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

Optimal. Leaf size=350 \[ \frac {4 a^{9/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {8 a^{9/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {8 a^{5/2} \sqrt [3]{x} \left (a x^{2/3}+b\right )}{5 b \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {a x+b \sqrt [3]{x}}}-\frac {8 a^2 \sqrt {a x+b \sqrt [3]{x}}}{5 b \sqrt [3]{x}}-\frac {2 \left (a x+b \sqrt [3]{x}\right )^{3/2}}{3 x^2}-\frac {4 a \sqrt {a x+b \sqrt [3]{x}}}{5 x} \]

[Out]

-2/3*(b*x^(1/3)+a*x)^(3/2)/x^2+8/5*a^(5/2)*(b+a*x^(2/3))*x^(1/3)/b/(x^(1/3)*a^(1/2)+b^(1/2))/(b*x^(1/3)+a*x)^(
1/2)-4/5*a*(b*x^(1/3)+a*x)^(1/2)/x-8/5*a^2*(b*x^(1/3)+a*x)^(1/2)/b/x^(1/3)-8/5*a^(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)/b^(3/4)/(
b*x^(1/3)+a*x)^(1/2)+4/5*a^(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)))*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^(3/4)/(b*x^(1/3)+a*x)^(1/2)

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Rubi [A]  time = 0.43, antiderivative size = 350, 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 = {2018, 2020, 2025, 2032, 329, 305, 220, 1196} \[ \frac {4 a^{9/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {8 a^{9/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{5 b^{3/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {8 a^{5/2} \sqrt [3]{x} \left (a x^{2/3}+b\right )}{5 b \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {a x+b \sqrt [3]{x}}}-\frac {8 a^2 \sqrt {a x+b \sqrt [3]{x}}}{5 b \sqrt [3]{x}}-\frac {2 \left (a x+b \sqrt [3]{x}\right )^{3/2}}{3 x^2}-\frac {4 a \sqrt {a x+b \sqrt [3]{x}}}{5 x} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^(1/3) + a*x)^(3/2)/x^3,x]

[Out]

(8*a^(5/2)*(b + a*x^(2/3))*x^(1/3))/(5*b*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[b*x^(1/3) + a*x]) - (4*a*Sqrt[b*x^(1
/3) + a*x])/(5*x) - (8*a^2*Sqrt[b*x^(1/3) + a*x])/(5*b*x^(1/3)) - (2*(b*x^(1/3) + a*x)^(3/2))/(3*x^2) - (8*a^(
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*ArcTa
n[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(5*b^(3/4)*Sqrt[b*x^(1/3) + a*x]) + (4*a^(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])/(5*b^(3/4)*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 2020

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

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 {\left (b \sqrt [3]{x}+a x\right )^{3/2}}{x^3} \, dx &=3 \operatorname {Subst}\left (\int \frac {\left (b x+a x^3\right )^{3/2}}{x^7} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{3 x^2}+(2 a) \operatorname {Subst}\left (\int \frac {\sqrt {b x+a x^3}}{x^4} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {4 a \sqrt {b \sqrt [3]{x}+a x}}{5 x}-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{3 x^2}+\frac {1}{5} \left (4 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {4 a \sqrt {b \sqrt [3]{x}+a x}}{5 x}-\frac {8 a^2 \sqrt {b \sqrt [3]{x}+a x}}{5 b \sqrt [3]{x}}-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{3 x^2}+\frac {\left (4 a^3\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{5 b}\\ &=-\frac {4 a \sqrt {b \sqrt [3]{x}+a x}}{5 x}-\frac {8 a^2 \sqrt {b \sqrt [3]{x}+a x}}{5 b \sqrt [3]{x}}-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{3 x^2}+\frac {\left (4 a^3 \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 )}{5 b \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {4 a \sqrt {b \sqrt [3]{x}+a x}}{5 x}-\frac {8 a^2 \sqrt {b \sqrt [3]{x}+a x}}{5 b \sqrt [3]{x}}-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{3 x^2}+\frac {\left (8 a^3 \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 )}{5 b \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {4 a \sqrt {b \sqrt [3]{x}+a x}}{5 x}-\frac {8 a^2 \sqrt {b \sqrt [3]{x}+a x}}{5 b \sqrt [3]{x}}-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{3 x^2}+\frac {\left (8 a^{5/2} \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 )}{5 \sqrt {b} \sqrt {b \sqrt [3]{x}+a x}}-\frac {\left (8 a^{5/2} \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 )}{5 \sqrt {b} \sqrt {b \sqrt [3]{x}+a x}}\\ &=\frac {8 a^{5/2} \left (b+a x^{2/3}\right ) \sqrt [3]{x}}{5 b \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {b \sqrt [3]{x}+a x}}-\frac {4 a \sqrt {b \sqrt [3]{x}+a x}}{5 x}-\frac {8 a^2 \sqrt {b \sqrt [3]{x}+a x}}{5 b \sqrt [3]{x}}-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{3 x^2}-\frac {8 a^{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 b^{3/4} \sqrt {b \sqrt [3]{x}+a x}}+\frac {4 a^{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 )}{5 b^{3/4} \sqrt {b \sqrt [3]{x}+a x}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 62, normalized size = 0.18 \[ -\frac {2 b \sqrt {a x+b \sqrt [3]{x}} \, _2F_1\left (-\frac {9}{4},-\frac {3}{2};-\frac {5}{4};-\frac {a x^{2/3}}{b}\right )}{3 x^{5/3} \sqrt {\frac {a x^{2/3}}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*Sqrt[b*x^(1/3) + a*x]*Hypergeometric2F1[-9/4, -3/2, -5/4, -((a*x^(2/3))/b)])/(3*Sqrt[1 + (a*x^(2/3))/b]*
x^(5/3))

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fricas [F]  time = 7.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a x + b x^{\frac {1}{3}}\right )}^{\frac {3}{2}}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*x + b*x^(1/3))^(3/2)/x^3, 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((b*x^(1/3)+a*x)^(3/2)/x^3,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: 8.232*(-60
*b^3/180/b^2/x^(1/3)/x^(1/3)-132*b^2*a/180/b^2)/x^(1/3)*sqrt(a/x^(1/3)+b/x)+integrate(144*b^2*a^2/180/b^2/3/(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 = 0.08, size = 339, normalized size = 0.97 \[ -\frac {2 \left (-12 \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {a \,x^{\frac {1}{3}}}{\sqrt {-a b}}}\, \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a^{2} b \,x^{\frac {8}{3}} \EllipticE \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+6 \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {a \,x^{\frac {1}{3}}}{\sqrt {-a b}}}\, \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a^{2} b \,x^{\frac {8}{3}} \EllipticF \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+12 \sqrt {a x +b \,x^{\frac {1}{3}}}\, a^{3} x^{\frac {10}{3}}+12 \sqrt {a x +b \,x^{\frac {1}{3}}}\, a^{2} b \,x^{\frac {8}{3}}+11 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a^{2} b \,x^{\frac {8}{3}}+16 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a \,b^{2} x^{2}+5 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, b^{3} x^{\frac {4}{3}}\right )}{15 \left (a \,x^{\frac {2}{3}}+b \right ) b \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(-12*a^2*b*((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
)*(-1/(-a*b)^(1/2)*a*x^(1/3))^(1/2)*x^(8/3)*((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))+6*a^2*b*((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)*(-1/(-a*b)^(1/2)*a*x^(1/3))^(1/2)*x^(8/3)*((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))+12*x^(10/3)*(a*x+b*x^(1/3))^(1/2)*a^3+12*x^(8/3)*(a*x
+b*x^(1/3))^(1/2)*a^2*b+16*x^2*((a*x^(2/3)+b)*x^(1/3))^(1/2)*a*b^2+11*x^(8/3)*((a*x^(2/3)+b)*x^(1/3))^(1/2)*a^
2*b+5*x^(4/3)*((a*x^(2/3)+b)*x^(1/3))^(1/2)*b^3)/b/x^3/(a*x^(2/3)+b)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + b x^{\frac {1}{3}}\right )}^{\frac {3}{2}}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^(1/3))^(3/2)/x^3,x)

[Out]

int((a*x + b*x^(1/3))^(3/2)/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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