\(\int \frac {(a+b x^2)^{5/6}}{x^2} \, dx\) [1086]

Optimal result

Integrand size = 15, antiderivative size = 574 \[ \int \frac {\left (a+b x^2\right )^{5/6}}{x^2} \, dx=-\frac {\left (a+b x^2\right )^{5/6}}{x}-\frac {5 \left (1+\sqrt {3}\right ) b x \sqrt [6]{a+b x^2}}{2 \left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )}-\frac {5 \sqrt [4]{3} \sqrt [3]{a} \sqrt [6]{a+b x^2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )^2}} E\left (\arccos \left (\frac {\sqrt [3]{a}-\left (1-\sqrt {3}\right ) \sqrt [3]{a+b x^2}}{\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}}\right )|\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 x \sqrt {-\frac {\sqrt [3]{a+b x^2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )^2}}}-\frac {5 \left (1-\sqrt {3}\right ) \sqrt [3]{a} \sqrt [6]{a+b x^2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}-\left (1-\sqrt {3}\right ) \sqrt [3]{a+b x^2}}{\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} x \sqrt {-\frac {\sqrt [3]{a+b x^2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )^2}}} \] Output:

-(b*x^2+a)^(5/6)/x-5*(1+3^(1/2))*b*x*(b*x^2+a)^(1/6)/(2*a^(1/3)-2*(1+3^(1/ 
2))*(b*x^2+a)^(1/3))-5/2*3^(1/4)*a^(1/3)*(b*x^2+a)^(1/6)*(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))/(a^(1/3)-(1+3^ 
(1/2))*(b*x^2+a)^(1/3))^2)^(1/2)*EllipticE((1-(a^(1/3)-(1-3^(1/2))*(b*x^2+ 
a)^(1/3))^2/(a^(1/3)-(1+3^(1/2))*(b*x^2+a)^(1/3))^2)^(1/2),1/4*6^(1/2)+1/4 
*2^(1/2))/x/(-(b*x^2+a)^(1/3)*(a^(1/3)-(b*x^2+a)^(1/3))/(a^(1/3)-(1+3^(1/2 
))*(b*x^2+a)^(1/3))^2)^(1/2)-5/12*(1-3^(1/2))*a^(1/3)*(b*x^2+a)^(1/6)*(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))/( 
a^(1/3)-(1+3^(1/2))*(b*x^2+a)^(1/3))^2)^(1/2)*InverseJacobiAM(arccos((a^(1 
/3)-(1-3^(1/2))*(b*x^2+a)^(1/3))/(a^(1/3)-(1+3^(1/2))*(b*x^2+a)^(1/3))),1/ 
4*6^(1/2)+1/4*2^(1/2))*3^(3/4)/x/(-(b*x^2+a)^(1/3)*(a^(1/3)-(b*x^2+a)^(1/3 
))/(a^(1/3)-(1+3^(1/2))*(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 = 8.55 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.09 \[ \int \frac {\left (a+b x^2\right )^{5/6}}{x^2} \, dx=-\frac {\left (a+b x^2\right )^{5/6} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},-\frac {1}{2},\frac {1}{2},-\frac {b x^2}{a}\right )}{x \left (1+\frac {b x^2}{a}\right )^{5/6}} \] Input:

Integrate[(a + b*x^2)^(5/6)/x^2,x]
 

Output:

-(((a + b*x^2)^(5/6)*Hypergeometric2F1[-5/6, -1/2, 1/2, -((b*x^2)/a)])/(x* 
(1 + (b*x^2)/a)^(5/6)))
 

Rubi [A] (warning: unable to verify)

Time = 0.43 (sec) , antiderivative size = 721, normalized size of antiderivative = 1.26, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {247, 235, 214, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{5/6}}{x^2} \, dx\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {5}{3} b \int \frac {1}{\sqrt [6]{b x^2+a}}dx-\frac {\left (a+b x^2\right )^{5/6}}{x}\)

\(\Big \downarrow \) 235

\(\displaystyle \frac {5}{3} b \left (\frac {3 x}{2 \sqrt [6]{a+b x^2}}-\frac {1}{2} a \int \frac {1}{\left (b x^2+a\right )^{7/6}}dx\right )-\frac {\left (a+b x^2\right )^{5/6}}{x}\)

\(\Big \downarrow \) 214

\(\displaystyle \frac {5}{3} b \left (\frac {3 x}{2 \sqrt [6]{a+b x^2}}-\frac {a \int \frac {1}{\sqrt [3]{1-\frac {b x^2}{b x^2+a}}}d\frac {x}{\sqrt {b x^2+a}}}{2 \left (\frac {a}{a+b x^2}\right )^{2/3} \left (a+b x^2\right )^{2/3}}\right )-\frac {\left (a+b x^2\right )^{5/6}}{x}\)

\(\Big \downarrow \) 233

\(\displaystyle \frac {5}{3} b \left (\frac {3 a \sqrt {-\frac {b x^2}{a+b x^2}} \int \frac {\sqrt [3]{1-\frac {b x^2}{b x^2+a}}}{\sqrt {\frac {x^3}{\left (b x^2+a\right )^{3/2}}-1}}d\sqrt [3]{1-\frac {b x^2}{b x^2+a}}}{4 b x \left (\frac {a}{a+b x^2}\right )^{2/3} \sqrt [6]{a+b x^2}}+\frac {3 x}{2 \sqrt [6]{a+b x^2}}\right )-\frac {\left (a+b x^2\right )^{5/6}}{x}\)

\(\Big \downarrow \) 833

\(\displaystyle \frac {5}{3} b \left (\frac {3 a \sqrt {-\frac {b x^2}{a+b x^2}} \left (\left (1+\sqrt {3}\right ) \int \frac {1}{\sqrt {\frac {x^3}{\left (b x^2+a\right )^{3/2}}-1}}d\sqrt [3]{1-\frac {b x^2}{b x^2+a}}-\int \frac {-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}+\sqrt {3}+1}{\sqrt {\frac {x^3}{\left (b x^2+a\right )^{3/2}}-1}}d\sqrt [3]{1-\frac {b x^2}{b x^2+a}}\right )}{4 b x \left (\frac {a}{a+b x^2}\right )^{2/3} \sqrt [6]{a+b x^2}}+\frac {3 x}{2 \sqrt [6]{a+b x^2}}\right )-\frac {\left (a+b x^2\right )^{5/6}}{x}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {5}{3} b \left (\frac {3 a \sqrt {-\frac {b x^2}{a+b x^2}} \left (-\int \frac {-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}+\sqrt {3}+1}{\sqrt {\frac {x^3}{\left (b x^2+a\right )^{3/2}}-1}}d\sqrt [3]{1-\frac {b x^2}{b x^2+a}}-\frac {2 \sqrt {2-\sqrt {3}} \left (1+\sqrt {3}\right ) \left (1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}\right ) \sqrt {\frac {\frac {x^2}{a+b x^2}+\sqrt [3]{1-\frac {b x^2}{a+b x^2}}+1}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}+\sqrt {3}+1}{-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x^3}{\left (a+b x^2\right )^{3/2}}-1} \sqrt {-\frac {1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}}}\right )}{4 b x \left (\frac {a}{a+b x^2}\right )^{2/3} \sqrt [6]{a+b x^2}}+\frac {3 x}{2 \sqrt [6]{a+b x^2}}\right )-\frac {\left (a+b x^2\right )^{5/6}}{x}\)

\(\Big \downarrow \) 2418

\(\displaystyle \frac {5}{3} b \left (\frac {3 a \sqrt {-\frac {b x^2}{a+b x^2}} \left (-\frac {2 \sqrt {2-\sqrt {3}} \left (1+\sqrt {3}\right ) \left (1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}\right ) \sqrt {\frac {\frac {x^2}{a+b x^2}+\sqrt [3]{1-\frac {b x^2}{a+b x^2}}+1}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}+\sqrt {3}+1}{-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {\frac {x^3}{\left (a+b x^2\right )^{3/2}}-1} \sqrt {-\frac {1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}}}+\frac {\sqrt [4]{3} \sqrt {2+\sqrt {3}} \left (1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}\right ) \sqrt {\frac {\frac {x^2}{a+b x^2}+\sqrt [3]{1-\frac {b x^2}{a+b x^2}}+1}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}} E\left (\arcsin \left (\frac {-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}+\sqrt {3}+1}{-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{\sqrt {\frac {x^3}{\left (a+b x^2\right )^{3/2}}-1} \sqrt {-\frac {1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}}}-\frac {2 \sqrt {\frac {x^3}{\left (a+b x^2\right )^{3/2}}-1}}{-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1}\right )}{4 b x \left (\frac {a}{a+b x^2}\right )^{2/3} \sqrt [6]{a+b x^2}}+\frac {3 x}{2 \sqrt [6]{a+b x^2}}\right )-\frac {\left (a+b x^2\right )^{5/6}}{x}\)

Input:

Int[(a + b*x^2)^(5/6)/x^2,x]
 

Output:

-((a + b*x^2)^(5/6)/x) + (5*b*((3*x)/(2*(a + b*x^2)^(1/6)) + (3*a*Sqrt[-(( 
b*x^2)/(a + b*x^2))]*((-2*Sqrt[-1 + x^3/(a + b*x^2)^(3/2)])/(1 - Sqrt[3] - 
 (1 - (b*x^2)/(a + b*x^2))^(1/3)) + (3^(1/4)*Sqrt[2 + Sqrt[3]]*(1 - (1 - ( 
b*x^2)/(a + b*x^2))^(1/3))*Sqrt[(1 + x^2/(a + b*x^2) + (1 - (b*x^2)/(a + b 
*x^2))^(1/3))/(1 - Sqrt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))^2]*EllipticE 
[ArcSin[(1 + Sqrt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (1 
- (b*x^2)/(a + b*x^2))^(1/3))], -7 + 4*Sqrt[3]])/(Sqrt[-1 + x^3/(a + b*x^2 
)^(3/2)]*Sqrt[-((1 - (1 - (b*x^2)/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (1 - 
(b*x^2)/(a + b*x^2))^(1/3))^2)]) - (2*Sqrt[2 - Sqrt[3]]*(1 + Sqrt[3])*(1 - 
 (1 - (b*x^2)/(a + b*x^2))^(1/3))*Sqrt[(1 + x^2/(a + b*x^2) + (1 - (b*x^2) 
/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))^2]*El 
lipticF[ArcSin[(1 + Sqrt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))/(1 - Sqrt[3 
] - (1 - (b*x^2)/(a + b*x^2))^(1/3))], -7 + 4*Sqrt[3]])/(3^(1/4)*Sqrt[-1 + 
 x^3/(a + b*x^2)^(3/2)]*Sqrt[-((1 - (1 - (b*x^2)/(a + b*x^2))^(1/3))/(1 - 
Sqrt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))^2)])))/(4*b*x*(a/(a + b*x^2))^( 
2/3)*(a + b*x^2)^(1/6))))/3
 

Defintions of rubi rules used

rule 214

Int[((a_) + (b_.)*(x_)^2)^(-7/6), x_Symbol] :> Simp[1/((a + b*x^2)^(2/3)*(a 
/(a + b*x^2))^(2/3))   Subst[Int[1/(1 - b*x^2)^(1/3), x], x, x/Sqrt[a + b*x 
^2]], x] /; FreeQ[{a, b}, x]
 

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 235

Int[((a_) + (b_.)*(x_)^2)^(-1/6), x_Symbol] :> Simp[3*(x/(2*(a + b*x^2)^(1/ 
6))), x] - Simp[a/2   Int[1/(a + b*x^2)^(7/6), x], 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]
 

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]

\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {5}{6}}}{x^{2}}d x\]

Input:

int((b*x^2+a)^(5/6)/x^2,x)
 

Output:

int((b*x^2+a)^(5/6)/x^2,x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^2\right )^{5/6}}{x^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {5}{6}}}{x^{2}} \,d x } \] Input:

integrate((b*x^2+a)^(5/6)/x^2,x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(5/6)/x^2, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.73 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.05 \[ \int \frac {\left (a+b x^2\right )^{5/6}}{x^2} \, dx=- \frac {a^{\frac {5}{6}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{6}, - \frac {1}{2} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} \] Input:

integrate((b*x**2+a)**(5/6)/x**2,x)
 

Output:

-a**(5/6)*hyper((-5/6, -1/2), (1/2,), b*x**2*exp_polar(I*pi)/a)/x
 

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^{5/6}}{x^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {5}{6}}}{x^{2}} \,d x } \] Input:

integrate((b*x^2+a)^(5/6)/x^2,x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(5/6)/x^2, x)
 

Giac [F]

\[ \int \frac {\left (a+b x^2\right )^{5/6}}{x^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {5}{6}}}{x^{2}} \,d x } \] Input:

integrate((b*x^2+a)^(5/6)/x^2,x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^(5/6)/x^2, x)
 

Mupad [B] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.07 \[ \int \frac {\left (a+b x^2\right )^{5/6}}{x^2} \, dx=\frac {3\,{\left (b\,x^2+a\right )}^{5/6}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{6},-\frac {1}{3};\ \frac {2}{3};\ -\frac {a}{b\,x^2}\right )}{2\,x\,{\left (\frac {a}{b\,x^2}+1\right )}^{5/6}} \] Input:

int((a + b*x^2)^(5/6)/x^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.08 \[ \int \frac {\left (a+b x^2\right )^{5/6}}{x^2} \, dx=\frac {\sqrt {b \,x^{2}+a}\, \left (-b^{2} x^{4}-2 a b \,x^{2}-a^{2}\right )}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} a x} \] Input:

int((b*x^2+a)^(5/6)/x^2,x)
 

Output:

(sqrt(a + b*x**2)*( - a**2 - 2*a*b*x**2 - b**2*x**4))/((a + b*x**2)**(2/3) 
*a*x)