\(\int \frac {x^6}{(a+b x^3)^{3/2}} \, dx\) [222]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 15, antiderivative size = 251 \[ \int \frac {x^6}{\left (a+b x^3\right )^{3/2}} \, dx=-\frac {2 x^4}{3 b \sqrt {a+b x^3}}+\frac {16 x \sqrt {a+b x^3}}{15 b^2}-\frac {32 \sqrt {2+\sqrt {3}} a \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} b^{7/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \] Output:

-2/3*x^4/b/(b*x^3+a)^(1/2)+16/15*x*(b*x^3+a)^(1/2)/b^2-32/45*(1/2*6^(1/2)+ 
1/2*2^(1/2))*a*(a^(1/3)+b^(1/3)*x)*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2 
)/((1+3^(1/2))*a^(1/3)+b^(1/3)*x)^2)^(1/2)*EllipticF(((1-3^(1/2))*a^(1/3)+ 
b^(1/3)*x)/((1+3^(1/2))*a^(1/3)+b^(1/3)*x),I*3^(1/2)+2*I)*3^(3/4)/b^(7/3)/ 
(a^(1/3)*(a^(1/3)+b^(1/3)*x)/((1+3^(1/2))*a^(1/3)+b^(1/3)*x)^2)^(1/2)/(b*x 
^3+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 4.98 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.26 \[ \int \frac {x^6}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {2 x \left (8 a+3 b x^3-8 a \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},-\frac {b x^3}{a}\right )\right )}{15 b^2 \sqrt {a+b x^3}} \] Input:

Integrate[x^6/(a + b*x^3)^(3/2),x]
 

Output:

(2*x*(8*a + 3*b*x^3 - 8*a*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/3, 1/2, 
4/3, -((b*x^3)/a)]))/(15*b^2*Sqrt[a + b*x^3])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {817, 843, 759}

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.

\(\displaystyle \int \frac {x^6}{\left (a+b x^3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {8 \int \frac {x^3}{\sqrt {b x^3+a}}dx}{3 b}-\frac {2 x^4}{3 b \sqrt {a+b x^3}}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {8 \left (\frac {2 x \sqrt {a+b x^3}}{5 b}-\frac {2 a \int \frac {1}{\sqrt {b x^3+a}}dx}{5 b}\right )}{3 b}-\frac {2 x^4}{3 b \sqrt {a+b x^3}}\)

\(\Big \downarrow \) 759

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

Input:

Int[x^6/(a + b*x^3)^(3/2),x]
 

Output:

(-2*x^4)/(3*b*Sqrt[a + b*x^3]) + (8*((2*x*Sqrt[a + b*x^3])/(5*b) - (4*Sqrt 
[2 + Sqrt[3]]*a*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + 
b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*EllipticF[ArcSin[((1 - 
 Sqrt[3])*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 
4*Sqrt[3]])/(5*3^(1/4)*b^(4/3)*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)*x))/((1 + 
Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[a + b*x^3])))/(3*b)
 

Defintions of rubi rules used

rule 759
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] & 
& PosQ[a]
 

rule 817
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( 
n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n 
*((m - n + 1)/(b*n*(p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x 
] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  ! 
ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 843
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n 
 - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 
Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.27

method result size
default \(\frac {2 x a}{3 b^{2} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 x \sqrt {b \,x^{3}+a}}{5 b^{2}}+\frac {32 i a \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}}{-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{b \left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right )}{45 b^{3} \sqrt {b \,x^{3}+a}}\) \(320\)
elliptic \(\frac {2 x a}{3 b^{2} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {2 x \sqrt {b \,x^{3}+a}}{5 b^{2}}+\frac {32 i a \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}}{-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{b \left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right )}{45 b^{3} \sqrt {b \,x^{3}+a}}\) \(320\)
risch \(\frac {2 x \sqrt {b \,x^{3}+a}}{5 b^{2}}-\frac {a \left (2 a \left (\frac {2 x}{3 a \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {2 i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}}{-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{b \left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right )}{9 a b \sqrt {b \,x^{3}+a}}\right )+7 b \left (-\frac {2 x}{3 b \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {4 i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{b}}{-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (-a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{b \left (-\frac {3 \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right )}{9 b^{2} \sqrt {b \,x^{3}+a}}\right )\right )}{5 b^{2}}\) \(637\)

Input:

int(x^6/(b*x^3+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/3/b^2*x*a/((x^3+a/b)*b)^(1/2)+2/5*x*(b*x^3+a)^(1/2)/b^2+32/45*I*a/b^3*3^ 
(1/2)*(-a*b^2)^(1/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^( 
1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3))/(-3/2/b*(-a* 
b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(1 
/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3 
+a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b 
*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/ 
3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.27 \[ \int \frac {x^6}{\left (a+b x^3\right )^{3/2}} \, dx=-\frac {2 \, {\left (16 \, {\left (a b x^{3} + a^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right ) - {\left (3 \, b^{2} x^{4} + 8 \, a b x\right )} \sqrt {b x^{3} + a}\right )}}{15 \, {\left (b^{4} x^{3} + a b^{3}\right )}} \] Input:

integrate(x^6/(b*x^3+a)^(3/2),x, algorithm="fricas")
 

Output:

-2/15*(16*(a*b*x^3 + a^2)*sqrt(b)*weierstrassPInverse(0, -4*a/b, x) - (3*b 
^2*x^4 + 8*a*b*x)*sqrt(b*x^3 + a))/(b^4*x^3 + a*b^3)
 

Sympy [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.15 \[ \int \frac {x^6}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {3}{2}} \Gamma \left (\frac {10}{3}\right )} \] Input:

integrate(x**6/(b*x**3+a)**(3/2),x)
 

Output:

x**7*gamma(7/3)*hyper((3/2, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*a* 
*(3/2)*gamma(10/3))
 

Maxima [F]

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

integrate(x^6/(b*x^3+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate(x^6/(b*x^3 + a)^(3/2), x)
 

Giac [F]

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

integrate(x^6/(b*x^3+a)^(3/2),x, algorithm="giac")
 

Output:

integrate(x^6/(b*x^3 + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\left (a+b x^3\right )^{3/2}} \, dx=\int \frac {x^6}{{\left (b\,x^3+a\right )}^{3/2}} \,d x \] Input:

int(x^6/(a + b*x^3)^(3/2),x)
 

Output:

int(x^6/(a + b*x^3)^(3/2), x)
 

Reduce [F]

\[ \int \frac {x^6}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {\frac {16 \sqrt {b \,x^{3}+a}\, a x}{5}+\frac {2 \sqrt {b \,x^{3}+a}\, b \,x^{4}}{5}-\frac {16 \left (\int \frac {\sqrt {b \,x^{3}+a}}{b^{2} x^{6}+2 a b \,x^{3}+a^{2}}d x \right ) a^{3}}{5}-\frac {16 \left (\int \frac {\sqrt {b \,x^{3}+a}}{b^{2} x^{6}+2 a b \,x^{3}+a^{2}}d x \right ) a^{2} b \,x^{3}}{5}}{b^{2} \left (b \,x^{3}+a \right )} \] Input:

int(x^6/(b*x^3+a)^(3/2),x)
                                                                                    
                                                                                    
 

Output:

(2*(8*sqrt(a + b*x**3)*a*x + sqrt(a + b*x**3)*b*x**4 - 8*int(sqrt(a + b*x* 
*3)/(a**2 + 2*a*b*x**3 + b**2*x**6),x)*a**3 - 8*int(sqrt(a + b*x**3)/(a**2 
 + 2*a*b*x**3 + b**2*x**6),x)*a**2*b*x**3))/(5*b**2*(a + b*x**3))