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

Optimal. Leaf size=149 \[ -\frac {5 b^{3/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 a^{9/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {5 \sqrt {a x+b \sqrt [3]{x}}}{a^2}-\frac {3 x}{a \sqrt {a x+b \sqrt [3]{x}}} \]

[Out]

-3*x/a/(b*x^(1/3)+a*x)^(1/2)+5*(b*x^(1/3)+a*x)^(1/2)/a^2-5/2*b^(3/4)*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)/a^(9/4)/(b*x^(1/3)+a*x)^(1/
2)

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Rubi [A]  time = 0.21, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2018, 2022, 2024, 2011, 329, 220} \[ -\frac {5 b^{3/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 a^{9/4} \sqrt {a x+b \sqrt [3]{x}}}+\frac {5 \sqrt {a x+b \sqrt [3]{x}}}{a^2}-\frac {3 x}{a \sqrt {a x+b \sqrt [3]{x}}} \]

Antiderivative was successfully verified.

[In]

Int[x/(b*x^(1/3) + a*x)^(3/2),x]

[Out]

(-3*x)/(a*Sqrt[b*x^(1/3) + a*x]) + (5*Sqrt[b*x^(1/3) + a*x])/a^2 - (5*b^(3/4)*(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])/(
2*a^(9/4)*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 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 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

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 2022

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 2024

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]

Rubi steps

\begin {align*} \int \frac {x}{\left (b \sqrt [3]{x}+a x\right )^{3/2}} \, dx &=3 \operatorname {Subst}\left (\int \frac {x^5}{\left (b x+a x^3\right )^{3/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {3 x}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {15 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{2 a}\\ &=-\frac {3 x}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {5 \sqrt {b \sqrt [3]{x}+a x}}{a^2}-\frac {(5 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{2 a^2}\\ &=-\frac {3 x}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {5 \sqrt {b \sqrt [3]{x}+a x}}{a^2}-\frac {\left (5 b \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {b+a x^2}} \, dx,x,\sqrt [3]{x}\right )}{2 a^2 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {3 x}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {5 \sqrt {b \sqrt [3]{x}+a x}}{a^2}-\frac {\left (5 b \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 )}{a^2 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {3 x}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {5 \sqrt {b \sqrt [3]{x}+a x}}{a^2}-\frac {5 b^{3/4} \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 )}{2 a^{9/4} \sqrt {b \sqrt [3]{x}+a x}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 82, normalized size = 0.55 \[ \frac {\sqrt {a x+b \sqrt [3]{x}} \left (-5 b \sqrt {\frac {a x^{2/3}}{b}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {a x^{2/3}}{b}\right )+2 a x^{2/3}+5 b\right )}{a^2 \left (a x^{2/3}+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{4} x^{3} + 3 \, a^{2} b^{2} x^{\frac {5}{3}} - 2 \, a b^{3} x - {\left (2 \, a^{3} b x^{2} - b^{4}\right )} x^{\frac {1}{3}}\right )} \sqrt {a x + b x^{\frac {1}{3}}}}{a^{6} x^{4} + 2 \, a^{3} b^{3} x^{2} + b^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(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^
4 + 2*a^3*b^3*x^2 + b^6), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 184, normalized size = 1.23 \[ \frac {4 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a^{2} x +4 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a b \,x^{\frac {1}{3}}+6 \sqrt {a x +b \,x^{\frac {1}{3}}}\, a b \,x^{\frac {1}{3}}-5 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, \sqrt {-a b}\, \sqrt {-\frac {a \,x^{\frac {1}{3}}}{\sqrt {-a b}}}\, \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, b \EllipticF \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 \left (a \,x^{\frac {2}{3}}+b \right ) a^{3} x^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{{\left (a\,x+b\,x^{1/3}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a*x + b*x^(1/3))^(3/2),x)

[Out]

int(x/(a*x + b*x^(1/3))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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