3.3055 \(\int \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} \, dx\)

Optimal. Leaf size=113 \[ \frac {\left (4 a c-b^2 d\right ) \tanh ^{-1}\left (\frac {2 a+b \sqrt {\frac {d}{x}}}{2 \sqrt {a} \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}\right )}{4 a^{3/2}}+\frac {x \left (2 a+b \sqrt {\frac {d}{x}}\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{2 a} \]

[Out]

1/4*(-b^2*d+4*a*c)*arctanh(1/2*(2*a+b*(d/x)^(1/2))/a^(1/2)/(a+c/x+b*(d/x)^(1/2))^(1/2))/a^(3/2)+1/2*x*(2*a+b*(
d/x)^(1/2))*(a+c/x+b*(d/x)^(1/2))^(1/2)/a

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Rubi [A]  time = 0.12, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1969, 1357, 720, 724, 206} \[ \frac {\left (4 a c-b^2 d\right ) \tanh ^{-1}\left (\frac {2 a+b \sqrt {\frac {d}{x}}}{2 \sqrt {a} \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}\right )}{4 a^{3/2}}+\frac {x \left (2 a+b \sqrt {\frac {d}{x}}\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[d/x] + c/x],x]

[Out]

((2*a + b*Sqrt[d/x])*Sqrt[a + b*Sqrt[d/x] + c/x]*x)/(2*a) + ((4*a*c - b^2*d)*ArcTanh[(2*a + b*Sqrt[d/x])/(2*Sq
rt[a]*Sqrt[a + b*Sqrt[d/x] + c/x])])/(4*a^(3/2))

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])

Rule 720

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

Rule 724

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

Rule 1357

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1969

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

Rubi steps

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

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Mathematica [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [B]  time = 0.99, size = 213, normalized size = 1.88 \[ \frac {{\left (2 \, \sqrt {a d^{2} x + \sqrt {d x} b d^{2} + c d^{2}} {\left (\frac {b d}{a} + 2 \, \sqrt {d x}\right )} + \frac {{\left (b^{2} d^{3} - 4 \, a c d^{2}\right )} \log \left ({\left | -b d^{2} - 2 \, \sqrt {a d} {\left (\sqrt {a d} \sqrt {d x} - \sqrt {a d^{2} x + \sqrt {d x} b d^{2} + c d^{2}}\right )} \right |}\right )}{\sqrt {a d} a} - \frac {b^{2} d^{3} \log \left ({\left | -b d^{2} + 2 \, \sqrt {c d^{2}} \sqrt {a d} \right |}\right ) - 4 \, a c d^{2} \log \left ({\left | -b d^{2} + 2 \, \sqrt {c d^{2}} \sqrt {a d} \right |}\right ) + 2 \, \sqrt {c d^{2}} \sqrt {a d} b d}{\sqrt {a d} a}\right )} \mathrm {sgn}\relax (x)}{4 \, d^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*sqrt(a*d^2*x + sqrt(d*x)*b*d^2 + c*d^2)*(b*d/a + 2*sqrt(d*x)) + (b^2*d^3 - 4*a*c*d^2)*log(abs(-b*d^2 -
2*sqrt(a*d)*(sqrt(a*d)*sqrt(d*x) - sqrt(a*d^2*x + sqrt(d*x)*b*d^2 + c*d^2))))/(sqrt(a*d)*a) - (b^2*d^3*log(abs
(-b*d^2 + 2*sqrt(c*d^2)*sqrt(a*d))) - 4*a*c*d^2*log(abs(-b*d^2 + 2*sqrt(c*d^2)*sqrt(a*d))) + 2*sqrt(c*d^2)*sqr
t(a*d)*b*d)/(sqrt(a*d)*a))*sgn(x)/d^(3/2)

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maple [B]  time = 0.14, size = 213, normalized size = 1.88 \[ \frac {\sqrt {\frac {a x +\sqrt {\frac {d}{x}}\, b x +c}{x}}\, \left (-a \,b^{2} d \ln \left (\frac {2 a \sqrt {x}+\sqrt {\frac {d}{x}}\, b \sqrt {x}+2 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, \sqrt {a}}{2 \sqrt {a}}\right )+4 a^{2} c \ln \left (\frac {2 a \sqrt {x}+\sqrt {\frac {d}{x}}\, b \sqrt {x}+2 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, \sqrt {a}}{2 \sqrt {a}}\right )+4 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, a^{\frac {5}{2}} \sqrt {x}+2 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, \sqrt {\frac {d}{x}}\, a^{\frac {3}{2}} b \sqrt {x}\right ) \sqrt {x}}{4 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((a*x+(d/x)^(1/2)*b*x+c)/x)^(1/2)*x^(1/2)*(2*a^(3/2)*(a*x+(d/x)^(1/2)*b*x+c)^(1/2)*(d/x)^(1/2)*x^(1/2)*b-l
n(1/2*(2*a*x^(1/2)+(d/x)^(1/2)*b*x^(1/2)+2*(a*x+(d/x)^(1/2)*b*x+c)^(1/2)*a^(1/2))/a^(1/2))*d*a*b^2+4*a^(5/2)*(
a*x+(d/x)^(1/2)*b*x+c)^(1/2)*x^(1/2)+4*ln(1/2*(2*a*x^(1/2)+(d/x)^(1/2)*b*x^(1/2)+2*(a*x+(d/x)^(1/2)*b*x+c)^(1/
2)*a^(1/2))/a^(1/2))*a^2*c)/a^(5/2)/(a*x+(d/x)^(1/2)*b*x+c)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {a+\frac {c}{x}+b\,\sqrt {\frac {d}{x}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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