3.21.19 \(\int \frac {1}{\sqrt {a+\frac {b}{x^3}} x} \, dx\) [2019]

Optimal. Leaf size=27 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{3 \sqrt {a}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 27, 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 = {272, 65, 214} \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{3 \sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b/x^3]*x),x]

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(59\) vs. \(2(27)=54\).
time = 0.14, size = 59, normalized size = 2.19 \begin {gather*} \frac {2 \sqrt {b+a x^3} \tanh ^{-1}\left (\frac {\sqrt {b+a x^3}}{\sqrt {a} x^{3/2}}\right )}{3 \sqrt {a} \sqrt {a+\frac {b}{x^3}} x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b/x^3]*x),x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.03, size = 480, normalized size = 17.78

method	result	size
default	\(-\frac {4 \left (a \,x^{3}+b \right ) \left (-1+i \sqrt {3}\right ) \sqrt {-\frac {\left (i \sqrt {3}-3\right ) x a}{\left (-1+i \sqrt {3}\right ) \left (-a x +\left (-a^{2} b \right )^{\frac {1}{3}}\right )}}\, \left (-a x +\left (-a^{2} b \right )^{\frac {1}{3}}\right )^{2} \sqrt {\frac {i \sqrt {3}\, \left (-a^{2} b \right )^{\frac {1}{3}}+2 a x +\left (-a^{2} b \right )^{\frac {1}{3}}}{\left (1+i \sqrt {3}\right ) \left (-a x +\left (-a^{2} b \right )^{\frac {1}{3}}\right )}}\, \sqrt {\frac {i \sqrt {3}\, \left (-a^{2} b \right )^{\frac {1}{3}}-2 a x -\left (-a^{2} b \right )^{\frac {1}{3}}}{\left (-1+i \sqrt {3}\right ) \left (-a x +\left (-a^{2} b \right )^{\frac {1}{3}}\right )}}\, \left (\EllipticF \left (\sqrt {-\frac {\left (i \sqrt {3}-3\right ) x a}{\left (-1+i \sqrt {3}\right ) \left (-a x +\left (-a^{2} b \right )^{\frac {1}{3}}\right )}}, \sqrt {\frac {\left (i \sqrt {3}+3\right ) \left (-1+i \sqrt {3}\right )}{\left (1+i \sqrt {3}\right ) \left (i \sqrt {3}-3\right )}}\right )-\EllipticPi \left (\sqrt {-\frac {\left (i \sqrt {3}-3\right ) x a}{\left (-1+i \sqrt {3}\right ) \left (-a x +\left (-a^{2} b \right )^{\frac {1}{3}}\right )}}, \frac {-1+i \sqrt {3}}{i \sqrt {3}-3}, \sqrt {\frac {\left (i \sqrt {3}+3\right ) \left (-1+i \sqrt {3}\right )}{\left (1+i \sqrt {3}\right ) \left (i \sqrt {3}-3\right )}}\right )\right )}{\sqrt {\frac {a \,x^{3}+b}{x^{3}}}\, x \,a^{2} \sqrt {x \left (a \,x^{3}+b \right )}\, \left (i \sqrt {3}-3\right ) \sqrt {\frac {x \left (-a x +\left (-a^{2} b \right )^{\frac {1}{3}}\right ) \left (i \sqrt {3}\, \left (-a^{2} b \right )^{\frac {1}{3}}+2 a x +\left (-a^{2} b \right )^{\frac {1}{3}}\right ) \left (i \sqrt {3}\, \left (-a^{2} b \right )^{\frac {1}{3}}-2 a x -\left (-a^{2} b \right )^{\frac {1}{3}}\right )}{a^{2}}}}\)	\(480\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(a+b/x^3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-4/((a*x^3+b)/x^3)^(1/2)/x*(a*x^3+b)*(-1+I*3^(1/2))*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^
(1/2)*(-a*x+(-a^2*b)^(1/3))^2*((I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1
/3)))^(1/2)*((I*3^(1/2)*(-a^2*b)^(1/3)-2*a*x-(-a^2*b)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)/a^2*(
EllipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3
^(1/2))/(I*3^(1/2)-3))^(1/2))-EllipticPi((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2),(-1+I
*3^(1/2))/(I*3^(1/2)-3),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2)))/(x*(a*x^3+b))^(1/2)
/(I*3^(1/2)-3)/(1/a^2*x*(-a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*(-a^2*b)^(1/3)+2*a*x+(-a^2*b)^(1/3))*(I*3^(1/2)*(-a^2
*b)^(1/3)-2*a*x-(-a^2*b)^(1/3)))^(1/2)

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Maxima [A]
time = 0.51, size = 37, normalized size = 1.37 \begin {gather*} -\frac {\log \left (\frac {\sqrt {a + \frac {b}{x^{3}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{3}}} + \sqrt {a}}\right )}{3 \, \sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (19) = 38\).
time = 0.47, size = 102, normalized size = 3.78 \begin {gather*} \left [\frac {\log \left (-8 \, a^{2} x^{6} - 8 \, a b x^{3} - b^{2} - 4 \, {\left (2 \, a x^{6} + b x^{3}\right )} \sqrt {a} \sqrt {\frac {a x^{3} + b}{x^{3}}}\right )}{6 \, \sqrt {a}}, -\frac {\sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} x^{3} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{2 \, a x^{3} + b}\right )}{3 \, a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*log(-8*a^2*x^6 - 8*a*b*x^3 - b^2 - 4*(2*a*x^6 + b*x^3)*sqrt(a)*sqrt((a*x^3 + b)/x^3))/sqrt(a), -1/3*sqrt(
-a)*arctan(2*sqrt(-a)*x^3*sqrt((a*x^3 + b)/x^3)/(2*a*x^3 + b))/a]

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Sympy [A]
time = 0.54, size = 24, normalized size = 0.89 \begin {gather*} \frac {2 \operatorname {asinh}{\left (\frac {\sqrt {a} x^{\frac {3}{2}}}{\sqrt {b}} \right )}}{3 \sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*asinh(sqrt(a)*x**(3/2)/sqrt(b))/(3*sqrt(a))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [B]
time = 1.31, size = 19, normalized size = 0.70 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{3\,\sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b/x^3)^(1/2)),x)

[Out]

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

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