\(\int x^4 (a+b x^2)^{4/3} \, dx\) [745]

Optimal result

Integrand size = 15, antiderivative size = 335 \[ \int x^4 \left (a+b x^2\right )^{4/3} \, dx=-\frac {432 a^3 x \sqrt [3]{a+b x^2}}{21505 b^2}+\frac {48 a^2 x^3 \sqrt [3]{a+b x^2}}{4301 b}+\frac {24}{391} a x^5 \sqrt [3]{a+b x^2}+\frac {3}{23} x^5 \left (a+b x^2\right )^{4/3}-\frac {432\ 3^{3/4} \sqrt {2-\sqrt {3}} a^4 \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{21505 b^3 x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:

-432/21505*a^3*x*(b*x^2+a)^(1/3)/b^2+48/4301*a^2*x^3*(b*x^2+a)^(1/3)/b+24/ 
391*a*x^5*(b*x^2+a)^(1/3)+3/23*x^5*(b*x^2+a)^(4/3)-432/21505*3^(3/4)*(1/2* 
6^(1/2)-1/2*2^(1/2))*a^4*(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))/b^3/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 = 6.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.24 \[ \int x^4 \left (a+b x^2\right )^{4/3} \, dx=\frac {3 x \sqrt [3]{a+b x^2} \left (-\left (\left (9 a-17 b x^2\right ) \left (a+b x^2\right )^2\right )+\frac {9 a^3 \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{\sqrt [3]{1+\frac {b x^2}{a}}}\right )}{391 b^2} \] Input:

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

Output:

(3*x*(a + b*x^2)^(1/3)*(-((9*a - 17*b*x^2)*(a + b*x^2)^2) + (9*a^3*Hyperge 
ometric2F1[-4/3, 1/2, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^(1/3)))/(391*b^2 
)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {248, 248, 262, 262, 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.

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

\(\Big \downarrow \) 248

\(\displaystyle \frac {8}{23} a \int x^4 \sqrt [3]{b x^2+a}dx+\frac {3}{23} x^5 \left (a+b x^2\right )^{4/3}\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {8}{23} a \left (\frac {2}{17} a \int \frac {x^4}{\left (b x^2+a\right )^{2/3}}dx+\frac {3}{17} x^5 \sqrt [3]{a+b x^2}\right )+\frac {3}{23} x^5 \left (a+b x^2\right )^{4/3}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {8}{23} a \left (\frac {2}{17} a \left (\frac {3 x^3 \sqrt [3]{a+b x^2}}{11 b}-\frac {9 a \int \frac {x^2}{\left (b x^2+a\right )^{2/3}}dx}{11 b}\right )+\frac {3}{17} x^5 \sqrt [3]{a+b x^2}\right )+\frac {3}{23} x^5 \left (a+b x^2\right )^{4/3}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {8}{23} a \left (\frac {2}{17} a \left (\frac {3 x^3 \sqrt [3]{a+b x^2}}{11 b}-\frac {9 a \left (\frac {3 x \sqrt [3]{a+b x^2}}{5 b}-\frac {3 a \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx}{5 b}\right )}{11 b}\right )+\frac {3}{17} x^5 \sqrt [3]{a+b x^2}\right )+\frac {3}{23} x^5 \left (a+b x^2\right )^{4/3}\)

\(\Big \downarrow \) 234

\(\displaystyle \frac {8}{23} a \left (\frac {2}{17} a \left (\frac {3 x^3 \sqrt [3]{a+b x^2}}{11 b}-\frac {9 a \left (\frac {3 x \sqrt [3]{a+b x^2}}{5 b}-\frac {9 a \sqrt {b x^2} \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{10 b^2 x}\right )}{11 b}\right )+\frac {3}{17} x^5 \sqrt [3]{a+b x^2}\right )+\frac {3}{23} x^5 \left (a+b x^2\right )^{4/3}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {8}{23} a \left (\frac {2}{17} a \left (\frac {3 x^3 \sqrt [3]{a+b x^2}}{11 b}-\frac {9 a \left (\frac {3\ 3^{3/4} \sqrt {2-\sqrt {3}} a \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{5 b^2 x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}+\frac {3 x \sqrt [3]{a+b x^2}}{5 b}\right )}{11 b}\right )+\frac {3}{17} x^5 \sqrt [3]{a+b x^2}\right )+\frac {3}{23} x^5 \left (a+b x^2\right )^{4/3}\)

Input:

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

Output:

(3*x^5*(a + b*x^2)^(4/3))/23 + (8*a*((3*x^5*(a + b*x^2)^(1/3))/17 + (2*a*( 
(3*x^3*(a + b*x^2)^(1/3))/(11*b) - (9*a*((3*x*(a + b*x^2)^(1/3))/(5*b) + ( 
3*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^2*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)])))/(11*b)))/17))/23
 

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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b*x^6 + a*x^4)*(b*x^2 + a)^(1/3), x)
 

Sympy [A] (verification not implemented)

Time = 0.64 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.09 \[ \int x^4 \left (a+b x^2\right )^{4/3} \, dx=\frac {a^{\frac {4}{3}} x^{5} {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} \] Input:

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

Output:

a**(4/3)*x**5*hyper((-4/3, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/5
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^4 \left (a+b x^2\right )^{4/3} \, dx=\frac {-\frac {432 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a^{3} x}{21505}+\frac {48 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a^{2} b \,x^{3}}{4301}+\frac {75 \left (b \,x^{2}+a \right )^{\frac {1}{3}} a \,b^{2} x^{5}}{391}+\frac {3 \left (b \,x^{2}+a \right )^{\frac {1}{3}} b^{3} x^{7}}{23}+\frac {432 \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \right ) a^{4}}{21505}}{b^{2}} \] Input:

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

Output:

(3*( - 144*(a + b*x**2)**(1/3)*a**3*x + 80*(a + b*x**2)**(1/3)*a**2*b*x**3 
 + 1375*(a + b*x**2)**(1/3)*a*b**2*x**5 + 935*(a + b*x**2)**(1/3)*b**3*x** 
7 + 144*int((a + b*x**2)**(1/3)/(a + b*x**2),x)*a**4))/(21505*b**2)