3.973 \(\int \frac{1}{x^3 \sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=91 \[ \frac{(a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{3/2} c^{3/2}}-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{2 a c x^2} \]

[Out]

-(Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(2*a*c*x^2) + ((b*c + a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[
c + d*x^2])])/(2*a^(3/2)*c^(3/2))

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Rubi [A]  time = 0.0785099, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {446, 96, 93, 208} \[ \frac{(a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{3/2} c^{3/2}}-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{2 a c x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]),x]

[Out]

-(Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(2*a*c*x^2) + ((b*c + a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[
c + d*x^2])])/(2*a^(3/2)*c^(3/2))

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 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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{1}{x^3 \sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{2 a c x^2}-\frac{(b c+a d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,x^2\right )}{4 a c}\\ &=-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{2 a c x^2}-\frac{(b c+a d) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x^2}}{\sqrt{c+d x^2}}\right )}{2 a c}\\ &=-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{2 a c x^2}+\frac{(b c+a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{3/2} c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0550414, size = 91, normalized size = 1. \[ \frac{(a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{3/2} c^{3/2}}-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{2 a c x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]),x]

[Out]

-(Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(2*a*c*x^2) + ((b*c + a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[
c + d*x^2])])/(2*a^(3/2)*c^(3/2))

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Maple [B]  time = 0.021, size = 209, normalized size = 2.3 \begin{align*}{\frac{1}{4\,ac{x}^{2}} \left ( \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 ){x}^{2}ad+\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 ){x}^{2}bc-2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac} \right ) \sqrt{d{x}^{2}+c}\sqrt{b{x}^{2}+a}{\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x)

[Out]

1/4/a/c*(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)*x^2*a*d+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)*x^2*b*c-2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+
b*c*x^2+a*c)^(1/2))*(d*x^2+c)^(1/2)*(b*x^2+a)^(1/2)/x^2/(a*c)^(1/2)/(b*d*x^4+a*d*x^2+b*c*x^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(1/x^3/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.49727, size = 635, normalized size = 6.98 \begin{align*} \left [\frac{\sqrt{a c}{\left (b c + a d\right )} x^{2} \log \left (\frac{{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \,{\left (a b c^{2} + a^{2} c d\right )} x^{2} + 4 \,{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} \sqrt{a c}}{x^{4}}\right ) - 4 \, \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} a c}{8 \, a^{2} c^{2} x^{2}}, -\frac{\sqrt{-a c}{\left (b c + a d\right )} x^{2} \arctan \left (\frac{{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} \sqrt{-a c}}{2 \,{\left (a b c d x^{4} + a^{2} c^{2} +{\left (a b c^{2} + a^{2} c d\right )} x^{2}\right )}}\right ) + 2 \, \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} a c}{4 \, a^{2} c^{2} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \sqrt{a + b x^{2}} \sqrt{c + d x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x**2+a)**(1/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(1/(x**3*sqrt(a + b*x**2)*sqrt(c + d*x**2)), x)

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Giac [B]  time = 1.3526, size = 558, normalized size = 6.13 \begin{align*} \frac{\sqrt{b d} b^{4} d{\left (\frac{{\left (b c + a d\right )} \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} a b^{3} c d} - \frac{2 \,{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2} -{\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} b c -{\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} a d\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \,{\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} b^{2} c - 2 \,{\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} 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 )}^{4}\right )} a b^{2} c d}\right )}}{2 \,{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(b*d)*b^4*d*((b*c + a*d)*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)*a*b^3*c*d) - 2*(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2 -
(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*b*c - (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt
(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a*d)/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*x^2 + a)*sqrt(b*d)
- sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*
d - a*b*d))^2*a*b*d + (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4)*a*b^2*c*d))/abs(b
)