqrt [3]{x^2+1} \int \frac {\left (x^2+1\right )^{2/3}}{x^{2/3} \left (1-x^2\right )}dx}{\sqrt [3]{x^4+x^2}}\) |
\(\Big \downarrow \) 368 |
\(\displaystyle -\frac {3 x^{2/3} \sqrt [3]{x^2+1} \int \frac {\left (x^2+1\right )^{2/3}}{1-x^2}d\sqrt [3]{x}}{\sqrt [3]{x^4+x^2}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle -\frac {3 x \sqrt [3]{x^2+1} \operatorname {AppellF1}\left (\frac {1}{6},1,-\frac {2}{3},\frac {7}{6},x^2,-x^2\right )}{\sqrt [3]{x^4+x^2}}\) |
3.23.37.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) ], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 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 = 18.94 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.20
method | result | size |
pseudoelliptic | \(-\frac {2^{\frac {2}{3}} \left (\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (-2^{\frac {2}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}+x \right )}{3 x}\right )-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}+x \right )}{3 x}\right )-\frac {\ln \left (\frac {2^{\frac {2}{3}} x^{2}-2^{\frac {1}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}+\frac {\ln \left (\frac {2^{\frac {2}{3}} x^{2}+2^{\frac {1}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}+\ln \left (\frac {2^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {-2^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{x}\right )\right )}{4}\) | \(201\) |
trager | \(\text {Expression too large to display}\) | \(3878\) |
-1/4*2^(2/3)*(3^(1/2)*arctan(1/3*3^(1/2)*(-2^(2/3)*(x^2*(x^2+1))^(1/3)+x)/ x)-3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)*(x^2*(x^2+1))^(1/3)+x)/x)-1/2*ln((2 ^(2/3)*x^2-2^(1/3)*(x^2*(x^2+1))^(1/3)*x+(x^2*(x^2+1))^(2/3))/x^2)+1/2*ln( (2^(2/3)*x^2+2^(1/3)*(x^2*(x^2+1))^(1/3)*x+(x^2*(x^2+1))^(2/3))/x^2)+ln((2 ^(1/3)*x+(x^2*(x^2+1))^(1/3))/x)-ln((-2^(1/3)*x+(x^2*(x^2+1))^(1/3))/x))
Time = 0.57 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.39 \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=-\frac {1}{4} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {1}{6}} {\left (2^{\frac {5}{6}} {\left (x^{5} + 8 \, x^{4} - 2 \, x^{3} + 8 \, x^{2} + x\right )} + 8 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} {\left (x^{3} + 2 \, x^{2} + x\right )} + 8 \cdot 2^{\frac {1}{6}} {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}} {\left (x^{2} - 2 \, x + 1\right )}\right )}}{6 \, {\left (x^{5} - 8 \, x^{4} - 2 \, x^{3} - 8 \, x^{2} + x\right )}}\right ) + \frac {1}{4} \cdot 2^{\frac {2}{3}} \log \left (-\frac {2^{\frac {2}{3}} {\left (x^{3} - 2 \, x^{2} + x\right )} + 4 \cdot 2^{\frac {1}{3}} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} x - 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}}}{x^{3} + 2 \, x^{2} + x}\right ) - \frac {1}{8} \cdot 2^{\frac {2}{3}} \log \left (\frac {2 \cdot 2^{\frac {2}{3}} {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (x^{3} + 2 \, x^{2} + x\right )} + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} x}{x^{3} + 2 \, x^{2} + x}\right ) \]
-1/4*sqrt(3)*2^(2/3)*arctan(1/6*sqrt(3)*2^(1/6)*(2^(5/6)*(x^5 + 8*x^4 - 2* x^3 + 8*x^2 + x) + 8*sqrt(2)*(x^4 + x^2)^(1/3)*(x^3 + 2*x^2 + x) + 8*2^(1/ 6)*(x^4 + x^2)^(2/3)*(x^2 - 2*x + 1))/(x^5 - 8*x^4 - 2*x^3 - 8*x^2 + x)) + 1/4*2^(2/3)*log(-(2^(2/3)*(x^3 - 2*x^2 + x) + 4*2^(1/3)*(x^4 + x^2)^(1/3) *x - 4*(x^4 + x^2)^(2/3))/(x^3 + 2*x^2 + x)) - 1/8*2^(2/3)*log((2*2^(2/3)* (x^4 + x^2)^(2/3) + 2^(1/3)*(x^3 + 2*x^2 + x) + 4*(x^4 + x^2)^(1/3)*x)/(x^ 3 + 2*x^2 + x))
\[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=\int \frac {x^{2} + 1}{\sqrt [3]{x^{2} \left (x^{2} + 1\right )} \left (x - 1\right ) \left (x + 1\right )}\, dx \]
\[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=\int { \frac {x^{2} + 1}{{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} - 1\right )}} \,d x } \]
\[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=\int { \frac {x^{2} + 1}{{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} - 1\right )}} \,d x } \]
Timed out. \[ \int \frac {1+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=\int \frac {x^2+1}{{\left (x^4+x^2\right )}^{1/3}\,\left (x^2-1\right )} \,d x \]