3.4.83 \(\int \frac {(a+b x)^{2/3}}{x^2} \, dx\) [383]

3.4.83.1 Optimal result
3.4.83.2 Mathematica [A] (verified)
3.4.83.3 Rubi [A] (verified)
3.4.83.4 Maple [A] (verified)
3.4.83.5 Fricas [A] (verification not implemented)
3.4.83.6 Sympy [C] (verification not implemented)
3.4.83.7 Maxima [A] (verification not implemented)
3.4.83.8 Giac [A] (verification not implemented)
3.4.83.9 Mupad [B] (verification not implemented)
3.4.83.10 Reduce [B] (verification not implemented)

3.4.83.1 Optimal result

Integrand size = 13, antiderivative size = 94 \[ \int \frac {(a+b x)^{2/3}}{x^2} \, dx=-\frac {(a+b x)^{2/3}}{x}+\frac {2 b \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a}}-\frac {b \log (x)}{3 \sqrt [3]{a}}+\frac {b \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{\sqrt [3]{a}} \]

output
-(b*x+a)^(2/3)/x-1/3*b*ln(x)/a^(1/3)+b*ln(a^(1/3)-(b*x+a)^(1/3))/a^(1/3)+2 
/3*b*arctan(1/3*(a^(1/3)+2*(b*x+a)^(1/3))/a^(1/3)*3^(1/2))/a^(1/3)*3^(1/2)
 
3.4.83.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.28 \[ \int \frac {(a+b x)^{2/3}}{x^2} \, dx=\frac {-3 \sqrt [3]{a} (a+b x)^{2/3}+2 \sqrt {3} b x \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+2 b x \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )-b x \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )}{3 \sqrt [3]{a} x} \]

input
Integrate[(a + b*x)^(2/3)/x^2,x]
 
output
(-3*a^(1/3)*(a + b*x)^(2/3) + 2*Sqrt[3]*b*x*ArcTan[(1 + (2*(a + b*x)^(1/3) 
)/a^(1/3))/Sqrt[3]] + 2*b*x*Log[a^(1/3) - (a + b*x)^(1/3)] - b*x*Log[a^(2/ 
3) + a^(1/3)*(a + b*x)^(1/3) + (a + b*x)^(2/3)])/(3*a^(1/3)*x)
 
3.4.83.3 Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {51, 67, 16, 1082, 217}

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.

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 67

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

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

3.4.83.3.1 Defintions of rubi rules used

rule 16
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 51
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 67
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x 
] + (Simp[3/(2*b)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], 
 x] - Simp[3/(2*b*q)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] / 
; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
 

rule 217
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1082
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 
3.4.83.4 Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.97

method result size
risch \(-\frac {\left (b x +a \right )^{\frac {2}{3}}}{x}+\frac {2 b \left (\frac {\ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{a^{\frac {1}{3}}}-\frac {\ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{2 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{a^{\frac {1}{3}}}\right )}{3}\) \(91\)
derivativedivides \(3 b \left (-\frac {\left (b x +a \right )^{\frac {2}{3}}}{3 b x}+\frac {2 \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{9 a^{\frac {1}{3}}}-\frac {\ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{9 a^{\frac {1}{3}}}+\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{9 a^{\frac {1}{3}}}\right )\) \(95\)
default \(3 b \left (-\frac {\left (b x +a \right )^{\frac {2}{3}}}{3 b x}+\frac {2 \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{9 a^{\frac {1}{3}}}-\frac {\ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{9 a^{\frac {1}{3}}}+\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{9 a^{\frac {1}{3}}}\right )\) \(95\)
pseudoelliptic \(\frac {2 \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b x +a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) \sqrt {3}\, b x +2 \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right ) b x -\ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right ) b x -3 \left (b x +a \right )^{\frac {2}{3}} a^{\frac {1}{3}}}{3 x \,a^{\frac {1}{3}}}\) \(96\)

input
int((b*x+a)^(2/3)/x^2,x,method=_RETURNVERBOSE)
 
output
-(b*x+a)^(2/3)/x+2/3*b*(1/a^(1/3)*ln((b*x+a)^(1/3)-a^(1/3))-1/2/a^(1/3)*ln 
((b*x+a)^(2/3)+a^(1/3)*(b*x+a)^(1/3)+a^(2/3))+3^(1/2)/a^(1/3)*arctan(1/3*3 
^(1/2)*(2/a^(1/3)*(b*x+a)^(1/3)+1)))
 
3.4.83.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.68 \[ \int \frac {(a+b x)^{2/3}}{x^2} \, dx=\left [\frac {3 \, \sqrt {\frac {1}{3}} a b x \sqrt {-\frac {1}{a^{\frac {2}{3}}}} \log \left (\frac {2 \, b x + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x + a\right )}^{\frac {2}{3}} a^{\frac {2}{3}} - {\left (b x + a\right )}^{\frac {1}{3}} a - a^{\frac {4}{3}}\right )} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} - 3 \, {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + 3 \, a}{x}\right ) - a^{\frac {2}{3}} b x \log \left ({\left (b x + a\right )}^{\frac {2}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + 2 \, a^{\frac {2}{3}} b x \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) - 3 \, {\left (b x + a\right )}^{\frac {2}{3}} a}{3 \, a x}, \frac {6 \, \sqrt {\frac {1}{3}} a^{\frac {2}{3}} b x \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{a^{\frac {1}{3}}}\right ) - a^{\frac {2}{3}} b x \log \left ({\left (b x + a\right )}^{\frac {2}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + 2 \, a^{\frac {2}{3}} b x \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) - 3 \, {\left (b x + a\right )}^{\frac {2}{3}} a}{3 \, a x}\right ] \]

input
integrate((b*x+a)^(2/3)/x^2,x, algorithm="fricas")
 
output
[1/3*(3*sqrt(1/3)*a*b*x*sqrt(-1/a^(2/3))*log((2*b*x + 3*sqrt(1/3)*(2*(b*x 
+ a)^(2/3)*a^(2/3) - (b*x + a)^(1/3)*a - a^(4/3))*sqrt(-1/a^(2/3)) - 3*(b* 
x + a)^(1/3)*a^(2/3) + 3*a)/x) - a^(2/3)*b*x*log((b*x + a)^(2/3) + (b*x + 
a)^(1/3)*a^(1/3) + a^(2/3)) + 2*a^(2/3)*b*x*log((b*x + a)^(1/3) - a^(1/3)) 
 - 3*(b*x + a)^(2/3)*a)/(a*x), 1/3*(6*sqrt(1/3)*a^(2/3)*b*x*arctan(sqrt(1/ 
3)*(2*(b*x + a)^(1/3) + a^(1/3))/a^(1/3)) - a^(2/3)*b*x*log((b*x + a)^(2/3 
) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3)) + 2*a^(2/3)*b*x*log((b*x + a)^(1/3) 
 - a^(1/3)) - 3*(b*x + a)^(2/3)*a)/(a*x)]
 
3.4.83.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.10 (sec) , antiderivative size = 643, normalized size of antiderivative = 6.84 \[ \int \frac {(a+b x)^{2/3}}{x^2} \, dx =\text {Too large to display} \]

input
integrate((b*x+a)**(2/3)/x**2,x)
 
output
10*a**(8/3)*b*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*ga 
mma(5/3)/(9*a**3*exp(2*I*pi/3)*gamma(8/3) - 9*a**2*b*(a/b + x)*exp(2*I*pi/ 
3)*gamma(8/3)) + 10*a**(8/3)*b*exp(-2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)** 
(1/3)*exp_polar(2*I*pi/3)/a**(1/3))*gamma(5/3)/(9*a**3*exp(2*I*pi/3)*gamma 
(8/3) - 9*a**2*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3)) + 10*a**(8/3)*b*log(1 
 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamma(5/3)/(9*a 
**3*exp(2*I*pi/3)*gamma(8/3) - 9*a**2*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3) 
) - 10*a**(5/3)*b**2*(a/b + x)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**( 
1/3)/a**(1/3))*gamma(5/3)/(9*a**3*exp(2*I*pi/3)*gamma(8/3) - 9*a**2*b*(a/b 
 + x)*exp(2*I*pi/3)*gamma(8/3)) - 10*a**(5/3)*b**2*(a/b + x)*exp(-2*I*pi/3 
)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(2*I*pi/3)/a**(1/3))*gamma(5/ 
3)/(9*a**3*exp(2*I*pi/3)*gamma(8/3) - 9*a**2*b*(a/b + x)*exp(2*I*pi/3)*gam 
ma(8/3)) - 10*a**(5/3)*b**2*(a/b + x)*log(1 - b**(1/3)*(a/b + x)**(1/3)*ex 
p_polar(4*I*pi/3)/a**(1/3))*gamma(5/3)/(9*a**3*exp(2*I*pi/3)*gamma(8/3) - 
9*a**2*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3)) + 15*a**2*b**(5/3)*(a/b + x)* 
*(2/3)*exp(2*I*pi/3)*gamma(5/3)/(9*a**3*exp(2*I*pi/3)*gamma(8/3) - 9*a**2* 
b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3))
 
3.4.83.7 Maxima [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99 \[ \int \frac {(a+b x)^{2/3}}{x^2} \, dx=\frac {2 \, \sqrt {3} b \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{3 \, a^{\frac {1}{3}}} - \frac {b \log \left ({\left (b x + a\right )}^{\frac {2}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{3 \, a^{\frac {1}{3}}} + \frac {2 \, b \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{3 \, a^{\frac {1}{3}}} - \frac {{\left (b x + a\right )}^{\frac {2}{3}}}{x} \]

input
integrate((b*x+a)^(2/3)/x^2,x, algorithm="maxima")
 
output
2/3*sqrt(3)*b*arctan(1/3*sqrt(3)*(2*(b*x + a)^(1/3) + a^(1/3))/a^(1/3))/a^ 
(1/3) - 1/3*b*log((b*x + a)^(2/3) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3))/a^( 
1/3) + 2/3*b*log((b*x + a)^(1/3) - a^(1/3))/a^(1/3) - (b*x + a)^(2/3)/x
 
3.4.83.8 Giac [A] (verification not implemented)

Time = 0.59 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.13 \[ \int \frac {(a+b x)^{2/3}}{x^2} \, dx=\frac {\frac {2 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {1}{3}}} - \frac {b^{2} \log \left ({\left (b x + a\right )}^{\frac {2}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {1}{3}}} + \frac {2 \, b^{2} \log \left ({\left | {\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {1}{3}}} - \frac {3 \, {\left (b x + a\right )}^{\frac {2}{3}} b}{x}}{3 \, b} \]

input
integrate((b*x+a)^(2/3)/x^2,x, algorithm="giac")
 
output
1/3*(2*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x + a)^(1/3) + a^(1/3))/a^(1/3 
))/a^(1/3) - b^2*log((b*x + a)^(2/3) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3))/ 
a^(1/3) + 2*b^2*log(abs((b*x + a)^(1/3) - a^(1/3)))/a^(1/3) - 3*(b*x + a)^ 
(2/3)*b/x)/b
 
3.4.83.9 Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.35 \[ \int \frac {(a+b x)^{2/3}}{x^2} \, dx=\frac {2\,b\,\ln \left (4\,a^{1/3}\,b^2-4\,b^2\,{\left (a+b\,x\right )}^{1/3}\right )}{3\,a^{1/3}}-\frac {{\left (a+b\,x\right )}^{2/3}}{x}-\frac {\ln \left (a^{1/3}\,{\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}^2-4\,b^2\,{\left (a+b\,x\right )}^{1/3}\right )\,\left (b-\sqrt {3}\,b\,1{}\mathrm {i}\right )}{3\,a^{1/3}}-\frac {\ln \left (a^{1/3}\,{\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}^2-4\,b^2\,{\left (a+b\,x\right )}^{1/3}\right )\,\left (b+\sqrt {3}\,b\,1{}\mathrm {i}\right )}{3\,a^{1/3}} \]

input
int((a + b*x)^(2/3)/x^2,x)
 
output
(2*b*log(4*a^(1/3)*b^2 - 4*b^2*(a + b*x)^(1/3)))/(3*a^(1/3)) - (a + b*x)^( 
2/3)/x - (log(a^(1/3)*(b - 3^(1/2)*b*1i)^2 - 4*b^2*(a + b*x)^(1/3))*(b - 3 
^(1/2)*b*1i))/(3*a^(1/3)) - (log(a^(1/3)*(b + 3^(1/2)*b*1i)^2 - 4*b^2*(a + 
 b*x)^(1/3))*(b + 3^(1/2)*b*1i))/(3*a^(1/3))
 
3.4.83.10 Reduce [B] (verification not implemented)

Time = 0.01 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.79 \[ \int \frac {(a+b x)^{2/3}}{x^2} \, dx=\frac {-2 \sqrt {3}\, \mathit {atan} \left (\frac {2 \left (b x +a \right )^{\frac {1}{6}}+a^{\frac {1}{6}}}{a^{\frac {1}{6}} \sqrt {3}}\right ) b x +2 \sqrt {3}\, \mathit {atan} \left (\frac {2 \left (b x +a \right )^{\frac {1}{6}}-a^{\frac {1}{6}}}{a^{\frac {1}{6}} \sqrt {3}}\right ) b x -3 a^{\frac {1}{3}} \left (b x +a \right )^{\frac {2}{3}}+2 \,\mathrm {log}\left (\left (b x +a \right )^{\frac {1}{6}}+a^{\frac {1}{6}}\right ) b x +2 \,\mathrm {log}\left (\left (b x +a \right )^{\frac {1}{6}}-a^{\frac {1}{6}}\right ) b x -\mathrm {log}\left (-a^{\frac {1}{6}} \left (b x +a \right )^{\frac {1}{6}}+\left (b x +a \right )^{\frac {1}{3}}+a^{\frac {1}{3}}\right ) b x -\mathrm {log}\left (a^{\frac {1}{6}} \left (b x +a \right )^{\frac {1}{6}}+\left (b x +a \right )^{\frac {1}{3}}+a^{\frac {1}{3}}\right ) b x}{3 a^{\frac {1}{3}} x} \]

input
int((a + b*x)**(2/3)/x**2,x)
 
output
( - 2*sqrt(3)*atan((2*(a + b*x)**(1/6) + a**(1/6))/(a**(1/6)*sqrt(3)))*b*x 
 + 2*sqrt(3)*atan((2*(a + b*x)**(1/6) - a**(1/6))/(a**(1/6)*sqrt(3)))*b*x 
- 3*a**(1/3)*(a + b*x)**(2/3) + 2*log((a + b*x)**(1/6) + a**(1/6))*b*x + 2 
*log((a + b*x)**(1/6) - a**(1/6))*b*x - log( - a**(1/6)*(a + b*x)**(1/6) + 
 (a + b*x)**(1/3) + a**(1/3))*b*x - log(a**(1/6)*(a + b*x)**(1/6) + (a + b 
*x)**(1/3) + a**(1/3))*b*x)/(3*a**(1/3)*x)