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

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

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Rubi [A]  time = 0.025524, antiderivative size = 43, normalized size of antiderivative = 1., 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 = {266, 50, 63, 208} \[ \frac{2}{3} \sqrt{a+b x^3}-\frac{2}{3} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right ) \]

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 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]]

Rule 50

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 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 208

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

Rubi steps

\begin{align*} \int \frac{\sqrt{a+b x^3}}{x} \, dx &=\frac{1}{3} \operatorname{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 \operatorname{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) \operatorname{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*}

Mathematica [A]  time = 0.0094017, size = 43, normalized size = 1. \[ \frac{2}{3} \sqrt{a+b x^3}-\frac{2}{3} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right ) \]

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 = 0.157, size = 32, normalized size = 0.7 \begin{align*} -{\frac{2}{3}{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ) \sqrt{a}}+{\frac{2}{3}\sqrt{b{x}^{3}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52955, size = 217, normalized size = 5.05 \begin{align*} \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{align*}

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]  time = 1.90964, size = 76, normalized size = 1.77 \begin{align*} - \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{align*}

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 = 1.12086, size = 49, normalized size = 1.14 \begin{align*} \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{align*}

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)