\(\int \frac {1}{x^2 \sqrt [3]{a+b x^2}} \, dx\) [760]

Optimal result
Mathematica [C] (verified)
Rubi [A] (warning: unable to verify)
Maple [F]
Fricas [F]
Sympy [A] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 15, antiderivative size = 546 \[ \int \frac {1}{x^2 \sqrt [3]{a+b x^2}} \, dx=-\frac {\left (a+b x^2\right )^{2/3}}{a x}-\frac {b x}{a \left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}+\frac {\sqrt [4]{3} \sqrt {2+\sqrt {3}} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} E\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right )|-7+4 \sqrt {3}\right )}{2 a^{2/3} x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}-\frac {\sqrt {2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} a^{2/3} x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:

-(b*x^2+a)^(2/3)/a/x-b*x/a/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))+1/2*3^(1/ 
4)*(1/2*6^(1/2)+1/2*2^(1/2))*(a^(1/3)-(b*x^2+a)^(1/3))*((a^(2/3)+a^(1/3)*( 
b*x^2+a)^(1/3)+(b*x^2+a)^(2/3))/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))^2)^( 
1/2)*EllipticE(((1+3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))/((1-3^(1/2))*a^(1/3)- 
(b*x^2+a)^(1/3)),2*I-I*3^(1/2))/a^(2/3)/x/(-a^(1/3)*(a^(1/3)-(b*x^2+a)^(1/ 
3))/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))^2)^(1/2)-1/3*2^(1/2)*(a^(1/3)-(b 
*x^2+a)^(1/3))*((a^(2/3)+a^(1/3)*(b*x^2+a)^(1/3)+(b*x^2+a)^(2/3))/((1-3^(1 
/2))*a^(1/3)-(b*x^2+a)^(1/3))^2)^(1/2)*EllipticF(((1+3^(1/2))*a^(1/3)-(b*x 
^2+a)^(1/3))/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3)),2*I-I*3^(1/2))*3^(3/4)/ 
a^(2/3)/x/(-a^(1/3)*(a^(1/3)-(b*x^2+a)^(1/3))/((1-3^(1/2))*a^(1/3)-(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 = 4.47 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.09 \[ \int \frac {1}{x^2 \sqrt [3]{a+b x^2}} \, dx=-\frac {\sqrt [3]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{3},\frac {1}{2},-\frac {b x^2}{a}\right )}{x \sqrt [3]{a+b x^2}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.45 (sec) , antiderivative size = 594, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {264, 233, 833, 760, 2418}

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.

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 233

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

\(\Big \downarrow \) 833

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

\(\Big \downarrow \) 760

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

\(\Big \downarrow \) 2418

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

Input:

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

Output:

-((a + b*x^2)^(2/3)/(a*x)) + (Sqrt[b*x^2]*((-2*Sqrt[b*x^2])/((1 - Sqrt[3]) 
*a^(1/3) - (a + b*x^2)^(1/3)) + (3^(1/4)*Sqrt[2 + Sqrt[3]]*a^(1/3)*(a^(1/3 
) - (a + b*x^2)^(1/3))*Sqrt[(a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b* 
x^2)^(2/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))^2]*EllipticE[ArcSi 
n[((1 + Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))/((1 - Sqrt[3])*a^(1/3) - (a 
+ b*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(Sqrt[b*x^2]*Sqrt[-((a^(1/3)*(a^(1/3) - 
 (a + b*x^2)^(1/3)))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))^2)]) - (2 
*Sqrt[2 - Sqrt[3]]*(1 + Sqrt[3])*a^(1/3)*(a^(1/3) - (a + b*x^2)^(1/3))*Sqr 
t[(a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3))/((1 - Sqrt[3]) 
*a^(1/3) - (a + b*x^2)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*a^(1/3) - 
 (a + b*x^2)^(1/3))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))], -7 + 4*S 
qrt[3]])/(3^(1/4)*Sqrt[b*x^2]*Sqrt[-((a^(1/3)*(a^(1/3) - (a + b*x^2)^(1/3) 
))/((1 - Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))^2)])))/(2*a*x)
 

Defintions of rubi rules used

rule 233
Int[((a_) + (b_.)*(x_)^2)^(-1/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) 
   Subst[Int[x/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 833
Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3] 
], s = Denom[Rt[b/a, 3]]}, Simp[(-(1 + Sqrt[3]))*(s/r)   Int[1/Sqrt[a + b*x 
^3], x], x] + Simp[1/r   Int[((1 + Sqrt[3])*s + r*x)/Sqrt[a + b*x^3], x], x 
]] /; FreeQ[{a, b}, x] && NegQ[a]
 

rule 2418
Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N 
umer[Simplify[(1 + Sqrt[3])*(d/c)]], s = Denom[Simplify[(1 + Sqrt[3])*(d/c) 
]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 - Sqrt[3])*s + r*x))), x] + S 
imp[3^(1/4)*Sqrt[2 + Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/( 
(1 - Sqrt[3])*s + r*x)^2]/(r^2*Sqrt[a + b*x^3]*Sqrt[(-s)*((s + r*x)/((1 - S 
qrt[3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 + Sqrt[3])*s + r*x)/((1 - Sqrt[ 
3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && NegQ[a] && 
 EqQ[b*c^3 - 2*(5 + 3*Sqrt[3])*a*d^3, 0]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.05 \[ \int \frac {1}{x^2 \sqrt [3]{a+b x^2}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{3} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{\sqrt [3]{a} x} \] Input:

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

Output:

-hyper((-1/2, 1/3), (1/2,), b*x**2*exp_polar(I*pi)/a)/(a**(1/3)*x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.07 \[ \int \frac {1}{x^2 \sqrt [3]{a+b x^2}} \, dx=-\frac {3\,{\left (\frac {a}{b\,x^2}+1\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {5}{6};\ \frac {11}{6};\ -\frac {a}{b\,x^2}\right )}{5\,x\,{\left (b\,x^2+a\right )}^{1/3}} \] Input:

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

Output:

-(3*(a/(b*x^2) + 1)^(1/3)*hypergeom([1/3, 5/6], 11/6, -a/(b*x^2)))/(5*x*(a 
 + b*x^2)^(1/3))
 

Reduce [F]

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

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

Output:

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