Integrand size = 18, antiderivative size = 88 \[ \int \frac {\sqrt {-2-3 x+5 x^2}}{x} \, dx=\sqrt {-2-3 x+5 x^2}+\sqrt {2} \arctan \left (\frac {4+3 x}{2 \sqrt {2} \sqrt {-2-3 x+5 x^2}}\right )+\frac {3 \text {arctanh}\left (\frac {3-10 x}{2 \sqrt {5} \sqrt {-2-3 x+5 x^2}}\right )}{2 \sqrt {5}} \]
arctan(1/4*(4+3*x)*2^(1/2)/(5*x^2-3*x-2)^(1/2))*2^(1/2)+3/10*arctanh(1/10* (3-10*x)*5^(1/2)/(5*x^2-3*x-2)^(1/2))*5^(1/2)+(5*x^2-3*x-2)^(1/2)
Time = 0.13 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {-2-3 x+5 x^2}}{x} \, dx=\sqrt {-2-3 x+5 x^2}+2 \sqrt {2} \arctan \left (\frac {\sqrt {-2-3 x+5 x^2}}{\sqrt {2} (-1+x)}\right )-\frac {3 \text {arctanh}\left (\frac {\sqrt {5} \sqrt {-2-3 x+5 x^2}}{2+5 x}\right )}{\sqrt {5}} \]
Sqrt[-2 - 3*x + 5*x^2] + 2*Sqrt[2]*ArcTan[Sqrt[-2 - 3*x + 5*x^2]/(Sqrt[2]* (-1 + x))] - (3*ArcTanh[(Sqrt[5]*Sqrt[-2 - 3*x + 5*x^2])/(2 + 5*x)])/Sqrt[ 5]
Time = 0.25 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1162, 1269, 1092, 219, 1154, 217}
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.
\(\displaystyle \int \frac {\sqrt {5 x^2-3 x-2}}{x} \, dx\) |
\(\Big \downarrow \) 1162 |
\(\displaystyle \sqrt {5 x^2-3 x-2}-\frac {1}{2} \int \frac {3 x+4}{x \sqrt {5 x^2-3 x-2}}dx\) |
\(\Big \downarrow \) 1269 |
\(\displaystyle \frac {1}{2} \left (-3 \int \frac {1}{\sqrt {5 x^2-3 x-2}}dx-4 \int \frac {1}{x \sqrt {5 x^2-3 x-2}}dx\right )+\sqrt {5 x^2-3 x-2}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{2} \left (-4 \int \frac {1}{x \sqrt {5 x^2-3 x-2}}dx-6 \int \frac {1}{20-\frac {(3-10 x)^2}{5 x^2-3 x-2}}d\left (-\frac {3-10 x}{\sqrt {5 x^2-3 x-2}}\right )\right )+\sqrt {5 x^2-3 x-2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {3 \text {arctanh}\left (\frac {3-10 x}{2 \sqrt {5} \sqrt {5 x^2-3 x-2}}\right )}{\sqrt {5}}-4 \int \frac {1}{x \sqrt {5 x^2-3 x-2}}dx\right )+\sqrt {5 x^2-3 x-2}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{2} \left (8 \int \frac {1}{-\frac {(3 x+4)^2}{5 x^2-3 x-2}-8}d\left (-\frac {3 x+4}{\sqrt {5 x^2-3 x-2}}\right )+\frac {3 \text {arctanh}\left (\frac {3-10 x}{2 \sqrt {5} \sqrt {5 x^2-3 x-2}}\right )}{\sqrt {5}}\right )+\sqrt {5 x^2-3 x-2}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (2 \sqrt {2} \arctan \left (\frac {3 x+4}{2 \sqrt {2} \sqrt {5 x^2-3 x-2}}\right )+\frac {3 \text {arctanh}\left (\frac {3-10 x}{2 \sqrt {5} \sqrt {5 x^2-3 x-2}}\right )}{\sqrt {5}}\right )+\sqrt {5 x^2-3 x-2}\) |
Sqrt[-2 - 3*x + 5*x^2] + (2*Sqrt[2]*ArcTan[(4 + 3*x)/(2*Sqrt[2]*Sqrt[-2 - 3*x + 5*x^2])] + (3*ArcTanh[(3 - 10*x)/(2*Sqrt[5]*Sqrt[-2 - 3*x + 5*x^2])] )/Sqrt[5])/2
3.24.64.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
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[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-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]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x ] - Simp[p/(e*(m + 2*p + 1)) Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x ] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || LtQ[m, 1]) && !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.81
method | result | size |
default | \(\sqrt {5 x^{2}-3 x -2}-\frac {3 \ln \left (\frac {\left (-\frac {3}{2}+5 x \right ) \sqrt {5}}{5}+\sqrt {5 x^{2}-3 x -2}\right ) \sqrt {5}}{10}-\sqrt {2}\, \arctan \left (\frac {\left (-3 x -4\right ) \sqrt {2}}{4 \sqrt {5 x^{2}-3 x -2}}\right )\) | \(71\) |
trager | \(\sqrt {5 x^{2}-3 x -2}-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) \ln \left (10 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right ) x +10 \sqrt {5 x^{2}-3 x -2}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-5\right )\right )}{10}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x +4 \sqrt {5 x^{2}-3 x -2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )}{x}\right )\) | \(99\) |
(5*x^2-3*x-2)^(1/2)-3/10*ln(1/5*(-3/2+5*x)*5^(1/2)+(5*x^2-3*x-2)^(1/2))*5^ (1/2)-2^(1/2)*arctan(1/4*(-3*x-4)*2^(1/2)/(5*x^2-3*x-2)^(1/2))
Time = 0.37 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {-2-3 x+5 x^2}}{x} \, dx=\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (3 \, x + 4\right )}}{4 \, \sqrt {5 \, x^{2} - 3 \, x - 2}}\right ) + \frac {3}{20} \, \sqrt {5} \log \left (-4 \, \sqrt {5} \sqrt {5 \, x^{2} - 3 \, x - 2} {\left (10 \, x - 3\right )} + 200 \, x^{2} - 120 \, x - 31\right ) + \sqrt {5 \, x^{2} - 3 \, x - 2} \]
sqrt(2)*arctan(1/4*sqrt(2)*(3*x + 4)/sqrt(5*x^2 - 3*x - 2)) + 3/20*sqrt(5) *log(-4*sqrt(5)*sqrt(5*x^2 - 3*x - 2)*(10*x - 3) + 200*x^2 - 120*x - 31) + sqrt(5*x^2 - 3*x - 2)
\[ \int \frac {\sqrt {-2-3 x+5 x^2}}{x} \, dx=\int \frac {\sqrt {\left (x - 1\right ) \left (5 x + 2\right )}}{x}\, dx \]
Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {-2-3 x+5 x^2}}{x} \, dx=\sqrt {2} \arcsin \left (\frac {3 \, x}{7 \, {\left | x \right |}} + \frac {4}{7 \, {\left | x \right |}}\right ) - \frac {3}{10} \, \sqrt {5} \log \left (2 \, \sqrt {5} \sqrt {5 \, x^{2} - 3 \, x - 2} + 10 \, x - 3\right ) + \sqrt {5 \, x^{2} - 3 \, x - 2} \]
sqrt(2)*arcsin(3/7*x/abs(x) + 4/7/abs(x)) - 3/10*sqrt(5)*log(2*sqrt(5)*sqr t(5*x^2 - 3*x - 2) + 10*x - 3) + sqrt(5*x^2 - 3*x - 2)
Time = 0.35 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {-2-3 x+5 x^2}}{x} \, dx=-2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {5} x - \sqrt {5 \, x^{2} - 3 \, x - 2}\right )}\right ) + \frac {3}{10} \, \sqrt {5} \log \left ({\left | -10 \, \sqrt {5} x + 3 \, \sqrt {5} + 10 \, \sqrt {5 \, x^{2} - 3 \, x - 2} \right |}\right ) + \sqrt {5 \, x^{2} - 3 \, x - 2} \]
-2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(5)*x - sqrt(5*x^2 - 3*x - 2))) + 3/10 *sqrt(5)*log(abs(-10*sqrt(5)*x + 3*sqrt(5) + 10*sqrt(5*x^2 - 3*x - 2))) + sqrt(5*x^2 - 3*x - 2)
Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {-2-3 x+5 x^2}}{x} \, dx=\sqrt {5\,x^2-3\,x-2}-\frac {3\,\sqrt {5}\,\ln \left (\sqrt {5\,x^2-3\,x-2}+\frac {\sqrt {5}\,\left (5\,x-\frac {3}{2}\right )}{5}\right )}{10}-\sqrt {2}\,\ln \left (-\frac {2}{x}-\frac {3}{2}+\frac {\sqrt {2}\,\sqrt {5\,x^2-3\,x-2}\,1{}\mathrm {i}}{x}\right )\,1{}\mathrm {i} \]