Integrand size = 26, antiderivative size = 298 \[ \int \frac {x^2}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {3 x \left (a+b x^2\right )}{5 b \sqrt [3]{a^2+2 a b x^2+b^2 x^4}}+\frac {3\ 3^{3/4} \sqrt {2-\sqrt {3}} a^2 \left (1+\frac {b x^2}{a}\right )^{2/3} \left (1-\sqrt [3]{1+\frac {b x^2}{a}}\right ) \sqrt {\frac {1+\sqrt [3]{1+\frac {b x^2}{a}}+\left (1+\frac {b x^2}{a}\right )^{2/3}}{\left (1-\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}}{1-\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}}\right ),-7+4 \sqrt {3}\right )}{5 b^2 x \sqrt [3]{a^2+2 a b x^2+b^2 x^4} \sqrt {-\frac {1-\sqrt [3]{1+\frac {b x^2}{a}}}{\left (1-\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}\right )^2}}} \]
3/5*x*(b*x^2+a)/b/(b^2*x^4+2*a*b*x^2+a^2)^(1/3)+3/5*3^(3/4)*a^2*(1+b*x^2/a )^(2/3)*(1-(1+b*x^2/a)^(1/3))*EllipticF((1-(1+b*x^2/a)^(1/3)+3^(1/2))/(1-( 1+b*x^2/a)^(1/3)-3^(1/2)),2*I-I*3^(1/2))*(1/2*6^(1/2)-1/2*2^(1/2))*((1+(1+ b*x^2/a)^(1/3)+(1+b*x^2/a)^(2/3))/(1-(1+b*x^2/a)^(1/3)-3^(1/2))^2)^(1/2)/b ^2/x/(b^2*x^4+2*a*b*x^2+a^2)^(1/3)/((-1+(1+b*x^2/a)^(1/3))/(1-(1+b*x^2/a)^ (1/3)-3^(1/2))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.21 \[ \int \frac {x^2}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\frac {3 x \left (a+b x^2-a \left (1+\frac {b x^2}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{5 b \sqrt [3]{\left (a+b x^2\right )^2}} \]
(3*x*(a + b*x^2 - a*(1 + (b*x^2)/a)^(2/3)*Hypergeometric2F1[1/2, 2/3, 3/2, -((b*x^2)/a)]))/(5*b*((a + b*x^2)^2)^(1/3))
Time = 0.30 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1385, 262, 234, 760}
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.
\(\displaystyle \int \frac {x^2}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx\) |
\(\Big \downarrow \) 1385 |
\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{2/3} \int \frac {x^2}{\left (\frac {b x^2}{a}+1\right )^{2/3}}dx}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{2/3} \left (\frac {3 a x \sqrt [3]{\frac {b x^2}{a}+1}}{5 b}-\frac {3 a \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{2/3}}dx}{5 b}\right )}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 234 |
\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{2/3} \left (\frac {3 a x \sqrt [3]{\frac {b x^2}{a}+1}}{5 b}-\frac {9 a^2 \sqrt {\frac {b x^2}{a}} \int \frac {1}{\sqrt {\frac {b x^2}{a}}}d\sqrt [3]{\frac {b x^2}{a}+1}}{10 b^2 x}\right )}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 760 |
\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{2/3} \left (\frac {3\ 3^{3/4} \sqrt {2-\sqrt {3}} a^2 \left (1-\sqrt [3]{\frac {b x^2}{a}+1}\right ) \sqrt {\frac {\left (\frac {b x^2}{a}+1\right )^{2/3}+\sqrt [3]{\frac {b x^2}{a}+1}+1}{\left (-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-\sqrt [3]{\frac {b x^2}{a}+1}+\sqrt {3}+1}{-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{5 b^2 x \sqrt {-\frac {1-\sqrt [3]{\frac {b x^2}{a}+1}}{\left (-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1\right )^2}}}+\frac {3 a x \sqrt [3]{\frac {b x^2}{a}+1}}{5 b}\right )}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}}\) |
((1 + (b*x^2)/a)^(2/3)*((3*a*x*(1 + (b*x^2)/a)^(1/3))/(5*b) + (3*3^(3/4)*S qrt[2 - Sqrt[3]]*a^2*(1 - (1 + (b*x^2)/a)^(1/3))*Sqrt[(1 + (1 + (b*x^2)/a) ^(1/3) + (1 + (b*x^2)/a)^(2/3))/(1 - Sqrt[3] - (1 + (b*x^2)/a)^(1/3))^2]*E llipticF[ArcSin[(1 + Sqrt[3] - (1 + (b*x^2)/a)^(1/3))/(1 - Sqrt[3] - (1 + (b*x^2)/a)^(1/3))], -7 + 4*Sqrt[3]])/(5*b^2*x*Sqrt[-((1 - (1 + (b*x^2)/a)^ (1/3))/(1 - Sqrt[3] - (1 + (b*x^2)/a)^(1/3))^2)])))/(a^2 + 2*a*b*x^2 + b^2 *x^4)^(1/3)
3.7.59.3.1 Defintions of rubi rules used
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, 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[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[a^IntPart[p]*((a + b*x^n + c*x^(2*n))^FracPart[p]/(1 + 2*c*(x^n/b))^(2* FracPart[p])) Int[u*(1 + 2*c*(x^n/b))^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[2*p] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)]
\[\int \frac {x^{2}}{\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {1}{3}}}d x\]
\[ \int \frac {x^2}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int { \frac {x^{2}}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {x^2}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {x^{2}}{\sqrt [3]{\left (a + b x^{2}\right )^{2}}}\, dx \]
\[ \int \frac {x^2}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int { \frac {x^{2}}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {x^2}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int { \frac {x^{2}}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {x^2}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {x^2}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{1/3}} \,d x \]