\(\int \frac {\sqrt [3]{a+b x^2}}{x^4} \, dx\) [725]

Optimal result

Integrand size = 15, antiderivative size = 290 \[ \int \frac {\sqrt [3]{a+b x^2}}{x^4} \, dx=-\frac {\sqrt [3]{a+b x^2}}{3 x^3}-\frac {2 b \sqrt [3]{a+b x^2}}{9 a x}+\frac {2 \sqrt {2-\sqrt {3}} b \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{9 \sqrt [4]{3} a x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:

-1/3*(b*x^2+a)^(1/3)/x^3-2/9*b*(b*x^2+a)^(1/3)/a/x+2/27*(1/2*6^(1/2)-1/2*2 
^(1/2))*b*(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/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 = 10.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.18 \[ \int \frac {\sqrt [3]{a+b x^2}}{x^4} \, dx=-\frac {\sqrt [3]{a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{3},-\frac {1}{2},-\frac {b x^2}{a}\right )}{3 x^3 \sqrt [3]{1+\frac {b x^2}{a}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {247, 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.

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

\(\Big \downarrow \) 247

\(\displaystyle \frac {2}{9} b \int \frac {1}{x^2 \left (b x^2+a\right )^{2/3}}dx-\frac {\sqrt [3]{a+b x^2}}{3 x^3}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {2}{9} b \left (-\frac {b \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x}\right )-\frac {\sqrt [3]{a+b x^2}}{3 x^3}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {2}{9} b \left (-\frac {\sqrt {b x^2} \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{2 a x}-\frac {\sqrt [3]{a+b x^2}}{a x}\right )-\frac {\sqrt [3]{a+b x^2}}{3 x^3}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {2}{9} b \left (\frac {\sqrt {2-\sqrt {3}} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} a x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}-\frac {\sqrt [3]{a+b x^2}}{a x}\right )-\frac {\sqrt [3]{a+b x^2}}{3 x^3}\)

Input:

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

Output:

-1/3*(a + b*x^2)^(1/3)/x^3 + (2*b*(-((a + b*x^2)^(1/3)/(a*x)) + (Sqrt[2 - 
Sqrt[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] 
*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*Sqrt[3]])/(3^(1/4)*a*x*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)])))/9
 

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 247

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1)))   Int[ 
(c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 
0] && LtQ[m, -1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, 
m, p, 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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.59 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.12 \[ \int \frac {\sqrt [3]{a+b x^2}}{x^4} \, dx=- \frac {\sqrt [3]{a} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {1}{3} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 x^{3}} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt [3]{a+b x^2}}{x^4} \, dx=\frac {-3 \left (b \,x^{2}+a \right )^{\frac {1}{3}}-2 \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{3}}}{b \,x^{6}+a \,x^{4}}d x \right ) a \,x^{3}}{7 x^{3}} \] Input:

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

Output:

( - 3*(a + b*x**2)**(1/3) - 2*int((a + b*x**2)**(1/3)/(a*x**4 + b*x**6),x) 
*a*x**3)/(7*x**3)