Integrand size = 24, antiderivative size = 208 \[ \int \frac {\sqrt [3]{-x^2+x^4}}{x \left (1+x^2\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x^2}{-x^2+\sqrt [3]{2} \left (-x^2+x^4\right )^{2/3}}\right )}{2\ 2^{2/3}}+\frac {\log \left (-2 x^2+2^{2/3} \sqrt {3} x \sqrt [3]{-x^2+x^4}-\sqrt [3]{2} \left (-x^2+x^4\right )^{2/3}\right )}{4\ 2^{2/3}}-\frac {\log \left (2 x^2+\sqrt [3]{2} \left (-x^2+x^4\right )^{2/3}\right )}{2\ 2^{2/3}}+\frac {\log \left (2 x^2+2^{2/3} \sqrt {3} x \sqrt [3]{-x^2+x^4}+\sqrt [3]{2} \left (-x^2+x^4\right )^{2/3}\right )}{4\ 2^{2/3}} \] Output:
-1/4*3^(1/2)*arctan(3^(1/2)*x^2/(-x^2+2^(1/3)*(x^4-x^2)^(2/3)))*2^(1/3)+1/ 8*ln(-2*x^2+2^(2/3)*3^(1/2)*x*(x^4-x^2)^(1/3)-2^(1/3)*(x^4-x^2)^(2/3))*2^( 1/3)-1/4*ln(2*x^2+2^(1/3)*(x^4-x^2)^(2/3))*2^(1/3)+1/8*ln(2*x^2+2^(2/3)*3^ (1/2)*x*(x^4-x^2)^(1/3)+2^(1/3)*(x^4-x^2)^(2/3))*2^(1/3)
Time = 0.48 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt [3]{-x^2+x^4}}{x \left (1+x^2\right )} \, dx=\frac {x^{4/3} \left (-1+x^2\right )^{2/3} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}-\sqrt [3]{2} \left (-1+x^2\right )^{2/3}}\right )+\log \left (-2 x^{2/3}+2^{2/3} \sqrt {3} \sqrt [3]{x} \sqrt [3]{-1+x^2}-\sqrt [3]{2} \left (-1+x^2\right )^{2/3}\right )-2 \log \left (2 x^{2/3}+\sqrt [3]{2} \left (-1+x^2\right )^{2/3}\right )+\log \left (2 x^{2/3}+2^{2/3} \sqrt {3} \sqrt [3]{x} \sqrt [3]{-1+x^2}+\sqrt [3]{2} \left (-1+x^2\right )^{2/3}\right )\right )}{4\ 2^{2/3} \left (x^2 \left (-1+x^2\right )\right )^{2/3}} \] Input:
Integrate[(-x^2 + x^4)^(1/3)/(x*(1 + x^2)),x]
Output:
(x^(4/3)*(-1 + x^2)^(2/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*x^(2/3))/(x^(2/3) - 2 ^(1/3)*(-1 + x^2)^(2/3))] + Log[-2*x^(2/3) + 2^(2/3)*Sqrt[3]*x^(1/3)*(-1 + x^2)^(1/3) - 2^(1/3)*(-1 + x^2)^(2/3)] - 2*Log[2*x^(2/3) + 2^(1/3)*(-1 + x^2)^(2/3)] + Log[2*x^(2/3) + 2^(2/3)*Sqrt[3]*x^(1/3)*(-1 + x^2)^(1/3) + 2 ^(1/3)*(-1 + x^2)^(2/3)]))/(4*2^(2/3)*(x^2*(-1 + x^2))^(2/3))
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 0.50 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1648, 1424, 1093, 1090, 234, 760, 1940, 1178, 150}
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.
\(\displaystyle \int \frac {\sqrt [3]{x^4-x^2}}{x \left (x^2+1\right )} \, dx\) |
\(\Big \downarrow \) 1648 |
\(\displaystyle \int \frac {x}{\left (x^4-x^2\right )^{2/3}}dx-2 \int \frac {x}{\left (x^2+1\right ) \left (x^4-x^2\right )^{2/3}}dx\) |
\(\Big \downarrow \) 1424 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\left (x^4-x^2\right )^{2/3}}dx^2-2 \int \frac {x}{\left (x^2+1\right ) \left (x^4-x^2\right )^{2/3}}dx\) |
\(\Big \downarrow \) 1093 |
\(\displaystyle \frac {\left (x^2-x^4\right )^{2/3} \int \frac {1}{\left (x^2-x^4\right )^{2/3}}dx^2}{2 \left (x^4-x^2\right )^{2/3}}-2 \int \frac {x}{\left (x^2+1\right ) \left (x^4-x^2\right )^{2/3}}dx\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle -\frac {\left (x^2-x^4\right )^{2/3} \int \frac {1}{\left (1-x^4\right )^{2/3}}d\left (1-2 x^2\right )}{2^{2/3} \left (x^4-x^2\right )^{2/3}}-2 \int \frac {x}{\left (x^2+1\right ) \left (x^4-x^2\right )^{2/3}}dx\) |
\(\Big \downarrow \) 234 |
\(\displaystyle \frac {3 \sqrt {-x^4} \left (x^2-x^4\right )^{2/3} \int \frac {1}{\sqrt {x^6-1}}d\sqrt [3]{1-x^4}}{2\ 2^{2/3} x^2 \left (x^4-x^2\right )^{2/3}}-2 \int \frac {x}{\left (x^2+1\right ) \left (x^4-x^2\right )^{2/3}}dx\) |
\(\Big \downarrow \) 760 |
\(\displaystyle -2 \int \frac {x}{\left (x^2+1\right ) \left (x^4-x^2\right )^{2/3}}dx-\frac {3^{3/4} \sqrt {2-\sqrt {3}} \sqrt {-x^4} \left (x^2-x^4\right )^{2/3} \sqrt {\frac {x^4-2 x^2+2}{\left (2 x^2-\sqrt {3}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2 x^2+\sqrt {3}}{2 x^2-\sqrt {3}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [6]{2} \sqrt {-\frac {x^2}{\left (2 x^2-\sqrt {3}\right )^2}} \left (x^4-x^2\right )^{2/3} \sqrt {x^6-1}}\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle -\int \frac {1}{\left (x^2+1\right ) \left (x^4-x^2\right )^{2/3}}dx^2-\frac {3^{3/4} \sqrt {2-\sqrt {3}} \sqrt {-x^4} \left (x^2-x^4\right )^{2/3} \sqrt {\frac {x^4-2 x^2+2}{\left (2 x^2-\sqrt {3}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2 x^2+\sqrt {3}}{2 x^2-\sqrt {3}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [6]{2} \sqrt {-\frac {x^2}{\left (2 x^2-\sqrt {3}\right )^2}} \left (x^4-x^2\right )^{2/3} \sqrt {x^6-1}}\) |
\(\Big \downarrow \) 1178 |
\(\displaystyle \frac {\left (\frac {x^2}{x^2+1}\right )^{2/3} \left (-\frac {1-x^2}{x^2+1}\right )^{2/3} \int \frac {\sqrt [3]{\frac {1}{x^2+1}}}{\left (1-\frac {2}{x^2+1}\right )^{2/3} \left (1-\frac {1}{x^2+1}\right )^{2/3}}d\frac {1}{x^2+1}}{\left (\frac {1}{x^2+1}\right )^{4/3} \left (x^4-x^2\right )^{2/3}}-\frac {3^{3/4} \sqrt {2-\sqrt {3}} \sqrt {-x^4} \left (x^2-x^4\right )^{2/3} \sqrt {\frac {x^4-2 x^2+2}{\left (2 x^2-\sqrt {3}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2 x^2+\sqrt {3}}{2 x^2-\sqrt {3}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [6]{2} \sqrt {-\frac {x^2}{\left (2 x^2-\sqrt {3}\right )^2}} \left (x^4-x^2\right )^{2/3} \sqrt {x^6-1}}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {3 \left (\frac {x^2}{x^2+1}\right )^{2/3} \left (-\frac {1-x^2}{x^2+1}\right )^{2/3} \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},\frac {2}{3},\frac {7}{3},\frac {2}{x^2+1},\frac {1}{x^2+1}\right )}{4 \left (x^4-x^2\right )^{2/3}}-\frac {3^{3/4} \sqrt {2-\sqrt {3}} \sqrt {-x^4} \left (x^2-x^4\right )^{2/3} \sqrt {\frac {x^4-2 x^2+2}{\left (2 x^2-\sqrt {3}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2 x^2+\sqrt {3}}{2 x^2-\sqrt {3}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [6]{2} \sqrt {-\frac {x^2}{\left (2 x^2-\sqrt {3}\right )^2}} \left (x^4-x^2\right )^{2/3} \sqrt {x^6-1}}\) |
Input:
Int[(-x^2 + x^4)^(1/3)/(x*(1 + x^2)),x]
Output:
(3*(x^2/(1 + x^2))^(2/3)*(-((1 - x^2)/(1 + x^2)))^(2/3)*AppellF1[4/3, 2/3, 2/3, 7/3, 2/(1 + x^2), (1 + x^2)^(-1)])/(4*(-x^2 + x^4)^(2/3)) - (3^(3/4) *Sqrt[2 - Sqrt[3]]*Sqrt[-x^4]*(x^2 - x^4)^(2/3)*Sqrt[(2 - 2*x^2 + x^4)/(-S qrt[3] + 2*x^2)^2]*EllipticF[ArcSin[(Sqrt[3] + 2*x^2)/(-Sqrt[3] + 2*x^2)], -7 + 4*Sqrt[3]])/(2^(1/6)*Sqrt[-(x^2/(-Sqrt[3] + 2*x^2)^2)]*(-x^2 + x^4)^ (2/3)*Sqrt[-1 + x^6])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b }, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) *s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x ] && NegQ[a]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
Int[((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b*x + c*x^2)^p/((- c)*((b*x + c*x^2)/b^2))^p Int[((-c)*(x/b) - c^2*(x^2/b^2))^p, x], x] /; F reeQ[{b, c}, x] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(-(1/(d + e*x))^(2*p))*((a + b*x + c*x^2)^p/(e*(e*((b - q + 2*c*x)/(2*c*(d + e*x))))^p*(e*((b + q + 2*c* x)/(2*c*(d + e*x))))^p)) Subst[Int[x^(-m - 2*(p + 1))*Simp[1 - (d - e*((b - q)/(2*c)))*x, x]^p*Simp[1 - (d - e*((b + q)/(2*c)))*x, x]^p, x], x, 1/(d + e*x)], x]] /; FreeQ[{a, b, c, d, e, p}, x] && ILtQ[m, 0]
Int[(x_)^(m_.)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{b, c, m, p}, x] && !IntegerQ[p] && IntegerQ[(m - 1)/2]
Int[(((f_.)*(x_))^(m_)*((a_.) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[1/(d*e) Int[(f*x)^m*(a*e + c*d*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] - Simp[(c*d^2 - b*d*e + a*e^2)/(d*e*f^2) Int[(f*x)^(m + 2)*((a + b*x^2 + c*x^4)^(p - 1)/(d + e*x^2)), x], x] /; Fre eQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && LtQ[m, 0]
Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) ^(n_))^(q_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1) *(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] && !IntegerQ[p] && NeQ[k, j] && I ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]
Time = 27.19 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.61
method | result | size |
pseudoelliptic | \(-\frac {\left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (-2^{\frac {1}{3}} \left (x^{4}-x^{2}\right )^{\frac {2}{3}}+x^{2}\right )}{3 x^{2}}\right )+2 \ln \left (\frac {2^{\frac {2}{3}} x^{2}+\left (x^{4}-x^{2}\right )^{\frac {2}{3}}}{x^{2}}\right )-\ln \left (\frac {-2^{\frac {2}{3}} \left (x^{4}-x^{2}\right )^{\frac {2}{3}}+2 \,2^{\frac {1}{3}} x^{2}+\left (x^{4}-x^{2}\right )^{\frac {1}{3}} x^{2}-\left (x^{4}-x^{2}\right )^{\frac {1}{3}}}{x^{2}}\right )\right ) 2^{\frac {1}{3}}}{8}\) | \(127\) |
trager | \(\text {Expression too large to display}\) | \(1129\) |
Input:
int((x^4-x^2)^(1/3)/x/(x^2+1),x,method=_RETURNVERBOSE)
Output:
-1/8*(2*3^(1/2)*arctan(1/3*3^(1/2)*(-2^(1/3)*(x^4-x^2)^(2/3)+x^2)/x^2)+2*l n((2^(2/3)*x^2+(x^4-x^2)^(2/3))/x^2)-ln((-2^(2/3)*(x^4-x^2)^(2/3)+2*2^(1/3 )*x^2+(x^4-x^2)^(1/3)*x^2-(x^4-x^2)^(1/3))/x^2))*2^(1/3)
Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (160) = 320\).
Time = 1.08 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.61 \[ \int \frac {\sqrt [3]{-x^2+x^4}}{x \left (1+x^2\right )} \, dx=-\frac {1}{12} \cdot 4^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {4^{\frac {1}{6}} \sqrt {3} {\left (6 \cdot 4^{\frac {2}{3}} {\left (x^{10} - 33 \, x^{8} + 110 \, x^{6} - 110 \, x^{4} + 33 \, x^{2} - 1\right )} {\left (x^{4} - x^{2}\right )}^{\frac {1}{3}} + 48 \, {\left (x^{8} - 2 \, x^{6} - 6 \, x^{4} - 2 \, x^{2} + 1\right )} {\left (x^{4} - x^{2}\right )}^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (x^{12} + 42 \, x^{10} - 417 \, x^{8} + 812 \, x^{6} - 417 \, x^{4} + 42 \, x^{2} + 1\right )}\right )}}{6 \, {\left (x^{12} - 102 \, x^{10} + 447 \, x^{8} - 628 \, x^{6} + 447 \, x^{4} - 102 \, x^{2} + 1\right )}}\right ) + \frac {1}{48} \cdot 4^{\frac {2}{3}} \log \left (\frac {24 \cdot 4^{\frac {1}{3}} {\left (x^{4} - x^{2}\right )}^{\frac {2}{3}} {\left (x^{4} - 4 \, x^{2} + 1\right )} + 4^{\frac {2}{3}} {\left (x^{8} - 32 \, x^{6} + 78 \, x^{4} - 32 \, x^{2} + 1\right )} - 12 \, {\left (x^{6} - 11 \, x^{4} + 11 \, x^{2} - 1\right )} {\left (x^{4} - x^{2}\right )}^{\frac {1}{3}}}{x^{8} + 4 \, x^{6} + 6 \, x^{4} + 4 \, x^{2} + 1}\right ) - \frac {1}{24} \cdot 4^{\frac {2}{3}} \log \left (\frac {3 \cdot 4^{\frac {2}{3}} {\left (x^{4} - x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} - 1\right )} + 4^{\frac {1}{3}} {\left (x^{4} + 2 \, x^{2} + 1\right )} + 12 \, {\left (x^{4} - x^{2}\right )}^{\frac {2}{3}}}{x^{4} + 2 \, x^{2} + 1}\right ) \] Input:
integrate((x^4-x^2)^(1/3)/x/(x^2+1),x, algorithm="fricas")
Output:
-1/12*4^(1/6)*sqrt(3)*arctan(1/6*4^(1/6)*sqrt(3)*(6*4^(2/3)*(x^10 - 33*x^8 + 110*x^6 - 110*x^4 + 33*x^2 - 1)*(x^4 - x^2)^(1/3) + 48*(x^8 - 2*x^6 - 6 *x^4 - 2*x^2 + 1)*(x^4 - x^2)^(2/3) + 4^(1/3)*(x^12 + 42*x^10 - 417*x^8 + 812*x^6 - 417*x^4 + 42*x^2 + 1))/(x^12 - 102*x^10 + 447*x^8 - 628*x^6 + 44 7*x^4 - 102*x^2 + 1)) + 1/48*4^(2/3)*log((24*4^(1/3)*(x^4 - x^2)^(2/3)*(x^ 4 - 4*x^2 + 1) + 4^(2/3)*(x^8 - 32*x^6 + 78*x^4 - 32*x^2 + 1) - 12*(x^6 - 11*x^4 + 11*x^2 - 1)*(x^4 - x^2)^(1/3))/(x^8 + 4*x^6 + 6*x^4 + 4*x^2 + 1)) - 1/24*4^(2/3)*log((3*4^(2/3)*(x^4 - x^2)^(1/3)*(x^2 - 1) + 4^(1/3)*(x^4 + 2*x^2 + 1) + 12*(x^4 - x^2)^(2/3))/(x^4 + 2*x^2 + 1))
\[ \int \frac {\sqrt [3]{-x^2+x^4}}{x \left (1+x^2\right )} \, dx=\int \frac {\sqrt [3]{x^{2} \left (x - 1\right ) \left (x + 1\right )}}{x \left (x^{2} + 1\right )}\, dx \] Input:
integrate((x**4-x**2)**(1/3)/x/(x**2+1),x)
Output:
Integral((x**2*(x - 1)*(x + 1))**(1/3)/(x*(x**2 + 1)), x)
\[ \int \frac {\sqrt [3]{-x^2+x^4}}{x \left (1+x^2\right )} \, dx=\int { \frac {{\left (x^{4} - x^{2}\right )}^{\frac {1}{3}}}{{\left (x^{2} + 1\right )} x} \,d x } \] Input:
integrate((x^4-x^2)^(1/3)/x/(x^2+1),x, algorithm="maxima")
Output:
integrate((x^4 - x^2)^(1/3)/((x^2 + 1)*x), x)
\[ \int \frac {\sqrt [3]{-x^2+x^4}}{x \left (1+x^2\right )} \, dx=\int { \frac {{\left (x^{4} - x^{2}\right )}^{\frac {1}{3}}}{{\left (x^{2} + 1\right )} x} \,d x } \] Input:
integrate((x^4-x^2)^(1/3)/x/(x^2+1),x, algorithm="giac")
Output:
integrate((x^4 - x^2)^(1/3)/((x^2 + 1)*x), x)
Timed out. \[ \int \frac {\sqrt [3]{-x^2+x^4}}{x \left (1+x^2\right )} \, dx=\int \frac {{\left (x^4-x^2\right )}^{1/3}}{x\,\left (x^2+1\right )} \,d x \] Input:
int((x^4 - x^2)^(1/3)/(x*(x^2 + 1)),x)
Output:
int((x^4 - x^2)^(1/3)/(x*(x^2 + 1)), x)
\[ \int \frac {\sqrt [3]{-x^2+x^4}}{x \left (1+x^2\right )} \, dx=\int \frac {\left (x^{2}-1\right )^{\frac {1}{3}}}{x^{\frac {7}{3}}+x^{\frac {1}{3}}}d x \] Input:
int((x^4-x^2)^(1/3)/x/(x^2+1),x)
Output:
int((x**2 - 1)**(1/3)/(x**(1/3)*x**2 + x**(1/3)),x)