\(\int \frac {(a+b x^2)^{4/3}}{x^2} \, dx\) [748]

Optimal result

Integrand size = 15, antiderivative size = 280 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^2} \, dx=\frac {8}{5} b x \sqrt [3]{a+b x^2}-\frac {\left (a+b x^2\right )^{4/3}}{x}-\frac {16 \sqrt {2-\sqrt {3}} a \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{5 \sqrt [4]{3} x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:

8/5*b*x*(b*x^2+a)^(1/3)-(b*x^2+a)^(4/3)/x-16/15*(1/2*6^(1/2)-1/2*2^(1/2))* 
a*(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)/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 = 5.91 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.18 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^2} \, dx=-\frac {a \sqrt [3]{a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-\frac {1}{2},\frac {1}{2},-\frac {b x^2}{a}\right )}{x \sqrt [3]{1+\frac {b x^2}{a}}} \] Input:

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

Output:

-((a*(a + b*x^2)^(1/3)*Hypergeometric2F1[-4/3, -1/2, 1/2, -((b*x^2)/a)])/( 
x*(1 + (b*x^2)/a)^(1/3)))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {247, 211, 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.

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

\(\Big \downarrow \) 247

\(\displaystyle \frac {8}{3} b \int \sqrt [3]{b x^2+a}dx-\frac {\left (a+b x^2\right )^{4/3}}{x}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {8}{3} b \left (\frac {2}{5} a \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx+\frac {3}{5} x \sqrt [3]{a+b x^2}\right )-\frac {\left (a+b x^2\right )^{4/3}}{x}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {8}{3} b \left (\frac {3 a \sqrt {b x^2} \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{5 b x}+\frac {3}{5} x \sqrt [3]{a+b x^2}\right )-\frac {\left (a+b x^2\right )^{4/3}}{x}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {8}{3} b \left (\frac {3}{5} x \sqrt [3]{a+b x^2}-\frac {2\ 3^{3/4} \sqrt {2-\sqrt {3}} a \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{5 b x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}\right )-\frac {\left (a+b x^2\right )^{4/3}}{x}\)

Input:

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

Output:

-((a + b*x^2)^(4/3)/x) + (8*b*((3*x*(a + b*x^2)^(1/3))/5 - (2*3^(3/4)*Sqrt 
[2 - Sqrt[3]]*a*(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]])/(5*b*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)])))/3
 

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.59 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.10 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^2} \, dx=- \frac {a^{\frac {4}{3}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, - \frac {1}{2} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} \] Input:

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

Output:

-a**(4/3)*hyper((-4/3, -1/2), (1/2,), b*x**2*exp_polar(I*pi)/a)/x
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.14 \[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^2} \, dx=\frac {3\,{\left (b\,x^2+a\right )}^{4/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {4}{3},-\frac {5}{6};\ \frac {1}{6};\ -\frac {a}{b\,x^2}\right )}{5\,x\,{\left (\frac {a}{b\,x^2}+1\right )}^{4/3}} \] Input:

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

Output:

(3*(a + b*x^2)^(4/3)*hypergeom([-4/3, -5/6], 1/6, -a/(b*x^2)))/(5*x*(a/(b* 
x^2) + 1)^(4/3))
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{4/3}}{x^2} \, dx=\frac {-21 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a +3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b \,x^{2}-16 \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{3}}}{b \,x^{4}+a \,x^{2}}d x \right ) a^{2} x}{5 x} \] Input:

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

Output:

( - 21*(a + b*x**2)**(1/3)*a + 3*(a + b*x**2)**(1/3)*b*x**2 - 16*int((a + 
b*x**2)**(1/3)/(a*x**2 + b*x**4),x)*a**2*x)/(5*x)