3.2945 \(\int \frac{\sqrt{a+b (c x^2)^{3/2}}}{x} \, dx\)

Optimal. Leaf size=55 \[ \frac{2}{3} \sqrt{a+b \left (c x^2\right )^{3/2}}-\frac{2}{3} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \left (c x^2\right )^{3/2}}}{\sqrt{a}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0429101, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {368, 266, 50, 63, 208} \[ \frac{2}{3} \sqrt{a+b \left (c x^2\right )^{3/2}}-\frac{2}{3} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \left (c x^2\right )^{3/2}}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*(c*x^2)^(3/2)]/x,x]

[Out]

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

Rule 368

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*((c*x^q
)^(1/q))^(m + 1)), Subst[Int[x^m*(a + b*x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q
}, x] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]

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

Mathematica [A]  time = 0.0239491, size = 55, normalized size = 1. \[ \frac{2}{3} \sqrt{a+b \left (c x^2\right )^{3/2}}-\frac{2}{3} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \left (c x^2\right )^{3/2}}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*(c*x^2)^(3/2)]/x,x]

[Out]

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

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Maple [F]  time = 0.048, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x}\sqrt{a+b \left ( c{x}^{2} \right ) ^{{\frac{3}{2}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.35816, size = 331, normalized size = 6.02 \begin{align*} \left [\frac{1}{3} \, \sqrt{a} \log \left (\frac{b c^{2} x^{4} - 2 \, \sqrt{\sqrt{c x^{2}} b c x^{2} + a} \sqrt{c x^{2}} \sqrt{a} + 2 \, \sqrt{c x^{2}} a}{x^{4}}\right ) + \frac{2}{3} \, \sqrt{\sqrt{c x^{2}} b c x^{2} + a}, \frac{2}{3} \, \sqrt{-a} \arctan \left (\frac{\sqrt{\sqrt{c x^{2}} b c x^{2} + a} \sqrt{-a}}{a}\right ) + \frac{2}{3} \, \sqrt{\sqrt{c x^{2}} b c x^{2} + a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + b \left (c x^{2}\right )^{\frac{3}{2}}}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*(c*x**2)**(3/2))/x, x)

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Giac [A]  time = 1.14873, size = 57, normalized size = 1.04 \begin{align*} \frac{2 \, a \arctan \left (\frac{\sqrt{b c^{\frac{3}{2}} x^{3} + a}}{\sqrt{-a}}\right )}{3 \, \sqrt{-a}} + \frac{2}{3} \, \sqrt{b c^{\frac{3}{2}} x^{3} + a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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