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

Optimal. Leaf size=54 \[ a^{3/2} \left (-\tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\right )+a \sqrt{a+b x^2}+\frac{1}{3} \left (a+b x^2\right )^{3/2} \]

[Out]

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

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Rubi [A]  time = 0.0356325, antiderivative size = 54, 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} \[ a^{3/2} \left (-\tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\right )+a \sqrt{a+b x^2}+\frac{1}{3} \left (a+b x^2\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/x,x]

[Out]

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

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

Mathematica [A]  time = 0.020053, size = 50, normalized size = 0.93 \[ \frac{1}{3} \sqrt{a+b x^2} \left (4 a+b x^2\right )-a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/x,x]

[Out]

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

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Maple [A]  time = 0.003, size = 52, normalized size = 1. \begin{align*}{\frac{1}{3} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) +a\sqrt{b{x}^{2}+a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58719, size = 251, normalized size = 4.65 \begin{align*} \left [\frac{1}{2} \, a^{\frac{3}{2}} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + \frac{1}{3} \,{\left (b x^{2} + 4 \, a\right )} \sqrt{b x^{2} + a}, \sqrt{-a} a \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) + \frac{1}{3} \,{\left (b x^{2} + 4 \, a\right )} \sqrt{b x^{2} + a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.86413, size = 78, normalized size = 1.44 \begin{align*} \frac{4 a^{\frac{3}{2}} \sqrt{1 + \frac{b x^{2}}{a}}}{3} + \frac{a^{\frac{3}{2}} \log{\left (\frac{b x^{2}}{a} \right )}}{2} - a^{\frac{3}{2}} \log{\left (\sqrt{1 + \frac{b x^{2}}{a}} + 1 \right )} + \frac{\sqrt{a} b x^{2} \sqrt{1 + \frac{b x^{2}}{a}}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.37338, size = 65, normalized size = 1.2 \begin{align*} \frac{a^{2} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + \frac{1}{3} \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} + \sqrt{b x^{2} + a} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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