\(\int \frac {x^3}{(a x^2+b x^3)^{4/3}} \, dx\) [380]

Optimal result

Integrand size = 19, antiderivative size = 198 \[ \int \frac {x^3}{\left (a x^2+b x^3\right )^{4/3}} \, dx=-\frac {3 x}{b \sqrt [3]{a x^2+b x^3}}-\frac {\sqrt {3} x^{2/3} \sqrt [3]{a+b x} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{x}}\right )}{b^{4/3} \sqrt [3]{a x^2+b x^3}}-\frac {x^{2/3} \sqrt [3]{a+b x} \log (x)}{2 b^{4/3} \sqrt [3]{a x^2+b x^3}}-\frac {3 x^{2/3} \sqrt [3]{a+b x} \log \left (1-\frac {\sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{x}}\right )}{2 b^{4/3} \sqrt [3]{a x^2+b x^3}} \] Output:

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.93 \[ \int \frac {x^3}{\left (a x^2+b x^3\right )^{4/3}} \, dx=\frac {x^{2/3} \left (-6 \sqrt [3]{b} \sqrt [3]{x}+2 \sqrt {3} \sqrt [3]{a+b x} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{b} \sqrt [3]{x}+2 \sqrt [3]{a+b x}}\right )-2 \sqrt [3]{a+b x} \log \left (-\sqrt [3]{b} \sqrt [3]{x}+\sqrt [3]{a+b x}\right )+\sqrt [3]{a+b x} \log \left (b^{2/3} x^{2/3}+\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )\right )}{2 b^{4/3} \sqrt [3]{x^2 (a+b x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.73, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1928, 1917, 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.

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

\(\Big \downarrow \) 1928

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

\(\Big \downarrow \) 1917

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

\(\Big \downarrow \) 71

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

Input:

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

Output:

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

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 1917

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + 
b*x^n)^FracPart[p]/(x^(j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])   Int[ 
x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !Integ 
erQ[p] && NeQ[n, j] && PosQ[n - j]
 

rule 1928

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*(n - j)*( 
p + 1))), x] - Simp[c^n*((m + j*p - n + j + 1)/(b*(n - j)*(p + 1)))   Int[( 
c*x)^(m - n)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c}, x] &&  !In 
tegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1] & 
& GtQ[m + j*p + 1, n - j]
 

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.80

method	result	size
pseudoelliptic	\(-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}}+\ln \left (\frac {-b^{\frac {1}{3}} x +\left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}}}{x}\right ) \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}}-\frac {\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}} x +\left (x^{2} \left (b x +a \right )\right )^{\frac {2}{3}}}{x^{2}}\right ) \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}}}{2}+3 b^{\frac {1}{3}} x}{b^{\frac {4}{3}} \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}}}\)	\(158\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 432, normalized size of antiderivative = 2.18 \[ \int \frac {x^3}{\left (a x^2+b x^3\right )^{4/3}} \, dx=\left [\frac {\sqrt {3} {\left (b^{2} x^{2} + a b x\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (\frac {3 \, b x^{2} + 2 \, a x - 3 \, {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} b^{\frac {2}{3}} x + \sqrt {3} {\left (b^{\frac {4}{3}} x^{2} + {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} b x - 2 \, {\left (b x^{3} + a x^{2}\right )}^{\frac {2}{3}} b^{\frac {2}{3}}\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}}}{x}\right ) - 2 \, {\left (b x^{2} + a x\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}}}{x}\right ) + {\left (b x^{2} + a x\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 6 \, {\left (b x^{3} + a x^{2}\right )}^{\frac {2}{3}} b}{2 \, {\left (b^{3} x^{2} + a b^{2} x\right )}}, -\frac {2 \, {\left (b x^{2} + a x\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - {\left (b x^{2} + a x\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) + \frac {2 \, \sqrt {3} {\left (b^{2} x^{2} + a b x\right )} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} x + 2 \, {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}} x}\right )}{b^{\frac {1}{3}}} + 6 \, {\left (b x^{3} + a x^{2}\right )}^{\frac {2}{3}} b}{2 \, {\left (b^{3} x^{2} + a b^{2} x\right )}}\right ] \] Input:

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

Output:

[1/2*(sqrt(3)*(b^2*x^2 + a*b*x)*sqrt(-1/b^(2/3))*log((3*b*x^2 + 2*a*x - 3* 
(b*x^3 + a*x^2)^(1/3)*b^(2/3)*x + sqrt(3)*(b^(4/3)*x^2 + (b*x^3 + a*x^2)^( 
1/3)*b*x - 2*(b*x^3 + a*x^2)^(2/3)*b^(2/3))*sqrt(-1/b^(2/3)))/x) - 2*(b*x^ 
2 + a*x)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a*x^2)^(1/3))/x) + (b*x^2 + a* 
x)*b^(2/3)*log((b^(2/3)*x^2 + (b*x^3 + a*x^2)^(1/3)*b^(1/3)*x + (b*x^3 + a 
*x^2)^(2/3))/x^2) - 6*(b*x^3 + a*x^2)^(2/3)*b)/(b^3*x^2 + a*b^2*x), -1/2*( 
2*(b*x^2 + a*x)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a*x^2)^(1/3))/x) - (b*x 
^2 + a*x)*b^(2/3)*log((b^(2/3)*x^2 + (b*x^3 + a*x^2)^(1/3)*b^(1/3)*x + (b* 
x^3 + a*x^2)^(2/3))/x^2) + 2*sqrt(3)*(b^2*x^2 + a*b*x)*arctan(1/3*sqrt(3)* 
(b^(1/3)*x + 2*(b*x^3 + a*x^2)^(1/3))/(b^(1/3)*x))/b^(1/3) + 6*(b*x^3 + a* 
x^2)^(2/3)*b)/(b^3*x^2 + a*b^2*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 8.32 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.51 \[ \int \frac {x^3}{\left (a x^2+b x^3\right )^{4/3}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {4}{3}}} + \frac {\log \left ({\left (b + \frac {a}{x}\right )}^{\frac {2}{3}} + {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} b^{\frac {1}{3}} + b^{\frac {2}{3}}\right )}{2 \, b^{\frac {4}{3}}} - \frac {\log \left ({\left | {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} - b^{\frac {1}{3}} \right |}\right )}{b^{\frac {4}{3}}} - \frac {3}{{\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} b} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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