3.9.91 \(\int \frac {1}{\sqrt {a+\frac {b}{x^2}} \sqrt {c+d x^2}} \, dx\) [891]

Optimal. Leaf size=70 \[ \frac {\sqrt {b+a x^2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {b+a x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} \sqrt {d} \sqrt {a+\frac {b}{x^2}} x} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {446, 455, 65, 223, 212} \begin {gather*} \frac {\sqrt {a x^2+b} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a x^2+b}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} \sqrt {d} x \sqrt {a+\frac {b}{x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 446

Int[((c_) + (d_.)*(x_)^(mn_.))^(q_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[x^(n*FracPart[q])*((c +
d/x^n)^FracPart[q]/(d + c*x^n)^FracPart[q]), Int[(a + b*x^n)^p*((d + c*x^n)^q/x^(n*q)), x], x] /; FreeQ[{a, b,
c, d, n, p, q}, x] && EqQ[mn, -n] &&  !IntegerQ[q] &&  !IntegerQ[p]

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps

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

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Mathematica [A]
time = 0.38, size = 70, normalized size = 1.00 \begin {gather*} \frac {\sqrt {b+a x^2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {b+a x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} \sqrt {d} \sqrt {a+\frac {b}{x^2}} x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.23, size = 103, normalized size = 1.47

method	result	size
default	\(\frac {\left (a \,x^{2}+b \right ) \ln \left (\frac {2 a d \,x^{2}+2 \sqrt {\left (a \,x^{2}+b \right ) \left (d \,x^{2}+c \right )}\, \sqrt {a d}+a c +b d}{2 \sqrt {a d}}\right ) \sqrt {d \,x^{2}+c}}{2 \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x \sqrt {a d}\, \sqrt {\left (a \,x^{2}+b \right ) \left (d \,x^{2}+c \right )}}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 208, normalized size = 2.97 \begin {gather*} \left [\frac {\sqrt {a d} \log \left (8 \, a^{2} d^{2} x^{4} + a^{2} c^{2} + 6 \, a b c d + b^{2} d^{2} + 8 \, {\left (a^{2} c d + a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, a d x^{3} + {\left (a c + b d\right )} x\right )} \sqrt {d x^{2} + c} \sqrt {a d} \sqrt {\frac {a x^{2} + b}{x^{2}}}\right )}{4 \, a d}, -\frac {\sqrt {-a d} \arctan \left (\frac {{\left (2 \, a d x^{3} + {\left (a c + b d\right )} x\right )} \sqrt {d x^{2} + c} \sqrt {-a d} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{2 \, {\left (a^{2} d^{2} x^{4} + a b c d + {\left (a^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right )}{2 \, a d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*sqrt(a*d)*log(8*a^2*d^2*x^4 + a^2*c^2 + 6*a*b*c*d + b^2*d^2 + 8*(a^2*c*d + a*b*d^2)*x^2 + 4*(2*a*d*x^3 +
(a*c + b*d)*x)*sqrt(d*x^2 + c)*sqrt(a*d)*sqrt((a*x^2 + b)/x^2))/(a*d), -1/2*sqrt(-a*d)*arctan(1/2*(2*a*d*x^3 +
(a*c + b*d)*x)*sqrt(d*x^2 + c)*sqrt(-a*d)*sqrt((a*x^2 + b)/x^2)/(a^2*d^2*x^4 + a*b*c*d + (a^2*c*d + a*b*d^2)*
x^2))/(a*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a + \frac {b}{x^{2}}} \sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.29, size = 92, normalized size = 1.31 \begin {gather*} \frac {a \log \left ({\left | -\sqrt {a d} \sqrt {b} + \sqrt {a^{2} c} \right |}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {a d} {\left | a \right |}} - \frac {a \log \left ({\left | -\sqrt {a x^{2} + b} \sqrt {a d} + \sqrt {a^{2} c + {\left (a x^{2} + b\right )} a d - a b d} \right |}\right )}{\sqrt {a d} {\left | a \right |} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {a+\frac {b}{x^2}}\,\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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