\(\int (a x^3+b x^6)^{4/3} \, dx\) [9]

Optimal result

Integrand size = 15, antiderivative size = 207 \[ \int \left (a x^3+b x^6\right )^{4/3} \, dx=\frac {2 a^2 x \sqrt [3]{a x^3+b x^6}}{81 b}+\frac {2}{27} a x^4 \sqrt [3]{a x^3+b x^6}+\frac {1}{9} x \left (a x^3+b x^6\right )^{4/3}+\frac {4 a^3 x^2 \left (a+b x^3\right )^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{81 \sqrt {3} b^{5/3} \left (a x^3+b x^6\right )^{2/3}}+\frac {2 a^3 x^2 \left (a+b x^3\right )^{2/3} \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{81 b^{5/3} \left (a x^3+b x^6\right )^{2/3}} \] Output:

2/81*a^2*x*(b*x^6+a*x^3)^(1/3)/b+2/27*a*x^4*(b*x^6+a*x^3)^(1/3)+1/9*x*(b*x 
^6+a*x^3)^(4/3)+4/243*a^3*x^2*(b*x^3+a)^(2/3)*arctan(1/3*(1+2*b^(1/3)*x/(b 
*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/b^(5/3)/(b*x^6+a*x^3)^(2/3)+2/81*a^3*x^2*( 
b*x^3+a)^(2/3)*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(5/3)/(b*x^6+a*x^3)^(2/3)
 

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.08 \[ \int \left (a x^3+b x^6\right )^{4/3} \, dx=\frac {x^2 \left (a+b x^3\right )^{2/3} \left (6 a^2 b^{2/3} x^2 \sqrt [3]{a+b x^3}+45 a b^{5/3} x^5 \sqrt [3]{a+b x^3}+27 b^{8/3} x^8 \sqrt [3]{a+b x^3}+4 \sqrt {3} a^3 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )+4 a^3 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )-2 a^3 \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )\right )}{243 b^{5/3} \left (x^3 \left (a+b x^3\right )\right )^{2/3}} \] Input:

Integrate[(a*x^3 + b*x^6)^(4/3),x]
 

Output:

(x^2*(a + b*x^3)^(2/3)*(6*a^2*b^(2/3)*x^2*(a + b*x^3)^(1/3) + 45*a*b^(5/3) 
*x^5*(a + b*x^3)^(1/3) + 27*b^(8/3)*x^8*(a + b*x^3)^(1/3) + 4*Sqrt[3]*a^3* 
ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))] + 4*a^3*Log[ 
-(b^(1/3)*x) + (a + b*x^3)^(1/3)] - 2*a^3*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + 
 b*x^3)^(1/3) + (a + b*x^3)^(2/3)]))/(243*b^(5/3)*(x^3*(a + b*x^3))^(2/3))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1910, 1927, 1930, 1938, 853}

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.

\(\displaystyle \int \left (a x^3+b x^6\right )^{4/3} \, dx\)

\(\Big \downarrow \) 1910

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

\(\Big \downarrow \) 1927

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

\(\Big \downarrow \) 1930

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

\(\Big \downarrow \) 1938

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

\(\Big \downarrow \) 853

\(\displaystyle \frac {4}{9} a \left (\frac {1}{6} a \left (\frac {x \sqrt [3]{a x^3+b x^6}}{3 b}-\frac {2 a x^2 \left (a+b x^3\right )^{2/3} \left (-\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {\log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}}\right )}{3 b \left (a x^3+b x^6\right )^{2/3}}\right )+\frac {1}{6} x^4 \sqrt [3]{a x^3+b x^6}\right )+\frac {1}{9} x \left (a x^3+b x^6\right )^{4/3}\)

Input:

Int[(a*x^3 + b*x^6)^(4/3),x]
 

Output:

(x*(a*x^3 + b*x^6)^(4/3))/9 + (4*a*((x^4*(a*x^3 + b*x^6)^(1/3))/6 + (a*((x 
*(a*x^3 + b*x^6)^(1/3))/(3*b) - (2*a*x^2*(a + b*x^3)^(2/3)*(-(ArcTan[(1 + 
(2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*b^(2/3))) - Log[b^(1/3) 
*x - (a + b*x^3)^(1/3)]/(2*b^(2/3))))/(3*b*(a*x^3 + b*x^6)^(2/3))))/6))/9
 

Defintions of rubi rules used

rule 853

Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Sim 
p[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp 
[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]
 

rule 1910

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[x*((a*x^j 
 + b*x^n)^p/(n*p + 1)), x] + Simp[a*(n - j)*(p/(n*p + 1))   Int[x^j*(a*x^j 
+ b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0, j, 
n] && GtQ[p, 0] && NeQ[n*p + 1, 0]
 

rule 1927

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a* 
(n - j)*(p/(c^j*(m + n*p + 1)))   Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) 
, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Int 
egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 0]
 

rule 1930

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p 
+ 1))), x] - Simp[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1)))   I 
nt[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, 
x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && Gt 
Q[m + j*p - n + j + 1, 0] && NeQ[m + n*p + 1, 0]
 

rule 1938

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F 
racPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]))   Int[x^(m + j* 
p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !Inte 
gerQ[p] && NeQ[n, j] && PosQ[n - j]
 

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.96

method	result	size
pseudoelliptic	\(\frac {27 \left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {1}{3}} b^{\frac {8}{3}} x^{7}+45 a \,b^{\frac {5}{3}} x^{4} \left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {1}{3}}+6 a^{2} x \left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {1}{3}} b^{\frac {2}{3}}-4 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x^{2}+2 \left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x^{2}}\right ) a^{3}+4 \ln \left (\frac {-b^{\frac {1}{3}} x^{2}+\left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {1}{3}}}{x^{2}}\right ) a^{3}-2 \ln \left (\frac {b^{\frac {2}{3}} x^{4}+b^{\frac {1}{3}} \left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {1}{3}} x^{2}+\left (x^{3} \left (b \,x^{3}+a \right )\right )^{\frac {2}{3}}}{x^{4}}\right ) a^{3}}{243 b^{\frac {5}{3}}}\)	\(198\)

Input:

int((b*x^6+a*x^3)^(4/3),x,method=_RETURNVERBOSE)
 

Output:

1/243*(27*(x^3*(b*x^3+a))^(1/3)*b^(8/3)*x^7+45*a*b^(5/3)*x^4*(x^3*(b*x^3+a 
))^(1/3)+6*a^2*x*(x^3*(b*x^3+a))^(1/3)*b^(2/3)-4*3^(1/2)*arctan(1/3*3^(1/2 
)*(b^(1/3)*x^2+2*(x^3*(b*x^3+a))^(1/3))/b^(1/3)/x^2)*a^3+4*ln((-b^(1/3)*x^ 
2+(x^3*(b*x^3+a))^(1/3))/x^2)*a^3-2*ln((b^(2/3)*x^4+b^(1/3)*(x^3*(b*x^3+a) 
)^(1/3)*x^2+(x^3*(b*x^3+a))^(2/3))/x^4)*a^3)/b^(5/3)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.03 \[ \int \left (a x^3+b x^6\right )^{4/3} \, dx=-\frac {12 \, \sqrt {\frac {1}{3}} a^{3} {\left (b^{2}\right )}^{\frac {1}{6}} b \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left ({\left (b^{2}\right )}^{\frac {1}{3}} b x^{2} + 2 \, {\left (b x^{6} + a x^{3}\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}}\right )} {\left (b^{2}\right )}^{\frac {1}{6}}}{b^{2} x^{2}}\right ) - 4 \, a^{3} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x^{2} - {\left (b x^{6} + a x^{3}\right )}^{\frac {1}{3}} b}{x^{2}}\right ) + 2 \, a^{3} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (b^{2}\right )}^{\frac {1}{3}} b x^{4} + {\left (b x^{6} + a x^{3}\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}} x^{2} + {\left (b x^{6} + a x^{3}\right )}^{\frac {2}{3}} b}{x^{4}}\right ) - 3 \, {\left (9 \, b^{4} x^{7} + 15 \, a b^{3} x^{4} + 2 \, a^{2} b^{2} x\right )} {\left (b x^{6} + a x^{3}\right )}^{\frac {1}{3}}}{243 \, b^{3}} \] Input:

integrate((b*x^6+a*x^3)^(4/3),x, algorithm="fricas")
 

Output:

-1/243*(12*sqrt(1/3)*a^3*(b^2)^(1/6)*b*arctan(sqrt(1/3)*((b^2)^(1/3)*b*x^2 
 + 2*(b*x^6 + a*x^3)^(1/3)*(b^2)^(2/3))*(b^2)^(1/6)/(b^2*x^2)) - 4*a^3*(b^ 
2)^(2/3)*log(-((b^2)^(2/3)*x^2 - (b*x^6 + a*x^3)^(1/3)*b)/x^2) + 2*a^3*(b^ 
2)^(2/3)*log(((b^2)^(1/3)*b*x^4 + (b*x^6 + a*x^3)^(1/3)*(b^2)^(2/3)*x^2 + 
(b*x^6 + a*x^3)^(2/3)*b)/x^4) - 3*(9*b^4*x^7 + 15*a*b^3*x^4 + 2*a^2*b^2*x) 
*(b*x^6 + a*x^3)^(1/3))/b^3
 

Sympy [F]

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

integrate((b*x**6+a*x**3)**(4/3),x)
 

Output:

Integral((a*x**3 + b*x**6)**(4/3), x)
 

Maxima [F]

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

integrate((b*x^6+a*x^3)^(4/3),x, algorithm="maxima")
 

Output:

integrate((b*x^6 + a*x^3)^(4/3), x)
 

Giac [A] (verification not implemented)

Time = 6.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.68 \[ \int \left (a x^3+b x^6\right )^{4/3} \, dx=\frac {1}{243} \, {\left (\frac {3 \, {\left (2 \, {\left (b + \frac {a}{x^{3}}\right )}^{\frac {7}{3}} + 11 \, {\left (b + \frac {a}{x^{3}}\right )}^{\frac {4}{3}} b - 4 \, {\left (b + \frac {a}{x^{3}}\right )}^{\frac {1}{3}} b^{2}\right )} x^{9}}{a^{3} b} - \frac {4 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b + \frac {a}{x^{3}}\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {5}{3}}} - \frac {2 \, \log \left ({\left (b + \frac {a}{x^{3}}\right )}^{\frac {2}{3}} + {\left (b + \frac {a}{x^{3}}\right )}^{\frac {1}{3}} b^{\frac {1}{3}} + b^{\frac {2}{3}}\right )}{b^{\frac {5}{3}}} + \frac {4 \, \log \left ({\left | {\left (b + \frac {a}{x^{3}}\right )}^{\frac {1}{3}} - b^{\frac {1}{3}} \right |}\right )}{b^{\frac {5}{3}}}\right )} a^{3} \] Input:

integrate((b*x^6+a*x^3)^(4/3),x, algorithm="giac")
 

Output:

1/243*(3*(2*(b + a/x^3)^(7/3) + 11*(b + a/x^3)^(4/3)*b - 4*(b + a/x^3)^(1/ 
3)*b^2)*x^9/(a^3*b) - 4*sqrt(3)*arctan(1/3*sqrt(3)*(2*(b + a/x^3)^(1/3) + 
b^(1/3))/b^(1/3))/b^(5/3) - 2*log((b + a/x^3)^(2/3) + (b + a/x^3)^(1/3)*b^ 
(1/3) + b^(2/3))/b^(5/3) + 4*log(abs((b + a/x^3)^(1/3) - b^(1/3)))/b^(5/3) 
)*a^3
 

Mupad [B] (verification not implemented)

Time = 19.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.20 \[ \int \left (a x^3+b x^6\right )^{4/3} \, dx=\frac {x\,{\left (b\,x^6+a\,x^3\right )}^{4/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {4}{3},\frac {5}{3};\ \frac {8}{3};\ -\frac {b\,x^3}{a}\right )}{5\,{\left (\frac {b\,x^3}{a}+1\right )}^{4/3}} \] Input:

int((a*x^3 + b*x^6)^(4/3),x)
 

Output:

(x*(a*x^3 + b*x^6)^(4/3)*hypergeom([-4/3, 5/3], 8/3, -(b*x^3)/a))/(5*((b*x 
^3)/a + 1)^(4/3))
 

Reduce [F]

\[ \int \left (a x^3+b x^6\right )^{4/3} \, dx=\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} x^{2}+15 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b \,x^{5}+9 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} x^{8}-4 \left (\int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a^{3}}{81 b} \] Input:

int((b*x^6+a*x^3)^(4/3),x)
 

Output:

(2*(a + b*x**3)**(1/3)*a**2*x**2 + 15*(a + b*x**3)**(1/3)*a*b*x**5 + 9*(a 
+ b*x**3)**(1/3)*b**2*x**8 - 4*int(((a + b*x**3)**(1/3)*x)/(a + b*x**3),x) 
*a**3)/(81*b)