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

Optimal. Leaf size=61 \[ \frac {3 a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {a x+b x^{2/3}}}\right )}{b^{3/2}}-\frac {3 \sqrt {a x+b x^{2/3}}}{b x^{2/3}} \]

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Rubi [A]  time = 0.09, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2025, 2029, 206} \begin {gather*} \frac {3 a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {a x+b x^{2/3}}}\right )}{b^{3/2}}-\frac {3 \sqrt {a x+b x^{2/3}}}{b x^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[b*x^(2/3) + a*x]),x]

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 90, normalized size = 1.48 \begin {gather*} \frac {6 a \sqrt [3]{x} \left (a \sqrt [3]{x}+b\right ) \left (\frac {\tanh ^{-1}\left (\sqrt {\frac {a \sqrt [3]{x}}{b}+1}\right )}{2 \sqrt {\frac {a \sqrt [3]{x}}{b}+1}}-\frac {b}{2 a \sqrt [3]{x}}\right )}{b^2 \sqrt {x^{2/3} \left (a \sqrt [3]{x}+b\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.14, size = 61, normalized size = 1.00 \begin {gather*} \frac {3 a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {a x+b x^{2/3}}}\right )}{b^{3/2}}-\frac {3 \sqrt {a x+b x^{2/3}}}{b x^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [A]  time = 0.22, size = 51, normalized size = 0.84 \begin {gather*} -\frac {3 \, {\left (\frac {a^{2} \arctan \left (\frac {\sqrt {a x^{\frac {1}{3}} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} + \frac {\sqrt {a x^{\frac {1}{3}} + b} a}{b x^{\frac {1}{3}}}\right )}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 61, normalized size = 1.00 \begin {gather*} \frac {3 \sqrt {a \,x^{\frac {1}{3}}+b}\, \left (a b \,x^{\frac {1}{3}} \arctanh \left (\frac {\sqrt {a \,x^{\frac {1}{3}}+b}}{\sqrt {b}}\right )-\sqrt {a \,x^{\frac {1}{3}}+b}\, b^{\frac {3}{2}}\right )}{\sqrt {a x +b \,x^{\frac {2}{3}}}\, b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a x + b x^{\frac {2}{3}}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{x\,\sqrt {a\,x+b\,x^{2/3}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a*x + b*x^(2/3))^(1/2)),x)

[Out]

int(1/(x*(a*x + b*x^(2/3))^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \sqrt {a x + b x^{\frac {2}{3}}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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