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

Optimal. Leaf size=163 \[ \frac{5 a^{7/4} \sqrt [6]{x} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{\frac{a x^{2/3}+b}{\left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{7 b^{9/4} \sqrt{a x+b \sqrt [3]{x}}}+\frac{10 a \sqrt{a x+b \sqrt [3]{x}}}{7 b^2 x^{2/3}}-\frac{6 \sqrt{a x+b \sqrt [3]{x}}}{7 b x^{4/3}} \]

[Out]

(-6*Sqrt[b*x^(1/3) + a*x])/(7*b*x^(4/3)) + (10*a*Sqrt[b*x^(1/3) + a*x])/(7*b^2*x^(2/3)) + (5*a^(7/4)*(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])/(7*b^(9/4)*Sqrt[b*x^(1/3) + a*x])

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Rubi [A]  time = 0.200723, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {2018, 2025, 2011, 329, 220} \[ \frac{5 a^{7/4} \sqrt [6]{x} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{\frac{a x^{2/3}+b}{\left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{7 b^{9/4} \sqrt{a x+b \sqrt [3]{x}}}+\frac{10 a \sqrt{a x+b \sqrt [3]{x}}}{7 b^2 x^{2/3}}-\frac{6 \sqrt{a x+b \sqrt [3]{x}}}{7 b x^{4/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[b*x^(1/3) + a*x]),x]

[Out]

(-6*Sqrt[b*x^(1/3) + a*x])/(7*b*x^(4/3)) + (10*a*Sqrt[b*x^(1/3) + a*x])/(7*b^2*x^(2/3)) + (5*a^(7/4)*(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])/(7*b^(9/4)*Sqrt[b*x^(1/3) + a*x])

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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{x^2 \sqrt{b \sqrt [3]{x}+a x}} \, dx &=3 \operatorname{Subst}\left (\int \frac{1}{x^4 \sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{6 \sqrt{b \sqrt [3]{x}+a x}}{7 b x^{4/3}}-\frac{(15 a) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{7 b}\\ &=-\frac{6 \sqrt{b \sqrt [3]{x}+a x}}{7 b x^{4/3}}+\frac{10 a \sqrt{b \sqrt [3]{x}+a x}}{7 b^2 x^{2/3}}+\frac{\left (5 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{7 b^2}\\ &=-\frac{6 \sqrt{b \sqrt [3]{x}+a x}}{7 b x^{4/3}}+\frac{10 a \sqrt{b \sqrt [3]{x}+a x}}{7 b^2 x^{2/3}}+\frac{\left (5 a^2 \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 )}{7 b^2 \sqrt{b \sqrt [3]{x}+a x}}\\ &=-\frac{6 \sqrt{b \sqrt [3]{x}+a x}}{7 b x^{4/3}}+\frac{10 a \sqrt{b \sqrt [3]{x}+a x}}{7 b^2 x^{2/3}}+\frac{\left (10 a^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 )}{7 b^2 \sqrt{b \sqrt [3]{x}+a x}}\\ &=-\frac{6 \sqrt{b \sqrt [3]{x}+a x}}{7 b x^{4/3}}+\frac{10 a \sqrt{b \sqrt [3]{x}+a x}}{7 b^2 x^{2/3}}+\frac{5 a^{7/4} \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 )}{7 b^{9/4} \sqrt{b \sqrt [3]{x}+a x}}\\ \end{align*}

Mathematica [C]  time = 0.0499106, size = 59, normalized size = 0.36 \[ -\frac{6 \sqrt{\frac{a x^{2/3}}{b}+1} \, _2F_1\left (-\frac{7}{4},\frac{1}{2};-\frac{3}{4};-\frac{a x^{2/3}}{b}\right )}{7 x \sqrt{a x+b \sqrt [3]{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*Sqrt[1 + (a*x^(2/3))/b]*Hypergeometric2F1[-7/4, 1/2, -3/4, -((a*x^(2/3))/b)])/(7*x*Sqrt[b*x^(1/3) + a*x])

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Maple [A]  time = 0.022, size = 142, normalized size = 0.9 \begin{align*}{\frac{1}{7\,{b}^{2}} \left ( 5\,a\sqrt{-ab}\sqrt{{\frac{a\sqrt [3]{x}+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-2\,{\frac{a\sqrt [3]{x}-\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{a\sqrt [3]{x}}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{a\sqrt [3]{x}+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){x}^{4/3}+4\,abx+10\,{x}^{5/3}{a}^{2}-6\,\sqrt [3]{x}{b}^{2} \right ){\frac{1}{\sqrt{\sqrt [3]{x} \left ( b+a{x}^{{\frac{2}{3}}} \right ) }}}{x}^{-{\frac{4}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*(5*a*(-a*b)^(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)*(-x^(1/3)*a/(-a*b)^(1/2))^(1/2)*EllipticF(((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*x^
(4/3)+4*a*b*x+10*x^(5/3)*a^2-6*x^(1/3)*b^2)/b^2/(x^(1/3)*(b+a*x^(2/3)))^(1/2)/x^(4/3)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a x + b x^{\frac{1}{3}}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{2} x^{2} - a b x^{\frac{4}{3}} + b^{2} x^{\frac{2}{3}}\right )} \sqrt{a x + b x^{\frac{1}{3}}}}{a^{3} x^{5} + b^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \sqrt{a x + b \sqrt [3]{x}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a x + b x^{\frac{1}{3}}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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