3.4.47 \(\int \frac {1}{x \sqrt {-b+a x^3}} \, dx\)

Optimal. Leaf size=29 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {a x^3-b}}{\sqrt {b}}\right )}{3 \sqrt {b}} \]

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Rubi [A]  time = 0.02, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {266, 63, 205} \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt {a x^3-b}}{\sqrt {b}}\right )}{3 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[-b + a*x^3]),x]

[Out]

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

Rule 63

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 205

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

Rule 266

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

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Mathematica [A]  time = 0.01, size = 29, normalized size = 1.00 \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt {a x^3-b}}{\sqrt {b}}\right )}{3 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[-b + a*x^3]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 29, normalized size = 1.00 \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt {-b+a x^3}}{\sqrt {b}}\right )}{3 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 64, normalized size = 2.21 \begin {gather*} \left [-\frac {\sqrt {-b} \log \left (\frac {a x^{3} - 2 \, \sqrt {a x^{3} - b} \sqrt {-b} - 2 \, b}{x^{3}}\right )}{3 \, b}, \frac {2 \, \arctan \left (\frac {\sqrt {a x^{3} - b}}{\sqrt {b}}\right )}{3 \, \sqrt {b}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.53, size = 21, normalized size = 0.72 \begin {gather*} \frac {2 \, \arctan \left (\frac {\sqrt {a x^{3} - b}}{\sqrt {b}}\right )}{3 \, \sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 26, normalized size = 0.90

method	result	size
default	\(-\frac {2 \arctanh \left (\frac {\sqrt {a \,x^{3}-b}}{\sqrt {-b}}\right )}{3 \sqrt {-b}}\)	\(26\)
elliptic	\(-\frac {2 \arctanh \left (\frac {\sqrt {a \,x^{3}-b}}{\sqrt {-b}}\right )}{3 \sqrt {-b}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.78, size = 21, normalized size = 0.72 \begin {gather*} \frac {2 \, \arctan \left (\frac {\sqrt {a x^{3} - b}}{\sqrt {b}}\right )}{3 \, \sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.08, size = 37, normalized size = 1.28 \begin {gather*} \frac {\ln \left (\frac {a\,x^3-2\,b+\sqrt {b}\,\sqrt {a\,x^3-b}\,2{}\mathrm {i}}{x^3}\right )\,1{}\mathrm {i}}{3\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a*x^3 - b)^(1/2)),x)

[Out]

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

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sympy [A]  time = 0.96, size = 60, normalized size = 2.07 \begin {gather*} \begin {cases} \frac {2 i \operatorname {acosh}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{\frac {3}{2}}} \right )}}{3 \sqrt {b}} & \text {for}\: \left |{\frac {b}{a x^{3}}}\right | > 1 \\- \frac {2 \operatorname {asin}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{\frac {3}{2}}} \right )}}{3 \sqrt {b}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*I*acosh(sqrt(b)/(sqrt(a)*x**(3/2)))/(3*sqrt(b)), Abs(b/(a*x**3)) > 1), (-2*asin(sqrt(b)/(sqrt(a)*
x**(3/2)))/(3*sqrt(b)), True))

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