3.561 \(\int \frac{A+B x^2}{x \sqrt{a+b x^2}} \, dx\)

Optimal. Leaf size=43 \[ \frac{B \sqrt{a+b x^2}}{b}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{\sqrt{a}} \]

[Out]

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

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Rubi [A]  time = 0.0331544, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {446, 80, 63, 208} \[ \frac{B \sqrt{a+b x^2}}{b}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x*Sqrt[a + b*x^2]),x]

[Out]

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

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

Mathematica [A]  time = 0.0196275, size = 43, normalized size = 1. \[ \frac{B \sqrt{a+b x^2}}{b}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x*Sqrt[a + b*x^2]),x]

[Out]

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

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Maple [A]  time = 0.008, size = 45, normalized size = 1.1 \begin{align*}{\frac{B}{b}\sqrt{b{x}^{2}+a}}-{A\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x/(b*x^2+a)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58653, size = 246, normalized size = 5.72 \begin{align*} \left [\frac{A \sqrt{a} b \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \, \sqrt{b x^{2} + a} B a}{2 \, a b}, \frac{A \sqrt{-a} b \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) + \sqrt{b x^{2} + a} B a}{a b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 6.52413, size = 61, normalized size = 1.42 \begin{align*} \frac{A \operatorname{atan}{\left (\frac{1}{\sqrt{- \frac{1}{a}} \sqrt{a + b x^{2}}} \right )}}{a \sqrt{- \frac{1}{a}}} - \frac{B \left (\begin{cases} - \frac{x^{2}}{\sqrt{a}} & \text{for}\: b = 0 \\- \frac{2 \sqrt{a + b x^{2}}}{b} & \text{otherwise} \end{cases}\right )}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x/(b*x**2+a)**(1/2),x)

[Out]

A*atan(1/(sqrt(-1/a)*sqrt(a + b*x**2)))/(a*sqrt(-1/a)) - B*Piecewise((-x**2/sqrt(a), Eq(b, 0)), (-2*sqrt(a + b
*x**2)/b, True))/2

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Giac [A]  time = 1.13029, size = 51, normalized size = 1.19 \begin{align*} \frac{A \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + \frac{\sqrt{b x^{2} + a} B}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + sqrt(b*x^2 + a)*B/b