3.156 \(\int \frac {1}{x^3 \sqrt {b \sqrt [3]{x}+a x}} \, dx\)

Optimal. Leaf size=388 \[ -\frac {77 a^{13/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{65 b^{15/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {154 a^{13/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{65 b^{15/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {154 a^{7/2} \sqrt [3]{x} \left (a x^{2/3}+b\right )}{65 b^4 \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {a x+b \sqrt [3]{x}}}+\frac {154 a^3 \sqrt {a x+b \sqrt [3]{x}}}{65 b^4 \sqrt [3]{x}}-\frac {154 a^2 \sqrt {a x+b \sqrt [3]{x}}}{195 b^3 x}+\frac {22 a \sqrt {a x+b \sqrt [3]{x}}}{39 b^2 x^{5/3}}-\frac {6 \sqrt {a x+b \sqrt [3]{x}}}{13 b x^{7/3}} \]

[Out]

-154/65*a^(7/2)*(b+a*x^(2/3))*x^(1/3)/b^4/(x^(1/3)*a^(1/2)+b^(1/2))/(b*x^(1/3)+a*x)^(1/2)-6/13*(b*x^(1/3)+a*x)
^(1/2)/b/x^(7/3)+22/39*a*(b*x^(1/3)+a*x)^(1/2)/b^2/x^(5/3)-154/195*a^2*(b*x^(1/3)+a*x)^(1/2)/b^3/x+154/65*a^3*
(b*x^(1/3)+a*x)^(1/2)/b^4/x^(1/3)+154/65*a^(13/4)*x^(1/6)*(cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))^2)^(1/2)/cos
(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))*EllipticE(sin(2*arctan(a^(1/4)*x^(1/6)/b^(1/4))),1/2*2^(1/2))*(x^(1/3)*a^(
1/2)+b^(1/2))*((b+a*x^(2/3))/(x^(1/3)*a^(1/2)+b^(1/2))^2)^(1/2)/b^(15/4)/(b*x^(1/3)+a*x)^(1/2)-77/65*a^(13/4)*
x^(1/6)*(cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))^2)^(1/2)/cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))*EllipticF(sin(
2*arctan(a^(1/4)*x^(1/6)/b^(1/4))),1/2*2^(1/2))*(x^(1/3)*a^(1/2)+b^(1/2))*((b+a*x^(2/3))/(x^(1/3)*a^(1/2)+b^(1
/2))^2)^(1/2)/b^(15/4)/(b*x^(1/3)+a*x)^(1/2)

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Rubi [A]  time = 0.48, antiderivative size = 388, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2018, 2025, 2032, 329, 305, 220, 1196} \[ -\frac {154 a^{7/2} \sqrt [3]{x} \left (a x^{2/3}+b\right )}{65 b^4 \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {a x+b \sqrt [3]{x}}}-\frac {77 a^{13/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{65 b^{15/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {154 a^{13/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{65 b^{15/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {154 a^3 \sqrt {a x+b \sqrt [3]{x}}}{65 b^4 \sqrt [3]{x}}-\frac {154 a^2 \sqrt {a x+b \sqrt [3]{x}}}{195 b^3 x}+\frac {22 a \sqrt {a x+b \sqrt [3]{x}}}{39 b^2 x^{5/3}}-\frac {6 \sqrt {a x+b \sqrt [3]{x}}}{13 b x^{7/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[b*x^(1/3) + a*x]),x]

[Out]

(-154*a^(7/2)*(b + a*x^(2/3))*x^(1/3))/(65*b^4*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[b*x^(1/3) + a*x]) - (6*Sqrt[b*
x^(1/3) + a*x])/(13*b*x^(7/3)) + (22*a*Sqrt[b*x^(1/3) + a*x])/(39*b^2*x^(5/3)) - (154*a^2*Sqrt[b*x^(1/3) + a*x
])/(195*b^3*x) + (154*a^3*Sqrt[b*x^(1/3) + a*x])/(65*b^4*x^(1/3)) + (154*a^(13/4)*(Sqrt[b] + Sqrt[a]*x^(1/3))*
Sqrt[(b + a*x^(2/3))/(Sqrt[b] + Sqrt[a]*x^(1/3))^2]*x^(1/6)*EllipticE[2*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2
])/(65*b^(15/4)*Sqrt[b*x^(1/3) + a*x]) - (77*a^(13/4)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[(b + a*x^(2/3))/(Sqrt[b
] + Sqrt[a]*x^(1/3))^2]*x^(1/6)*EllipticF[2*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(65*b^(15/4)*Sqrt[b*x^(1/
3) + a*x])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

\begin {align*} \int \frac {1}{x^3 \sqrt {b \sqrt [3]{x}+a x}} \, dx &=3 \operatorname {Subst}\left (\int \frac {1}{x^7 \sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{13 b x^{7/3}}-\frac {(33 a) \operatorname {Subst}\left (\int \frac {1}{x^5 \sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{13 b}\\ &=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{13 b x^{7/3}}+\frac {22 a \sqrt {b \sqrt [3]{x}+a x}}{39 b^2 x^{5/3}}+\frac {\left (77 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{39 b^2}\\ &=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{13 b x^{7/3}}+\frac {22 a \sqrt {b \sqrt [3]{x}+a x}}{39 b^2 x^{5/3}}-\frac {154 a^2 \sqrt {b \sqrt [3]{x}+a x}}{195 b^3 x}-\frac {\left (77 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{65 b^3}\\ &=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{13 b x^{7/3}}+\frac {22 a \sqrt {b \sqrt [3]{x}+a x}}{39 b^2 x^{5/3}}-\frac {154 a^2 \sqrt {b \sqrt [3]{x}+a x}}{195 b^3 x}+\frac {154 a^3 \sqrt {b \sqrt [3]{x}+a x}}{65 b^4 \sqrt [3]{x}}-\frac {\left (77 a^4\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{65 b^4}\\ &=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{13 b x^{7/3}}+\frac {22 a \sqrt {b \sqrt [3]{x}+a x}}{39 b^2 x^{5/3}}-\frac {154 a^2 \sqrt {b \sqrt [3]{x}+a x}}{195 b^3 x}+\frac {154 a^3 \sqrt {b \sqrt [3]{x}+a x}}{65 b^4 \sqrt [3]{x}}-\frac {\left (77 a^4 \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{\sqrt {b+a x^2}} \, dx,x,\sqrt [3]{x}\right )}{65 b^4 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{13 b x^{7/3}}+\frac {22 a \sqrt {b \sqrt [3]{x}+a x}}{39 b^2 x^{5/3}}-\frac {154 a^2 \sqrt {b \sqrt [3]{x}+a x}}{195 b^3 x}+\frac {154 a^3 \sqrt {b \sqrt [3]{x}+a x}}{65 b^4 \sqrt [3]{x}}-\frac {\left (154 a^4 \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{65 b^4 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{13 b x^{7/3}}+\frac {22 a \sqrt {b \sqrt [3]{x}+a x}}{39 b^2 x^{5/3}}-\frac {154 a^2 \sqrt {b \sqrt [3]{x}+a x}}{195 b^3 x}+\frac {154 a^3 \sqrt {b \sqrt [3]{x}+a x}}{65 b^4 \sqrt [3]{x}}-\frac {\left (154 a^{7/2} \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{65 b^{7/2} \sqrt {b \sqrt [3]{x}+a x}}+\frac {\left (154 a^{7/2} \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {a} x^2}{\sqrt {b}}}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{65 b^{7/2} \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {154 a^{7/2} \left (b+a x^{2/3}\right ) \sqrt [3]{x}}{65 b^4 \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {b \sqrt [3]{x}+a x}}-\frac {6 \sqrt {b \sqrt [3]{x}+a x}}{13 b x^{7/3}}+\frac {22 a \sqrt {b \sqrt [3]{x}+a x}}{39 b^2 x^{5/3}}-\frac {154 a^2 \sqrt {b \sqrt [3]{x}+a x}}{195 b^3 x}+\frac {154 a^3 \sqrt {b \sqrt [3]{x}+a x}}{65 b^4 \sqrt [3]{x}}+\frac {154 a^{13/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{65 b^{15/4} \sqrt {b \sqrt [3]{x}+a x}}-\frac {77 a^{13/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{65 b^{15/4} \sqrt {b \sqrt [3]{x}+a x}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 59, normalized size = 0.15 \[ -\frac {6 \sqrt {\frac {a x^{2/3}}{b}+1} \, _2F_1\left (-\frac {13}{4},\frac {1}{2};-\frac {9}{4};-\frac {a x^{2/3}}{b}\right )}{13 x^2 \sqrt {a x+b \sqrt [3]{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[b*x^(1/3) + a*x]),x]

[Out]

(-6*Sqrt[1 + (a*x^(2/3))/b]*Hypergeometric2F1[-13/4, 1/2, -9/4, -((a*x^(2/3))/b)])/(13*x^2*Sqrt[b*x^(1/3) + a*
x])

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fricas [F]  time = 7.39, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} x^{2} - a b x^{\frac {4}{3}} + b^{2} x^{\frac {2}{3}}\right )} \sqrt {a x + b x^{\frac {1}{3}}}}{a^{3} x^{6} + b^{3} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^2*x^2 - a*b*x^(4/3) + b^2*x^(2/3))*sqrt(a*x + b*x^(1/3))/(a^3*x^6 + b^3*x^4), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(x)]sym2poly
/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument ValueEvaluation time: 1.416*((-3
780*b^4/49140/b^5/x^(1/3)/x^(1/3)+4620*b^3*a/49140/b^5)/x^(1/3)/x^(1/3)-6468*b^2*a^2/49140/b^5)/x^(1/3)*sqrt(a
/x^(1/3)+b/x)+integrate(-58212*b^2*a^3/49140/b^5/3/(x^(1/3)*x^(1/3)*sqrt(x)*sqrt(a*(x^(1/3))^2+b)*sign(x)),x)

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maple [A]  time = 0.09, size = 365, normalized size = 0.94 \[ -\frac {462 \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {a \,x^{\frac {1}{3}}}{\sqrt {-a b}}}\, \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a^{3} b \,x^{\frac {10}{3}} \EllipticE \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-231 \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {a \,x^{\frac {1}{3}}}{\sqrt {-a b}}}\, \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a^{3} b \,x^{\frac {10}{3}} \EllipticF \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-462 \sqrt {a x +b \,x^{\frac {1}{3}}}\, a^{4} x^{4}-462 \sqrt {a x +b \,x^{\frac {1}{3}}}\, a^{3} b \,x^{\frac {10}{3}}+154 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a^{3} b \,x^{\frac {10}{3}}+44 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a^{2} b^{2} x^{\frac {8}{3}}-20 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a \,b^{3} x^{2}+90 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, b^{4} x^{\frac {4}{3}}}{195 \left (a \,x^{\frac {2}{3}}+b \right ) b^{4} x^{\frac {11}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/195*(462*a^3*b*((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2
))^(1/2)*(-1/(-a*b)^(1/2)*a*x^(1/3))^(1/2)*x^(10/3)*((a*x^(2/3)+b)*x^(1/3))^(1/2)*EllipticE(((a*x^(1/3)+(-a*b)
^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))-231*a^3*b*((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-a*
x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-1/(-a*b)^(1/2)*a*x^(1/3))^(1/2)*x^(10/3)*((a*x^(2/3)+b)*x^(1/3))^(
1/2)*EllipticF(((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))-462*x^(10/3)*(a*x+b*x^(1/3))^(1/2)*a
^3*b+44*x^(8/3)*((a*x^(2/3)+b)*x^(1/3))^(1/2)*a^2*b^2+154*x^(10/3)*((a*x^(2/3)+b)*x^(1/3))^(1/2)*a^3*b-20*x^2*
((a*x^(2/3)+b)*x^(1/3))^(1/2)*a*b^3-462*x^4*(a*x+b*x^(1/3))^(1/2)*a^4+90*x^(4/3)*((a*x^(2/3)+b)*x^(1/3))^(1/2)
*b^4)/x^(11/3)/(a*x^(2/3)+b)/b^4

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a x + b x^{\frac {1}{3}}} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \sqrt {a x + b \sqrt [3]{x}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**3*sqrt(a*x + b*x**(1/3))), x)

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