\(\int \frac {1}{x^4 (a+b x^2)^{5/6}} \, dx\) [1110]

Optimal result

Integrand size = 15, antiderivative size = 293 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=-\frac {\sqrt [6]{a+b x^2}}{3 a x^3}+\frac {8 b \sqrt [6]{a+b x^2}}{9 a^2 x}+\frac {8 b \sqrt [6]{a+b x^2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}-\left (1-\sqrt {3}\right ) \sqrt [3]{a+b x^2}}{\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{9 \sqrt [4]{3} a^{7/3} x \sqrt {-\frac {\sqrt [3]{a+b x^2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )^2}}} \] Output:

-1/3*(b*x^2+a)^(1/6)/a/x^3+8/9*b*(b*x^2+a)^(1/6)/a^2/x+8/27*b*(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(arcc 
os((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)/a^(7/3)/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 = 10.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.17 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=-\frac {\left (1+\frac {b x^2}{a}\right )^{5/6} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{6},-\frac {1}{2},-\frac {b x^2}{a}\right )}{3 x^3 \left (a+b x^2\right )^{5/6}} \] Input:

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

Output:

-1/3*((1 + (b*x^2)/a)^(5/6)*Hypergeometric2F1[-3/2, 5/6, -1/2, -((b*x^2)/a 
)])/(x^3*(a + b*x^2)^(5/6))
 

Rubi [A] (warning: unable to verify)

Time = 0.29 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.31, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {264, 264, 236, 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.

\(\displaystyle \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {8 b \int \frac {1}{x^2 \left (b x^2+a\right )^{5/6}}dx}{9 a}-\frac {\sqrt [6]{a+b x^2}}{3 a x^3}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {8 b \left (-\frac {2 b \int \frac {1}{\left (b x^2+a\right )^{5/6}}dx}{3 a}-\frac {\sqrt [6]{a+b x^2}}{a x}\right )}{9 a}-\frac {\sqrt [6]{a+b x^2}}{3 a x^3}\)

\(\Big \downarrow \) 236

\(\displaystyle -\frac {8 b \left (-\frac {2 b \int \frac {1}{\left (1-\frac {b x^2}{b x^2+a}\right )^{2/3}}d\frac {x}{\sqrt {b x^2+a}}}{3 a \sqrt [3]{\frac {a}{a+b x^2}} \sqrt [3]{a+b x^2}}-\frac {\sqrt [6]{a+b x^2}}{a x}\right )}{9 a}-\frac {\sqrt [6]{a+b x^2}}{3 a x^3}\)

\(\Big \downarrow \) 234

\(\displaystyle -\frac {8 b \left (\frac {\sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2} \int \frac {1}{\sqrt {\frac {x^3}{\left (b x^2+a\right )^{3/2}}-1}}d\sqrt [3]{1-\frac {b x^2}{b x^2+a}}}{a x \sqrt [3]{\frac {a}{a+b x^2}}}-\frac {\sqrt [6]{a+b x^2}}{a x}\right )}{9 a}-\frac {\sqrt [6]{a+b x^2}}{3 a x^3}\)

\(\Big \downarrow \) 760

\(\displaystyle -\frac {8 b \left (-\frac {2 \sqrt {2-\sqrt {3}} \sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}\right ) \sqrt {\frac {\frac {x^2}{a+b x^2}+\sqrt [3]{1-\frac {b x^2}{a+b x^2}}+1}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}+\sqrt {3}+1}{-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} a x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {\frac {x^3}{\left (a+b x^2\right )^{3/2}}-1} \sqrt {-\frac {1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}}}-\frac {\sqrt [6]{a+b x^2}}{a x}\right )}{9 a}-\frac {\sqrt [6]{a+b x^2}}{3 a x^3}\)

Input:

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

Output:

-1/3*(a + b*x^2)^(1/6)/(a*x^3) - (8*b*(-((a + b*x^2)^(1/6)/(a*x)) - (2*Sqr 
t[2 - Sqrt[3]]*Sqrt[-((b*x^2)/(a + b*x^2))]*(a + b*x^2)^(1/6)*(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]*EllipticF[ 
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)*a*x*(a/(a + b*x^2 
))^(1/3)*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)])))/(9*a)
 

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 236

Int[((a_) + (b_.)*(x_)^2)^(-5/6), x_Symbol] :> Simp[1/((a/(a + b*x^2))^(1/3 
)*(a + b*x^2)^(1/3))   Subst[Int[1/(1 - b*x^2)^(2/3), x], x, x/Sqrt[a + b*x 
^2]], 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]
 

Maple [F]

\[\int \frac {1}{x^{4} \left (b \,x^{2}+a \right )^{\frac {5}{6}}}d x\]

Input:

int(1/x^4/(b*x^2+a)^(5/6),x)
 

Output:

int(1/x^4/(b*x^2+a)^(5/6),x)
 

Fricas [F]

\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{6}} x^{4}} \,d x } \] Input:

integrate(1/x^4/(b*x^2+a)^(5/6),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(1/6)/(b*x^6 + a*x^4), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.63 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.11 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {5}{6} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {5}{6}} x^{3}} \] Input:

integrate(1/x**4/(b*x**2+a)**(5/6),x)
 

Output:

-hyper((-3/2, 5/6), (-1/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(5/6)*x**3)
 

Maxima [F]

\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{6}} x^{4}} \,d x } \] Input:

integrate(1/x^4/(b*x^2+a)^(5/6),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^(5/6)*x^4), x)
 

Giac [F]

\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{6}} x^{4}} \,d x } \] Input:

integrate(1/x^4/(b*x^2+a)^(5/6),x, algorithm="giac")
 

Output:

integrate(1/((b*x^2 + a)^(5/6)*x^4), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=\int \frac {1}{x^4\,{\left (b\,x^2+a\right )}^{5/6}} \,d x \] Input:

int(1/(x^4*(a + b*x^2)^(5/6)),x)
 

Output:

int(1/(x^4*(a + b*x^2)^(5/6)), x)
 

Reduce [F]

\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{5/6}} \, dx=\frac {-\left (b \,x^{2}+a \right )^{\frac {5}{6}}-4 \left (b \,x^{2}+a \right )^{\frac {2}{3}} \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {7}{6}}}{b^{2} x^{6}+2 a b \,x^{4}+a^{2} x^{2}}d x \right ) b \,x^{3}}{3 \left (b \,x^{2}+a \right )^{\frac {2}{3}} a \,x^{3}} \] Input:

int(1/x^4/(b*x^2+a)^(5/6),x)
 

Output:

( - (a + b*x**2)**(5/6) - 4*(a + b*x**2)**(2/3)*int((a + b*x**2)**(7/6)/(a 
**2*x**2 + 2*a*b*x**4 + b**2*x**6),x)*b*x**3)/(3*(a + b*x**2)**(2/3)*a*x** 
3)