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

Optimal. Leaf size=232 \[ \frac{b \sqrt{c} \sqrt{a+b x^2} \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right ),1-\frac{b c}{a d}\right )}{a \sqrt{d} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}+\frac{d x \sqrt{a+b x^2}}{c \sqrt{c+d x^2}}-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{c x}-\frac{\sqrt{d} \sqrt{a+b x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{\sqrt{c} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

[Out]

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

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Rubi [A]  time = 0.148864, antiderivative size = 232, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {475, 21, 422, 418, 492, 411} \[ \frac{d x \sqrt{a+b x^2}}{c \sqrt{c+d x^2}}-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{c x}+\frac{b \sqrt{c} \sqrt{a+b x^2} F\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{a \sqrt{d} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}}-\frac{\sqrt{d} \sqrt{a+b x^2} E\left (\tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )|1-\frac{b c}{a d}\right )}{\sqrt{c} \sqrt{c+d x^2} \sqrt{\frac{c \left (a+b x^2\right )}{a \left (c+d x^2\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 475

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)/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*
x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x
], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] &&
IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 422

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

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*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[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

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

Mathematica [A]  time = 0.265678, size = 111, normalized size = 0.48 \[ \frac{\frac{b c x \sqrt{\frac{b x^2}{a}+1} \sqrt{\frac{d x^2}{c}+1} E\left (\sin ^{-1}\left (\sqrt{-\frac{b}{a}} x\right )|\frac{a d}{b c}\right )}{\sqrt{-\frac{b}{a}}}-\left (a+b x^2\right ) \left (c+d x^2\right )}{c x \sqrt{a+b x^2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.023, size = 168, normalized size = 0.7 \begin{align*}{\frac{1}{ \left ( bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac \right ) cx}\sqrt{b{x}^{2}+a}\sqrt{d{x}^{2}+c} \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 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 ){\frac{1}{\sqrt{-{\frac{b}{a}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(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}}{d x^{4} + c x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Integral(sqrt(a + b*x**2)/(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{\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((b*x^2+a)^(1/2)/x^2/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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