\(\int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx\) [668]

Optimal result

Integrand size = 26, antiderivative size = 289 \[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {a+b x^2}{a x \sqrt [3]{a^2+2 a b x^2+b^2 x^4}}+\frac {\sqrt {2-\sqrt {3}} \left (1+\frac {b x^2}{a}\right )^{2/3} \left (1-\sqrt [3]{1+\frac {b x^2}{a}}\right ) \sqrt {\frac {1+\sqrt [3]{1+\frac {b x^2}{a}}+\left (1+\frac {b x^2}{a}\right )^{2/3}}{\left (1-\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}}{1-\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} x \sqrt [3]{a^2+2 a b x^2+b^2 x^4} \sqrt {-\frac {1-\sqrt [3]{1+\frac {b x^2}{a}}}{\left (1-\sqrt {3}-\sqrt [3]{1+\frac {b x^2}{a}}\right )^2}}} \] Output:

-(b*x^2+a)/a/x/(b^2*x^4+2*a*b*x^2+a^2)^(1/3)+1/3*(1/2*6^(1/2)-1/2*2^(1/2)) 
*(1+b*x^2/a)^(2/3)*(1-(1+b*x^2/a)^(1/3))*((1+(1+b*x^2/a)^(1/3)+(1+b*x^2/a) 
^(2/3))/(1-3^(1/2)-(1+b*x^2/a)^(1/3))^2)^(1/2)*EllipticF((1+3^(1/2)-(1+b*x 
^2/a)^(1/3))/(1-3^(1/2)-(1+b*x^2/a)^(1/3)),2*I-I*3^(1/2))*3^(3/4)/x/(b^2*x 
^4+2*a*b*x^2+a^2)^(1/3)/(-(1-(1+b*x^2/a)^(1/3))/(1-3^(1/2)-(1+b*x^2/a)^(1/ 
3))^2)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 5.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.18 \[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=-\frac {\left (1+\frac {b x^2}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {2}{3},\frac {1}{2},-\frac {b x^2}{a}\right )}{x \sqrt [3]{\left (a+b x^2\right )^2}} \] Input:

Integrate[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/3)),x]
 

Output:

-(((1 + (b*x^2)/a)^(2/3)*Hypergeometric2F1[-1/2, 2/3, 1/2, -((b*x^2)/a)])/ 
(x*((a + b*x^2)^2)^(1/3)))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1385, 264, 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.

\(\displaystyle \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx\)

\(\Big \downarrow \) 1385

\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{2/3} \int \frac {1}{x^2 \left (\frac {b x^2}{a}+1\right )^{2/3}}dx}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{2/3} \left (-\frac {b \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{2/3}}dx}{3 a}-\frac {\sqrt [3]{\frac {b x^2}{a}+1}}{x}\right )}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{2/3} \left (-\frac {\sqrt {\frac {b x^2}{a}} \int \frac {1}{\sqrt {\frac {b x^2}{a}}}d\sqrt [3]{\frac {b x^2}{a}+1}}{2 x}-\frac {\sqrt [3]{\frac {b x^2}{a}+1}}{x}\right )}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{2/3} \left (\frac {\sqrt {2-\sqrt {3}} \left (1-\sqrt [3]{\frac {b x^2}{a}+1}\right ) \sqrt {\frac {\left (\frac {b x^2}{a}+1\right )^{2/3}+\sqrt [3]{\frac {b x^2}{a}+1}+1}{\left (-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-\sqrt [3]{\frac {b x^2}{a}+1}+\sqrt {3}+1}{-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} x \sqrt {-\frac {1-\sqrt [3]{\frac {b x^2}{a}+1}}{\left (-\sqrt [3]{\frac {b x^2}{a}+1}-\sqrt {3}+1\right )^2}}}-\frac {\sqrt [3]{\frac {b x^2}{a}+1}}{x}\right )}{\sqrt [3]{a^2+2 a b x^2+b^2 x^4}}\)

Input:

Int[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/3)),x]
 

Output:

((1 + (b*x^2)/a)^(2/3)*(-((1 + (b*x^2)/a)^(1/3)/x) + (Sqrt[2 - Sqrt[3]]*(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]*EllipticF[ArcSin[(1 + Sq 
rt[3] - (1 + (b*x^2)/a)^(1/3))/(1 - Sqrt[3] - (1 + (b*x^2)/a)^(1/3))], -7 
+ 4*Sqrt[3]])/(3^(1/4)*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)
 

Defintions of rubi rules used

rule 234

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]
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 760

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]
 

rule 1385

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)]
 

Maple [F]

\[\int \frac {1}{x^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {1}{3}}}d x\]

Input:

int(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(1/3),x)
 

Output:

int(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(1/3),x)
 

Fricas [F]

\[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int { \frac {1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{3}} x^{2}} \,d x } \] Input:

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(1/3),x, algorithm="fricas")
 

Output:

integral((b^2*x^4 + 2*a*b*x^2 + a^2)^(2/3)/(b^2*x^6 + 2*a*b*x^4 + a^2*x^2) 
, x)
 

Sympy [F]

\[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {1}{x^{2} \sqrt [3]{\left (a + b x^{2}\right )^{2}}}\, dx \] Input:

integrate(1/x**2/(b**2*x**4+2*a*b*x**2+a**2)**(1/3),x)
 

Output:

Integral(1/(x**2*((a + b*x**2)**2)**(1/3)), x)
 

Maxima [F]

\[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int { \frac {1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{3}} x^{2}} \,d x } \] Input:

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(1/3),x, algorithm="maxima")
 

Output:

integrate(1/((b^2*x^4 + 2*a*b*x^2 + a^2)^(1/3)*x^2), x)
 

Giac [F]

\[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int { \frac {1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{3}} x^{2}} \,d x } \] Input:

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(1/3),x, algorithm="giac")
 

Output:

integrate(1/((b^2*x^4 + 2*a*b*x^2 + a^2)^(1/3)*x^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {1}{x^2\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{1/3}} \,d x \] Input:

int(1/(x^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/3)),x)
 

Output:

int(1/(x^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/3)), x)
 

Reduce [F]

\[ \int \frac {1}{x^2 \sqrt [3]{a^2+2 a b x^2+b^2 x^4}} \, dx=\int \frac {1}{\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {1}{3}} x^{2}}d x \] Input:

int(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(1/3),x)
 

Output:

int(1/((a**2 + 2*a*b*x**2 + b**2*x**4)**(1/3)*x**2),x)