\(\int x^4 \sqrt [6]{a+b x^2} \, dx\) [1075]

Optimal result

Integrand size = 15, antiderivative size = 314 \[ \int x^4 \sqrt [6]{a+b x^2} \, dx=-\frac {27 a^2 x \sqrt [6]{a+b x^2}}{640 b^2}+\frac {3 a x^3 \sqrt [6]{a+b x^2}}{160 b}+\frac {3}{16} x^5 \sqrt [6]{a+b x^2}+\frac {27\ 3^{3/4} a^{8/3} \sqrt [6]{a+b x^2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}-\left (1-\sqrt {3}\right ) \sqrt [3]{a+b x^2}}{\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{1280 b^3 x \sqrt {-\frac {\sqrt [3]{a+b x^2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )^2}}} \] Output:

-27/640*a^2*x*(b*x^2+a)^(1/6)/b^2+3/160*a*x^3*(b*x^2+a)^(1/6)/b+3/16*x^5*( 
b*x^2+a)^(1/6)+27/1280*3^(3/4)*a^(8/3)*(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))/b^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 = 7.17 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.30 \[ \int x^4 \sqrt [6]{a+b x^2} \, dx=\frac {3 x \sqrt [6]{a+b x^2} \left (\sqrt [6]{1+\frac {b x^2}{a}} \left (-9 a^2+a b x^2+10 b^2 x^4\right )+9 a^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{160 b^2 \sqrt [6]{1+\frac {b x^2}{a}}} \] Input:

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

Output:

(3*x*(a + b*x^2)^(1/6)*((1 + (b*x^2)/a)^(1/6)*(-9*a^2 + a*b*x^2 + 10*b^2*x 
^4) + 9*a^2*Hypergeometric2F1[-1/6, 1/2, 3/2, -((b*x^2)/a)]))/(160*b^2*(1 
+ (b*x^2)/a)^(1/6))
 

Rubi [A] (warning: unable to verify)

Time = 0.29 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.31, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {248, 262, 262, 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.

\(\displaystyle \int x^4 \sqrt [6]{a+b x^2} \, dx\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {1}{16} a \int \frac {x^4}{\left (b x^2+a\right )^{5/6}}dx+\frac {3}{16} x^5 \sqrt [6]{a+b x^2}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{16} a \left (\frac {3 x^3 \sqrt [6]{a+b x^2}}{10 b}-\frac {9 a \int \frac {x^2}{\left (b x^2+a\right )^{5/6}}dx}{10 b}\right )+\frac {3}{16} x^5 \sqrt [6]{a+b x^2}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{16} a \left (\frac {3 x^3 \sqrt [6]{a+b x^2}}{10 b}-\frac {9 a \left (\frac {3 x \sqrt [6]{a+b x^2}}{4 b}-\frac {3 a \int \frac {1}{\left (b x^2+a\right )^{5/6}}dx}{4 b}\right )}{10 b}\right )+\frac {3}{16} x^5 \sqrt [6]{a+b x^2}\)

\(\Big \downarrow \) 236

\(\displaystyle \frac {1}{16} a \left (\frac {3 x^3 \sqrt [6]{a+b x^2}}{10 b}-\frac {9 a \left (\frac {3 x \sqrt [6]{a+b x^2}}{4 b}-\frac {3 a \int \frac {1}{\left (1-\frac {b x^2}{b x^2+a}\right )^{2/3}}d\frac {x}{\sqrt {b x^2+a}}}{4 b \sqrt [3]{\frac {a}{a+b x^2}} \sqrt [3]{a+b x^2}}\right )}{10 b}\right )+\frac {3}{16} x^5 \sqrt [6]{a+b x^2}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {1}{16} a \left (\frac {3 x^3 \sqrt [6]{a+b x^2}}{10 b}-\frac {9 a \left (\frac {9 a \sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2} \int \frac {1}{\sqrt {\frac {x^3}{\left (b x^2+a\right )^{3/2}}-1}}d\sqrt [3]{1-\frac {b x^2}{b x^2+a}}}{8 b^2 x \sqrt [3]{\frac {a}{a+b x^2}}}+\frac {3 x \sqrt [6]{a+b x^2}}{4 b}\right )}{10 b}\right )+\frac {3}{16} x^5 \sqrt [6]{a+b x^2}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {1}{16} a \left (\frac {3 x^3 \sqrt [6]{a+b x^2}}{10 b}-\frac {9 a \left (\frac {3 x \sqrt [6]{a+b x^2}}{4 b}-\frac {3\ 3^{3/4} \sqrt {2-\sqrt {3}} a \sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}\right ) \sqrt {\frac {\frac {x^2}{a+b x^2}+\sqrt [3]{1-\frac {b x^2}{a+b x^2}}+1}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}+\sqrt {3}+1}{-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{4 b^2 x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {\frac {x^3}{\left (a+b x^2\right )^{3/2}}-1} \sqrt {-\frac {1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}}}\right )}{10 b}\right )+\frac {3}{16} x^5 \sqrt [6]{a+b x^2}\)

Input:

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

Output:

(3*x^5*(a + b*x^2)^(1/6))/16 + (a*((3*x^3*(a + b*x^2)^(1/6))/(10*b) - (9*a 
*((3*x*(a + b*x^2)^(1/6))/(4*b) - (3*3^(3/4)*Sqrt[2 - Sqrt[3]]*a*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]])/(4*b^2*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)])))/(10*b)))/16
 

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 248

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

rule 262

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.55 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.09 \[ \int x^4 \sqrt [6]{a+b x^2} \, dx=\frac {\sqrt [6]{a} x^{5} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{6}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} \] Input:

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

Output:

a**(1/6)*x**5*hyper((-1/6, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/5
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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