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

Optimal. Leaf size=60 \[ \frac {6 \sqrt [3]{x}}{b \sqrt {a x+b x^{2/3}}}-\frac {6 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {a x+b x^{2/3}}}\right )}{b^{3/2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2006, 2029, 206} \begin {gather*} \frac {6 \sqrt [3]{x}}{b \sqrt {a x+b x^{2/3}}}-\frac {6 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {a x+b x^{2/3}}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^(2/3) + a*x)^(-3/2),x]

[Out]

(6*x^(1/3))/(b*Sqrt[b*x^(2/3) + a*x]) - (6*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 2006

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1)*
x^(j - 1)), x] + Dist[(n*p + n - j + 1)/(a*(n - j)*(p + 1)), Int[(a*x^j + b*x^n)^(p + 1)/x^j, x], x] /; FreeQ[
{a, b}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && LtQ[p, -1]

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}{\left (b x^{2/3}+a x\right )^{3/2}} \, dx &=\frac {6 \sqrt [3]{x}}{b \sqrt {b x^{2/3}+a x}}+\frac {\int \frac {1}{x^{2/3} \sqrt {b x^{2/3}+a x}} \, dx}{b}\\ &=\frac {6 \sqrt [3]{x}}{b \sqrt {b x^{2/3}+a x}}-\frac {6 \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 {6 \sqrt [3]{x}}{b \sqrt {b x^{2/3}+a x}}-\frac {6 \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 [C]  time = 0.04, size = 45, normalized size = 0.75 \begin {gather*} \frac {6 \sqrt [3]{x} \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {\sqrt [3]{x} a}{b}+1\right )}{b \sqrt {a x+b x^{2/3}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^(2/3) + a*x)^(-3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 2.65, size = 71, normalized size = 1.18 \begin {gather*} \frac {6 \sqrt {a x+b x^{2/3}}}{b \sqrt [3]{x} \left (a \sqrt [3]{x}+b\right )}-\frac {6 \tanh ^{-1}\left (\frac {\sqrt {a x+b x^{2/3}}}{\sqrt {b} \sqrt [3]{x}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Sqrt[b*x^(2/3) + a*x])/(b*(b + a*x^(1/3))*x^(1/3)) - (6*ArcTanh[Sqrt[b*x^(2/3) + a*x]/(Sqrt[b]*x^(1/3))])/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/(b*x^(2/3)+a*x)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.21, size = 71, normalized size = 1.18 \begin {gather*} \frac {6 \, \arctan \left (\frac {\sqrt {a x^{\frac {1}{3}} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} - \frac {6 \, {\left (\sqrt {b} \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + \sqrt {-b}\right )}}{\sqrt {-b} b^{\frac {3}{2}}} + \frac {6}{\sqrt {a x^{\frac {1}{3}} + b} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*arctan(sqrt(a*x^(1/3) + b)/sqrt(-b))/(sqrt(-b)*b) - 6*(sqrt(b)*arctan(sqrt(b)/sqrt(-b)) + sqrt(-b))/(sqrt(-b
)*b^(3/2)) + 6/(sqrt(a*x^(1/3) + b)*b)

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maple [A]  time = 0.05, size = 56, normalized size = 0.93 \begin {gather*} -\frac {6 \left (a \,x^{\frac {1}{3}}+b \right ) \left (\sqrt {a \,x^{\frac {1}{3}}+b}\, b \arctanh \left (\frac {\sqrt {a \,x^{\frac {1}{3}}+b}}{\sqrt {b}}\right )-b^{\frac {3}{2}}\right ) x}{\left (a x +b \,x^{\frac {2}{3}}\right )^{\frac {3}{2}} b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.36, size = 40, normalized size = 0.67 \begin {gather*} -\frac {2\,x\,{\left (\frac {b}{a\,x^{1/3}}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {3}{2};\ \frac {5}{2};\ -\frac {b}{a\,x^{1/3}}\right )}{{\left (a\,x+b\,x^{2/3}\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x + b*x^(2/3))^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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