Integrand size = 26, antiderivative size = 93 \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^2} \, dx=-\frac {2 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}{c}+\frac {b \sqrt {d} \text {arctanh}\left (\frac {b d+2 c \sqrt {\frac {d}{x}}}{2 \sqrt {c} \sqrt {d} \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}\right )}{c^{3/2}} \]
b*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^(3/2)-2*(a+c/x+b*(d/x)^(1/2))^(1/2)/c
Time = 0.52 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.57 \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^2} \, dx=\frac {\sqrt {\frac {d \left (c+\left (a+b \sqrt {\frac {d}{x}}\right ) x\right )}{x}} \left (-2 \sqrt {c} \sqrt {\frac {d \left (c+a x+b \sqrt {\frac {d}{x}} x\right )}{x}}+b d \text {arctanh}\left (\frac {b d+2 c \sqrt {\frac {d}{x}}}{2 \sqrt {c} \sqrt {\frac {d \left (c+\left (a+b \sqrt {\frac {d}{x}}\right ) x\right )}{x}}}\right )\right )}{c^{3/2} d \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}} \]
(Sqrt[(d*(c + (a + b*Sqrt[d/x])*x))/x]*(-2*Sqrt[c]*Sqrt[(d*(c + a*x + b*Sq rt[d/x]*x))/x] + b*d*ArcTanh[(b*d + 2*c*Sqrt[d/x])/(2*Sqrt[c]*Sqrt[(d*(c + (a + b*Sqrt[d/x])*x))/x])]))/(c^(3/2)*d*Sqrt[a + b*Sqrt[d/x] + c/x])
Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.72, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2066, 1680, 1160, 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.
\(\displaystyle \int \frac {1}{x^2 \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}} \, dx\) |
\(\Big \downarrow \) 2066 |
\(\displaystyle -\frac {\int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}}}d\frac {d}{x}}{d}\) |
\(\Big \downarrow \) 1680 |
\(\displaystyle -\frac {2 \int \frac {\sqrt {\frac {d}{x}}}{\sqrt {a+\frac {b d}{x}+\frac {c d}{x^2}}}d\sqrt {\frac {d}{x}}}{d}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle -\frac {2 \left (\frac {d \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}{c}-\frac {b d \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}d\sqrt {\frac {d}{x}}}{2 c}\right )}{d}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle -\frac {2 \left (\frac {d \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}{c}-\frac {b d \int \frac {1}{\frac {4 c}{d}-\frac {d^2}{x^2}}d\frac {2 \sqrt {\frac {d}{x}} c+b d}{d \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}}{c}\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2 \left (\frac {d \sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c d}{x^2}}}{c}-\frac {b d^{3/2} \text {arctanh}\left (\frac {d^{3/2}}{2 \sqrt {c} x}\right )}{2 c^{3/2}}\right )}{d}\) |
(-2*((d*Sqrt[a + b*Sqrt[d/x] + (c*d)/x^2])/c - (b*d^(3/2)*ArcTanh[d^(3/2)/ (2*Sqrt[c]*x)])/(2*c^(3/2))))/d
3.31.65.3.1 Defintions of rubi rules used
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])
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]
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] + Simp[(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[p, -1]
Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[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] && Fr actionQ[n]
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]
Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.29
method | result | size |
default | \(-\frac {\sqrt {\frac {b \sqrt {\frac {d}{x}}\, x +a x +c}{x}}\, \left (-\sqrt {\frac {d}{x}}\, x b \ln \left (\frac {2 c +b \sqrt {\frac {d}{x}}\, x +2 \sqrt {c}\, \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}}{\sqrt {x}}\right ) c +2 \sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, c^{\frac {3}{2}}\right )}{\sqrt {b \sqrt {\frac {d}{x}}\, x +a x +c}\, c^{\frac {5}{2}}}\) | \(120\) |
-((b*(d/x)^(1/2)*x+a*x+c)/x)^(1/2)*(-(d/x)^(1/2)*x*b*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))*c+2*(b*(d/x)^(1/2)*x+ a*x+c)^(1/2)*c^(3/2))/(b*(d/x)^(1/2)*x+a*x+c)^(1/2)/c^(5/2)
Timed out. \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^2} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^2} \, dx=\int \frac {1}{x^{2} \sqrt {a + b \sqrt {\frac {d}{x}} + \frac {c}{x}}}\, dx \]
\[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^2} \, dx=\int { \frac {1}{\sqrt {b \sqrt {\frac {d}{x}} + a + \frac {c}{x}} x^{2}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{\sqrt {a+b \sqrt {\frac {d}{x}}+\frac {c}{x}} x^2} \, dx=\int \frac {1}{x^2\,\sqrt {a+\frac {c}{x}+b\,\sqrt {\frac {d}{x}}}} \,d x \]