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

Optimal. Leaf size=80 \[ \frac{\sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a \sqrt{b}}-\frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.0733895, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {446, 83, 63, 208} \[ \frac{\sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a \sqrt{b}}-\frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x^2]/(x*(a + b*x^2)),x]

[Out]

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

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 83

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

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

Mathematica [A]  time = 0.0343538, size = 78, normalized size = 0.98 \[ \frac{\frac{\sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{\sqrt{b}}-\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.01, size = 984, normalized size = 12.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.82432, size = 1261, normalized size = 15.76 \begin{align*} \left [\frac{\sqrt{\frac{b c - a d}{b}} \log \left (\frac{b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \,{\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \,{\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt{d x^{2} + c} \sqrt{\frac{b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 2 \, \sqrt{c} \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right )}{4 \, a}, \frac{4 \, \sqrt{-c} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) + \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \,{\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \,{\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt{d x^{2} + c} \sqrt{\frac{b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \, a}, \frac{\sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt{d x^{2} + c} \sqrt{-\frac{b c - a d}{b}}}{2 \,{\left (b c^{2} - a c d +{\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + \sqrt{c} \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right )}{2 \, a}, \frac{\sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt{d x^{2} + c} \sqrt{-\frac{b c - a d}{b}}}{2 \,{\left (b c^{2} - a c d +{\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + 2 \, \sqrt{-c} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right )}{2 \, a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt((b*c - a*d)/b)*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^2 +
4*(b^2*d*x^2 + 2*b^2*c - a*b*d)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/b))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 2*sqrt(c)
*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2))/a, 1/4*(4*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + sq
rt((b*c - a*d)/b)*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^2 + 4*(b^2*
d*x^2 + 2*b^2*c - a*b*d)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/b))/(b^2*x^4 + 2*a*b*x^2 + a^2)))/a, 1/2*(sqrt(-(b*c
- a*d)/b)*arctan(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/b)/(b*c^2 - a*c*d + (b*c*d -
a*d^2)*x^2)) + sqrt(c)*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2))/a, 1/2*(sqrt(-(b*c - a*d)/b)*arcta
n(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/b)/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)) + 2
*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)))/a]

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Sympy [A]  time = 6.25726, size = 78, normalized size = 0.98 \begin{align*} \frac{2 \left (\frac{c d \operatorname{atan}{\left (\frac{\sqrt{c + d x^{2}}}{\sqrt{- c}} \right )}}{2 a \sqrt{- c}} + \frac{d \left (a d - b c\right ) \operatorname{atan}{\left (\frac{\sqrt{c + d x^{2}}}{\sqrt{\frac{a d - b c}{b}}} \right )}}{2 a b \sqrt{\frac{a d - b c}{b}}}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(c*d*atan(sqrt(c + d*x**2)/sqrt(-c))/(2*a*sqrt(-c)) + d*(a*d - b*c)*atan(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b
))/(2*a*b*sqrt((a*d - b*c)/b)))/d

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Giac [A]  time = 1.13864, size = 117, normalized size = 1.46 \begin{align*} -d{\left (\frac{{\left (b c - a d\right )} \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} a d} - \frac{c \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{a \sqrt{-c} d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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