3.2031 \(\int \frac {x^3}{\sqrt {a+\frac {b}{x^3}}} \, dx\)

Optimal. Leaf size=542 \[ \frac {5 b^{4/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{4 \sqrt {2} \sqrt [4]{3} a^{5/3} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}-\frac {5 \sqrt [4]{3} \sqrt {2-\sqrt {3}} b^{4/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} E\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{16 a^{5/3} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}+\frac {5 b^{4/3} \sqrt {a+\frac {b}{x^3}}}{8 a^2 \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}-\frac {5 b x \sqrt {a+\frac {b}{x^3}}}{8 a^2}+\frac {x^4 \sqrt {a+\frac {b}{x^3}}}{4 a} \]

[Out]

-5/8*b*x*(a+b/x^3)^(1/2)/a^2+1/4*x^4*(a+b/x^3)^(1/2)/a+5/8*b^(4/3)*(a+b/x^3)^(1/2)/a^2/(b^(1/3)/x+a^(1/3)*(1+3
^(1/2)))+5/24*b^(4/3)*(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)*((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)*3^(3/4)
/a^(5/3)*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)-5/16*3^
(1/4)*b^(4/3)*(a^(1/3)+b^(1/3)/x)*EllipticE((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^(5/3)/(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 = 0.29, antiderivative size = 542, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {335, 325, 303, 218, 1877} \[ \frac {5 b^{4/3} \sqrt {a+\frac {b}{x^3}}}{8 a^2 \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}+\frac {5 b^{4/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{4 \sqrt {2} \sqrt [4]{3} a^{5/3} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}-\frac {5 \sqrt [4]{3} \sqrt {2-\sqrt {3}} b^{4/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} E\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{16 a^{5/3} \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}-\frac {5 b x \sqrt {a+\frac {b}{x^3}}}{8 a^2}+\frac {x^4 \sqrt {a+\frac {b}{x^3}}}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[a + b/x^3],x]

[Out]

(5*b^(4/3)*Sqrt[a + b/x^3])/(8*a^2*((1 + Sqrt[3])*a^(1/3) + b^(1/3)/x)) - (5*b*Sqrt[a + b/x^3]*x)/(8*a^2) + (S
qrt[a + b/x^3]*x^4)/(4*a) - (5*3^(1/4)*Sqrt[2 - Sqrt[3]]*b^(4/3)*(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]*EllipticE[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]])/(16*a^(5/3)*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]) + (5*b^(4/3)*(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]])/(4*Sqrt[2]*3^(1/4)*a^(5/3)*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 303

Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Dist[(Sq
rt[2]*s)/(Sqrt[2 + Sqrt[3]]*r), Int[1/Sqrt[a + b*x^3], x], x] + Dist[1/r, Int[((1 - Sqrt[3])*s + r*x)/Sqrt[a +
b*x^3], x], x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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]

Rule 1877

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Simplify[((1 - Sqrt[3])*d)/c]]
, s = Denom[Simplify[((1 - Sqrt[3])*d)/c]]}, Simp[(2*d*s^3*Sqrt[a + b*x^3])/(a*r^2*((1 + Sqrt[3])*s + r*x)), x
] - Simp[(3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*Elli
pticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(r^2*Sqrt[a + b*x^3]*Sqrt[(s*(
s + r*x))/((1 + Sqrt[3])*s + r*x)^2]), x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && EqQ[b*c^3 - 2*(5 - 3*Sqrt[3
])*a*d^3, 0]

Rubi steps

\begin {align*} \int \frac {x^3}{\sqrt {a+\frac {b}{x^3}}} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{x^5 \sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {\sqrt {a+\frac {b}{x^3}} x^4}{4 a}+\frac {(5 b) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{8 a}\\ &=-\frac {5 b \sqrt {a+\frac {b}{x^3}} x}{8 a^2}+\frac {\sqrt {a+\frac {b}{x^3}} x^4}{4 a}+\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{16 a^2}\\ &=-\frac {5 b \sqrt {a+\frac {b}{x^3}} x}{8 a^2}+\frac {\sqrt {a+\frac {b}{x^3}} x^4}{4 a}+\frac {\left (5 b^{5/3}\right ) \operatorname {Subst}\left (\int \frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{16 a^2}+\frac {\left (5 \sqrt {\frac {1}{2} \left (2-\sqrt {3}\right )} b^{5/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{8 a^{5/3}}\\ &=\frac {5 b^{4/3} \sqrt {a+\frac {b}{x^3}}}{8 a^2 \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}-\frac {5 b \sqrt {a+\frac {b}{x^3}} x}{8 a^2}+\frac {\sqrt {a+\frac {b}{x^3}} x^4}{4 a}-\frac {5 \sqrt [4]{3} \sqrt {2-\sqrt {3}} b^{4/3} \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}} E\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 )}{16 a^{5/3} \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}}}+\frac {5 b^{4/3} \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 )}{4 \sqrt {2} \sqrt [4]{3} a^{5/3} \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 = 0.02, size = 62, normalized size = 0.11 \[ \frac {x \left (-b \sqrt {\frac {a x^3}{b}+1} \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};-\frac {a x^3}{b}\right )+a x^3+b\right )}{4 a \sqrt {a+\frac {b}{x^3}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[a + b/x^3],x]

[Out]

(x*(b + a*x^3 - b*Sqrt[1 + (a*x^3)/b]*Hypergeometric2F1[1/2, 5/6, 11/6, -((a*x^3)/b)]))/(4*a*Sqrt[a + b/x^3])

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{6} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{a x^{3} + b}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b/x^3)^(1/2),x, algorithm="fricas")

[Out]

integral(x^6*sqrt((a*x^3 + b)/x^3)/(a*x^3 + b), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {a + \frac {b}{x^{3}}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b/x^3)^(1/2),x, algorithm="giac")

[Out]

integrate(x^3/sqrt(a + b/x^3), x)

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maple [B]  time = 0.01, size = 2586, normalized size = 4.77 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a+b/x^3)^(1/2),x)

[Out]

1/4/((a*x^3+b)/x^3)^(1/2)/x*(a*x^3+b)/a^3*(5*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)*EllipticE((-(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*b^2+10*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)*EllipticE((-(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*b-10*(-(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^2*a*b+15*(-(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)*EllipticE((-
(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^2*a*b-5*I*3^(1/2)*x^3*a^2*b+20*(-(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*b-30*(-(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)*EllipticE((-(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*b-5*I
*(-a^2*b)^(2/3)*3^(1/2)*x*b-5*I*(-a^2*b)^(1/3)*3^(1/2)*x^2*a*b+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)*a^2*x^2+10*(-(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*b^2-15*(-(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)*El
lipticE((-(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*b^2-5*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)*EllipticE((-(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*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)*a^2*x^2+15*a^2*b*x^3+15*(-a^2*b)^(1/3)*a*b*x^2+15*(-a^
2*b)^(2/3)*b*x)/((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 \[ \int \frac {x^{3}}{\sqrt {a + \frac {b}{x^{3}}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b/x^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^3/sqrt(a + b/x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3}{\sqrt {a+\frac {b}{x^3}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a + b/x^3)^(1/2),x)

[Out]

int(x^3/(a + b/x^3)^(1/2), x)

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sympy [A]  time = 1.14, size = 46, normalized size = 0.08 \[ - \frac {x^{4} \Gamma \left (- \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {1}{2} \\ - \frac {1}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 \sqrt {a} \Gamma \left (- \frac {1}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(a+b/x**3)**(1/2),x)

[Out]

-x**4*gamma(-4/3)*hyper((-4/3, 1/2), (-1/3,), b*exp_polar(I*pi)/(a*x**3))/(3*sqrt(a)*gamma(-1/3))

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