3.2.60 \(\int \frac {x^2}{(b \sqrt [3]{x}+a x)^{3/2}} \, dx\) [160]

Optimal. Leaf size=349 \[ \frac {77 b^2 \left (b+a x^{2/3}\right ) \sqrt [3]{x}}{5 a^{7/2} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {b \sqrt [3]{x}+a x}}-\frac {3 x^2}{a \sqrt {b \sqrt [3]{x}+a x}}-\frac {77 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a^3}+\frac {11 x \sqrt {b \sqrt [3]{x}+a x}}{3 a^2}-\frac {77 b^{9/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 a^{15/4} \sqrt {b \sqrt [3]{x}+a x}}+\frac {77 b^{9/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{10 a^{15/4} \sqrt {b \sqrt [3]{x}+a x}} \]

[Out]

-3*x^2/a/(b*x^(1/3)+a*x)^(1/2)+77/5*b^2*(b+a*x^(2/3))*x^(1/3)/a^(7/2)/(x^(1/3)*a^(1/2)+b^(1/2))/(b*x^(1/3)+a*x
)^(1/2)-77/15*b*x^(1/3)*(b*x^(1/3)+a*x)^(1/2)/a^3+11/3*x*(b*x^(1/3)+a*x)^(1/2)/a^2-77/5*b^(9/4)*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)/
a^(15/4)/(b*x^(1/3)+a*x)^(1/2)+77/10*b^(9/4)*x^(1/6)*(cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))^2)^(1/2)/cos(2*ar
ctan(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)/a^(15/4)/(b*x^(1/3)+a*x)^(1/2)

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Rubi [A]
time = 0.31, antiderivative size = 349, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {2043, 2047, 2049, 2057, 335, 311, 226, 1210} \begin {gather*} \frac {77 b^{9/4} \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 \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{10 a^{15/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {77 b^{9/4} \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 \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 a^{15/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {77 b^2 \sqrt [3]{x} \left (a x^{2/3}+b\right )}{5 a^{7/2} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {a x+b \sqrt [3]{x}}}-\frac {77 b \sqrt [3]{x} \sqrt {a x+b \sqrt [3]{x}}}{15 a^3}+\frac {11 x \sqrt {a x+b \sqrt [3]{x}}}{3 a^2}-\frac {3 x^2}{a \sqrt {a x+b \sqrt [3]{x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(b*x^(1/3) + a*x)^(3/2),x]

[Out]

(77*b^2*(b + a*x^(2/3))*x^(1/3))/(5*a^(7/2)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[b*x^(1/3) + a*x]) - (3*x^2)/(a*Sq
rt[b*x^(1/3) + a*x]) - (77*b*x^(1/3)*Sqrt[b*x^(1/3) + a*x])/(15*a^3) + (11*x*Sqrt[b*x^(1/3) + a*x])/(3*a^2) -
(77*b^(9/4)*(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])/(5*a^(15/4)*Sqrt[b*x^(1/3) + a*x]) + (77*b^(9/4)*(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])/(10*a^(15/4)*Sqrt[b*x^(1/3) + a*x])

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 311

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 335

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 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 2043

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 2047

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

Rule 2049

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n +
1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*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]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[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 {x^2}{\left (b \sqrt [3]{x}+a x\right )^{3/2}} \, dx &=3 \text {Subst}\left (\int \frac {x^8}{\left (b x+a x^3\right )^{3/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {3 x^2}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {33 \text {Subst}\left (\int \frac {x^5}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{2 a}\\ &=-\frac {3 x^2}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {11 x \sqrt {b \sqrt [3]{x}+a x}}{3 a^2}-\frac {(77 b) \text {Subst}\left (\int \frac {x^3}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{6 a^2}\\ &=-\frac {3 x^2}{a \sqrt {b \sqrt [3]{x}+a x}}-\frac {77 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a^3}+\frac {11 x \sqrt {b \sqrt [3]{x}+a x}}{3 a^2}+\frac {\left (77 b^2\right ) \text {Subst}\left (\int \frac {x}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{10 a^3}\\ &=-\frac {3 x^2}{a \sqrt {b \sqrt [3]{x}+a x}}-\frac {77 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a^3}+\frac {11 x \sqrt {b \sqrt [3]{x}+a x}}{3 a^2}+\frac {\left (77 b^2 \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{\sqrt {b+a x^2}} \, dx,x,\sqrt [3]{x}\right )}{10 a^3 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {3 x^2}{a \sqrt {b \sqrt [3]{x}+a x}}-\frac {77 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a^3}+\frac {11 x \sqrt {b \sqrt [3]{x}+a x}}{3 a^2}+\frac {\left (77 b^2 \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{5 a^3 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {3 x^2}{a \sqrt {b \sqrt [3]{x}+a x}}-\frac {77 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a^3}+\frac {11 x \sqrt {b \sqrt [3]{x}+a x}}{3 a^2}+\frac {\left (77 b^{5/2} \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{5 a^{7/2} \sqrt {b \sqrt [3]{x}+a x}}-\frac {\left (77 b^{5/2} \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {a} x^2}{\sqrt {b}}}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{5 a^{7/2} \sqrt {b \sqrt [3]{x}+a x}}\\ &=\frac {77 b^2 \left (b+a x^{2/3}\right ) \sqrt [3]{x}}{5 a^{7/2} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {b \sqrt [3]{x}+a x}}-\frac {3 x^2}{a \sqrt {b \sqrt [3]{x}+a x}}-\frac {77 b \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x}}{15 a^3}+\frac {11 x \sqrt {b \sqrt [3]{x}+a x}}{3 a^2}-\frac {77 b^{9/4} \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 )}{5 a^{15/4} \sqrt {b \sqrt [3]{x}+a x}}+\frac {77 b^{9/4} \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 )}{10 a^{15/4} \sqrt {b \sqrt [3]{x}+a x}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.06, size = 94, normalized size = 0.27 \begin {gather*} \frac {2 x^{2/3} \left (77 b^2-11 a b x^{2/3}+5 a^2 x^{4/3}-77 b^2 \sqrt {1+\frac {a x^{2/3}}{b}} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {a x^{2/3}}{b}\right )\right )}{15 a^3 \sqrt {b \sqrt [3]{x}+a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(b*x^(1/3) + a*x)^(3/2),x]

[Out]

(2*x^(2/3)*(77*b^2 - 11*a*b*x^(2/3) + 5*a^2*x^(4/3) - 77*b^2*Sqrt[1 + (a*x^(2/3))/b]*Hypergeometric2F1[3/4, 3/
2, 7/4, -((a*x^(2/3))/b)]))/(15*a^3*Sqrt[b*x^(1/3) + a*x])

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Maple [A]
time = 0.35, size = 312, normalized size = 0.89

method	result	size
derivativedivides	\(-\frac {3 x^{\frac {2}{3}} b^{2}}{a^{3} \sqrt {\left (x^{\frac {2}{3}}+\frac {b}{a}\right ) x^{\frac {1}{3}} a}}+\frac {2 x \sqrt {b \,x^{\frac {1}{3}}+a x}}{3 a^{2}}-\frac {32 b \,x^{\frac {1}{3}} \sqrt {b \,x^{\frac {1}{3}}+a x}}{15 a^{3}}+\frac {77 b^{2} \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{a}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{a}\right )}{10 a^{4} \sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(237\)
default	\(-\frac {-462 b^{3} \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, \EllipticE \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+231 b^{3} \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, \EllipticF \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+90 \sqrt {b \,x^{\frac {1}{3}}+a x}\, x^{\frac {2}{3}} a \,b^{2}+64 x^{\frac {2}{3}} \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a \,b^{2}+44 x^{\frac {4}{3}} \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a^{2} b -20 \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\, a^{3} x^{2}}{30 a^{4} x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}\)	\(312\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30/a^4*(-462*b^3*((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-2*(a*x^(1/3)-(-a*b)^(1/2))/(-a*b)^(1/2))^(
1/2)*(-x^(1/3)*a/(-a*b)^(1/2))^(1/2)*(x^(1/3)*(b+a*x^(2/3)))^(1/2)*EllipticE(((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^
(1/2))^(1/2),1/2*2^(1/2))+231*b^3*((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-2*(a*x^(1/3)-(-a*b)^(1/2))/(
-a*b)^(1/2))^(1/2)*(-x^(1/3)*a/(-a*b)^(1/2))^(1/2)*(x^(1/3)*(b+a*x^(2/3)))^(1/2)*EllipticF(((a*x^(1/3)+(-a*b)^
(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))+90*(b*x^(1/3)+a*x)^(1/2)*x^(2/3)*a*b^2+64*x^(2/3)*(x^(1/3)*(b+a*x^(2/3
)))^(1/2)*a*b^2+44*x^(4/3)*(x^(1/3)*(b+a*x^(2/3)))^(1/2)*a^2*b-20*(x^(1/3)*(b+a*x^(2/3)))^(1/2)*a^3*x^2)/x^(1/
3)/(b+a*x^(2/3))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^4*x^4 + 3*a^2*b^2*x^(8/3) - 2*a*b^3*x^2 - (2*a^3*b*x^3 - b^4*x)*x^(1/3))*sqrt(a*x + b*x^(1/3))/(a^
6*x^4 + 2*a^3*b^3*x^2 + b^6), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\left (a x + b \sqrt [3]{x}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2}{{\left (a\,x+b\,x^{1/3}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x + b*x^(1/3))^(3/2),x)

[Out]

int(x^2/(a*x + b*x^(1/3))^(3/2), x)

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