3.2939 \(\int \frac{\sqrt{a+b \sqrt{c x^2}}}{x^2} \, dx\)

Optimal. Leaf size=67 \[ -\frac{\sqrt{a+b \sqrt{c x^2}}}{x}-\frac{b \sqrt{c x^2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c x^2}}}{\sqrt{a}}\right )}{\sqrt{a} x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0289001, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {368, 47, 63, 208} \[ -\frac{\sqrt{a+b \sqrt{c x^2}}}{x}-\frac{b \sqrt{c x^2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c x^2}}}{\sqrt{a}}\right )}{\sqrt{a} x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c*x^2]]/x^2,x]

[Out]

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

Rule 368

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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{\sqrt{a+b \sqrt{c x^2}}}{x^2} \, dx &=\frac{\sqrt{c x^2} \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^2} \, dx,x,\sqrt{c x^2}\right )}{x}\\ &=-\frac{\sqrt{a+b \sqrt{c x^2}}}{x}+\frac{\left (b \sqrt{c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sqrt{c x^2}\right )}{2 x}\\ &=-\frac{\sqrt{a+b \sqrt{c x^2}}}{x}+\frac{\sqrt{c x^2} \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sqrt{c x^2}}\right )}{x}\\ &=-\frac{\sqrt{a+b \sqrt{c x^2}}}{x}-\frac{b \sqrt{c x^2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c x^2}}}{\sqrt{a}}\right )}{\sqrt{a} x}\\ \end{align*}

Mathematica [A]  time = 0.0484649, size = 87, normalized size = 1.3 \[ -\frac{b \sqrt{c x^2} \sqrt{\frac{b \sqrt{c x^2}}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{b \sqrt{c x^2}}{a}+1}\right )+a+b \sqrt{c x^2}}{x \sqrt{a+b \sqrt{c x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sqrt[c*x^2]]/x^2,x]

[Out]

-((a + b*Sqrt[c*x^2] + b*Sqrt[c*x^2]*Sqrt[1 + (b*Sqrt[c*x^2])/a]*ArcTanh[Sqrt[1 + (b*Sqrt[c*x^2])/a]])/(x*Sqrt
[a + b*Sqrt[c*x^2]]))

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 54, normalized size = 0.8 \begin{align*} -{\frac{1}{x} \left ({\it Artanh} \left ({\sqrt{a+b\sqrt{c{x}^{2}}}{\frac{1}{\sqrt{a}}}} \right ) b\sqrt{c{x}^{2}}+\sqrt{a+b\sqrt{c{x}^{2}}}\sqrt{a} \right ){\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(arctanh((a+b*(c*x^2)^(1/2))^(1/2)/a^(1/2))*b*(c*x^2)^(1/2)+(a+b*(c*x^2)^(1/2))^(1/2)*a^(1/2))/x/a^(1/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\sqrt{c x^{2}} b + a}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x^2)^(1/2))^(1/2)/x^2,x, algorithm="maxima")

[Out]

integrate(sqrt(sqrt(c*x^2)*b + a)/x^2, x)

________________________________________________________________________________________

Fricas [A]  time = 1.56047, size = 392, normalized size = 5.85 \begin{align*} \left [\frac{b x \sqrt{\frac{c}{a}} \log \left (\frac{b c x^{2} - 2 \, \sqrt{\sqrt{c x^{2}} b + a} a x \sqrt{\frac{c}{a}} + 2 \, \sqrt{c x^{2}} a}{x^{2}}\right ) - 2 \, \sqrt{\sqrt{c x^{2}} b + a}}{2 \, x}, -\frac{b x \sqrt{-\frac{c}{a}} \arctan \left (-\frac{{\left (a b c x^{2} \sqrt{-\frac{c}{a}} - \sqrt{c x^{2}} a^{2} \sqrt{-\frac{c}{a}}\right )} \sqrt{\sqrt{c x^{2}} b + a}}{b^{2} c^{2} x^{3} - a^{2} c x}\right ) + \sqrt{\sqrt{c x^{2}} b + a}}{x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + b \sqrt{c x^{2}}}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*sqrt(c*x**2))/x**2, x)

________________________________________________________________________________________

Giac [A]  time = 1.15603, size = 73, normalized size = 1.09 \begin{align*} \frac{\frac{b^{2} c \arctan \left (\frac{\sqrt{b \sqrt{c} x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{\sqrt{b \sqrt{c} x + a} b \sqrt{c}}{x}}{b \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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