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

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

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {375, 475, 21, 422, 418, 492, 411} \[ -\frac {d \sqrt {a+\frac {b}{x^2}}}{c x \sqrt {c+\frac {d}{x^2}}}+\frac {x \sqrt {a+\frac {b}{x^2}} \sqrt {c+\frac {d}{x^2}}}{c}-\frac {b \sqrt {c} \sqrt {a+\frac {b}{x^2}} F\left (\cot ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {d}}\right )|1-\frac {b c}{a d}\right )}{a \sqrt {d} \sqrt {c+\frac {d}{x^2}} \sqrt {\frac {c \left (a+\frac {b}{x^2}\right )}{a \left (c+\frac {d}{x^2}\right )}}}+\frac {\sqrt {d} \sqrt {a+\frac {b}{x^2}} E\left (\cot ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {d}}\right )|1-\frac {b c}{a d}\right )}{\sqrt {c} \sqrt {c+\frac {d}{x^2}} \sqrt {\frac {c \left (a+\frac {b}{x^2}\right )}{a \left (c+\frac {d}{x^2}\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

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]

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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 86, normalized size = 0.37 \[ \frac {\sqrt {a+\frac {b}{x^2}} \sqrt {\frac {c x^2+d}{d}} E\left (\sin ^{-1}\left (\sqrt {-\frac {c}{d}} x\right )|\frac {a d}{b c}\right )}{\sqrt {-\frac {c}{d}} \sqrt {\frac {a x^2+b}{b}} \sqrt {c+\frac {d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + \frac {b}{x^{2}}}}{\sqrt {c + \frac {d}{x^{2}}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 94, normalized size = 0.41 \[ \frac {\sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, \sqrt {\frac {a \,x^{2}+b}{b}}\, \sqrt {\frac {c \,x^{2}+d}{d}}\, b \EllipticE \left (\sqrt {-\frac {c}{d}}\, x , \sqrt {\frac {a d}{b c}}\right )}{\left (a \,x^{2}+b \right ) \sqrt {-\frac {c}{d}}\, \sqrt {\frac {c \,x^{2}+d}{x^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + \frac {b}{x^{2}}}}{\sqrt {c + \frac {d}{x^{2}}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {c+\frac {d}{x^2}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + \frac {b}{x^{2}}}}{\sqrt {c + \frac {d}{x^{2}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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