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

Optimal. Leaf size=92 \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{\sqrt{d}}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{c}} \]

[Out]

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

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Rubi [A]  time = 0.0953195, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {446, 105, 63, 217, 206, 93, 208} \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{\sqrt{d}}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -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 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])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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 x^2}}{x \sqrt{c+d x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x \sqrt{c+d x}} \, dx,x,x^2\right )\\ &=\frac{1}{2} a \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,x^2\right )+\frac{1}{2} b \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx,x,x^2\right )\\ &=a \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x^2}}{\sqrt{c+d x^2}}\right )+\operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x^2}\right )\\ &=-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{c}}+\operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x^2}}{\sqrt{c+d x^2}}\right )\\ &=-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{c}}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{\sqrt{d}}\\ \end{align*}

Mathematica [A]  time = 0.135443, size = 129, normalized size = 1.4 \[ \frac{\sqrt{b c-a d} \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )}{\sqrt{d} \sqrt{c+d x^2}}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.026, size = 177, normalized size = 1.9 \begin{align*}{\frac{1}{2}\sqrt{b{x}^{2}+a}\sqrt{d{x}^{2}+c} \left ( b\ln \left ({\frac{1}{2} \left ( 2\,d{x}^{2}b+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ) \sqrt{ac}-a\ln \left ({\frac{1}{{x}^{2}} \left ( ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac \right ) } \right ) \sqrt{bd} \right ){\frac{1}{\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(b*ln(1/2*(2*d*x^2*b+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d
+b*c)/(b*d)^(1/2))*(a*c)^(1/2)-a*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+2*a*c)/
x^2)*(b*d)^(1/2))/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/(b*d)^(1/2)/(a*c)^(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)^(1/2)/x/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.9764, size = 1708, normalized size = 18.57 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*sqrt(b/d)*log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 + 4*(2*b*d^2*x^2
+ b*c*d + a*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b/d)) + 1/4*sqrt(a/c)*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^
2)*x^4 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*x^2 - 4*(2*a*c^2 + (b*c^2 + a*c*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2
+ c)*sqrt(a/c))/x^4), -1/2*sqrt(-b/d)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(
-b/d)/(b^2*d*x^4 + a*b*c + (b^2*c + a*b*d)*x^2)) + 1/4*sqrt(a/c)*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*
a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*x^2 - 4*(2*a*c^2 + (b*c^2 + a*c*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(a
/c))/x^4), 1/2*sqrt(-a/c)*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-a/c)/(a*b
*d*x^4 + a^2*c + (a*b*c + a^2*d)*x^2)) + 1/4*sqrt(b/d)*log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(
b^2*c*d + a*b*d^2)*x^2 + 4*(2*b*d^2*x^2 + b*c*d + a*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b/d)), 1/2*sqrt(
-a/c)*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-a/c)/(a*b*d*x^4 + a^2*c + (a*
b*c + a^2*d)*x^2)) - 1/2*sqrt(-b/d)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b
/d)/(b^2*d*x^4 + a*b*c + (b^2*c + a*b*d)*x^2))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.20248, size = 208, normalized size = 2.26 \begin{align*} -\frac{b^{2}{\left (\frac{2 \, \sqrt{b d} a \arctan \left (-\frac{b^{2} c + a b d -{\left (\sqrt{b x^{2} + a} \sqrt{b d} - \sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt{-a b c d} b}\right )}{\sqrt{-a b c d} b} + \frac{\sqrt{b d} \log \left ({\left (\sqrt{b x^{2} + a} \sqrt{b d} - \sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}\right )}{b d}\right )}}{2 \,{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b^2*(2*sqrt(b*d)*a*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d
- a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*b) + sqrt(b*d)*log((sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c +
(b*x^2 + a)*b*d - a*b*d))^2)/(b*d))/abs(b)