3.3057 \(\int \frac{\sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}}}{x^2} \, dx\)

Optimal. Leaf size=155 \[ \frac{b \sqrt{d} \left (4 a c-b^2 d\right ) \tanh ^{-1}\left (\frac{b d+2 c \sqrt{\frac{d}{x}}}{2 \sqrt{c} \sqrt{d} \sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}}}\right )}{8 c^{5/2}}+\frac{b \left (b d+2 c \sqrt{\frac{d}{x}}\right ) \sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}}}{4 c^2}-\frac{2 \left (a+b \sqrt{\frac{d}{x}}+\frac{c}{x}\right )^{3/2}}{3 c} \]

[Out]

(b*(b*d + 2*c*Sqrt[d/x])*Sqrt[a + b*Sqrt[d/x] + c/x])/(4*c^2) - (2*(a + b*Sqrt[d/x] + c/x)^(3/2))/(3*c) + (b*S
qrt[d]*(4*a*c - b^2*d)*ArcTanh[(b*d + 2*c*Sqrt[d/x])/(2*Sqrt[c]*Sqrt[d]*Sqrt[a + b*Sqrt[d/x] + c/x])])/(8*c^(5
/2))

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Rubi [A]  time = 0.183611, antiderivative size = 155, 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 = {1970, 1341, 640, 612, 621, 206} \[ \frac{b \sqrt{d} \left (4 a c-b^2 d\right ) \tanh ^{-1}\left (\frac{b d+2 c \sqrt{\frac{d}{x}}}{2 \sqrt{c} \sqrt{d} \sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}}}\right )}{8 c^{5/2}}+\frac{b \left (b d+2 c \sqrt{\frac{d}{x}}\right ) \sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}}}{4 c^2}-\frac{2 \left (a+b \sqrt{\frac{d}{x}}+\frac{c}{x}\right )^{3/2}}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(b*d + 2*c*Sqrt[d/x])*Sqrt[a + b*Sqrt[d/x] + c/x])/(4*c^2) - (2*(a + b*Sqrt[d/x] + c/x)^(3/2))/(3*c) + (b*S
qrt[d]*(4*a*c - b^2*d)*ArcTanh[(b*d + 2*c*Sqrt[d/x])/(2*Sqrt[c]*Sqrt[d]*Sqrt[a + b*Sqrt[d/x] + c/x])])/(8*c^(5
/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 612

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

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{\sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}}}{x^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \sqrt{a+b \sqrt{x}+\frac{c x}{d}} \, dx,x,\frac{d}{x}\right )}{d}\\ &=-\frac{2 \operatorname{Subst}\left (\int x \sqrt{a+b x+\frac{c x^2}{d}} \, dx,x,\sqrt{\frac{d}{x}}\right )}{d}\\ &=-\frac{2 \left (a+b \sqrt{\frac{d}{x}}+\frac{c}{x}\right )^{3/2}}{3 c}+\frac{b \operatorname{Subst}\left (\int \sqrt{a+b x+\frac{c x^2}{d}} \, dx,x,\sqrt{\frac{d}{x}}\right )}{c}\\ &=\frac{b \left (b d+2 c \sqrt{\frac{d}{x}}\right ) \sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}}}{4 c^2}-\frac{2 \left (a+b \sqrt{\frac{d}{x}}+\frac{c}{x}\right )^{3/2}}{3 c}+\frac{\left (b \left (4 a c-b^2 d\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+\frac{c x^2}{d}}} \, dx,x,\sqrt{\frac{d}{x}}\right )}{8 c^2}\\ &=\frac{b \left (b d+2 c \sqrt{\frac{d}{x}}\right ) \sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}}}{4 c^2}-\frac{2 \left (a+b \sqrt{\frac{d}{x}}+\frac{c}{x}\right )^{3/2}}{3 c}+\frac{\left (b \left (4 a c-b^2 d\right )\right ) \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 )}{4 c^2}\\ &=\frac{b \left (b d+2 c \sqrt{\frac{d}{x}}\right ) \sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}}}{4 c^2}-\frac{2 \left (a+b \sqrt{\frac{d}{x}}+\frac{c}{x}\right )^{3/2}}{3 c}+\frac{b \sqrt{d} \left (4 a c-b^2 d\right ) \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 )}{8 c^{5/2}}\\ \end{align*}

Mathematica [F]  time = 0.121034, size = 0, normalized size = 0. \[ \int \frac{\sqrt{a+b \sqrt{\frac{d}{x}}+\frac{c}{x}}}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [B]  time = 0.134, size = 331, normalized size = 2.1 \begin{align*} -{\frac{1}{24\,x{c}^{3}}\sqrt{{\frac{1}{x} \left ( b\sqrt{{\frac{d}{x}}}x+ax+c \right ) }} \left ( 3\,\sqrt{c}\ln \left ({\frac{1}{\sqrt{x}} \left ( 2\,c+b\sqrt{{\frac{d}{x}}}x+2\,\sqrt{c}\sqrt{b\sqrt{{\frac{d}{x}}}x+ax+c} \right ) } \right ) \left ({\frac{d}{x}} \right ) ^{3/2}{x}^{3}{b}^{3}-6\,\sqrt{b\sqrt{{\frac{d}{x}}}x+ax+c} \left ({\frac{d}{x}} \right ) ^{3/2}{x}^{3}{b}^{3}-12\,{c}^{3/2}a\ln \left ({\frac{1}{\sqrt{x}} \left ( 2\,c+b\sqrt{{\frac{d}{x}}}x+2\,\sqrt{c}\sqrt{b\sqrt{{\frac{d}{x}}}x+ax+c} \right ) } \right ) \sqrt{{\frac{d}{x}}}{x}^{2}b-12\, \left ( b\sqrt{{\frac{d}{x}}}x+ax+c \right ) ^{3/2}\sqrt{{\frac{d}{x}}}xbc+12\,a\sqrt{b\sqrt{{\frac{d}{x}}}x+ax+c}\sqrt{{\frac{d}{x}}}{x}^{2}bc+6\, \left ( b\sqrt{{\frac{d}{x}}}x+ax+c \right ) ^{3/2}dx{b}^{2}-6\,a\sqrt{b\sqrt{{\frac{d}{x}}}x+ax+c}d{x}^{2}{b}^{2}+16\, \left ( b\sqrt{{\frac{d}{x}}}x+ax+c \right ) ^{3/2}{c}^{2} \right ){\frac{1}{\sqrt{b\sqrt{{\frac{d}{x}}}x+ax+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*((b*(d/x)^(1/2)*x+a*x+c)/x)^(1/2)*(3*c^(1/2)*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))*(d/x)^(3/2)*x^3*b^3-6*(b*(d/x)^(1/2)*x+a*x+c)^(1/2)*(d/x)^(3/2)*x^3*b^3-12*c^(3/2)*a*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))*(d/x)^(1/2)*x^2*b-12*(b*(d/x)^(1/2)*x+a*x+c)^(
3/2)*(d/x)^(1/2)*x*b*c+12*a*(b*(d/x)^(1/2)*x+a*x+c)^(1/2)*(d/x)^(1/2)*x^2*b*c+6*(b*(d/x)^(1/2)*x+a*x+c)^(3/2)*
d*x*b^2-6*a*(b*(d/x)^(1/2)*x+a*x+c)^(1/2)*d*x^2*b^2+16*(b*(d/x)^(1/2)*x+a*x+c)^(3/2)*c^2)/x/(b*(d/x)^(1/2)*x+a
*x+c)^(1/2)/c^3

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out