3.3066 \(\int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^3} \, dx\)

Optimal. Leaf size=165 \[ -\frac {b \sqrt {d} \left (12 a c-5 b^2 d\right ) \tanh ^{-1}\left (\frac {b d+2 c \sqrt {\frac {d}{x}}}{2 \sqrt {c} \sqrt {d} \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}\right )}{8 c^{7/2}}+\frac {\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} \left (16 a c-15 b^2 d+10 b c \sqrt {\frac {d}{x}}\right )}{12 c^3}-\frac {2 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{3 c x} \]

[Out]

-1/8*b*(-5*b^2*d+12*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^(7/2)-2/3*(a+c/x+b*(d/x)^(1/2))^(1/2)/c/x+1/12*(16*a*c-15*b^2*d+10*b*c*(d/x)^(1/2))*(a+c/x+b*(d/x)^(1/2))^
(1/2)/c^3

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Rubi [A]  time = 0.23, antiderivative size = 165, 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 = {1970, 1357, 742, 779, 621, 206} \[ \frac {\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} \left (16 a c-15 b^2 d+10 b c \sqrt {\frac {d}{x}}\right )}{12 c^3}-\frac {b \sqrt {d} \left (12 a c-5 b^2 d\right ) \tanh ^{-1}\left (\frac {b d+2 c \sqrt {\frac {d}{x}}}{2 \sqrt {c} \sqrt {d} \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}\right )}{8 c^{7/2}}-\frac {2 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{3 c x} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b*Sqrt[d/x] + c/x]*x^3),x]

[Out]

((16*a*c - 15*b^2*d + 10*b*c*Sqrt[d/x])*Sqrt[a + b*Sqrt[d/x] + c/x])/(12*c^3) - (2*Sqrt[a + b*Sqrt[d/x] + c/x]
)/(3*c*x) - (b*Sqrt[d]*(12*a*c - 5*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^(7/2))

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 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 742

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

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

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 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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^3} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x}{\sqrt {a+b \sqrt {x}+\frac {c x}{d}}} \, dx,x,\frac {d}{x}\right )}{d^2}\\ &=-\frac {2 \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {a+b x+\frac {c x^2}{d}}} \, dx,x,\sqrt {\frac {d}{x}}\right )}{d^2}\\ &=-\frac {2 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{3 c x}-\frac {2 \operatorname {Subst}\left (\int \frac {x \left (-2 a-\frac {5 b x}{2}\right )}{\sqrt {a+b x+\frac {c x^2}{d}}} \, dx,x,\sqrt {\frac {d}{x}}\right )}{3 c d}\\ &=\frac {\left (16 a c-5 b \left (3 b d-2 c \sqrt {\frac {d}{x}}\right )\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{12 c^3}-\frac {2 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{3 c x}-\frac {\left (b \left (12 a c-5 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^3}\\ &=\frac {\left (16 a c-5 b \left (3 b d-2 c \sqrt {\frac {d}{x}}\right )\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{12 c^3}-\frac {2 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{3 c x}-\frac {\left (b \left (12 a c-5 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^3}\\ &=\frac {\left (16 a c-5 b \left (3 b d-2 c \sqrt {\frac {d}{x}}\right )\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{12 c^3}-\frac {2 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{3 c x}-\frac {b \sqrt {d} \left (12 a c-5 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^{7/2}}\\ \end {align*}

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Mathematica [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^3} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[1/(Sqrt[a + b*Sqrt[d/x] + c/x]*x^3),x]

[Out]

Integrate[1/(Sqrt[a + b*Sqrt[d/x] + c/x]*x^3), 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(1/x^3/(a+c/x+b*(d/x)^(1/2))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+c/x+b*(d/x)^(1/2))^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, checking for positivity of a ro
ot depending of parameters might return wrong sign: -dWarning, checking for positivity of a root depending of
parameters might return wrong sign: d-1/d/d/abs(d)*d^2*2*(2*((16*c^2/96/d/c^3*sqrt(d/x)-20*d*c*b/96/d/c^3)*sqr
t(d/x)+(30*d^2*b^2-32*d*c*a)/96/d/c^3)*sqrt(a*d^2+b*d^2*sqrt(d/x)+c*d*d/x)+2*(-12*a*b*c*d^2+5*b^3*d^3)/32/c^3/
sqrt(c*d)*ln(abs(2*sqrt(c*d)*(sqrt(a*d^2+b*d^2*sqrt(d/x)+c*d*d/x)-sqrt(c*d)*sqrt(d/x))-d^2*b)))

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maple [A]  time = 0.16, size = 267, normalized size = 1.62 \[ -\frac {\sqrt {\frac {a x +\sqrt {\frac {d}{x}}\, b x +c}{x}}\, \left (-15 \left (\frac {d}{x}\right )^{\frac {3}{2}} b^{3} c \,x^{3} \ln \left (\frac {\sqrt {\frac {d}{x}}\, b x +2 c +2 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, \sqrt {c}}{\sqrt {x}}\right )+36 \sqrt {\frac {d}{x}}\, a b \,c^{2} x^{2} \ln \left (\frac {\sqrt {\frac {d}{x}}\, b x +2 c +2 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, \sqrt {c}}{\sqrt {x}}\right )+30 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, b^{2} c^{\frac {3}{2}} d x -32 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, a \,c^{\frac {5}{2}} x -20 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, \sqrt {\frac {d}{x}}\, b \,c^{\frac {5}{2}} x +16 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, c^{\frac {7}{2}}\right )}{24 \sqrt {a x +\sqrt {\frac {d}{x}}\, b x +c}\, c^{\frac {9}{2}} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(a+(d/x)^(1/2)*b+c/x)^(1/2),x)

[Out]

-1/24*((a*x+(d/x)^(1/2)*b*x+c)/x)^(1/2)/x*(-15*ln(((d/x)^(1/2)*b*x+2*c+2*(a*x+(d/x)^(1/2)*b*x+c)^(1/2)*c^(1/2)
)/x^(1/2))*(d/x)^(3/2)*x^3*b^3*c+30*c^(3/2)*(a*x+(d/x)^(1/2)*b*x+c)^(1/2)*d*x*b^2+36*ln(((d/x)^(1/2)*b*x+2*c+2
*(a*x+(d/x)^(1/2)*b*x+c)^(1/2)*c^(1/2))/x^(1/2))*(d/x)^(1/2)*x^2*a*b*c^2-20*c^(5/2)*(a*x+(d/x)^(1/2)*b*x+c)^(1
/2)*(d/x)^(1/2)*x*b-32*c^(5/2)*(a*x+(d/x)^(1/2)*b*x+c)^(1/2)*x*a+16*(a*x+(d/x)^(1/2)*b*x+c)^(1/2)*c^(7/2))/(a*
x+(d/x)^(1/2)*b*x+c)^(1/2)/c^(9/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \sqrt {\frac {d}{x}} + a + \frac {c}{x}} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(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^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^3\,\sqrt {a+\frac {c}{x}+b\,\sqrt {\frac {d}{x}}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + c/x + b*(d/x)^(1/2))^(1/2)),x)

[Out]

int(1/(x^3*(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 \[ \int \frac {1}{x^{3} \sqrt {a + b \sqrt {\frac {d}{x}} + \frac {c}{x}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(a+c/x+b*(d/x)**(1/2))**(1/2),x)

[Out]

Integral(1/(x**3*sqrt(a + b*sqrt(d/x) + c/x)), x)

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