3.31.57 \(\int \frac {\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{x^2} \, dx\) [3057]

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

[Out]

1/8*b*(-b^2*d+4*a*c)*arctanh(1/2*(b*d+2*c*(d/x)^(1/2))/c^(1/2)/d^(1/2)/(a+c/x+b*(d/x)^(1/2))^(1/2))*d^(1/2)/c^
(5/2)-2/3*(a+c/x+b*(d/x)^(1/2))^(3/2)/c+1/4*b*(b*d+2*c*(d/x)^(1/2))*(a+c/x+b*(d/x)^(1/2))^(1/2)/c^2

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 155, normalized size of antiderivative = 1.00, 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 = {1994, 1355, 654, 626, 635, 212} \begin {gather*} \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} \end {gather*}

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 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 626

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 635

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 654

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 1355

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 1994

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/d^(2*n))*x^(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]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{x^2} \, dx &=-\frac {\text {Subst}\left (\int \sqrt {a+b \sqrt {x}+\frac {c x}{d}} \, dx,x,\frac {d}{x}\right )}{d}\\ &=-\frac {2 \text {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 \text {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 ) \text {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 ) \text {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 [A]
time = 0.64, size = 168, normalized size = 1.08 \begin {gather*} \frac {\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} \left (-\frac {2 \sqrt {c} \left (8 c^2-3 b^2 d x+2 c \left (4 a+b \sqrt {\frac {d}{x}}\right ) x\right )}{x}+\frac {3 b d \left (-4 a c+b^2 d\right ) \log \left (c^2 \left (b d+2 c \sqrt {\frac {d}{x}}-2 \sqrt {c} \sqrt {\frac {d \left (c+a x+b \sqrt {\frac {d}{x}} x\right )}{x}}\right )\right )}{\sqrt {\frac {d \left (c+\left (a+b \sqrt {\frac {d}{x}}\right ) x\right )}{x}}}\right )}{24 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*Sqrt[d/x] + c/x]*((-2*Sqrt[c]*(8*c^2 - 3*b^2*d*x + 2*c*(4*a + b*Sqrt[d/x])*x))/x + (3*b*d*(-4*a*c
+ b^2*d)*Log[c^2*(b*d + 2*c*Sqrt[d/x] - 2*Sqrt[c]*Sqrt[(d*(c + a*x + b*Sqrt[d/x]*x))/x])])/Sqrt[(d*(c + (a + b
*Sqrt[d/x])*x))/x]))/(24*c^(5/2))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(330\) vs. \(2(123)=246\).
time = 0.07, size = 331, normalized size = 2.14

method	result	size
default	\(\frac {\sqrt {\frac {b \sqrt {\frac {d}{x}}\, x +a x +c}{x}}\, \left (-3 \sqrt {c}\, \ln \left (\frac {2 c +b \sqrt {\frac {d}{x}}\, x +2 \sqrt {c}\, \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}}{\sqrt {x}}\right ) \left (\frac {d}{x}\right )^{\frac {3}{2}} x^{3} b^{3}+6 \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, \left (\frac {d}{x}\right )^{\frac {3}{2}} x^{3} b^{3}+12 c^{\frac {3}{2}} a \ln \left (\frac {2 c +b \sqrt {\frac {d}{x}}\, x +2 \sqrt {c}\, \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}}{\sqrt {x}}\right ) \sqrt {\frac {d}{x}}\, x^{2} b +6 a \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, d \,x^{2} b^{2}-12 a \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, \sqrt {\frac {d}{x}}\, x^{2} b c -6 \left (b \sqrt {\frac {d}{x}}\, x +a x +c \right )^{\frac {3}{2}} d x \,b^{2}+12 \left (b \sqrt {\frac {d}{x}}\, x +a x +c \right )^{\frac {3}{2}} \sqrt {\frac {d}{x}}\, x b c -16 \left (b \sqrt {\frac {d}{x}}\, x +a x +c \right )^{\frac {3}{2}} c^{2}\right )}{24 x \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, c^{3}}\)	\(331\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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((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)

________________________________________________________________________________________

Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

________________________________________________________________________________________

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

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]

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

________________________________________________________________________________________