3.5.25 \(\int \frac {1}{x (a+b x^3)^{3/2}} \, dx\) [425]

Optimal. Leaf size=46 \[ \frac {2}{3 a \sqrt {a+b x^3}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {272, 53, 65, 214} \begin {gather*} \frac {2}{3 a \sqrt {a+b x^3}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^3)^(3/2)),x]

[Out]

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

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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

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Mathematica [A]
time = 0.05, size = 46, normalized size = 1.00 \begin {gather*} \frac {2}{3 a \sqrt {a+b x^3}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^3)^(3/2)),x]

[Out]

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

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Maple [A]
time = 0.13, size = 39, normalized size = 0.85

method	result	size
default	\(\frac {2}{3 a \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {2 \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {3}{2}}}\)	\(39\)
elliptic	\(\frac {2}{3 a \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {2 \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {3}{2}}}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 52, normalized size = 1.13 \begin {gather*} \frac {\log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{3 \, a^{\frac {3}{2}}} + \frac {2}{3 \, \sqrt {b x^{3} + a} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 129, normalized size = 2.80 \begin {gather*} \left [\frac {{\left (b x^{3} + a\right )} \sqrt {a} \log \left (\frac {b x^{3} - 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + 2 \, \sqrt {b x^{3} + a} a}{3 \, {\left (a^{2} b x^{3} + a^{3}\right )}}, \frac {2 \, {\left ({\left (b x^{3} + a\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + \sqrt {b x^{3} + a} a\right )}}{3 \, {\left (a^{2} b x^{3} + a^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (39) = 78\).
time = 0.85, size = 184, normalized size = 4.00 \begin {gather*} \frac {2 a^{3} \sqrt {1 + \frac {b x^{3}}{a}}}{3 a^{\frac {9}{2}} + 3 a^{\frac {7}{2}} b x^{3}} + \frac {a^{3} \log {\left (\frac {b x^{3}}{a} \right )}}{3 a^{\frac {9}{2}} + 3 a^{\frac {7}{2}} b x^{3}} - \frac {2 a^{3} \log {\left (\sqrt {1 + \frac {b x^{3}}{a}} + 1 \right )}}{3 a^{\frac {9}{2}} + 3 a^{\frac {7}{2}} b x^{3}} + \frac {a^{2} b x^{3} \log {\left (\frac {b x^{3}}{a} \right )}}{3 a^{\frac {9}{2}} + 3 a^{\frac {7}{2}} b x^{3}} - \frac {2 a^{2} b x^{3} \log {\left (\sqrt {1 + \frac {b x^{3}}{a}} + 1 \right )}}{3 a^{\frac {9}{2}} + 3 a^{\frac {7}{2}} b x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a**3*sqrt(1 + b*x**3/a)/(3*a**(9/2) + 3*a**(7/2)*b*x**3) + a**3*log(b*x**3/a)/(3*a**(9/2) + 3*a**(7/2)*b*x**
3) - 2*a**3*log(sqrt(1 + b*x**3/a) + 1)/(3*a**(9/2) + 3*a**(7/2)*b*x**3) + a**2*b*x**3*log(b*x**3/a)/(3*a**(9/
2) + 3*a**(7/2)*b*x**3) - 2*a**2*b*x**3*log(sqrt(1 + b*x**3/a) + 1)/(3*a**(9/2) + 3*a**(7/2)*b*x**3)

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Giac [A]
time = 1.04, size = 41, normalized size = 0.89 \begin {gather*} \frac {2 \, \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{3 \, \sqrt {-a} a} + \frac {2}{3 \, \sqrt {b x^{3} + a} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.23, size = 55, normalized size = 1.20 \begin {gather*} \frac {\ln \left (\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3\,\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}{x^6}\right )}{3\,a^{3/2}}+\frac {2}{3\,a\,\sqrt {b\,x^3+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x^3)^(3/2)),x)

[Out]

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

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