3.20.91 \(\int \sqrt {a+\frac {b}{x^3}} x^5 \, dx\) [1991]
Optimal. Leaf size=71 \[ \frac {b \sqrt {a+\frac {b}{x^3}} x^3}{12 a}+\frac {1}{6} \sqrt {a+\frac {b}{x^3}} x^6-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{12 a^{3/2}} \]
[Out]
-1/12*b^2*arctanh((a+b/x^3)^(1/2)/a^(1/2))/a^(3/2)+1/12*b*x^3*(a+b/x^3)^(1/2)/a+1/6*x^6*(a+b/x^3)^(1/2)
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Rubi [A]
time = 0.03, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 43, 44, 65,
214} \begin {gather*} -\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{12 a^{3/2}}+\frac {b x^3 \sqrt {a+\frac {b}{x^3}}}{12 a}+\frac {1}{6} x^6 \sqrt {a+\frac {b}{x^3}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + b/x^3]*x^5,x]
[Out]
(b*Sqrt[a + b/x^3]*x^3)/(12*a) + (Sqrt[a + b/x^3]*x^6)/6 - (b^2*ArcTanh[Sqrt[a + b/x^3]/Sqrt[a]])/(12*a^(3/2))
Rule 43
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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] && !IntegerQ[n] && GtQ[n, 0]
Rule 44
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] && ILtQ[m, -1] && !IntegerQ[n] && LtQ[n, 0]
Rule 65
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 214
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]
Rule 272
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 \sqrt {a+\frac {b}{x^3}} x^5 \, dx &=-\left (\frac {1}{3} \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x^3} \, dx,x,\frac {1}{x^3}\right )\right )\\ &=\frac {1}{6} \sqrt {a+\frac {b}{x^3}} x^6-\frac {1}{12} b \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{x^3}\right )\\ &=\frac {b \sqrt {a+\frac {b}{x^3}} x^3}{12 a}+\frac {1}{6} \sqrt {a+\frac {b}{x^3}} x^6+\frac {b^2 \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x^3}\right )}{24 a}\\ &=\frac {b \sqrt {a+\frac {b}{x^3}} x^3}{12 a}+\frac {1}{6} \sqrt {a+\frac {b}{x^3}} x^6+\frac {b \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x^3}}\right )}{12 a}\\ &=\frac {b \sqrt {a+\frac {b}{x^3}} x^3}{12 a}+\frac {1}{6} \sqrt {a+\frac {b}{x^3}} x^6-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{12 a^{3/2}}\\ \end {align*}
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Mathematica [A]
time = 0.16, size = 92, normalized size = 1.30 \begin {gather*} \frac {\sqrt {a+\frac {b}{x^3}} x^3 \left (b+2 a x^3\right )}{12 a}-\frac {b^2 \sqrt {a+\frac {b}{x^3}} x^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b+a x^3}}{\sqrt {a} x^{3/2}}\right )}{12 a^{3/2} \sqrt {b+a x^3}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + b/x^3]*x^5,x]
[Out]
(Sqrt[a + b/x^3]*x^3*(b + 2*a*x^3))/(12*a) - (b^2*Sqrt[a + b/x^3]*x^(3/2)*ArcTanh[Sqrt[b + a*x^3]/(Sqrt[a]*x^(
3/2))])/(12*a^(3/2)*Sqrt[b + a*x^3])
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 1.38, size = 3560, normalized size = 50.14
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method |
result |
size |
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risch |
\(\text {Expression too large to display}\) |
\(1045\) |
default |
\(\text {Expression too large to display}\) |
\(3560\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^5*(a+b/x^3)^(1/2),x,method=_RETURNVERBOSE)
[Out]
1/12*((a*x^3+b)/x^3)^(1/2)*x^2/a^3*(-6*I*3^(1/2)*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/
2)*((I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a
^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)*x*a/(
-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2),(-1+I*3^(1/2))/(I*3^(1/2)-3),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(
1/2))/(I*3^(1/2)-3))^(1/2))*a^2*b^2*x^2+I*3^(1/2)*(a*x^4+b*x)^(1/2)*(1/a^2*x*(-a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*
(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3)))^(1/2)*a^2*b*x+2*I*3^(1/2
)*(a*x^4+b*x)^(1/2)*(1/a^2*x*(-a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*
(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3)))^(1/2)*a^3*x^4+6*I*3^(1/2)*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2
*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*
((I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^
(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2
)-3))^(1/2))*(-a^2*b)^(2/3)*b^2+6*I*3^(1/2)*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((
I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)
^(1/3)-2*a*x-(-a^2*b)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3
^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*a^2*b^2
*x^2+12*I*3^(1/2)*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)+2
*a*x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3
))/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1
/3)))^(1/2),(-1+I*3^(1/2))/(I*3^(1/2)-3),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a
^2*b)^(1/3)*a*b^2*x-12*I*3^(1/2)*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(
-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)-2*a*
x-(-a^2*b)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a
*x+(-a^2*b)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(1/3)*a*b
^2*x-6*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*
b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3))/(-1+I*3^
(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2),
((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*a^2*b^2*x^2+6*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1
/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b
)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*E
llipticPi((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2),(-1+I*3^(1/2))/(I*3^(1/2)-3),((I*3^(
1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*a^2*b^2*x^2-6*x^4*(a*x^4+b*x)^(1/2)*a^3*(1/a^2*x*(-
a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(
1/3)))^(1/2)+12*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)+2*a
*x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3))
/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)
))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(1/3)*a*b^2*x-12*(-(I*3^(1
/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))/(1+I*3
^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-a
^2*b)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2),(-1+I*3^(1/2))/
(I*3^(1/2)-3),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(1/3)*a*b^2*x-6*I*3^(
1/2)*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)
^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3))/(-1+I*3^(1
/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2),(
-1+I*3^(1/2))/(I*3^(1/2)-3),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(2/3)*b
^2-6*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)
^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1...
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Maxima [A]
time = 0.52, size = 100, normalized size = 1.41 \begin {gather*} \frac {b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x^{3}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{3}}} + \sqrt {a}}\right )}{24 \, a^{\frac {3}{2}}} + \frac {{\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} b^{2} + \sqrt {a + \frac {b}{x^{3}}} a b^{2}}{12 \, {\left ({\left (a + \frac {b}{x^{3}}\right )}^{2} a - 2 \, {\left (a + \frac {b}{x^{3}}\right )} a^{2} + a^{3}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^5*(a+b/x^3)^(1/2),x, algorithm="maxima")
[Out]
1/24*b^2*log((sqrt(a + b/x^3) - sqrt(a))/(sqrt(a + b/x^3) + sqrt(a)))/a^(3/2) + 1/12*((a + b/x^3)^(3/2)*b^2 +
sqrt(a + b/x^3)*a*b^2)/((a + b/x^3)^2*a - 2*(a + b/x^3)*a^2 + a^3)
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Fricas [A]
time = 0.48, size = 175, normalized size = 2.46 \begin {gather*} \left [\frac {\sqrt {a} b^{2} \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 (2 \, a^{2} x^{6} + a b x^{3}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{48 \, a^{2}}, \frac {\sqrt {-a} b^{2} \arctan \left (\frac {2 \, \sqrt {-a} x^{3} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{2 \, a x^{3} + b}\right ) + 2 \, {\left (2 \, a^{2} x^{6} + a b x^{3}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{24 \, a^{2}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^5*(a+b/x^3)^(1/2),x, algorithm="fricas")
[Out]
[1/48*(sqrt(a)*b^2*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*(
2*a^2*x^6 + a*b*x^3)*sqrt((a*x^3 + b)/x^3))/a^2, 1/24*(sqrt(-a)*b^2*arctan(2*sqrt(-a)*x^3*sqrt((a*x^3 + b)/x^3
)/(2*a*x^3 + b)) + 2*(2*a^2*x^6 + a*b*x^3)*sqrt((a*x^3 + b)/x^3))/a^2]
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Sympy [A]
time = 2.15, size = 100, normalized size = 1.41 \begin {gather*} \frac {a x^{\frac {15}{2}}}{6 \sqrt {b} \sqrt {\frac {a x^{3}}{b} + 1}} + \frac {\sqrt {b} x^{\frac {9}{2}}}{4 \sqrt {\frac {a x^{3}}{b} + 1}} + \frac {b^{\frac {3}{2}} x^{\frac {3}{2}}}{12 a \sqrt {\frac {a x^{3}}{b} + 1}} - \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} x^{\frac {3}{2}}}{\sqrt {b}} \right )}}{12 a^{\frac {3}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**5*(a+b/x**3)**(1/2),x)
[Out]
a*x**(15/2)/(6*sqrt(b)*sqrt(a*x**3/b + 1)) + sqrt(b)*x**(9/2)/(4*sqrt(a*x**3/b + 1)) + b**(3/2)*x**(3/2)/(12*a
*sqrt(a*x**3/b + 1)) - b**2*asinh(sqrt(a)*x**(3/2)/sqrt(b))/(12*a**(3/2))
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Giac [A]
time = 1.04, size = 55, normalized size = 0.77 \begin {gather*} \frac {1}{12} \, \sqrt {a x^{4} + b x} {\left (2 \, x^{3} + \frac {b}{a}\right )} x + \frac {b^{2} \arctan \left (\frac {\sqrt {a + \frac {b}{x^{3}}}}{\sqrt {-a}}\right )}{12 \, \sqrt {-a} a} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^5*(a+b/x^3)^(1/2),x, algorithm="giac")
[Out]
1/12*sqrt(a*x^4 + b*x)*(2*x^3 + b/a)*x + 1/12*b^2*arctan(sqrt(a + b/x^3)/sqrt(-a))/(sqrt(-a)*a)
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Mupad [B]
time = 1.77, size = 76, normalized size = 1.07 \begin {gather*} \frac {x^6\,\sqrt {a+\frac {b}{x^3}}}{6}+\frac {b^2\,\ln \left (x^6\,{\left (\sqrt {a+\frac {b}{x^3}}-\sqrt {a}\right )}^3\,\left (\sqrt {a+\frac {b}{x^3}}+\sqrt {a}\right )\right )}{24\,a^{3/2}}+\frac {b\,x^3\,\sqrt {a+\frac {b}{x^3}}}{12\,a} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^5*(a + b/x^3)^(1/2),x)
[Out]
(x^6*(a + b/x^3)^(1/2))/6 + (b^2*log(x^6*((a + b/x^3)^(1/2) - a^(1/2))^3*((a + b/x^3)^(1/2) + a^(1/2))))/(24*a
^(3/2)) + (b*x^3*(a + b/x^3)^(1/2))/(12*a)
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