3.391 \(\int \frac{(a+b x^2)^{5/2}}{x} \, dx\)

Optimal. Leaf size=72 \[ a^2 \sqrt{a+b x^2}+a^{5/2} \left (-\tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\right )+\frac{1}{3} a \left (a+b x^2\right )^{3/2}+\frac{1}{5} \left (a+b x^2\right )^{5/2} \]

[Out]

a^2*Sqrt[a + b*x^2] + (a*(a + b*x^2)^(3/2))/3 + (a + b*x^2)^(5/2)/5 - a^(5/2)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]

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Rubi [A]  time = 0.0434574, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 50, 63, 208} \[ a^2 \sqrt{a+b x^2}+a^{5/2} \left (-\tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\right )+\frac{1}{3} a \left (a+b x^2\right )^{3/2}+\frac{1}{5} \left (a+b x^2\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/x,x]

[Out]

a^2*Sqrt[a + b*x^2] + (a*(a + b*x^2)^(3/2))/3 + (a + b*x^2)^(5/2)/5 - a^(5/2)*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 )^{5/2}}{x} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{5} \left (a+b x^2\right )^{5/2}+\frac{1}{2} a \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{3} a \left (a+b x^2\right )^{3/2}+\frac{1}{5} \left (a+b x^2\right )^{5/2}+\frac{1}{2} a^2 \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^2\right )\\ &=a^2 \sqrt{a+b x^2}+\frac{1}{3} a \left (a+b x^2\right )^{3/2}+\frac{1}{5} \left (a+b x^2\right )^{5/2}+\frac{1}{2} a^3 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=a^2 \sqrt{a+b x^2}+\frac{1}{3} a \left (a+b x^2\right )^{3/2}+\frac{1}{5} \left (a+b x^2\right )^{5/2}+\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{b}\\ &=a^2 \sqrt{a+b x^2}+\frac{1}{3} a \left (a+b x^2\right )^{3/2}+\frac{1}{5} \left (a+b x^2\right )^{5/2}-a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0267292, size = 62, normalized size = 0.86 \[ \frac{1}{15} \sqrt{a+b x^2} \left (23 a^2+11 a b x^2+3 b^2 x^4\right )-a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x,x]

[Out]

(Sqrt[a + b*x^2]*(23*a^2 + 11*a*b*x^2 + 3*b^2*x^4))/15 - a^(5/2)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]

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Maple [A]  time = 0.005, size = 66, normalized size = 0.9 \begin{align*}{\frac{1}{5} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{a}{3} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{a}^{{\frac{5}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) +{a}^{2}\sqrt{b{x}^{2}+a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)/x,x)

[Out]

1/5*(b*x^2+a)^(5/2)+1/3*a*(b*x^2+a)^(3/2)-a^(5/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+a^2*(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)^(5/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.589, size = 311, normalized size = 4.32 \begin{align*} \left [\frac{1}{2} \, a^{\frac{5}{2}} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + \frac{1}{15} \,{\left (3 \, b^{2} x^{4} + 11 \, a b x^{2} + 23 \, a^{2}\right )} \sqrt{b x^{2} + a}, \sqrt{-a} a^{2} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) + \frac{1}{15} \,{\left (3 \, b^{2} x^{4} + 11 \, a b x^{2} + 23 \, a^{2}\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)^(5/2)/x,x, algorithm="fricas")

[Out]

[1/2*a^(5/2)*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 1/15*(3*b^2*x^4 + 11*a*b*x^2 + 23*a^2)*sqrt
(b*x^2 + a), sqrt(-a)*a^2*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + 1/15*(3*b^2*x^4 + 11*a*b*x^2 + 23*a^2)*sqrt(b*x^2
+ a)]

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Sympy [A]  time = 3.40694, size = 105, normalized size = 1.46 \begin{align*} \frac{23 a^{\frac{5}{2}} \sqrt{1 + \frac{b x^{2}}{a}}}{15} + \frac{a^{\frac{5}{2}} \log{\left (\frac{b x^{2}}{a} \right )}}{2} - a^{\frac{5}{2}} \log{\left (\sqrt{1 + \frac{b x^{2}}{a}} + 1 \right )} + \frac{11 a^{\frac{3}{2}} b x^{2} \sqrt{1 + \frac{b x^{2}}{a}}}{15} + \frac{\sqrt{a} b^{2} x^{4} \sqrt{1 + \frac{b x^{2}}{a}}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)/x,x)

[Out]

23*a**(5/2)*sqrt(1 + b*x**2/a)/15 + a**(5/2)*log(b*x**2/a)/2 - a**(5/2)*log(sqrt(1 + b*x**2/a) + 1) + 11*a**(3
/2)*b*x**2*sqrt(1 + b*x**2/a)/15 + sqrt(a)*b**2*x**4*sqrt(1 + b*x**2/a)/5

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Giac [A]  time = 2.772, size = 84, normalized size = 1.17 \begin{align*} \frac{a^{3} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + \frac{1}{5} \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} + \frac{1}{3} \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a + \sqrt{b x^{2} + a} a^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x,x, algorithm="giac")

[Out]

a^3*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 1/5*(b*x^2 + a)^(5/2) + 1/3*(b*x^2 + a)^(3/2)*a + sqrt(b*x^2 +
a)*a^2