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

Optimal. Leaf size=47 \[ -\frac{A \sqrt{a+c x^2}}{a x}-\frac{B \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a}} \]

[Out]

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

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Rubi [A]  time = 0.0308582, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {807, 266, 63, 208} \[ -\frac{A \sqrt{a+c x^2}}{a x}-\frac{B \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

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

Mathematica [A]  time = 0.0157645, size = 47, normalized size = 1. \[ -\frac{A \sqrt{a+c x^2}}{a x}-\frac{B \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 49, normalized size = 1. \begin{align*} -{B\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}-{\frac{A}{ax}\sqrt{c{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.7426, size = 240, normalized size = 5.11 \begin{align*} \left [\frac{B \sqrt{a} x \log \left (-\frac{c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \, \sqrt{c x^{2} + a} A}{2 \, a x}, \frac{B \sqrt{-a} x \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) - \sqrt{c x^{2} + a} A}{a x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.9938, size = 41, normalized size = 0.87 \begin{align*} - \frac{A \sqrt{c} \sqrt{\frac{a}{c x^{2}} + 1}}{a} - \frac{B \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{c} x} \right )}}{\sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*sqrt(c)*sqrt(a/(c*x**2) + 1)/a - B*asinh(sqrt(a)/(sqrt(c)*x))/sqrt(a)

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Giac [A]  time = 1.13986, size = 88, normalized size = 1.87 \begin{align*} \frac{2 \, B \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + \frac{2 \, A \sqrt{c}}{{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} - a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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