3.31.66 \(\int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^3} \, dx\) [3066]

3.31.66.1 Optimal result

Integrand size = 26, antiderivative size = 165 \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^3} \, dx=\frac {\left (16 a c-15 b^2 d+10 b c \sqrt {\frac {d}{x}}\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{12 c^3}-\frac {2 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{3 c x}-\frac {b \sqrt {d} \left (12 a c-5 b^2 d\right ) \text {arctanh}\left (\frac {b d+2 c \sqrt {\frac {d}{x}}}{2 \sqrt {c} \sqrt {d} \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}\right )}{8 c^{7/2}} \]

output

-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
 

3.31.66.2 Mathematica [A] (verified)

Time = 1.00 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.32 \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^3} \, dx=\frac {2 \sqrt {c} \left (-8 c^3+2 c^2 \left (4 a+b \sqrt {\frac {d}{x}}\right ) x-15 b^2 d \left (a+b \sqrt {\frac {d}{x}}\right ) x^2+c x \left (-5 b^2 d+16 a^2 x+26 a b \sqrt {\frac {d}{x}} x\right )\right )+3 b \left (12 a c-5 b^2 d\right ) x^2 \sqrt {\frac {d \left (c+\left (a+b \sqrt {\frac {d}{x}}\right ) x\right )}{x}} \log \left (c^3 \left (b d+2 c \sqrt {\frac {d}{x}}-2 \sqrt {c} \sqrt {\frac {d \left (c+a x+b \sqrt {\frac {d}{x}} x\right )}{x}}\right )\right )}{24 c^{7/2} \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^2} \]

input

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

output

(2*Sqrt[c]*(-8*c^3 + 2*c^2*(4*a + b*Sqrt[d/x])*x - 15*b^2*d*(a + b*Sqrt[d/ 
x])*x^2 + c*x*(-5*b^2*d + 16*a^2*x + 26*a*b*Sqrt[d/x]*x)) + 3*b*(12*a*c - 
5*b^2*d)*x^2*Sqrt[(d*(c + (a + b*Sqrt[d/x])*x))/x]*Log[c^3*(b*d + 2*c*Sqrt 
[d/x] - 2*Sqrt[c]*Sqrt[(d*(c + a*x + b*Sqrt[d/x]*x))/x])])/(24*c^(7/2)*Sqr 
t[a + b*Sqrt[d/x] + c/x]*x^2)
 

3.31.66.3 Rubi [A] (warning: unable to verify)

Time = 0.39 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {2066, 1693, 1166, 27, 1225, 1092, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2066

\(\displaystyle -\frac {\int \frac {d}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x}d\frac {d}{x}}{d^2}\)

\(\Big \downarrow \) 1693

\(\displaystyle -\frac {2 \int \frac {d^3}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}} x^3}d\sqrt {\frac {d}{x}}}{d^2}\)

\(\Big \downarrow \) 1166

\(\displaystyle -\frac {2 \left (\frac {d \int -\frac {\left (4 a+\frac {5 b d}{x}\right ) \sqrt {\frac {d}{x}}}{2 \sqrt {a+\frac {b d}{x}+\frac {c d}{x^2}}}d\sqrt {\frac {d}{x}}}{3 c}+\frac {d^3 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}{3 c x^2}\right )}{d^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2 \left (\frac {d^3 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}{3 c x^2}-\frac {d \int \frac {\left (4 a+\frac {5 b d}{x}\right ) \sqrt {\frac {d}{x}}}{\sqrt {a+\frac {b d}{x}+\frac {c d}{x^2}}}d\sqrt {\frac {d}{x}}}{6 c}\right )}{d^2}\)

\(\Big \downarrow \) 1225

\(\displaystyle -\frac {2 \left (\frac {d^3 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}{3 c x^2}-\frac {d \left (-\frac {3 b d \left (12 a c-5 b^2 d\right ) \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}d\sqrt {\frac {d}{x}}}{8 c^2}-\frac {d \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}} \left (d \left (15 b^2-\frac {16 a c}{d}\right )-10 b c \sqrt {\frac {d}{x}}\right )}{4 c^2}\right )}{6 c}\right )}{d^2}\)

\(\Big \downarrow \) 1092

\(\displaystyle -\frac {2 \left (\frac {d^3 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}{3 c x^2}-\frac {d \left (-\frac {3 b d \left (12 a c-5 b^2 d\right ) \int \frac {1}{\frac {4 c}{d}-\frac {d^2}{x^2}}d\frac {2 \sqrt {\frac {d}{x}} c+b d}{d \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}}{4 c^2}-\frac {d \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}} \left (d \left (15 b^2-\frac {16 a c}{d}\right )-10 b c \sqrt {\frac {d}{x}}\right )}{4 c^2}\right )}{6 c}\right )}{d^2}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {2 \left (\frac {d^3 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}{3 c x^2}-\frac {d \left (-\frac {3 b d^{3/2} \left (12 a c-5 b^2 d\right ) \text {arctanh}\left (\frac {d^{3/2}}{2 \sqrt {c} x}\right )}{8 c^{5/2}}-\frac {d \left (d \left (15 b^2-\frac {16 a c}{d}\right )-10 b c \sqrt {\frac {d}{x}}\right ) \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}{4 c^2}\right )}{6 c}\right )}{d^2}\)

input

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

output

(-2*((d^3*Sqrt[a + b*Sqrt[d/x] + (c*d)/x^2])/(3*c*x^2) - (d*(-1/4*(d*((15* 
b^2 - (16*a*c)/d)*d - 10*b*c*Sqrt[d/x])*Sqrt[a + b*Sqrt[d/x] + (c*d)/x^2]) 
/c^2 - (3*b*d^(3/2)*(12*a*c - 5*b^2*d)*ArcTanh[d^(3/2)/(2*Sqrt[c]*x)])/(8* 
c^(5/2))))/(6*c)))/d^2
 

3.31.66.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1092

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

rule 1166

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 
 1))), x] + Simp[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] && If[Ration 
alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat 
icQ[a, b, c, d, e, m, p, x]
 

rule 1225

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] + Simp[(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] &&  !LeQ[p, -1]
 

rule 1693

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(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] && IntegerQ 
[Simplify[(m + 1)/n]]
 

rule 2066

Int[(x_)^(m_.)*((a_) + (b_.)*((d_.)/(x_))^(n_) + (c_.)*(x_)^(n2_.))^(p_), x 
_Symbol] :> Simp[-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]
 

3.31.66.4 Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.62

method	result	size
default	\(\frac {\sqrt {\frac {b \sqrt {\frac {d}{x}}\, x +a x +c}{x}}\, \left (15 \ln \left (\frac {2 c +b \sqrt {\frac {d}{x}}\, x +2 \sqrt {c}\, \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}}{\sqrt {x}}\right ) \left (\frac {d}{x}\right )^{\frac {3}{2}} x^{3} b^{3} c -30 d \,c^{\frac {3}{2}} \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, x \,b^{2}-36 \ln \left (\frac {2 c +b \sqrt {\frac {d}{x}}\, x +2 \sqrt {c}\, \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}}{\sqrt {x}}\right ) \sqrt {\frac {d}{x}}\, x^{2} a b \,c^{2}+20 c^{\frac {5}{2}} \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, \sqrt {\frac {d}{x}}\, x b +32 c^{\frac {5}{2}} \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, a x -16 \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, c^{\frac {7}{2}}\right )}{24 x \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, c^{\frac {9}{2}}}\)	\(267\)

input

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

output

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

3.31.66.5 Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^3} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.31.66.6 Sympy [F]

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

input

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

output

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

3.31.66.7 Maxima [F]

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

input

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

output

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

3.31.66.8 Giac [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^3} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.31.66.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^3} \, dx=\int \frac {1}{x^3\,\sqrt {a+\frac {c}{x}+b\,\sqrt {\frac {d}{x}}}} \,d x \]

input

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

output

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