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

Optimal. Leaf size=63 \[ -\frac{\left (a+b x^2\right )^{3/2}}{x}+\frac{3}{2} b x \sqrt{a+b x^2}+\frac{3}{2} a \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0177573, antiderivative size = 63, 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 = {277, 195, 217, 206} \[ -\frac{\left (a+b x^2\right )^{3/2}}{x}+\frac{3}{2} b x \sqrt{a+b x^2}+\frac{3}{2} a \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/x^2,x]

[Out]

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

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.0079266, size = 50, normalized size = 0.79 \[ -\frac{a \sqrt{a+b x^2} \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};-\frac{b x^2}{a}\right )}{x \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/x^2,x]

[Out]

-((a*Sqrt[a + b*x^2]*Hypergeometric2F1[-3/2, -1/2, 1/2, -((b*x^2)/a)])/(x*Sqrt[1 + (b*x^2)/a]))

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Maple [A]  time = 0.004, size = 69, normalized size = 1.1 \begin{align*} -{\frac{1}{ax} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{bx}{a} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,bx}{2}\sqrt{b{x}^{2}+a}}+{\frac{3\,a}{2}\sqrt{b}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.56303, size = 271, normalized size = 4.3 \begin{align*} \left [\frac{3 \, a \sqrt{b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \, \sqrt{b x^{2} + a}{\left (b x^{2} - 2 \, a\right )}}{4 \, x}, -\frac{3 \, a \sqrt{-b} x \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) - \sqrt{b x^{2} + a}{\left (b x^{2} - 2 \, a\right )}}{2 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.29281, size = 88, normalized size = 1.4 \begin{align*} - \frac{a^{\frac{3}{2}}}{x \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{\sqrt{a} b x}{2 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 a \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{2} + \frac{b^{2} x^{3}}{2 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.47025, size = 99, normalized size = 1.57 \begin{align*} \frac{1}{2} \, \sqrt{b x^{2} + a} b x - \frac{3}{4} \, a \sqrt{b} \log \left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2}\right ) + \frac{2 \, a^{2} \sqrt{b}}{{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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