\(\int \frac {1}{\sqrt [3]{a x^2+b x^3}} \, dx\) [109]

Optimal result

Integrand size = 15, antiderivative size = 177 \[ \int \frac {1}{\sqrt [3]{a x^2+b x^3}} \, dx=-\frac {\sqrt {3} x^{2/3} \sqrt [3]{a+b x} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{x}}\right )}{\sqrt [3]{b} \sqrt [3]{a x^2+b x^3}}-\frac {x^{2/3} \sqrt [3]{a+b x} \log (x)}{2 \sqrt [3]{b} \sqrt [3]{a x^2+b x^3}}-\frac {3 x^{2/3} \sqrt [3]{a+b x} \log \left (1-\frac {\sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{x}}\right )}{2 \sqrt [3]{b} \sqrt [3]{a x^2+b x^3}} \] Output:

-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^(1/3)/(b*x^3+a*x^2)^(1/3)-1/2*x^(2/3)*(b*x+a)^(1/3)*l 
n(x)/b^(1/3)/(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^(1/3)/(b*x^3+a*x^2)^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\sqrt [3]{a x^2+b x^3}} \, dx=\frac {x^{2/3} \sqrt [3]{a+b x} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{b} \sqrt [3]{x}+2 \sqrt [3]{a+b x}}\right )-2 \log \left (-\sqrt [3]{b} \sqrt [3]{x}+\sqrt [3]{a+b x}\right )+\log \left (b^{2/3} x^{2/3}+\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )\right )}{2 \sqrt [3]{b} \sqrt [3]{x^2 (a+b x)}} \] Input:

Integrate[(a*x^2 + b*x^3)^(-1/3),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.67, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {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.

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

\(\Big \downarrow \) 1917

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

\(\Big \downarrow \) 71

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

Input:

Int[(a*x^2 + b*x^3)^(-1/3),x]
 

Output:

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

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.60

method	result	size
pseudoelliptic	\(-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+\ln \left (\frac {-b^{\frac {1}{3}} x +\left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (x^{2} \left (b x +a \right )\right )^{\frac {1}{3}} x +\left (x^{2} \left (b x +a \right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}}{b^{\frac {1}{3}}}\)	\(107\)

Input:

int(1/(b*x^3+a*x^2)^(1/3),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.08 \[ \int \frac {1}{\sqrt [3]{a x^2+b x^3}} \, dx=\left [\frac {\sqrt {3} b \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {3 \, b x^{2} + 2 \, a x - 3 \, {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {2}{3}} x - \sqrt {3} {\left (\left (-b\right )^{\frac {1}{3}} b x^{2} - {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} b x + 2 \, {\left (b x^{3} + a x^{2}\right )}^{\frac {2}{3}} \left (-b\right )^{\frac {2}{3}}\right )} \sqrt {\frac {\left (-b\right )^{\frac {1}{3}}}{b}}}{x}\right ) - 2 \, \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}}}{x}\right ) + \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right )}{2 \, b}, -\frac {2 \, \sqrt {3} b \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}} \arctan \left (-\frac {\sqrt {3} {\left (\left (-b\right )^{\frac {1}{3}} x - 2 \, {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-b\right )^{\frac {1}{3}}}{b}}}{3 \, x}\right ) + 2 \, \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - \left (-b\right )^{\frac {2}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a x^{2}\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right )}{2 \, b}\right ] \] Input:

integrate(1/(b*x^3+a*x^2)^(1/3),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {1}{\sqrt [3]{a x^2+b x^3}} \, dx=\int \frac {1}{\sqrt [3]{a x^{2} + b x^{3}}}\, dx \] Input:

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

Output:

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

Maxima [F]

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

integrate(1/(b*x^3+a*x^2)^(1/3),x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a*x^2)^(-1/3), x)
 

Giac [A] (verification not implemented)

Time = 3.56 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.49 \[ \int \frac {1}{\sqrt [3]{a x^2+b x^3}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} + b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} + \frac {\log \left ({\left (b + \frac {a}{x}\right )}^{\frac {2}{3}} + {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} b^{\frac {1}{3}} + b^{\frac {2}{3}}\right )}{2 \, b^{\frac {1}{3}}} - \frac {\log \left ({\left | {\left (b + \frac {a}{x}\right )}^{\frac {1}{3}} - b^{\frac {1}{3}} \right |}\right )}{b^{\frac {1}{3}}} \] Input:

integrate(1/(b*x^3+a*x^2)^(1/3),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 9.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.21 \[ \int \frac {1}{\sqrt [3]{a x^2+b x^3}} \, dx=\frac {3\,x\,{\left (\frac {b\,x}{a}+1\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{3},\frac {1}{3};\ \frac {4}{3};\ -\frac {b\,x}{a}\right )}{{\left (b\,x^3+a\,x^2\right )}^{1/3}} \] Input:

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

Output:

(3*x*((b*x)/a + 1)^(1/3)*hypergeom([1/3, 1/3], 4/3, -(b*x)/a))/(a*x^2 + b* 
x^3)^(1/3)
 

Reduce [F]

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

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

Output:

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