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

Optimal. Leaf size=158 \[ -\frac {5 a^{3/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 b^{9/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {5 \sqrt {a x+b \sqrt [3]{x}}}{b^2 x^{2/3}}+\frac {3}{b \sqrt [3]{x} \sqrt {a x+b \sqrt [3]{x}}} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2018, 2023, 2025, 2011, 329, 220} \[ -\frac {5 a^{3/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{2 b^{9/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {5 \sqrt {a x+b \sqrt [3]{x}}}{b^2 x^{2/3}}+\frac {3}{b \sqrt [3]{x} \sqrt {a x+b \sqrt [3]{x}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(b*x^(1/3) + a*x)^(3/2)),x]

[Out]

3/(b*x^(1/3)*Sqrt[b*x^(1/3) + a*x]) - (5*Sqrt[b*x^(1/3) + a*x])/(b^2*x^(2/3)) - (5*a^(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*b^(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 2023

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*(n - j)*(p + 1)), x] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] &
& (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -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]

Rubi steps

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

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Mathematica [C]  time = 0.07, size = 62, normalized size = 0.39 \[ -\frac {2 \sqrt {\frac {a x^{2/3}}{b}+1} \, _2F_1\left (-\frac {3}{4},\frac {3}{2};\frac {1}{4};-\frac {a x^{2/3}}{b}\right )}{b \sqrt [3]{x} \sqrt {a x+b \sqrt [3]{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{4} x^{3} + 3 \, a^{2} b^{2} x^{\frac {5}{3}} - 2 \, a b^{3} x - {\left (2 \, a^{3} b x^{2} - b^{4}\right )} x^{\frac {1}{3}}\right )} \sqrt {a x + b x^{\frac {1}{3}}}}{a^{6} x^{6} + 2 \, a^{3} b^{3} x^{4} + b^{6} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/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^
6 + 2*a^3*b^3*x^4 + b^6*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 181, normalized size = 1.15 \[ -\frac {6 \sqrt {a x +b \,x^{\frac {1}{3}}}\, a x +4 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a x +5 \sqrt {-a b}\, \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {a \,x^{\frac {1}{3}}}{\sqrt {-a b}}}\, \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, x^{\frac {2}{3}} \EllipticF \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+4 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, b \,x^{\frac {1}{3}}}{2 \left (a \,x^{\frac {2}{3}}+b \right ) b^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a*x + b*x^(1/3))^(3/2)),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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