\(\int \frac {(c x)^{2/3}}{\sqrt [3]{a x+b x^2}} \, dx\) [128]

Optimal result

Integrand size = 21, antiderivative size = 235 \[ \int \frac {(c x)^{2/3}}{\sqrt [3]{a x+b x^2}} \, dx=\frac {c \left (a x+b x^2\right )^{2/3}}{b \sqrt [3]{c x}}+\frac {a \sqrt [3]{c} \sqrt [3]{c x} \sqrt [3]{a+b x} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{c} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c x}}\right )}{\sqrt {3} b^{4/3} \sqrt [3]{a x+b x^2}}+\frac {a \sqrt [3]{c} \sqrt [3]{c x} \sqrt [3]{a+b x} \log (c x)}{6 b^{4/3} \sqrt [3]{a x+b x^2}}+\frac {a \sqrt [3]{c} \sqrt [3]{c x} \sqrt [3]{a+b x} \log \left (1-\frac {\sqrt [3]{c} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c x}}\right )}{2 b^{4/3} \sqrt [3]{a x+b x^2}} \] Output:

c*(b*x^2+a*x)^(2/3)/b/(c*x)^(1/3)+1/3*a*c^(1/3)*(c*x)^(1/3)*(b*x+a)^(1/3)* 
arctan(1/3*3^(1/2)+2/3*c^(1/3)*(b*x+a)^(1/3)*3^(1/2)/b^(1/3)/(c*x)^(1/3))* 
3^(1/2)/b^(4/3)/(b*x^2+a*x)^(1/3)+1/6*a*c^(1/3)*(c*x)^(1/3)*(b*x+a)^(1/3)* 
ln(c*x)/b^(4/3)/(b*x^2+a*x)^(1/3)+1/2*a*c^(1/3)*(c*x)^(1/3)*(b*x+a)^(1/3)* 
ln(1-c^(1/3)*(b*x+a)^(1/3)/b^(1/3)/(c*x)^(1/3))/b^(4/3)/(b*x^2+a*x)^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.88 \[ \int \frac {(c x)^{2/3}}{\sqrt [3]{a x+b x^2}} \, dx=\frac {(c x)^{2/3} \left (6 a \sqrt [3]{b} \sqrt [3]{x}+6 b^{4/3} x^{4/3}-2 \sqrt {3} a \sqrt [3]{a+b x} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{b} \sqrt [3]{x}+2 \sqrt [3]{a+b x}}\right )+2 a \sqrt [3]{a+b x} \log \left (-\sqrt [3]{b} \sqrt [3]{x}+\sqrt [3]{a+b x}\right )-a \sqrt [3]{a+b x} \log \left (b^{2/3} x^{2/3}+\sqrt [3]{b} \sqrt [3]{x} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )\right )}{6 b^{4/3} \sqrt [3]{x} \sqrt [3]{x (a+b x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.64, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1137, 60, 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.

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

\(\Big \downarrow \) 1137

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 71

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

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 1137

Int[((e_.)*(x_))^(m_)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[( 
e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*(b + c*x)^p))   Int[x^(m + p)*(b + c*x)^ 
p, x], x] /; FreeQ[{b, c, e, m}, x]
 

Maple [F]

\[\int \frac {\left (c x \right )^{\frac {2}{3}}}{\left (b \,x^{2}+a x \right )^{\frac {1}{3}}}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.14 \[ \int \frac {(c x)^{2/3}}{\sqrt [3]{a x+b x^2}} \, dx=-\frac {2 \, \sqrt {3} a \left (\frac {c^{2}}{b}\right )^{\frac {1}{3}} x \arctan \left (\frac {2 \, \sqrt {3} {\left (b x^{2} + a x\right )}^{\frac {2}{3}} \left (c x\right )^{\frac {2}{3}} b \left (\frac {c^{2}}{b}\right )^{\frac {2}{3}} + \sqrt {3} {\left (b c^{2} x^{2} + a c^{2} x\right )}}{3 \, {\left (b c^{2} x^{2} + a c^{2} x\right )}}\right ) + a \left (\frac {c^{2}}{b}\right )^{\frac {1}{3}} x \log \left (\frac {{\left (b x^{2} + a x\right )}^{\frac {1}{3}} \left (c x\right )^{\frac {1}{3}} c x + {\left (b x^{2} + a x\right )}^{\frac {2}{3}} \left (c x\right )^{\frac {2}{3}} \left (\frac {c^{2}}{b}\right )^{\frac {1}{3}} + {\left (b x^{2} + a x\right )} \left (\frac {c^{2}}{b}\right )^{\frac {2}{3}}}{b x^{2} + a x}\right ) - 2 \, a \left (\frac {c^{2}}{b}\right )^{\frac {1}{3}} x \log \left (\frac {{\left (b x^{2} + a x\right )}^{\frac {2}{3}} \left (c x\right )^{\frac {2}{3}} - {\left (b x^{2} + a x\right )} \left (\frac {c^{2}}{b}\right )^{\frac {1}{3}}}{b x^{2} + a x}\right ) - 6 \, {\left (b x^{2} + a x\right )}^{\frac {2}{3}} \left (c x\right )^{\frac {2}{3}}}{6 \, b x} \] Input:

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

Output:

-1/6*(2*sqrt(3)*a*(c^2/b)^(1/3)*x*arctan(1/3*(2*sqrt(3)*(b*x^2 + a*x)^(2/3 
)*(c*x)^(2/3)*b*(c^2/b)^(2/3) + sqrt(3)*(b*c^2*x^2 + a*c^2*x))/(b*c^2*x^2 
+ a*c^2*x)) + a*(c^2/b)^(1/3)*x*log(((b*x^2 + a*x)^(1/3)*(c*x)^(1/3)*c*x + 
 (b*x^2 + a*x)^(2/3)*(c*x)^(2/3)*(c^2/b)^(1/3) + (b*x^2 + a*x)*(c^2/b)^(2/ 
3))/(b*x^2 + a*x)) - 2*a*(c^2/b)^(1/3)*x*log(((b*x^2 + a*x)^(2/3)*(c*x)^(2 
/3) - (b*x^2 + a*x)*(c^2/b)^(1/3))/(b*x^2 + a*x)) - 6*(b*x^2 + a*x)^(2/3)* 
(c*x)^(2/3))/(b*x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.69 \[ \int \frac {(c x)^{2/3}}{\sqrt [3]{a x+b x^2}} \, dx=\frac {1}{6} \, a c^{2} {\left (\frac {2 \, \sqrt {3} \left (b c\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b c + \frac {a c}{x}\right )}^{\frac {1}{3}} + \left (b c\right )^{\frac {1}{3}}\right )}}{3 \, \left (b c\right )^{\frac {1}{3}}}\right )}{b^{2} c^{2}} + \frac {6 \, {\left (b c + \frac {a c}{x}\right )}^{\frac {2}{3}} x}{a b c^{2}} - \frac {\left (b c\right )^{\frac {2}{3}} \log \left ({\left (b c + \frac {a c}{x}\right )}^{\frac {2}{3}} + {\left (b c + \frac {a c}{x}\right )}^{\frac {1}{3}} \left (b c\right )^{\frac {1}{3}} + \left (b c\right )^{\frac {2}{3}}\right )}{b^{2} c^{2}} + \frac {2 \, \left (b c\right )^{\frac {2}{3}} \log \left ({\left | {\left (b c + \frac {a c}{x}\right )}^{\frac {1}{3}} - \left (b c\right )^{\frac {1}{3}} \right |}\right )}{b^{2} c^{2}}\right )} \] Input:

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

Output:

1/6*a*c^2*(2*sqrt(3)*(b*c)^(2/3)*arctan(1/3*sqrt(3)*(2*(b*c + a*c/x)^(1/3) 
 + (b*c)^(1/3))/(b*c)^(1/3))/(b^2*c^2) + 6*(b*c + a*c/x)^(2/3)*x/(a*b*c^2) 
 - (b*c)^(2/3)*log((b*c + a*c/x)^(2/3) + (b*c + a*c/x)^(1/3)*(b*c)^(1/3) + 
 (b*c)^(2/3))/(b^2*c^2) + 2*(b*c)^(2/3)*log(abs((b*c + a*c/x)^(1/3) - (b*c 
)^(1/3)))/(b^2*c^2))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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