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

Optimal. Leaf size=90 \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [3]{x}}{\sqrt{a x+b x^{2/3}}}\right )}{4 b^{3/2}}-\frac{3 a \sqrt{a x+b x^{2/3}}}{4 b x^{2/3}}-\frac{3 \sqrt{a x+b x^{2/3}}}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.138582, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2020, 2025, 2029, 206} \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [3]{x}}{\sqrt{a x+b x^{2/3}}}\right )}{4 b^{3/2}}-\frac{3 a \sqrt{a x+b x^{2/3}}}{4 b x^{2/3}}-\frac{3 \sqrt{a x+b x^{2/3}}}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^(2/3) + a*x]/x^2,x]

[Out]

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

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 2029

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.0415882, size = 57, normalized size = 0.63 \[ -\frac{2 a^2 \left (a \sqrt [3]{x}+b\right ) \sqrt{a x+b x^{2/3}} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{\sqrt [3]{x} a}{b}+1\right )}{b^3 \sqrt [3]{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 80, normalized size = 0.9 \begin{align*}{\frac{3}{4\,x}\sqrt{b{x}^{{\frac{2}{3}}}+ax} \left ({\it Artanh} \left ({\sqrt{b+a\sqrt [3]{x}}{\frac{1}{\sqrt{b}}}} \right ) b{a}^{2}{x}^{{\frac{2}{3}}}-{b}^{{\frac{3}{2}}} \left ( b+a\sqrt [3]{x} \right ) ^{{\frac{3}{2}}}-{b}^{{\frac{5}{2}}}\sqrt{b+a\sqrt [3]{x}} \right ){b}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{b+a\sqrt [3]{x}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21738, size = 97, normalized size = 1.08 \begin{align*} -\frac{3 \,{\left (\frac{a^{3} \arctan \left (\frac{\sqrt{a x^{\frac{1}{3}} + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b} + \frac{{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{3}{2}} a^{3} + \sqrt{a x^{\frac{1}{3}} + b} a^{3} b}{a^{2} b x^{\frac{2}{3}}}\right )}}{4 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*(a^3*arctan(sqrt(a*x^(1/3) + b)/sqrt(-b))/(sqrt(-b)*b) + ((a*x^(1/3) + b)^(3/2)*a^3 + sqrt(a*x^(1/3) + b)
*a^3*b)/(a^2*b*x^(2/3)))/a