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

Optimal. Leaf size=153 \[ \frac{d x \sqrt{a+b x^2}}{a c \sqrt{c+d x^2}}-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{a c x}-\frac{\sqrt{d} \sqrt{a+b x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{a \sqrt{c} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

[Out]

(d*x*Sqrt[a + b*x^2])/(a*c*Sqrt[c + d*x^2]) - (Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(a*c*x) - (Sqrt[d]*Sqrt[a + b*
x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(a*Sqrt[c]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]
*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.0948289, antiderivative size = 153, 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 = {480, 12, 492, 411} \[ \frac{d x \sqrt{a+b x^2}}{a c \sqrt{c+d x^2}}-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{a c x}-\frac{\sqrt{d} \sqrt{a+b x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{a \sqrt{c} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x*Sqrt[a + b*x^2])/(a*c*Sqrt[c + d*x^2]) - (Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(a*c*x) - (Sqrt[d]*Sqrt[a + b*
x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(a*Sqrt[c]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]
*Sqrt[c + d*x^2])

Rule 480

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*e*(m + 1)), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rubi steps

\begin{align*} \int \frac{1}{x^2 \sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx &=-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{a c x}+\frac{\int \frac{b d x^2}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{a c}\\ &=-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{a c x}+\frac{(b d) \int \frac{x^2}{\sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx}{a c}\\ &=\frac{d x \sqrt{a+b x^2}}{a c \sqrt{c+d x^2}}-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{a c x}-\frac{d \int \frac{\sqrt{a+b x^2}}{\left (c+d x^2\right )^{3/2}} \, dx}{a}\\ &=\frac{d x \sqrt{a+b x^2}}{a c \sqrt{c+d x^2}}-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{a c x}-\frac{\sqrt{d} \sqrt{a+b x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{a \sqrt{c} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.324959, size = 146, normalized size = 0.95 \[ \frac{-\frac{\left (a+b x^2\right ) \left (c+d x^2\right )}{c x}-i a \sqrt{\frac{b}{a}} \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} \left (E\left (i \sinh ^{-1}\left (\sqrt{\frac{b}{a}} x\right )|\frac{a d}{b c}\right )-\text{EllipticF}\left (i \sinh ^{-1}\left (x \sqrt{\frac{b}{a}}\right ),\frac{a d}{b c}\right )\right )}{a \sqrt{a+b x^2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(((a + b*x^2)*(c + d*x^2))/(c*x)) - I*a*Sqrt[b/a]*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*(EllipticE[I*ArcSi
nh[Sqrt[b/a]*x], (a*d)/(b*c)] - EllipticF[I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)]))/(a*Sqrt[a + b*x^2]*Sqrt[c + d
*x^2])

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Maple [A]  time = 0.019, size = 224, normalized size = 1.5 \begin{align*}{\frac{1}{axc \left ( bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac \right ) } \left ( -\sqrt{-{\frac{b}{a}}}{x}^{4}bd-bc\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}x{\it EllipticF} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ) +bc\sqrt{{\frac{b{x}^{2}+a}{a}}}\sqrt{{\frac{d{x}^{2}+c}{c}}}x{\it EllipticE} \left ( x\sqrt{-{\frac{b}{a}}},\sqrt{{\frac{ad}{bc}}} \right ) -\sqrt{-{\frac{b}{a}}}{x}^{2}ad-\sqrt{-{\frac{b}{a}}}{x}^{2}bc-\sqrt{-{\frac{b}{a}}}ac \right ) \sqrt{d{x}^{2}+c}\sqrt{b{x}^{2}+a}{\frac{1}{\sqrt{-{\frac{b}{a}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-b/a)^(1/2)*x^4*b*d-b*c*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*x*EllipticF(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))
+b*c*((b*x^2+a)/a)^(1/2)*((d*x^2+c)/c)^(1/2)*x*EllipticE(x*(-b/a)^(1/2),(a*d/b/c)^(1/2))-(-b/a)^(1/2)*x^2*a*d-
(-b/a)^(1/2)*x^2*b*c-(-b/a)^(1/2)*a*c)*(d*x^2+c)^(1/2)*(b*x^2+a)^(1/2)/x/a/(-b/a)^(1/2)/c/(b*d*x^4+a*d*x^2+b*c
*x^2+a*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{b d x^{6} +{\left (b c + a d\right )} x^{4} + a c x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)/(b*d*x^6 + (b*c + a*d)*x^4 + a*c*x^2), x)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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