\(\int \frac {x^2}{(a x^2+b x^3)^{2/3}} \, dx\) [373]

Optimal result

Integrand size = 19, antiderivative size = 200 \[ \int \frac {x^2}{\left (a x^2+b x^3\right )^{2/3}} \, dx=\frac {\sqrt [3]{a x^2+b x^3}}{b}+\frac {2 a x^{4/3} (a+b x)^{2/3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a+b x}}\right )}{\sqrt {3} b^{5/3} \left (a x^2+b x^3\right )^{2/3}}+\frac {a x^{4/3} (a+b x)^{2/3} \log (a+b x)}{3 b^{5/3} \left (a x^2+b x^3\right )^{2/3}}+\frac {a x^{4/3} (a+b x)^{2/3} \log \left (1-\frac {\sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a+b x}}\right )}{b^{5/3} \left (a x^2+b x^3\right )^{2/3}} \] Output:

(b*x^3+a*x^2)^(1/3)/b+2/3*a*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^(5/3)/(b*x^3+a*x^2)^(2/3)+ 
1/3*a*x^(4/3)*(b*x+a)^(2/3)*ln(b*x+a)/b^(5/3)/(b*x^3+a*x^2)^(2/3)+a*x^(4/3 
)*(b*x+a)^(2/3)*ln(1-b^(1/3)*x^(1/3)/(b*x+a)^(1/3))/b^(5/3)/(b*x^3+a*x^2)^ 
(2/3)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{\left (a x^2+b x^3\right )^{2/3}} \, dx=\frac {x^{4/3} \left (3 a b^{2/3} x^{2/3}+3 b^{5/3} x^{5/3}+2 \sqrt {3} a (a+b x)^{2/3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{b} \sqrt [3]{x}+2 \sqrt [3]{a+b x}}\right )+2 a (a+b x)^{2/3} \log \left (-\sqrt [3]{b} \sqrt [3]{x}+\sqrt [3]{a+b x}\right )-a (a+b x)^{2/3} \log \left (b^{2/3} x^{2/3}+\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )\right )}{3 b^{5/3} \left (x^2 (a+b x)\right )^{2/3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.75, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {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.

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

\(\Big \downarrow \) 1930

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

\(\Big \downarrow \) 1938

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

\(\Big \downarrow \) 71

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

Input:

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

Output:

(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))/(Sqrt[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))
 

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 = 0.48 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.64

method	result	size
pseudoelliptic	\(\frac {-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) a +3 \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}} b^{\frac {2}{3}}+2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}}}{x}\right ) a -\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}} x +\left (x^{2} \left (b x +a \right )\right )^{\frac {2}{3}}}{x^{2}}\right ) a}{3 b^{\frac {5}{3}}}\)	\(129\)

Input:

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

Output:

1/3*(-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)*a+3*(x^2*(b*x+a))^(1/3)*b^(2/3)+2*ln((-b^(1/3)*x+(x^2*(b*x+a))^(1/3) 
)/x)*a-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)*a)/b^(5/3)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.88 \[ \int \frac {x^2}{\left (a x^2+b x^3\right )^{2/3}} \, dx=-\frac {6 \, \sqrt {\frac {1}{3}} a {\left (b^{2}\right )}^{\frac {1}{6}} b \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left ({\left (b^{2}\right )}^{\frac {1}{3}} b x + 2 \, {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}}\right )} {\left (b^{2}\right )}^{\frac {1}{6}}}{b^{2} x}\right ) - 2 \, a {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x - {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} b}{x}\right ) + a {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (b^{2}\right )}^{\frac {1}{3}} b x^{2} + {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}} x + {\left (b x^{3} + a x^{2}\right )}^{\frac {2}{3}} b}{x^{2}}\right ) - 3 \, {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} b^{2}}{3 \, b^{3}} \] Input:

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

Output:

-1/3*(6*sqrt(1/3)*a*(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)) - 2*a*(b^2)^(2/3)*lo 
g(-((b^2)^(2/3)*x - (b*x^3 + a*x^2)^(1/3)*b)/x) + a*(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*(b*x^3 + a*x^2)^(1/3)*b^2)/b^3
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 4.16 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.54 \[ \int \frac {x^2}{\left (a x^2+b x^3\right )^{2/3}} \, dx=-\frac {1}{3} \, a {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {5}{3}}} - \frac {3 \, {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} x}{a b} + \frac {\log \left ({\left (b + \frac {a}{x}\right )}^{\frac {2}{3}} + {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} b^{\frac {1}{3}} + b^{\frac {2}{3}}\right )}{b^{\frac {5}{3}}} - \frac {2 \, \log \left ({\left | {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} - b^{\frac {1}{3}} \right |}\right )}{b^{\frac {5}{3}}}\right )} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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