3.3065 \(\int \frac{1}{\sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}} x^2} \, dx\)

Optimal. Leaf size=93 \[ \frac{b \sqrt{d} \tanh ^{-1}\left (\frac{b d+2 c \sqrt{\frac{d}{x}}}{2 \sqrt{c} \sqrt{d} \sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}}}\right )}{c^{3/2}}-\frac{2 \sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}}}{c} \]

[Out]

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

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Rubi [A]  time = 0.122088, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {1970, 1341, 640, 621, 206} \[ \frac{b \sqrt{d} \tanh ^{-1}\left (\frac{b d+2 c \sqrt{\frac{d}{x}}}{2 \sqrt{c} \sqrt{d} \sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}}}\right )}{c^{3/2}}-\frac{2 \sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}}}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1970

Int[(x_)^(m_.)*((a_) + (b_.)*((d_.)/(x_))^(n_) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> -Dist[d^(m + 1), Subst
[Int[(a + b*x^n + (c*x^(2*n))/d^(2*n))^p/x^(m + 2), x], x, d/x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[n2,
-2*n] && IntegerQ[2*n] && IntegerQ[m]

Rule 1341

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[I
nt[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] &
& FractionQ[n]

Rule 640

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

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

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

Rubi steps

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

Mathematica [F]  time = 0.189127, size = 0, normalized size = 0. \[ \int \frac{1}{\sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}} x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 0.143, size = 118, normalized size = 1.3 \begin{align*}{\sqrt{{\frac{1}{x} \left ( b\sqrt{{\frac{d}{x}}}x+ax+c \right ) }} \left ( b\sqrt{{\frac{d}{x}}}x\ln \left ({ \left ( 2\,c+b\sqrt{{\frac{d}{x}}}x+2\,\sqrt{c}\sqrt{b\sqrt{{\frac{d}{x}}}x+ax+c} \right ){\frac{1}{\sqrt{x}}}} \right ) c-2\,\sqrt{b\sqrt{{\frac{d}{x}}}x+ax+c}{c}^{3/2} \right ){\frac{1}{\sqrt{b\sqrt{{\frac{d}{x}}}x+ax+c}}}{c}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*(d/x)^(1/2)*x+a*x+c)/x)^(1/2)*(b*(d/x)^(1/2)*x*ln((2*c+b*(d/x)^(1/2)*x+2*c^(1/2)*(b*(d/x)^(1/2)*x+a*x+c)^(
1/2))/x^(1/2))*c-2*(b*(d/x)^(1/2)*x+a*x+c)^(1/2)*c^(3/2))/(b*(d/x)^(1/2)*x+a*x+c)^(1/2)/c^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.64431, size = 123, normalized size = 1.32 \begin{align*} -\frac{{\left (\frac{b d \log \left ({\left | -b d - 2 \, \sqrt{c}{\left (\sqrt{c} \sqrt{\frac{d}{x}} - \sqrt{b d \sqrt{\frac{d}{x}} + a d + \frac{c d}{x}}\right )} \right |}\right )}{c^{\frac{3}{2}}} + \frac{2 \, \sqrt{b d \sqrt{\frac{d}{x}} + a d + \frac{c d}{x}}}{c}\right )} \sqrt{d}}{{\left | d \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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