\(\int \frac {(a+b x^2)^{2/3}}{x^5} \, dx\) [732]

Optimal result

Integrand size = 15, antiderivative size = 135 \[ \int \frac {\left (a+b x^2\right )^{2/3}}{x^5} \, dx=-\frac {\left (a+b x^2\right )^{2/3}}{4 x^4}-\frac {b \left (a+b x^2\right )^{2/3}}{6 a x^2}-\frac {b^2 \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x^2}}{\sqrt {3} \sqrt [3]{a}}\right )}{6 \sqrt {3} a^{4/3}}+\frac {b^2 \log (x)}{18 a^{4/3}}-\frac {b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{12 a^{4/3}} \] Output:

-1/4*(b*x^2+a)^(2/3)/x^4-1/6*b*(b*x^2+a)^(2/3)/a/x^2-1/18*b^2*arctan(1/3*( 
a^(1/3)+2*(b*x^2+a)^(1/3))*3^(1/2)/a^(1/3))*3^(1/2)/a^(4/3)+1/18*b^2*ln(x) 
/a^(4/3)-1/12*b^2*ln(a^(1/3)-(b*x^2+a)^(1/3))/a^(4/3)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a+b x^2\right )^{2/3}}{x^5} \, dx=\frac {\left (-3 a-2 b x^2\right ) \left (a+b x^2\right )^{2/3}}{12 a x^4}-\frac {b^2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt {3} \sqrt [3]{a}}\right )}{6 \sqrt {3} a^{4/3}}-\frac {b^2 \log \left (-\sqrt [3]{a}+\sqrt [3]{a+b x^2}\right )}{18 a^{4/3}}+\frac {b^2 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )}{36 a^{4/3}} \] Input:

Integrate[(a + b*x^2)^(2/3)/x^5,x]
 

Output:

((-3*a - 2*b*x^2)*(a + b*x^2)^(2/3))/(12*a*x^4) - (b^2*ArcTan[1/Sqrt[3] + 
(2*(a + b*x^2)^(1/3))/(Sqrt[3]*a^(1/3))])/(6*Sqrt[3]*a^(4/3)) - (b^2*Log[- 
a^(1/3) + (a + b*x^2)^(1/3)])/(18*a^(4/3)) + (b^2*Log[a^(2/3) + a^(1/3)*(a 
 + b*x^2)^(1/3) + (a + b*x^2)^(2/3)])/(36*a^(4/3))
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {243, 51, 52, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{2/3}}{x^5} \, dx\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{2/3}}{x^6}dx^2\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} b \int \frac {1}{x^4 \sqrt [3]{b x^2+a}}dx^2-\frac {\left (a+b x^2\right )^{2/3}}{2 x^4}\right )\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} b \left (-\frac {b \int \frac {1}{x^2 \sqrt [3]{b x^2+a}}dx^2}{3 a}-\frac {\left (a+b x^2\right )^{2/3}}{a x^2}\right )-\frac {\left (a+b x^2\right )^{2/3}}{2 x^4}\right )\)

\(\Big \downarrow \) 67

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} b \left (-\frac {b \left (\frac {3}{2} \int \frac {1}{x^4+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}-\frac {3 \int \frac {1}{\sqrt [3]{a}-\sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}}{2 \sqrt [3]{a}}-\frac {\log \left (x^2\right )}{2 \sqrt [3]{a}}\right )}{3 a}-\frac {\left (a+b x^2\right )^{2/3}}{a x^2}\right )-\frac {\left (a+b x^2\right )^{2/3}}{2 x^4}\right )\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} b \left (-\frac {b \left (\frac {3}{2} \int \frac {1}{x^4+a^{2/3}+\sqrt [3]{a} \sqrt [3]{b x^2+a}}d\sqrt [3]{b x^2+a}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 \sqrt [3]{a}}-\frac {\log \left (x^2\right )}{2 \sqrt [3]{a}}\right )}{3 a}-\frac {\left (a+b x^2\right )^{2/3}}{a x^2}\right )-\frac {\left (a+b x^2\right )^{2/3}}{2 x^4}\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} b \left (-\frac {b \left (-\frac {3 \int \frac {1}{-x^4-3}d\left (\frac {2 \sqrt [3]{b x^2+a}}{\sqrt [3]{a}}+1\right )}{\sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 \sqrt [3]{a}}-\frac {\log \left (x^2\right )}{2 \sqrt [3]{a}}\right )}{3 a}-\frac {\left (a+b x^2\right )^{2/3}}{a x^2}\right )-\frac {\left (a+b x^2\right )^{2/3}}{2 x^4}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} b \left (-\frac {b \left (\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{\sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{2 \sqrt [3]{a}}-\frac {\log \left (x^2\right )}{2 \sqrt [3]{a}}\right )}{3 a}-\frac {\left (a+b x^2\right )^{2/3}}{a x^2}\right )-\frac {\left (a+b x^2\right )^{2/3}}{2 x^4}\right )\)

Input:

Int[(a + b*x^2)^(2/3)/x^5,x]
 

Output:

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

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 52

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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[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 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

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]
 

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00

method	result	size
pseudoelliptic	\(\frac {-2 b^{2} \sqrt {3}\, \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b \,x^{2}+a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) x^{4}-2 b^{2} \ln \left (\left (b \,x^{2}+a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right ) x^{4}+b^{2} \ln \left (a^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b \,x^{2}+a \right )^{\frac {1}{3}}+\left (b \,x^{2}+a \right )^{\frac {2}{3}}\right ) x^{4}-6 b \,x^{2} \left (b \,x^{2}+a \right )^{\frac {2}{3}} a^{\frac {1}{3}}-9 \left (b \,x^{2}+a \right )^{\frac {2}{3}} a^{\frac {4}{3}}}{36 a^{\frac {4}{3}} x^{4}}\)	\(135\)

Input:

int((b*x^2+a)^(2/3)/x^5,x,method=_RETURNVERBOSE)
 

Output:

1/36*(-2*b^2*3^(1/2)*arctan(1/3*(a^(1/3)+2*(b*x^2+a)^(1/3))*3^(1/2)/a^(1/3 
))*x^4-2*b^2*ln((b*x^2+a)^(1/3)-a^(1/3))*x^4+b^2*ln(a^(2/3)+a^(1/3)*(b*x^2 
+a)^(1/3)+(b*x^2+a)^(2/3))*x^4-6*b*x^2*(b*x^2+a)^(2/3)*a^(1/3)-9*(b*x^2+a) 
^(2/3)*a^(4/3))/a^(4/3)/x^4
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.41 \[ \int \frac {\left (a+b x^2\right )^{2/3}}{x^5} \, dx=\left [\frac {3 \, \sqrt {\frac {1}{3}} a b^{2} x^{4} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} \log \left (\frac {2 \, b x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {2}{3}} a^{\frac {2}{3}} - {\left (b x^{2} + a\right )}^{\frac {1}{3}} a - a^{\frac {4}{3}}\right )} \sqrt {-\frac {1}{a^{\frac {2}{3}}}} - 3 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + 3 \, a}{x^{2}}\right ) + a^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) - 2 \, a^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) - 3 \, {\left (2 \, a b x^{2} + 3 \, a^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {2}{3}}}{36 \, a^{2} x^{4}}, -\frac {6 \, \sqrt {\frac {1}{3}} a^{\frac {2}{3}} b^{2} x^{4} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{a^{\frac {1}{3}}}\right ) - a^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + 2 \, a^{\frac {2}{3}} b^{2} x^{4} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + 3 \, {\left (2 \, a b x^{2} + 3 \, a^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {2}{3}}}{36 \, a^{2} x^{4}}\right ] \] Input:

integrate((b*x^2+a)^(2/3)/x^5,x, algorithm="fricas")
 

Output:

[1/36*(3*sqrt(1/3)*a*b^2*x^4*sqrt(-1/a^(2/3))*log((2*b*x^2 - 3*sqrt(1/3)*( 
2*(b*x^2 + a)^(2/3)*a^(2/3) - (b*x^2 + a)^(1/3)*a - a^(4/3))*sqrt(-1/a^(2/ 
3)) - 3*(b*x^2 + a)^(1/3)*a^(2/3) + 3*a)/x^2) + a^(2/3)*b^2*x^4*log((b*x^2 
 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3)) - 2*a^(2/3)*b^2*x^4*log 
((b*x^2 + a)^(1/3) - a^(1/3)) - 3*(2*a*b*x^2 + 3*a^2)*(b*x^2 + a)^(2/3))/( 
a^2*x^4), -1/36*(6*sqrt(1/3)*a^(2/3)*b^2*x^4*arctan(sqrt(1/3)*(2*(b*x^2 + 
a)^(1/3) + a^(1/3))/a^(1/3)) - a^(2/3)*b^2*x^4*log((b*x^2 + a)^(2/3) + (b* 
x^2 + a)^(1/3)*a^(1/3) + a^(2/3)) + 2*a^(2/3)*b^2*x^4*log((b*x^2 + a)^(1/3 
) - a^(1/3)) + 3*(2*a*b*x^2 + 3*a^2)*(b*x^2 + a)^(2/3))/(a^2*x^4)]
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.31 \[ \int \frac {\left (a+b x^2\right )^{2/3}}{x^5} \, dx=- \frac {b^{\frac {2}{3}} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 x^{\frac {8}{3}} \Gamma \left (\frac {7}{3}\right )} \] Input:

integrate((b*x**2+a)**(2/3)/x**5,x)
 

Output:

-b**(2/3)*gamma(4/3)*hyper((-2/3, 4/3), (7/3,), a*exp_polar(I*pi)/(b*x**2) 
)/(2*x**(8/3)*gamma(7/3))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x^2\right )^{2/3}}{x^5} \, dx=-\frac {\sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{18 \, a^{\frac {4}{3}}} + \frac {b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{36 \, a^{\frac {4}{3}}} - \frac {b^{2} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{18 \, a^{\frac {4}{3}}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {5}{3}} b^{2} + {\left (b x^{2} + a\right )}^{\frac {2}{3}} a b^{2}}{12 \, {\left ({\left (b x^{2} + a\right )}^{2} a - 2 \, {\left (b x^{2} + a\right )} a^{2} + a^{3}\right )}} \] Input:

integrate((b*x^2+a)^(2/3)/x^5,x, algorithm="maxima")
 

Output:

-1/18*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/ 
3))/a^(4/3) + 1/36*b^2*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + 
 a^(2/3))/a^(4/3) - 1/18*b^2*log((b*x^2 + a)^(1/3) - a^(1/3))/a^(4/3) - 1/ 
12*(2*(b*x^2 + a)^(5/3)*b^2 + (b*x^2 + a)^(2/3)*a*b^2)/((b*x^2 + a)^2*a - 
2*(b*x^2 + a)*a^2 + a^3)
 

Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^{2/3}}{x^5} \, dx=-\frac {\frac {2 \, \sqrt {3} b^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {4}{3}}} - \frac {b^{3} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {4}{3}}} + \frac {2 \, b^{3} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {4}{3}}} + \frac {3 \, {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {5}{3}} b^{3} + {\left (b x^{2} + a\right )}^{\frac {2}{3}} a b^{3}\right )}}{a b^{2} x^{4}}}{36 \, b} \] Input:

integrate((b*x^2+a)^(2/3)/x^5,x, algorithm="giac")
 

Output:

-1/36*(2*sqrt(3)*b^3*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^ 
(1/3))/a^(4/3) - b^3*log((b*x^2 + a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a 
^(2/3))/a^(4/3) + 2*b^3*log(abs((b*x^2 + a)^(1/3) - a^(1/3)))/a^(4/3) + 3* 
(2*(b*x^2 + a)^(5/3)*b^3 + (b*x^2 + a)^(2/3)*a*b^3)/(a*b^2*x^4))/b
 

Mupad [B] (verification not implemented)

Time = 0.79 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.57 \[ \int \frac {\left (a+b x^2\right )^{2/3}}{x^5} \, dx=\frac {{\left (-1\right )}^{1/3}\,b^2\,\ln \left ({\left (b\,x^2+a\right )}^{1/3}-{\left (-1\right )}^{2/3}\,a^{1/3}\right )}{18\,a^{4/3}}-\frac {\frac {b^2\,{\left (b\,x^2+a\right )}^{2/3}}{6}+\frac {b^2\,{\left (b\,x^2+a\right )}^{5/3}}{3\,a}}{2\,{\left (b\,x^2+a\right )}^2-4\,a\,\left (b\,x^2+a\right )+2\,a^2}+\frac {{\left (-1\right )}^{1/3}\,b^2\,\ln \left (\frac {b^4\,{\left (b\,x^2+a\right )}^{1/3}}{36\,a^2}-\frac {{\left (-1\right )}^{2/3}\,b^4\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{36\,a^{5/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{18\,a^{4/3}}-\frac {{\left (-1\right )}^{1/3}\,b^2\,\ln \left (\frac {b^4\,{\left (b\,x^2+a\right )}^{1/3}}{36\,a^2}-\frac {{\left (-1\right )}^{2/3}\,b^4\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{36\,a^{5/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{18\,a^{4/3}} \] Input:

int((a + b*x^2)^(2/3)/x^5,x)
 

Output:

((-1)^(1/3)*b^2*log((a + b*x^2)^(1/3) - (-1)^(2/3)*a^(1/3)))/(18*a^(4/3)) 
- ((b^2*(a + b*x^2)^(2/3))/6 + (b^2*(a + b*x^2)^(5/3))/(3*a))/(2*(a + b*x^ 
2)^2 - 4*a*(a + b*x^2) + 2*a^2) + ((-1)^(1/3)*b^2*log((b^4*(a + b*x^2)^(1/ 
3))/(36*a^2) - ((-1)^(2/3)*b^4*((3^(1/2)*1i)/2 - 1/2)^2)/(36*a^(5/3)))*((3 
^(1/2)*1i)/2 - 1/2))/(18*a^(4/3)) - ((-1)^(1/3)*b^2*log((b^4*(a + b*x^2)^( 
1/3))/(36*a^2) - ((-1)^(2/3)*b^4*((3^(1/2)*1i)/2 + 1/2)^2)/(36*a^(5/3)))*( 
(3^(1/2)*1i)/2 + 1/2))/(18*a^(4/3))
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{2/3}}{x^5} \, dx=\frac {-9 \left (b \,x^{2}+a \right )^{\frac {2}{3}} a -6 \left (b \,x^{2}+a \right )^{\frac {2}{3}} b \,x^{2}-4 \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {2}{3}}}{b \,x^{3}+a x}d x \right ) b^{2} x^{4}}{36 a \,x^{4}} \] Input:

int((b*x^2+a)^(2/3)/x^5,x)
 

Output:

( - 9*(a + b*x**2)**(2/3)*a - 6*(a + b*x**2)**(2/3)*b*x**2 - 4*int((a + b* 
x**2)**(2/3)/(a*x + b*x**3),x)*b**2*x**4)/(36*a*x**4)