3.2000 \(\int \sqrt {a+\frac {b}{x^3}} x \, dx\)
Optimal. Leaf size=242 \[ \frac {1}{2} x^2 \sqrt {a+\frac {b}{x^3}}-\frac {3^{3/4} \sqrt {2+\sqrt {3}} b^{2/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{2 \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}} \]
[Out]
1/2*x^2*(a+b/x^3)^(1/2)-1/2*3^(3/4)*b^(2/3)*(a^(1/3)+b^(1/3)/x)*EllipticF((b^(1/3)/x+a^(1/3)*(1-3^(1/2)))/(b^(
1/3)/x+a^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*((a^(2/3)+b^(2/3)/x^2-a^(1/3)*b^(1/3)/x)/
(b^(1/3)/x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)/(a+b/x^3)^(1/2)/(a^(1/3)*(a^(1/3)+b^(1/3)/x)/(b^(1/3)/x+a^(1/3)*(1+3^
(1/2)))^2)^(1/2)
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Rubi [A] time = 0.09, antiderivative size = 242, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used =
{335, 277, 218} \[ \frac {1}{2} x^2 \sqrt {a+\frac {b}{x^3}}-\frac {3^{3/4} \sqrt {2+\sqrt {3}} b^{2/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{2 \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + b/x^3]*x,x]
[Out]
(Sqrt[a + b/x^3]*x^2)/2 - (3^(3/4)*Sqrt[2 + Sqrt[3]]*b^(2/3)*(a^(1/3) + b^(1/3)/x)*Sqrt[(a^(2/3) + b^(2/3)/x^2
- (a^(1/3)*b^(1/3))/x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)/x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/
3)/x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)/x)], -7 - 4*Sqrt[3]])/(2*Sqrt[a + b/x^3]*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/
3)/x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)/x)^2])
Rule 218
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]
Rule 277
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Rule 335
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]
Rubi steps
\begin {align*} \int \sqrt {a+\frac {b}{x^3}} x \, dx &=-\operatorname {Subst}\left (\int \frac {\sqrt {a+b x^3}}{x^3} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{2} \sqrt {a+\frac {b}{x^3}} x^2-\frac {1}{4} (3 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{2} \sqrt {a+\frac {b}{x^3}} x^2-\frac {3^{3/4} \sqrt {2+\sqrt {3}} b^{2/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}+\frac {b^{2/3}}{x^2}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{2 \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}\\ \end {align*}
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Mathematica [C] time = 0.01, size = 49, normalized size = 0.20 \[ \frac {2 x^2 \sqrt {a+\frac {b}{x^3}} \, _2F_1\left (-\frac {1}{2},\frac {1}{6};\frac {7}{6};-\frac {a x^3}{b}\right )}{\sqrt {\frac {a x^3}{b}+1}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + b/x^3]*x,x]
[Out]
(2*Sqrt[a + b/x^3]*x^2*Hypergeometric2F1[-1/2, 1/6, 7/6, -((a*x^3)/b)])/Sqrt[1 + (a*x^3)/b]
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fricas [F] time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \sqrt {\frac {a x^{3} + b}{x^{3}}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(a+b/x^3)^(1/2),x, algorithm="fricas")
[Out]
integral(x*sqrt((a*x^3 + b)/x^3), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + \frac {b}{x^{3}}} x\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(a+b/x^3)^(1/2),x, algorithm="giac")
[Out]
integrate(sqrt(a + b/x^3)*x, x)
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maple [B] time = 0.03, size = 1786, normalized size = 7.38 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*(a+b/x^3)^(1/2),x)
[Out]
-1/2*((a*x^3+b)/x^3)^(1/2)*x^2/(-a^2*b)^(1/3)/a*(6*I*3^(1/2)*(-(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^2*a^2*b-12*I*(-a^2*b)^(1/3)*3^(1/2)*(-(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*a*b+6*I
*(-a^2*b)^(2/3)*3^(1/2)*(-(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))*b-6*(-(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^2*a^2*b+12*(-a^2*b)^(1/3)*(-(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*a*b-6*(-a^2*b)^(2/3)*(-(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))*b-I*(-a^2*b)^(1/3)*3^(1/2)*
(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)*a+3*(a*x^4+b*x)^(1/2)*a*(-a^2*b)^(1/3)*((-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))/((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 [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + \frac {b}{x^{3}}} x\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(a+b/x^3)^(1/2),x, algorithm="maxima")
[Out]
integrate(sqrt(a + b/x^3)*x, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int x\,\sqrt {a+\frac {b}{x^3}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*(a + b/x^3)^(1/2),x)
[Out]
int(x*(a + b/x^3)^(1/2), x)
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sympy [A] time = 1.09, size = 44, normalized size = 0.18 \[ - \frac {\sqrt {a} x^{2} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {1}{2} \\ \frac {1}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 \Gamma \left (\frac {1}{3}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(a+b/x**3)**(1/2),x)
[Out]
-sqrt(a)*x**2*gamma(-2/3)*hyper((-2/3, -1/2), (1/3,), b*exp_polar(I*pi)/(a*x**3))/(3*gamma(1/3))
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