\(\int \frac {x^3}{(a x^2+b x^3)^{2/3}} \, dx\) [372]

Optimal result

Integrand size = 19, antiderivative size = 238 \[ \int \frac {x^3}{\left (a x^2+b x^3\right )^{2/3}} \, dx=-\frac {5 a \sqrt [3]{a x^2+b x^3}}{6 b^2}+\frac {x \sqrt [3]{a x^2+b x^3}}{2 b}-\frac {5 a^2 x^{4/3} (a+b x)^{2/3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a+b x}}\right )}{3 \sqrt {3} b^{8/3} \left (a x^2+b x^3\right )^{2/3}}-\frac {5 a^2 x^{4/3} (a+b x)^{2/3} \log (a+b x)}{18 b^{8/3} \left (a x^2+b x^3\right )^{2/3}}-\frac {5 a^2 x^{4/3} (a+b x)^{2/3} \log \left (1-\frac {\sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a+b x}}\right )}{6 b^{8/3} \left (a x^2+b x^3\right )^{2/3}} \] Output:

-5/6*a*(b*x^3+a*x^2)^(1/3)/b^2+1/2*x*(b*x^3+a*x^2)^(1/3)/b-5/9*a^2*x^(4/3) 
*(b*x+a)^(2/3)*arctan(1/3*3^(1/2)+2/3*b^(1/3)*x^(1/3)*3^(1/2)/(b*x+a)^(1/3 
))*3^(1/2)/b^(8/3)/(b*x^3+a*x^2)^(2/3)-5/18*a^2*x^(4/3)*(b*x+a)^(2/3)*ln(b 
*x+a)/b^(8/3)/(b*x^3+a*x^2)^(2/3)-5/6*a^2*x^(4/3)*(b*x+a)^(2/3)*ln(1-b^(1/ 
3)*x^(1/3)/(b*x+a)^(1/3))/b^(8/3)/(b*x^3+a*x^2)^(2/3)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.93 \[ \int \frac {x^3}{\left (a x^2+b x^3\right )^{2/3}} \, dx=\frac {x^{4/3} \left (-15 a^2 b^{2/3} x^{2/3}-6 a b^{5/3} x^{5/3}+9 b^{8/3} x^{8/3}-10 \sqrt {3} a^2 (a+b x)^{2/3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{b} \sqrt [3]{x}+2 \sqrt [3]{a+b x}}\right )-10 a^2 (a+b x)^{2/3} \log \left (-\sqrt [3]{b} \sqrt [3]{x}+\sqrt [3]{a+b x}\right )+5 a^2 (a+b x)^{2/3} \log \left (b^{2/3} x^{2/3}+\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )\right )}{18 b^{8/3} \left (x^2 (a+b x)\right )^{2/3}} \] Input:

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

Output:

(x^(4/3)*(-15*a^2*b^(2/3)*x^(2/3) - 6*a*b^(5/3)*x^(5/3) + 9*b^(8/3)*x^(8/3 
) - 10*Sqrt[3]*a^2*(a + b*x)^(2/3)*ArcTan[(Sqrt[3]*b^(1/3)*x^(1/3))/(b^(1/ 
3)*x^(1/3) + 2*(a + b*x)^(1/3))] - 10*a^2*(a + b*x)^(2/3)*Log[-(b^(1/3)*x^ 
(1/3)) + (a + b*x)^(1/3)] + 5*a^2*(a + b*x)^(2/3)*Log[b^(2/3)*x^(2/3) + b^ 
(1/3)*x^(1/3)*(a + b*x)^(1/3) + (a + b*x)^(2/3)]))/(18*b^(8/3)*(x^2*(a + b 
*x))^(2/3))
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1930, 1930, 1938, 71}

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.

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

\(\Big \downarrow \) 1930

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

\(\Big \downarrow \) 1930

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

\(\Big \downarrow \) 1938

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

\(\Big \downarrow \) 71

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

Input:

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

Output:

(x*(a*x^2 + b*x^3)^(1/3))/(2*b) - (5*a*((a*x^2 + b*x^3)^(1/3)/b - (2*a*x^( 
4/3)*(a + b*x)^(2/3)*(-((Sqrt[3]*ArcTan[1/Sqrt[3] + (2*b^(1/3)*x^(1/3))/(S 
qrt[3]*(a + b*x)^(1/3))])/b^(2/3)) - Log[a + b*x]/(2*b^(2/3)) - (3*Log[-1 
+ (b^(1/3)*x^(1/3))/(a + b*x)^(1/3)])/(2*b^(2/3))))/(3*b*(a*x^2 + b*x^3)^( 
2/3))))/(6*b)
 

Defintions of rubi rules used

rule 71

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

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 = 0.51 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.64

method	result	size
pseudoelliptic	\(\frac {9 x \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}} b^{\frac {5}{3}}+10 a^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )-15 a \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}} b^{\frac {2}{3}}-10 a^{2} \ln \left (\frac {-b^{\frac {1}{3}} x +\left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}}}{x}\right )+5 a^{2} \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}} x +\left (x^{2} \left (b x +a \right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{18 b^{\frac {8}{3}}}\)	\(153\)

Input:

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

Output:

1/18/b^(8/3)*(9*x*(x^2*(b*x+a))^(1/3)*b^(5/3)+10*a^2*3^(1/2)*arctan(1/3*3^ 
(1/2)*(b^(1/3)*x+2*(x^2*(b*x+a))^(1/3))/b^(1/3)/x)-15*a*(x^2*(b*x+a))^(1/3 
)*b^(2/3)-10*a^2*ln((-b^(1/3)*x+(x^2*(b*x+a))^(1/3))/x)+5*a^2*ln((b^(2/3)* 
x^2+b^(1/3)*(x^2*(b*x+a))^(1/3)*x+(x^2*(b*x+a))^(2/3))/x^2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.82 \[ \int \frac {x^3}{\left (a x^2+b x^3\right )^{2/3}} \, dx=\frac {30 \, \sqrt {\frac {1}{3}} a^{2} {\left (b^{2}\right )}^{\frac {1}{6}} b \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left ({\left (b^{2}\right )}^{\frac {1}{3}} b x + 2 \, {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}}\right )} {\left (b^{2}\right )}^{\frac {1}{6}}}{b^{2} x}\right ) - 10 \, a^{2} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x - {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} b}{x}\right ) + 5 \, a^{2} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (b^{2}\right )}^{\frac {1}{3}} b x^{2} + {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}} x + {\left (b x^{3} + a x^{2}\right )}^{\frac {2}{3}} b}{x^{2}}\right ) + 3 \, {\left (3 \, b^{3} x - 5 \, a b^{2}\right )} {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}}}{18 \, b^{4}} \] Input:

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

Output:

1/18*(30*sqrt(1/3)*a^2*(b^2)^(1/6)*b*arctan(sqrt(1/3)*((b^2)^(1/3)*b*x + 2 
*(b*x^3 + a*x^2)^(1/3)*(b^2)^(2/3))*(b^2)^(1/6)/(b^2*x)) - 10*a^2*(b^2)^(2 
/3)*log(-((b^2)^(2/3)*x - (b*x^3 + a*x^2)^(1/3)*b)/x) + 5*a^2*(b^2)^(2/3)* 
log(((b^2)^(1/3)*b*x^2 + (b*x^3 + a*x^2)^(1/3)*(b^2)^(2/3)*x + (b*x^3 + a* 
x^2)^(2/3)*b)/x^2) + 3*(3*b^3*x - 5*a*b^2)*(b*x^3 + a*x^2)^(1/3))/b^4
 

Sympy [F]

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

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

Output:

Integral(x**3/(x**2*(a + b*x))**(2/3), x)
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 4.29 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.60 \[ \int \frac {x^3}{\left (a x^2+b x^3\right )^{2/3}} \, dx=\frac {\frac {10 \, \sqrt {3} a^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {8}{3}}} + \frac {5 \, a^{3} \log \left ({\left (b + \frac {a}{x}\right )}^{\frac {2}{3}} + {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} b^{\frac {1}{3}} + b^{\frac {2}{3}}\right )}{b^{\frac {8}{3}}} - \frac {10 \, a^{3} \log \left ({\left | {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} - b^{\frac {1}{3}} \right |}\right )}{b^{\frac {8}{3}}} - \frac {3 \, {\left (5 \, a^{3} {\left (b + \frac {a}{x}\right )}^{\frac {4}{3}} - 8 \, a^{3} {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} b\right )} x^{2}}{a^{2} b^{2}}}{18 \, a} \] Input:

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

Output:

1/18*(10*sqrt(3)*a^3*arctan(1/3*sqrt(3)*(2*(b + a/x)^(1/3) + b^(1/3))/b^(1 
/3))/b^(8/3) + 5*a^3*log((b + a/x)^(2/3) + (b + a/x)^(1/3)*b^(1/3) + b^(2/ 
3))/b^(8/3) - 10*a^3*log(abs((b + a/x)^(1/3) - b^(1/3)))/b^(8/3) - 3*(5*a^ 
3*(b + a/x)^(4/3) - 8*a^3*(b + a/x)^(1/3)*b)*x^2/(a^2*b^2))/a
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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