3.391 \(\int \frac{(a+b x^3)^{3/2}}{x} \, dx\)

Optimal. Leaf size=59 \[ -\frac{2}{3} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )+\frac{2}{3} a \sqrt{a+b x^3}+\frac{2}{9} \left (a+b x^3\right )^{3/2} \]

[Out]

(2*a*Sqrt[a + b*x^3])/3 + (2*(a + b*x^3)^(3/2))/9 - (2*a^(3/2)*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/3

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Rubi [A]  time = 0.0343941, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 50, 63, 208} \[ -\frac{2}{3} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )+\frac{2}{3} a \sqrt{a+b x^3}+\frac{2}{9} \left (a+b x^3\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(3/2)/x,x]

[Out]

(2*a*Sqrt[a + b*x^3])/3 + (2*(a + b*x^3)^(3/2))/9 - (2*a^(3/2)*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{\left (a+b x^3\right )^{3/2}}{x} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,x^3\right )\\ &=\frac{2}{9} \left (a+b x^3\right )^{3/2}+\frac{1}{3} a \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^3\right )\\ &=\frac{2}{3} a \sqrt{a+b x^3}+\frac{2}{9} \left (a+b x^3\right )^{3/2}+\frac{1}{3} a^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^3\right )\\ &=\frac{2}{3} a \sqrt{a+b x^3}+\frac{2}{9} \left (a+b x^3\right )^{3/2}+\frac{\left (2 a^2\right ) \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} a \sqrt{a+b x^3}+\frac{2}{9} \left (a+b x^3\right )^{3/2}-\frac{2}{3} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [A]  time = 0.030422, size = 51, normalized size = 0.86 \[ \frac{2}{9} \left (\sqrt{a+b x^3} \left (4 a+b x^3\right )-3 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(3/2)/x,x]

[Out]

(2*(Sqrt[a + b*x^3]*(4*a + b*x^3) - 3*a^(3/2)*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]]))/9

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Maple [A]  time = 0.011, size = 48, normalized size = 0.8 \begin{align*}{\frac{2\,b{x}^{3}}{9}\sqrt{b{x}^{3}+a}}+{\frac{8\,a}{9}\sqrt{b{x}^{3}+a}}-{\frac{2}{3}{a}^{{\frac{3}{2}}}{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*b*x^3*(b*x^3+a)^(1/2)+8/9*a*(b*x^3+a)^(1/2)-2/3*a^(3/2)*arctanh((b*x^3+a)^(1/2)/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)^(3/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.5212, size = 258, normalized size = 4.37 \begin{align*} \left [\frac{1}{3} \, a^{\frac{3}{2}} \log \left (\frac{b x^{3} - 2 \, \sqrt{b x^{3} + a} \sqrt{a} + 2 \, a}{x^{3}}\right ) + \frac{2}{9} \,{\left (b x^{3} + 4 \, a\right )} \sqrt{b x^{3} + a}, \frac{2}{3} \, \sqrt{-a} a \arctan \left (\frac{\sqrt{b x^{3} + a} \sqrt{-a}}{a}\right ) + \frac{2}{9} \,{\left (b x^{3} + 4 \, a\right )} \sqrt{b x^{3} + a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*a^(3/2)*log((b*x^3 - 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) + 2/9*(b*x^3 + 4*a)*sqrt(b*x^3 + a), 2/3*sqrt(
-a)*a*arctan(sqrt(b*x^3 + a)*sqrt(-a)/a) + 2/9*(b*x^3 + 4*a)*sqrt(b*x^3 + a)]

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Sympy [A]  time = 2.30488, size = 83, normalized size = 1.41 \begin{align*} \frac{8 a^{\frac{3}{2}} \sqrt{1 + \frac{b x^{3}}{a}}}{9} + \frac{a^{\frac{3}{2}} \log{\left (\frac{b x^{3}}{a} \right )}}{3} - \frac{2 a^{\frac{3}{2}} \log{\left (\sqrt{1 + \frac{b x^{3}}{a}} + 1 \right )}}{3} + \frac{2 \sqrt{a} b x^{3} \sqrt{1 + \frac{b x^{3}}{a}}}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1304, size = 68, normalized size = 1.15 \begin{align*} \frac{2 \, a^{2} \arctan \left (\frac{\sqrt{b x^{3} + a}}{\sqrt{-a}}\right )}{3 \, \sqrt{-a}} + \frac{2}{9} \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} + \frac{2}{3} \, \sqrt{b x^{3} + a} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*a^2*arctan(sqrt(b*x^3 + a)/sqrt(-a))/sqrt(-a) + 2/9*(b*x^3 + a)^(3/2) + 2/3*sqrt(b*x^3 + a)*a