3.162 \(\int \frac {1}{(b \sqrt [3]{x}+a x)^{3/2}} \, dx\)

Optimal. Leaf size=296 \[ -\frac {3 \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 a^{3/4} b^{3/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {3 \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{a^{3/4} b^{3/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {3 \sqrt [3]{x} \left (a x^{2/3}+b\right )}{\sqrt {a} b \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {a x+b \sqrt [3]{x}}}+\frac {3 x^{2/3}}{b \sqrt {a x+b \sqrt [3]{x}}} \]

[Out]

3*x^(2/3)/b/(b*x^(1/3)+a*x)^(1/2)-3*(b+a*x^(2/3))*x^(1/3)/b/a^(1/2)/(x^(1/3)*a^(1/2)+b^(1/2))/(b*x^(1/3)+a*x)^
(1/2)+3*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)))*Ellipt
icE(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^(3/4)/b^(3/4)/(b*x^(1/3)+a*x)^(1/2)-3/2*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)/a^(3/4)/b^(3/4)/(b*x^(1/3)+a*x)^
(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {2006, 2018, 2032, 329, 305, 220, 1196} \[ -\frac {3 \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 a^{3/4} b^{3/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {3 \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{a^{3/4} b^{3/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {3 \sqrt [3]{x} \left (a x^{2/3}+b\right )}{\sqrt {a} b \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {a x+b \sqrt [3]{x}}}+\frac {3 x^{2/3}}{b \sqrt {a x+b \sqrt [3]{x}}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^(1/3) + a*x)^(-3/2),x]

[Out]

(-3*(b + a*x^(2/3))*x^(1/3))/(Sqrt[a]*b*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[b*x^(1/3) + a*x]) + (3*x^(2/3))/(b*Sq
rt[b*x^(1/3) + a*x]) + (3*(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])/(a^(3/4)*b^(3/4)*Sqrt[b*x^(1/3) + a*x]) - (3*(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])/(2*a^(3/4)*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 2006

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

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

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Mathematica [C]  time = 0.04, size = 62, normalized size = 0.21 \[ \frac {2 x^{2/3} \sqrt {\frac {a x^{2/3}}{b}+1} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {a x^{2/3}}{b}\right )}{b \sqrt {a x+b \sqrt [3]{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^(1/3) + a*x)^(-3/2),x]

[Out]

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

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fricas [F]  time = 7.07, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{4} x^{3} + 3 \, a^{2} b^{2} x^{\frac {5}{3}} - 2 \, a b^{3} x - {\left (2 \, a^{3} b x^{2} - b^{4}\right )} x^{\frac {1}{3}}\right )} \sqrt {a x + b x^{\frac {1}{3}}}}{a^{6} x^{5} + 2 \, a^{3} b^{3} x^{3} + b^{6} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^4*x^3 + 3*a^2*b^2*x^(5/3) - 2*a*b^3*x - (2*a^3*b*x^2 - b^4)*x^(1/3))*sqrt(a*x + b*x^(1/3))/(a^6*x^
5 + 2*a^3*b^3*x^3 + b^6*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 242, normalized size = 0.82 \[ \frac {-3 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {a \,x^{\frac {1}{3}}}{\sqrt {-a b}}}\, b \EllipticE \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+\frac {3 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {a \,x^{\frac {1}{3}}}{\sqrt {-a b}}}\, b \EllipticF \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2}+3 \sqrt {a x +b \,x^{\frac {1}{3}}}\, a \,x^{\frac {2}{3}}}{\left (a \,x^{\frac {2}{3}}+b \right ) a b \,x^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2/a*(-2*((a*x^(2/3)+b)*x^(1/3))^(1/2)*((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)*EllipticE(((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(
1/2),1/2*2^(1/2))*b+((a*x^(2/3)+b)*x^(1/3))^(1/2)*((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)*EllipticF(((a*x^(1/3)+(-a*b)^(1/2))/(-a*b
)^(1/2))^(1/2),1/2*2^(1/2))*b+2*(a*x+b*x^(1/3))^(1/2)*a*x^(2/3))/x^(1/3)/(a*x^(2/3)+b)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.35, size = 40, normalized size = 0.14 \[ \frac {2\,x\,{\left (\frac {a\,x^{2/3}}{b}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{4},\frac {3}{2};\ \frac {7}{4};\ -\frac {a\,x^{2/3}}{b}\right )}{{\left (a\,x+b\,x^{1/3}\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x + b*x^(1/3))^(3/2),x)

[Out]

(2*x*((a*x^(2/3))/b + 1)^(3/2)*hypergeom([3/4, 3/2], 7/4, -(a*x^(2/3))/b))/(a*x + b*x^(1/3))^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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