Integrand size = 21, antiderivative size = 62 \[ \int \frac {x}{\left (-1+x^2\right ) \sqrt {4+2 x+x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {5+2 x}{\sqrt {7} \sqrt {4+2 x+x^2}}\right )}{2 \sqrt {7}}-\frac {\text {arctanh}\left (\frac {\sqrt {4+2 x+x^2}}{\sqrt {3}}\right )}{2 \sqrt {3}} \] Output:
-1/6*arctanh(1/3*(x^2+2*x+4)^(1/2)*3^(1/2))*3^(1/2)-1/14*arctanh(1/7*(5+2* x)*7^(1/2)/(x^2+2*x+4)^(1/2))*7^(1/2)
Time = 0.14 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\left (-1+x^2\right ) \sqrt {4+2 x+x^2}} \, dx=\frac {\text {arctanh}\left (\frac {1+x-\sqrt {4+2 x+x^2}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\text {arctanh}\left (\frac {1-x+\sqrt {4+2 x+x^2}}{\sqrt {7}}\right )}{\sqrt {7}} \] Input:
Integrate[x/((-1 + x^2)*Sqrt[4 + 2*x + x^2]),x]
Output:
ArcTanh[(1 + x - Sqrt[4 + 2*x + x^2])/Sqrt[3]]/Sqrt[3] - ArcTanh[(1 - x + Sqrt[4 + 2*x + x^2])/Sqrt[7]]/Sqrt[7]
Time = 0.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1366, 25, 1112, 220, 1154, 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.
\(\displaystyle \int \frac {x}{\left (x^2-1\right ) \sqrt {x^2+2 x+4}} \, dx\) |
\(\Big \downarrow \) 1366 |
\(\displaystyle \frac {1}{2} \int -\frac {1}{(1-x) \sqrt {x^2+2 x+4}}dx+\frac {1}{2} \int \frac {1}{(x+1) \sqrt {x^2+2 x+4}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \int \frac {1}{(x+1) \sqrt {x^2+2 x+4}}dx-\frac {1}{2} \int \frac {1}{(1-x) \sqrt {x^2+2 x+4}}dx\) |
\(\Big \downarrow \) 1112 |
\(\displaystyle 2 \int \frac {1}{4 \left (x^2+2 x+4\right )-12}d\sqrt {x^2+2 x+4}-\frac {1}{2} \int \frac {1}{(1-x) \sqrt {x^2+2 x+4}}dx\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {1}{2} \int \frac {1}{(1-x) \sqrt {x^2+2 x+4}}dx-\frac {\text {arctanh}\left (\frac {\sqrt {x^2+2 x+4}}{\sqrt {3}}\right )}{2 \sqrt {3}}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \int \frac {1}{28-\frac {4 (2 x+5)^2}{x^2+2 x+4}}d\left (-\frac {2 (2 x+5)}{\sqrt {x^2+2 x+4}}\right )-\frac {\text {arctanh}\left (\frac {\sqrt {x^2+2 x+4}}{\sqrt {3}}\right )}{2 \sqrt {3}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\text {arctanh}\left (\frac {2 x+5}{\sqrt {7} \sqrt {x^2+2 x+4}}\right )}{2 \sqrt {7}}-\frac {\text {arctanh}\left (\frac {\sqrt {x^2+2 x+4}}{\sqrt {3}}\right )}{2 \sqrt {3}}\) |
Input:
Int[x/((-1 + x^2)*Sqrt[4 + 2*x + x^2]),x]
Output:
-1/2*ArcTanh[(5 + 2*x)/(Sqrt[7]*Sqrt[4 + 2*x + x^2])]/Sqrt[7] - ArcTanh[Sq rt[4 + 2*x + x^2]/Sqrt[3]]/(2*Sqrt[3])
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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symb ol] :> Simp[4*c Subst[Int[1/(b^2*e - 4*a*c*e + 4*c*e*x^2), x], x, Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
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[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + ( f_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Simp[(h/2 + c*(g/(2*q ))) Int[1/((-q + c*x)*Sqrt[d + e*x + f*x^2]), x], x] + Simp[(h/2 - c*(g/( 2*q))) Int[1/((q + c*x)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d , e, f, g, h}, x] && NeQ[e^2 - 4*d*f, 0] && PosQ[(-a)*c]
Time = 0.34 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.79
method | result | size |
default | \(-\frac {\sqrt {7}\, \operatorname {arctanh}\left (\frac {\left (10+4 x \right ) \sqrt {7}}{14 \sqrt {\left (-1+x \right )^{2}+3+4 x}}\right )}{14}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {3}}{\sqrt {\left (1+x \right )^{2}+3}}\right )}{6}\) | \(49\) |
trager | \(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-7\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-7\right ) x +7 \sqrt {x^{2}+2 x +4}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-7\right )}{-1+x}\right )}{14}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (\frac {\sqrt {x^{2}+2 x +4}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )}{1+x}\right )}{6}\) | \(82\) |
Input:
int(x/(x^2-1)/(x^2+2*x+4)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/14*7^(1/2)*arctanh(1/14*(10+4*x)*7^(1/2)/((-1+x)^2+3+4*x)^(1/2))-1/6*3^ (1/2)*arctanh(3^(1/2)/((1+x)^2+3)^(1/2))
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.19 \[ \int \frac {x}{\left (-1+x^2\right ) \sqrt {4+2 x+x^2}} \, dx=\frac {1}{14} \, \sqrt {7} \log \left (\frac {\sqrt {7} {\left (2 \, x + 5\right )} + \sqrt {x^{2} + 2 \, x + 4} {\left (2 \, \sqrt {7} - 7\right )} - 4 \, x - 10}{x - 1}\right ) + \frac {1}{6} \, \sqrt {3} \log \left (-\frac {\sqrt {3} - \sqrt {x^{2} + 2 \, x + 4}}{x + 1}\right ) \] Input:
integrate(x/(x^2-1)/(x^2+2*x+4)^(1/2),x, algorithm="fricas")
Output:
1/14*sqrt(7)*log((sqrt(7)*(2*x + 5) + sqrt(x^2 + 2*x + 4)*(2*sqrt(7) - 7) - 4*x - 10)/(x - 1)) + 1/6*sqrt(3)*log(-(sqrt(3) - sqrt(x^2 + 2*x + 4))/(x + 1))
\[ \int \frac {x}{\left (-1+x^2\right ) \sqrt {4+2 x+x^2}} \, dx=\int \frac {x}{\left (x - 1\right ) \left (x + 1\right ) \sqrt {x^{2} + 2 x + 4}}\, dx \] Input:
integrate(x/(x**2-1)/(x**2+2*x+4)**(1/2),x)
Output:
Integral(x/((x - 1)*(x + 1)*sqrt(x**2 + 2*x + 4)), x)
Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.87 \[ \int \frac {x}{\left (-1+x^2\right ) \sqrt {4+2 x+x^2}} \, dx=-\frac {1}{14} \, \sqrt {7} \operatorname {arsinh}\left (\frac {4 \, \sqrt {3} x}{3 \, {\left | 2 \, x - 2 \right |}} + \frac {10 \, \sqrt {3}}{3 \, {\left | 2 \, x - 2 \right |}}\right ) - \frac {1}{6} \, \sqrt {3} \operatorname {arsinh}\left (\frac {2 \, \sqrt {3}}{{\left | 2 \, x + 2 \right |}}\right ) \] Input:
integrate(x/(x^2-1)/(x^2+2*x+4)^(1/2),x, algorithm="maxima")
Output:
-1/14*sqrt(7)*arcsinh(4/3*sqrt(3)*x/abs(2*x - 2) + 10/3*sqrt(3)/abs(2*x - 2)) - 1/6*sqrt(3)*arcsinh(2*sqrt(3)/abs(2*x + 2))
Leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (48) = 96\).
Time = 0.16 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.76 \[ \int \frac {x}{\left (-1+x^2\right ) \sqrt {4+2 x+x^2}} \, dx=\frac {1}{14} \, \sqrt {7} \log \left (\frac {{\left | -2 \, x - 2 \, \sqrt {7} + 2 \, \sqrt {x^{2} + 2 \, x + 4} + 2 \right |}}{{\left | -2 \, x + 2 \, \sqrt {7} + 2 \, \sqrt {x^{2} + 2 \, x + 4} + 2 \right |}}\right ) + \frac {1}{6} \, \sqrt {3} \log \left (-\frac {{\left | -2 \, x - 2 \, \sqrt {3} + 2 \, \sqrt {x^{2} + 2 \, x + 4} - 2 \right |}}{2 \, {\left (x - \sqrt {3} - \sqrt {x^{2} + 2 \, x + 4} + 1\right )}}\right ) \] Input:
integrate(x/(x^2-1)/(x^2+2*x+4)^(1/2),x, algorithm="giac")
Output:
1/14*sqrt(7)*log(abs(-2*x - 2*sqrt(7) + 2*sqrt(x^2 + 2*x + 4) + 2)/abs(-2* x + 2*sqrt(7) + 2*sqrt(x^2 + 2*x + 4) + 2)) + 1/6*sqrt(3)*log(-1/2*abs(-2* x - 2*sqrt(3) + 2*sqrt(x^2 + 2*x + 4) - 2)/(x - sqrt(3) - sqrt(x^2 + 2*x + 4) + 1))
Timed out. \[ \int \frac {x}{\left (-1+x^2\right ) \sqrt {4+2 x+x^2}} \, dx=\int \frac {x}{\left (x^2-1\right )\,\sqrt {x^2+2\,x+4}} \,d x \] Input:
int(x/((x^2 - 1)*(2*x + x^2 + 4)^(1/2)),x)
Output:
int(x/((x^2 - 1)*(2*x + x^2 + 4)^(1/2)), x)
Time = 0.16 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.08 \[ \int \frac {x}{\left (-1+x^2\right ) \sqrt {4+2 x+x^2}} \, dx=\frac {\sqrt {7}\, \mathrm {log}\left (\sqrt {x^{2}+2 x +4}\, \sqrt {7}-2 x -5\right )}{14}-\frac {\sqrt {7}\, \mathrm {log}\left (x -1\right )}{14}+\frac {\sqrt {3}\, \mathrm {log}\left (\sqrt {x^{2}+2 x +4}-\sqrt {3}\right )}{12}-\frac {\sqrt {3}\, \mathrm {log}\left (\sqrt {x^{2}+2 x +4}+\sqrt {3}\right )}{12} \] Input:
int(x/(x^2-1)/(x^2+2*x+4)^(1/2),x)
Output:
(6*sqrt(7)*log(sqrt(x**2 + 2*x + 4)*sqrt(7) - 2*x - 5) - 6*sqrt(7)*log(x - 1) + 7*sqrt(3)*log(sqrt(x**2 + 2*x + 4) - sqrt(3)) - 7*sqrt(3)*log(sqrt(x **2 + 2*x + 4) + sqrt(3)))/84