3.4.72 \(\int \frac {\sqrt {a+b x^3}}{x} \, dx\) [372]

Optimal. Leaf size=43 \[ \frac {2}{3} \sqrt {a+b x^3}-\frac {2}{3} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 43, 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, 52, 65, 214} \begin {gather*} \frac {2}{3} \sqrt {a+b x^3}-\frac {2}{3} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^3]/x,x]

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 {\sqrt {a+b x^3}}{x} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^3\right )\\ &=\frac {2}{3} \sqrt {a+b x^3}+\frac {1}{3} a \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^3\right )\\ &=\frac {2}{3} \sqrt {a+b x^3}+\frac {(2 a) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^3}\right )}{3 b}\\ &=\frac {2}{3} \sqrt {a+b x^3}-\frac {2}{3} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 43, normalized size = 1.00 \begin {gather*} \frac {2}{3} \sqrt {a+b x^3}-\frac {2}{3} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^3]/x,x]

[Out]

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

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Maple [A]
time = 0.13, size = 32, normalized size = 0.74

method result size
default \(-\frac {2 \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right ) \sqrt {a}}{3}+\frac {2 \sqrt {b \,x^{3}+a}}{3}\) \(32\)
elliptic \(-\frac {2 \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right ) \sqrt {a}}{3}+\frac {2 \sqrt {b \,x^{3}+a}}{3}\) \(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 49, normalized size = 1.14 \begin {gather*} \frac {1}{3} \, \sqrt {a} \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right ) + \frac {2}{3} \, \sqrt {b x^{3} + a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 84, normalized size = 1.95 \begin {gather*} \left [\frac {1}{3} \, \sqrt {a} \log \left (\frac {b x^{3} - 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + \frac {2}{3} \, \sqrt {b x^{3} + a}, \frac {2}{3} \, \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + \frac {2}{3} \, \sqrt {b x^{3} + a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (37) = 74\).
time = 0.68, size = 76, normalized size = 1.77 \begin {gather*} - \frac {2 \sqrt {a} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{3} + \frac {2 a}{3 \sqrt {b} x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {2 \sqrt {b} x^{\frac {3}{2}}}{3 \sqrt {\frac {a}{b x^{3}} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.72, size = 36, normalized size = 0.84 \begin {gather*} \frac {2 \, a \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{3 \, \sqrt {-a}} + \frac {2}{3} \, \sqrt {b x^{3} + a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.21, size = 52, normalized size = 1.21 \begin {gather*} \frac {\sqrt {a}\,\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}+\frac {2\,\sqrt {b\,x^3+a}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(1/2)/x,x)

[Out]

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

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