3.23.72 \(\int \frac {1}{(a+b x^{3/2})^{2/3}} \, dx\) [2272]

Optimal. Leaf size=82 \[ -\frac {2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} \sqrt {x}}{\sqrt [3]{a+b x^{3/2}}}}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {\log \left (\sqrt [3]{b} \sqrt {x}-\sqrt [3]{a+b x^{3/2}}\right )}{b^{2/3}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {249, 337} \begin {gather*} -\frac {2 \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b} \sqrt {x}}{\sqrt [3]{a+b x^{3/2}}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {\log \left (\sqrt [3]{b} \sqrt {x}-\sqrt [3]{a+b x^{3/2}}\right )}{b^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(3/2))^(-2/3),x]

[Out]

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

Rule 249

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k - 1)*(a + b*
x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, p}, x] && FractionQ[n]

Rule 337

Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Simp[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(
1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]

Rubi steps

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

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Mathematica [A]
time = 0.19, size = 138, normalized size = 1.68 \begin {gather*} \frac {-2 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt {x}}{\sqrt [3]{b} \sqrt {x}+2 \sqrt [3]{a+b x^{3/2}}}\right )-2 \log \left (-\sqrt [3]{b} \sqrt {x}+\sqrt [3]{a+b x^{3/2}}\right )+\log \left (b^{2/3} x+\sqrt [3]{b} \sqrt {x} \sqrt [3]{a+b x^{3/2}}+\left (a+b x^{3/2}\right )^{2/3}\right )}{3 b^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(3/2))^(-2/3),x]

[Out]

(-2*Sqrt[3]*ArcTan[(Sqrt[3]*b^(1/3)*Sqrt[x])/(b^(1/3)*Sqrt[x] + 2*(a + b*x^(3/2))^(1/3))] - 2*Log[-(b^(1/3)*Sq
rt[x]) + (a + b*x^(3/2))^(1/3)] + Log[b^(2/3)*x + b^(1/3)*Sqrt[x]*(a + b*x^(3/2))^(1/3) + (a + b*x^(3/2))^(2/3
)])/(3*b^(2/3))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \,x^{\frac {3}{2}}\right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 100, normalized size = 1.22 \begin {gather*} \frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}}}{\sqrt {x}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{3 \, b^{\frac {2}{3}}} + \frac {\log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{\sqrt {x}} + \frac {{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}}}{x}\right )}{3 \, b^{\frac {2}{3}}} - \frac {2 \, \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}}}{\sqrt {x}}\right )}{3 \, b^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
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/(a+b*x^(3/2))^(2/3),x, algorithm="fricas")

[Out]

Timed out

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Sympy [C] Result contains complex when optimal does not.
time = 0.46, size = 39, normalized size = 0.48 \begin {gather*} \frac {2 x \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{\frac {3}{2}} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {5}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x*gamma(2/3)*hyper((2/3, 2/3), (5/3,), b*x**(3/2)*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(5/3))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^(3/2) + a)^(-2/3), x)

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Mupad [B]
time = 1.21, size = 37, normalized size = 0.45 \begin {gather*} \frac {x\,{\left (\frac {b\,x^{3/2}}{a}+1\right )}^{2/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {2}{3},\frac {2}{3};\ \frac {5}{3};\ -\frac {b\,x^{3/2}}{a}\right )}{{\left (a+b\,x^{3/2}\right )}^{2/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x^(3/2))^(2/3),x)

[Out]

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

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