\(\int \frac {\sqrt [3]{a+b x^2}}{(c x)^{8/3}} \, dx\) [797]

Optimal result

Integrand size = 19, antiderivative size = 391 \[ \int \frac {\sqrt [3]{a+b x^2}}{(c x)^{8/3}} \, dx=-\frac {3 \sqrt [3]{a+b x^2}}{5 c (c x)^{5/3}}+\frac {3^{3/4} b \sqrt [3]{c x} \sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right ) \sqrt {\frac {c^{4/3}+\frac {b^{2/3} (c x)^{4/3}}{\left (a+b x^2\right )^{2/3}}+\frac {\sqrt [3]{b} c^{2/3} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{\left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {c^{2/3}-\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}{c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{5 a c^{11/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\frac {\sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )}{\sqrt [3]{a+b x^2} \left (c^{2/3}-\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{\sqrt [3]{a+b x^2}}\right )^2}}} \] Output:

-3/5*(b*x^2+a)^(1/3)/c/(c*x)^(5/3)+1/5*3^(3/4)*b*(c*x)^(1/3)*(b*x^2+a)^(1/ 
3)*(c^(2/3)-b^(1/3)*(c*x)^(2/3)/(b*x^2+a)^(1/3))*((c^(4/3)+b^(2/3)*(c*x)^( 
4/3)/(b*x^2+a)^(2/3)+b^(1/3)*c^(2/3)*(c*x)^(2/3)/(b*x^2+a)^(1/3))/(c^(2/3) 
-(1+3^(1/2))*b^(1/3)*(c*x)^(2/3)/(b*x^2+a)^(1/3))^2)^(1/2)*InverseJacobiAM 
(arccos((c^(2/3)-(1-3^(1/2))*b^(1/3)*(c*x)^(2/3)/(b*x^2+a)^(1/3))/(c^(2/3) 
-(1+3^(1/2))*b^(1/3)*(c*x)^(2/3)/(b*x^2+a)^(1/3))),1/4*6^(1/2)+1/4*2^(1/2) 
)/a/c^(11/3)/(-b^(1/3)*(c*x)^(2/3)*(c^(2/3)-b^(1/3)*(c*x)^(2/3)/(b*x^2+a)^ 
(1/3))/(b*x^2+a)^(1/3)/(c^(2/3)-(1+3^(1/2))*b^(1/3)*(c*x)^(2/3)/(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 = 56, normalized size of antiderivative = 0.14 \[ \int \frac {\sqrt [3]{a+b x^2}}{(c x)^{8/3}} \, dx=-\frac {3 x \sqrt [3]{a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},-\frac {1}{3},\frac {1}{6},-\frac {b x^2}{a}\right )}{5 (c x)^{8/3} \sqrt [3]{1+\frac {b x^2}{a}}} \] Input:

Integrate[(a + b*x^2)^(1/3)/(c*x)^(8/3),x]
 

Output:

(-3*x*(a + b*x^2)^(1/3)*Hypergeometric2F1[-5/6, -1/3, 1/6, -((b*x^2)/a)])/ 
(5*(c*x)^(8/3)*(1 + (b*x^2)/a)^(1/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.34 (sec) , antiderivative size = 326, normalized size of antiderivative = 0.83, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {247, 266, 771, 766}

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.

\(\displaystyle \int \frac {\sqrt [3]{a+b x^2}}{(c x)^{8/3}} \, dx\)

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 771

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

\(\Big \downarrow \) 766

\(\displaystyle \frac {3^{3/4} b \sqrt [3]{c x} \left (c^{2/3}-\sqrt [3]{b} (c x)^{2/3}\right ) \sqrt {\frac {b^{2/3} (c x)^{4/3}+\sqrt [3]{b} c^{2/3} (c x)^{2/3}+c^{4/3}}{\left (c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {c^{2/3}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}{c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{5 c^{11/3} \sqrt {1-b x^2} \left (a+b x^2\right )^{2/3} \sqrt {-\frac {\sqrt [3]{b} (c x)^{2/3} \left (c^{2/3}-\sqrt [3]{b} (c x)^{2/3}\right )}{\left (c^{2/3}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} (c x)^{2/3}\right )^2}} \sqrt {\frac {a c^2}{a c^2+b c^2 x^2}}}-\frac {3 \sqrt [3]{a+b x^2}}{5 c (c x)^{5/3}}\)

Input:

Int[(a + b*x^2)^(1/3)/(c*x)^(8/3),x]
 

Output:

(-3*(a + b*x^2)^(1/3))/(5*c*(c*x)^(5/3)) + (3^(3/4)*b*(c*x)^(1/3)*(c^(2/3) 
 - b^(1/3)*(c*x)^(2/3))*Sqrt[(c^(4/3) + b^(1/3)*c^(2/3)*(c*x)^(2/3) + b^(2 
/3)*(c*x)^(4/3))/(c^(2/3) - (1 + Sqrt[3])*b^(1/3)*(c*x)^(2/3))^2]*Elliptic 
F[ArcCos[(c^(2/3) - (1 - Sqrt[3])*b^(1/3)*(c*x)^(2/3))/(c^(2/3) - (1 + Sqr 
t[3])*b^(1/3)*(c*x)^(2/3))], (2 + Sqrt[3])/4])/(5*c^(11/3)*Sqrt[1 - b*x^2] 
*(a + b*x^2)^(2/3)*Sqrt[(a*c^2)/(a*c^2 + b*c^2*x^2)]*Sqrt[-((b^(1/3)*(c*x) 
^(2/3)*(c^(2/3) - b^(1/3)*(c*x)^(2/3)))/(c^(2/3) - (1 + Sqrt[3])*b^(1/3)*( 
c*x)^(2/3))^2)])
 

Defintions of rubi rules used

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 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 766

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[x*(s + r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/ 
(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*((s + 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])* 
r*x^2)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x 
]
 

rule 771

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a/(a + b*x^n))^(p + 1 
/n)*(a + b*x^n)^(p + 1/n)   Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x 
/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] 
&& NeQ[p, -2^(-1)] && LtQ[Denominator[p + 1/n], Denominator[p]]
 

Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{3}}}{\left (c x \right )^{\frac {8}{3}}}d x\]

Input:

int((b*x^2+a)^(1/3)/(c*x)^(8/3),x)
 

Output:

int((b*x^2+a)^(1/3)/(c*x)^(8/3),x)
 

Fricas [F]

\[ \int \frac {\sqrt [3]{a+b x^2}}{(c x)^{8/3}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{3}}}{\left (c x\right )^{\frac {8}{3}}} \,d x } \] Input:

integrate((b*x^2+a)^(1/3)/(c*x)^(8/3),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 6.39 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.08 \[ \int \frac {\sqrt [3]{a+b x^2}}{(c x)^{8/3}} \, dx=- \frac {\sqrt [3]{b} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{c^{\frac {8}{3}} x} \] Input:

integrate((b*x**2+a)**(1/3)/(c*x)**(8/3),x)
 

Output:

-b**(1/3)*hyper((-1/3, 1/2), (3/2,), a*exp_polar(I*pi)/(b*x**2))/(c**(8/3) 
*x)
 

Maxima [F]

\[ \int \frac {\sqrt [3]{a+b x^2}}{(c x)^{8/3}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{3}}}{\left (c x\right )^{\frac {8}{3}}} \,d x } \] Input:

integrate((b*x^2+a)^(1/3)/(c*x)^(8/3),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(1/3)/(c*x)^(8/3), x)
 

Giac [F]

\[ \int \frac {\sqrt [3]{a+b x^2}}{(c x)^{8/3}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {1}{3}}}{\left (c x\right )^{\frac {8}{3}}} \,d x } \] Input:

integrate((b*x^2+a)^(1/3)/(c*x)^(8/3),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^(1/3)/(c*x)^(8/3), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{a+b x^2}}{(c x)^{8/3}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{1/3}}{{\left (c\,x\right )}^{8/3}} \,d x \] Input:

int((a + b*x^2)^(1/3)/(c*x)^(8/3),x)
 

Output:

int((a + b*x^2)^(1/3)/(c*x)^(8/3), x)
 

Reduce [F]

\[ \int \frac {\sqrt [3]{a+b x^2}}{(c x)^{8/3}} \, dx=\frac {-3 \left (b \,x^{2}+a \right )^{\frac {1}{3}}+2 x^{\frac {5}{3}} \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{3}}}{x^{\frac {2}{3}} a +x^{\frac {8}{3}} b}d x \right ) b}{5 x^{\frac {5}{3}} c^{\frac {8}{3}}} \] Input:

int((b*x^2+a)^(1/3)/(c*x)^(8/3),x)
 

Output:

( - 3*(a + b*x**2)**(1/3) + 2*x**(2/3)*int((a + b*x**2)**(1/3)/(x**(2/3)*a 
 + x**(2/3)*b*x**2),x)*b*x)/(5*x**(2/3)*c**(2/3)*c**2*x)