3.2012 \(\int \frac {(a+\frac {b}{x^3})^{3/2}}{x} \, dx\)

Optimal. Leaf size=59 \[ \frac {2}{3} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )-\frac {2}{3} a \sqrt {a+\frac {b}{x^3}}-\frac {2}{9} \left (a+\frac {b}{x^3}\right )^{3/2} \]

[Out]

-2/9*(a+b/x^3)^(3/2)+2/3*a^(3/2)*arctanh((a+b/x^3)^(1/2)/a^(1/2))-2/3*a*(a+b/x^3)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 50, 63, 208} \[ \frac {2}{3} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )-\frac {2}{3} a \sqrt {a+\frac {b}{x^3}}-\frac {2}{9} \left (a+\frac {b}{x^3}\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^3)^(3/2)/x,x]

[Out]

(-2*a*Sqrt[a + b/x^3])/3 - (2*(a + b/x^3)^(3/2))/9 + (2*a^(3/2)*ArcTanh[Sqrt[a + b/x^3]/Sqrt[a]])/3

Rule 50

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] + Dist[(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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\left (a+\frac {b}{x^3}\right )^{3/2}}{x} \, dx &=-\left (\frac {1}{3} \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,\frac {1}{x^3}\right )\right )\\ &=-\frac {2}{9} \left (a+\frac {b}{x^3}\right )^{3/2}-\frac {1}{3} a \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\frac {1}{x^3}\right )\\ &=-\frac {2}{3} a \sqrt {a+\frac {b}{x^3}}-\frac {2}{9} \left (a+\frac {b}{x^3}\right )^{3/2}-\frac {1}{3} a^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x^3}\right )\\ &=-\frac {2}{3} a \sqrt {a+\frac {b}{x^3}}-\frac {2}{9} \left (a+\frac {b}{x^3}\right )^{3/2}-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x^3}}\right )}{3 b}\\ &=-\frac {2}{3} a \sqrt {a+\frac {b}{x^3}}-\frac {2}{9} \left (a+\frac {b}{x^3}\right )^{3/2}+\frac {2}{3} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.88 \[ -\frac {2 b \sqrt {a+\frac {b}{x^3}} \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};-\frac {a x^3}{b}\right )}{9 x^3 \sqrt {\frac {a x^3}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^3)^(3/2)/x,x]

[Out]

(-2*b*Sqrt[a + b/x^3]*Hypergeometric2F1[-3/2, -3/2, -1/2, -((a*x^3)/b)])/(9*x^3*Sqrt[1 + (a*x^3)/b])

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fricas [A]  time = 1.08, size = 164, normalized size = 2.78 \[ \left [\frac {3 \, a^{\frac {3}{2}} x^{3} \log \left (-8 \, a^{2} x^{6} - 8 \, a b x^{3} - b^{2} - 4 \, {\left (2 \, a x^{6} + b x^{3}\right )} \sqrt {a} \sqrt {\frac {a x^{3} + b}{x^{3}}}\right ) - 4 \, {\left (4 \, a x^{3} + b\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{18 \, x^{3}}, -\frac {3 \, \sqrt {-a} a x^{3} \arctan \left (\frac {2 \, \sqrt {-a} x^{3} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{2 \, a x^{3} + b}\right ) + 2 \, {\left (4 \, a x^{3} + b\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{9 \, x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/18*(3*a^(3/2)*x^3*log(-8*a^2*x^6 - 8*a*b*x^3 - b^2 - 4*(2*a*x^6 + b*x^3)*sqrt(a)*sqrt((a*x^3 + b)/x^3)) - 4
*(4*a*x^3 + b)*sqrt((a*x^3 + b)/x^3))/x^3, -1/9*(3*sqrt(-a)*a*x^3*arctan(2*sqrt(-a)*x^3*sqrt((a*x^3 + b)/x^3)/
(2*a*x^3 + b)) + 2*(4*a*x^3 + b)*sqrt((a*x^3 + b)/x^3))/x^3]

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giac [A]  time = 0.28, size = 50, normalized size = 0.85 \[ -\frac {2 \, a^{2} \arctan \left (\frac {\sqrt {a + \frac {b}{x^{3}}}}{\sqrt {-a}}\right )}{3 \, \sqrt {-a}} - \frac {2}{9} \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} - \frac {2}{3} \, \sqrt {a + \frac {b}{x^{3}}} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*a^2*arctan(sqrt(a + b/x^3)/sqrt(-a))/sqrt(-a) - 2/9*(a + b/x^3)^(3/2) - 2/3*sqrt(a + b/x^3)*a

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maple [C]  time = 0.05, size = 3535, normalized size = 59.92 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*((a*x^3+b)/x^3)^(3/2)*(-36*I*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(
1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/
3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-
a^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(1/3)*3^(1/
2)*x^6*a-18*I*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+
(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(
I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^
(1/2),(I*3^(1/2)-1)/(I*3^(1/2)-3),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(2
/3)*3^(1/2)*x^5+18*I*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)
^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(
1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))
*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(2/3)*3^(1/2)*x^5+36*I*(
-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))
/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-
a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),(I*3^(1/2
)-1)/(I*3^(1/2)-3),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(1/3)*3^(1/2)*x^6
*a+18*I*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*
b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1
/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),(
(I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*3^(1/2)*x^7*a^2-18*(-(I*3^(1/2)-3)/(I*3^(1/2)-
1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2
*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)
*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^
(1/2))/(I*3^(1/2)-3))^(1/2))*x^7*a^2+4*I*(a*x^4+b*x)^(1/2)*((-a*x+(-a^2*b)^(1/3))*(2*a*x+I*3^(1/2)*(-a^2*b)^(1
/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)*3^(1/2)*x^3*a+18*(-(I*3^(1/2
)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1
/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*
b)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),(I*3^(1/2)-1)/(I*3^
(1/2)-3),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x^7*a^2+36*(-(I*3^(1/2)-3)/(I*3^(1/2
)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a
^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/
2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*
3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(1/3)*x^6*a-36*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x
)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I
*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)/
(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),(I*3^(1/2)-1)/(I*3^(1/2)-3),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3
^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(1/3)*x^6*a-18*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)
^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*
3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I
*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*
(-a^2*b)^(2/3)*x^5+18*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b
)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^
(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3
))*a*x)^(1/2),(I*3^(1/2)-1)/(I*3^(1/2)-3),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a
^2*b)^(2/3)*x^5-18*I*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)
^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(
1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)
)*a*x)^(1/2),(I*3^(1/2)-1)/(I*3^(1/2)-3),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*3^(1
/2)*x^7*a^2-12*(a*x^4+b*x)^(1/2)*((-a*x+(-a^2*b)^(1/3))*(2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*
x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)*x^3*a+I*(a*x^4+b*x)^(1/2)*((-a*x+(-a^2*b)^(1/3))*(2*a*
x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)*3^(1/
2)*b-3*(a*x^4+b*x)^(1/2)*((-a*x+(-a^2*b)^(1/3))*(2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1
/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)*b)/(a*x^3+b)/((a*x^3+b)*x)^(1/2)/(I*3^(1/2)-3)/((-a*x+(-a^2*b)
^(1/3))*(2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x
)^(1/2)

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maxima [A]  time = 1.97, size = 61, normalized size = 1.03 \[ -\frac {1}{3} \, a^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {b}{x^{3}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{3}}} + \sqrt {a}}\right ) - \frac {2}{9} \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} - \frac {2}{3} \, \sqrt {a + \frac {b}{x^{3}}} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^(3/2)*log((sqrt(a + b/x^3) - sqrt(a))/(sqrt(a + b/x^3) + sqrt(a))) - 2/9*(a + b/x^3)^(3/2) - 2/3*sqrt(a
+ b/x^3)*a

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mupad [B]  time = 1.62, size = 43, normalized size = 0.73 \[ \frac {2\,a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{3}-\frac {2\,a\,\sqrt {a+\frac {b}{x^3}}}{3}-\frac {2\,{\left (a+\frac {b}{x^3}\right )}^{3/2}}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^3)^(3/2)/x,x)

[Out]

(2*a^(3/2)*atanh((a + b/x^3)^(1/2)/a^(1/2)))/3 - (2*a*(a + b/x^3)^(1/2))/3 - (2*(a + b/x^3)^(3/2))/9

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sympy [A]  time = 2.38, size = 83, normalized size = 1.41 \[ - \frac {8 a^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x^{3}}}}{9} - \frac {a^{\frac {3}{2}} \log {\left (\frac {b}{a x^{3}} \right )}}{3} + \frac {2 a^{\frac {3}{2}} \log {\left (\sqrt {1 + \frac {b}{a x^{3}}} + 1 \right )}}{3} - \frac {2 \sqrt {a} b \sqrt {1 + \frac {b}{a x^{3}}}}{9 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*a**(3/2)*sqrt(1 + b/(a*x**3))/9 - a**(3/2)*log(b/(a*x**3))/3 + 2*a**(3/2)*log(sqrt(1 + b/(a*x**3)) + 1)/3 -
2*sqrt(a)*b*sqrt(1 + b/(a*x**3))/(9*x**3)

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