3.384 \(\int \frac{(a+b x)^{2/3}}{x^3} \, dx\)
Optimal. Leaf size=127 \[ \frac{b^2 \log (x)}{18 a^{4/3}}-\frac{b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{6 a^{4/3}}-\frac{b^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{4/3}}-\frac{(a+b x)^{2/3}}{2 x^2}-\frac{b (a+b x)^{2/3}}{3 a x} \]
[Out]
-(a + b*x)^(2/3)/(2*x^2) - (b*(a + b*x)^(2/3))/(3*a*x) - (b^2*ArcTan[(a^(1/3) + 2*(a + b*x)^(1/3))/(Sqrt[3]*a^
(1/3))])/(3*Sqrt[3]*a^(4/3)) + (b^2*Log[x])/(18*a^(4/3)) - (b^2*Log[a^(1/3) - (a + b*x)^(1/3)])/(6*a^(4/3))
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Rubi [A] time = 0.0470729, antiderivative size = 127, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used =
{47, 51, 55, 617, 204, 31} \[ \frac{b^2 \log (x)}{18 a^{4/3}}-\frac{b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{6 a^{4/3}}-\frac{b^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{4/3}}-\frac{(a+b x)^{2/3}}{2 x^2}-\frac{b (a+b x)^{2/3}}{3 a x} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*x)^(2/3)/x^3,x]
[Out]
-(a + b*x)^(2/3)/(2*x^2) - (b*(a + b*x)^(2/3))/(3*a*x) - (b^2*ArcTan[(a^(1/3) + 2*(a + b*x)^(1/3))/(Sqrt[3]*a^
(1/3))])/(3*Sqrt[3]*a^(4/3)) + (b^2*Log[x])/(18*a^(4/3)) - (b^2*Log[a^(1/3) - (a + b*x)^(1/3)])/(6*a^(4/3))
Rule 47
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] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]
Rule 51
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]
Rule 55
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[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 617
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Rule 204
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rubi steps
\begin{align*} \int \frac{(a+b x)^{2/3}}{x^3} \, dx &=-\frac{(a+b x)^{2/3}}{2 x^2}+\frac{1}{3} b \int \frac{1}{x^2 \sqrt [3]{a+b x}} \, dx\\ &=-\frac{(a+b x)^{2/3}}{2 x^2}-\frac{b (a+b x)^{2/3}}{3 a x}-\frac{b^2 \int \frac{1}{x \sqrt [3]{a+b x}} \, dx}{9 a}\\ &=-\frac{(a+b x)^{2/3}}{2 x^2}-\frac{b (a+b x)^{2/3}}{3 a x}+\frac{b^2 \log (x)}{18 a^{4/3}}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x}\right )}{6 a^{4/3}}-\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x}\right )}{6 a}\\ &=-\frac{(a+b x)^{2/3}}{2 x^2}-\frac{b (a+b x)^{2/3}}{3 a x}+\frac{b^2 \log (x)}{18 a^{4/3}}-\frac{b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{6 a^{4/3}}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}\right )}{3 a^{4/3}}\\ &=-\frac{(a+b x)^{2/3}}{2 x^2}-\frac{b (a+b x)^{2/3}}{3 a x}-\frac{b^2 \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{3 \sqrt{3} a^{4/3}}+\frac{b^2 \log (x)}{18 a^{4/3}}-\frac{b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{6 a^{4/3}}\\ \end{align*}
Mathematica [C] time = 0.0135362, size = 35, normalized size = 0.28 \[ -\frac{3 b^2 (a+b x)^{5/3} \, _2F_1\left (\frac{5}{3},3;\frac{8}{3};\frac{b x}{a}+1\right )}{5 a^3} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*x)^(2/3)/x^3,x]
[Out]
(-3*b^2*(a + b*x)^(5/3)*Hypergeometric2F1[5/3, 3, 8/3, 1 + (b*x)/a])/(5*a^3)
________________________________________________________________________________________
Maple [A] time = 0.011, size = 113, normalized size = 0.9 \begin{align*} -{\frac{1}{3\,a{x}^{2}} \left ( bx+a \right ) ^{{\frac{5}{3}}}}-{\frac{1}{6\,{x}^{2}} \left ( bx+a \right ) ^{{\frac{2}{3}}}}-{\frac{{b}^{2}}{9}\ln \left ( \sqrt [3]{bx+a}-\sqrt [3]{a} \right ){a}^{-{\frac{4}{3}}}}+{\frac{{b}^{2}}{18}\ln \left ( \left ( bx+a \right ) ^{{\frac{2}{3}}}+\sqrt [3]{a}\sqrt [3]{bx+a}+{a}^{{\frac{2}{3}}} \right ){a}^{-{\frac{4}{3}}}}-{\frac{{b}^{2}\sqrt{3}}{9}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{\sqrt [3]{bx+a}}{\sqrt [3]{a}}}+1 \right ) } \right ){a}^{-{\frac{4}{3}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x+a)^(2/3)/x^3,x)
[Out]
-1/3/x^2/a*(b*x+a)^(5/3)-1/6*(b*x+a)^(2/3)/x^2-1/9*b^2/a^(4/3)*ln((b*x+a)^(1/3)-a^(1/3))+1/18*b^2/a^(4/3)*ln((
b*x+a)^(2/3)+a^(1/3)*(b*x+a)^(1/3)+a^(2/3))-1/9*b^2/a^(4/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/a^(1/3)*(b*x+a)^(1/3
)+1))
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(2/3)/x^3,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.58869, size = 950, normalized size = 7.48 \begin{align*} \left [\frac{3 \, \sqrt{\frac{1}{3}} a b^{2} x^{2} \sqrt{\frac{\left (-a\right )^{\frac{1}{3}}}{a}} \log \left (\frac{2 \, b x - 3 \, \sqrt{\frac{1}{3}}{\left (2 \,{\left (b x + a\right )}^{\frac{2}{3}} \left (-a\right )^{\frac{2}{3}} -{\left (b x + a\right )}^{\frac{1}{3}} a + \left (-a\right )^{\frac{1}{3}} a\right )} \sqrt{\frac{\left (-a\right )^{\frac{1}{3}}}{a}} - 3 \,{\left (b x + a\right )}^{\frac{1}{3}} \left (-a\right )^{\frac{2}{3}} + 3 \, a}{x}\right ) + \left (-a\right )^{\frac{2}{3}} b^{2} x^{2} \log \left ({\left (b x + a\right )}^{\frac{2}{3}} -{\left (b x + a\right )}^{\frac{1}{3}} \left (-a\right )^{\frac{1}{3}} + \left (-a\right )^{\frac{2}{3}}\right ) - 2 \, \left (-a\right )^{\frac{2}{3}} b^{2} x^{2} \log \left ({\left (b x + a\right )}^{\frac{1}{3}} + \left (-a\right )^{\frac{1}{3}}\right ) - 3 \,{\left (2 \, a b x + 3 \, a^{2}\right )}{\left (b x + a\right )}^{\frac{2}{3}}}{18 \, a^{2} x^{2}}, -\frac{6 \, \sqrt{\frac{1}{3}} a b^{2} x^{2} \sqrt{-\frac{\left (-a\right )^{\frac{1}{3}}}{a}} \arctan \left (\sqrt{\frac{1}{3}}{\left (2 \,{\left (b x + a\right )}^{\frac{1}{3}} - \left (-a\right )^{\frac{1}{3}}\right )} \sqrt{-\frac{\left (-a\right )^{\frac{1}{3}}}{a}}\right ) - \left (-a\right )^{\frac{2}{3}} b^{2} x^{2} \log \left ({\left (b x + a\right )}^{\frac{2}{3}} -{\left (b x + a\right )}^{\frac{1}{3}} \left (-a\right )^{\frac{1}{3}} + \left (-a\right )^{\frac{2}{3}}\right ) + 2 \, \left (-a\right )^{\frac{2}{3}} b^{2} x^{2} \log \left ({\left (b x + a\right )}^{\frac{1}{3}} + \left (-a\right )^{\frac{1}{3}}\right ) + 3 \,{\left (2 \, a b x + 3 \, a^{2}\right )}{\left (b x + a\right )}^{\frac{2}{3}}}{18 \, a^{2} x^{2}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(2/3)/x^3,x, algorithm="fricas")
[Out]
[1/18*(3*sqrt(1/3)*a*b^2*x^2*sqrt((-a)^(1/3)/a)*log((2*b*x - 3*sqrt(1/3)*(2*(b*x + a)^(2/3)*(-a)^(2/3) - (b*x
+ a)^(1/3)*a + (-a)^(1/3)*a)*sqrt((-a)^(1/3)/a) - 3*(b*x + a)^(1/3)*(-a)^(2/3) + 3*a)/x) + (-a)^(2/3)*b^2*x^2*
log((b*x + a)^(2/3) - (b*x + a)^(1/3)*(-a)^(1/3) + (-a)^(2/3)) - 2*(-a)^(2/3)*b^2*x^2*log((b*x + a)^(1/3) + (-
a)^(1/3)) - 3*(2*a*b*x + 3*a^2)*(b*x + a)^(2/3))/(a^2*x^2), -1/18*(6*sqrt(1/3)*a*b^2*x^2*sqrt(-(-a)^(1/3)/a)*a
rctan(sqrt(1/3)*(2*(b*x + a)^(1/3) - (-a)^(1/3))*sqrt(-(-a)^(1/3)/a)) - (-a)^(2/3)*b^2*x^2*log((b*x + a)^(2/3)
- (b*x + a)^(1/3)*(-a)^(1/3) + (-a)^(2/3)) + 2*(-a)^(2/3)*b^2*x^2*log((b*x + a)^(1/3) + (-a)^(1/3)) + 3*(2*a*
b*x + 3*a^2)*(b*x + a)^(2/3))/(a^2*x^2)]
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Sympy [C] time = 3.93751, size = 2266, normalized size = 17.84 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)**(2/3)/x**3,x)
[Out]
-10*a**(17/3)*b**2*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(5/3)/(54*a**7*exp(2*I*pi/3)
*gamma(8/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(8
/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(8/3)) - 10*a**(17/3)*b**2*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)/(54*a**7*exp(2*I*pi/3)*gamma(8/3) - 162*a**6*b*(a/b
+ x)*exp(2*I*pi/3)*gamma(8/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(8/3) - 54*a**4*b**3*(a/b + x)**
3*exp(2*I*pi/3)*gamma(8/3)) - 10*a**(17/3)*b**2*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3)
)*gamma(5/3)/(54*a**7*exp(2*I*pi/3)*gamma(8/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3) + 162*a**5*b**2
*(a/b + x)**2*exp(2*I*pi/3)*gamma(8/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(8/3)) + 30*a**(14/3)*b*
*3*(a/b + x)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(5/3)/(54*a**7*exp(2*I*pi/3)*gamma
(8/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(8/3) -
54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(8/3)) + 30*a**(14/3)*b**3*(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)/(54*a**7*exp(2*I*pi/3)*gamma(8/3) - 162*a**6*b*(
a/b + x)*exp(2*I*pi/3)*gamma(8/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(8/3) - 54*a**4*b**3*(a/b +
x)**3*exp(2*I*pi/3)*gamma(8/3)) + 30*a**(14/3)*b**3*(a/b + x)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*
pi/3)/a**(1/3))*gamma(5/3)/(54*a**7*exp(2*I*pi/3)*gamma(8/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3) +
162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(8/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(8/3)) - 3
0*a**(11/3)*b**4*(a/b + x)**2*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(5/3)/(54*a**7*ex
p(2*I*pi/3)*gamma(8/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi
/3)*gamma(8/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(8/3)) - 30*a**(11/3)*b**4*(a/b + x)**2*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)/(54*a**7*exp(2*I*pi/3)*gamma
(8/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(8/3) -
54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(8/3)) - 30*a**(11/3)*b**4*(a/b + x)**2*log(1 - b**(1/3)*(a/b + x
)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamma(5/3)/(54*a**7*exp(2*I*pi/3)*gamma(8/3) - 162*a**6*b*(a/b + x)*exp
(2*I*pi/3)*gamma(8/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(8/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*
I*pi/3)*gamma(8/3)) + 10*a**(8/3)*b**5*(a/b + x)**3*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*
gamma(5/3)/(54*a**7*exp(2*I*pi/3)*gamma(8/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3) + 162*a**5*b**2*(
a/b + x)**2*exp(2*I*pi/3)*gamma(8/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(8/3)) + 10*a**(8/3)*b**5*
(a/b + x)**3*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)/(54*a**
7*exp(2*I*pi/3)*gamma(8/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*
I*pi/3)*gamma(8/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(8/3)) + 10*a**(8/3)*b**5*(a/b + x)**3*log(1
- b**(1/3)*(a/b + x)**(1/3)*exp_polar(4*I*pi/3)/a**(1/3))*gamma(5/3)/(54*a**7*exp(2*I*pi/3)*gamma(8/3) - 162*
a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(8/3) - 54*a**4*b**3
*(a/b + x)**3*exp(2*I*pi/3)*gamma(8/3)) - 15*a**5*b**(8/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*gamma(5/3)/(54*a**7*
exp(2*I*pi/3)*gamma(8/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*
pi/3)*gamma(8/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(8/3)) - 15*a**4*b**(11/3)*(a/b + x)**(5/3)*ex
p(2*I*pi/3)*gamma(5/3)/(54*a**7*exp(2*I*pi/3)*gamma(8/3) - 162*a**6*b*(a/b + x)*exp(2*I*pi/3)*gamma(8/3) + 162
*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(8/3) - 54*a**4*b**3*(a/b + x)**3*exp(2*I*pi/3)*gamma(8/3)) + 30*a*
*3*b**(14/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(5/3)/(54*a**7*exp(2*I*pi/3)*gamma(8/3) - 162*a**6*b*(a/b + x
)*exp(2*I*pi/3)*gamma(8/3) + 162*a**5*b**2*(a/b + x)**2*exp(2*I*pi/3)*gamma(8/3) - 54*a**4*b**3*(a/b + x)**3*e
xp(2*I*pi/3)*gamma(8/3))
________________________________________________________________________________________
Giac [A] time = 1.79425, size = 174, normalized size = 1.37 \begin{align*} -\frac{\frac{2 \, \sqrt{3} b^{3} \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{4}{3}}} - \frac{b^{3} \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{4}{3}}} + \frac{2 \, b^{3} \log \left ({\left |{\left (b x + a\right )}^{\frac{1}{3}} - a^{\frac{1}{3}} \right |}\right )}{a^{\frac{4}{3}}} + \frac{3 \,{\left (2 \,{\left (b x + a\right )}^{\frac{5}{3}} b^{3} +{\left (b x + a\right )}^{\frac{2}{3}} a b^{3}\right )}}{a b^{2} x^{2}}}{18 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(2/3)/x^3,x, algorithm="giac")
[Out]
-1/18*(2*sqrt(3)*b^3*arctan(1/3*sqrt(3)*(2*(b*x + a)^(1/3) + a^(1/3))/a^(1/3))/a^(4/3) - b^3*log((b*x + a)^(2/
3) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3))/a^(4/3) + 2*b^3*log(abs((b*x + a)^(1/3) - a^(1/3)))/a^(4/3) + 3*(2*(b*
x + a)^(5/3)*b^3 + (b*x + a)^(2/3)*a*b^3)/(a*b^2*x^2))/b