Integrand size = 14, antiderivative size = 79 \[ \int \frac {1}{x^3 \left (1+x+x^2\right )^{3/2}} \, dx=\frac {2 (1-x)}{3 x^2 \sqrt {1+x+x^2}}-\frac {7 \sqrt {1+x+x^2}}{6 x^2}+\frac {37 \sqrt {1+x+x^2}}{12 x}-\frac {3}{8} \text {arctanh}\left (\frac {2+x}{2 \sqrt {1+x+x^2}}\right ) \]
-3/8*arctanh(1/2*(2+x)/(x^2+x+1)^(1/2))+2/3*(1-x)/x^2/(x^2+x+1)^(1/2)-7/6* (x^2+x+1)^(1/2)/x^2+37/12*(x^2+x+1)^(1/2)/x
Time = 0.20 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.66 \[ \int \frac {1}{x^3 \left (1+x+x^2\right )^{3/2}} \, dx=\frac {-6+15 x+23 x^2+37 x^3}{12 x^2 \sqrt {1+x+x^2}}+\frac {3}{4} \text {arctanh}\left (x-\sqrt {1+x+x^2}\right ) \]
(-6 + 15*x + 23*x^2 + 37*x^3)/(12*x^2*Sqrt[1 + x + x^2]) + (3*ArcTanh[x - Sqrt[1 + x + x^2]])/4
Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1165, 27, 1237, 27, 1228, 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.
\(\displaystyle \int \frac {1}{x^3 \left (x^2+x+1\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1165 |
\(\displaystyle \frac {2}{3} \int \frac {7-4 x}{2 x^3 \sqrt {x^2+x+1}}dx+\frac {2 (1-x)}{3 x^2 \sqrt {x^2+x+1}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {7-4 x}{x^3 \sqrt {x^2+x+1}}dx+\frac {2 (1-x)}{3 x^2 \sqrt {x^2+x+1}}\) |
\(\Big \downarrow \) 1237 |
\(\displaystyle \frac {1}{3} \left (-\frac {1}{2} \int \frac {14 x+37}{2 x^2 \sqrt {x^2+x+1}}dx-\frac {7 \sqrt {x^2+x+1}}{2 x^2}\right )+\frac {2 (1-x)}{3 x^2 \sqrt {x^2+x+1}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (-\frac {1}{4} \int \frac {14 x+37}{x^2 \sqrt {x^2+x+1}}dx-\frac {7 \sqrt {x^2+x+1}}{2 x^2}\right )+\frac {2 (1-x)}{3 x^2 \sqrt {x^2+x+1}}\) |
\(\Big \downarrow \) 1228 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{4} \left (\frac {9}{2} \int \frac {1}{x \sqrt {x^2+x+1}}dx+\frac {37 \sqrt {x^2+x+1}}{x}\right )-\frac {7 \sqrt {x^2+x+1}}{2 x^2}\right )+\frac {2 (1-x)}{3 x^2 \sqrt {x^2+x+1}}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{4} \left (\frac {37 \sqrt {x^2+x+1}}{x}-9 \int \frac {1}{4-\frac {(x+2)^2}{x^2+x+1}}d\frac {x+2}{\sqrt {x^2+x+1}}\right )-\frac {7 \sqrt {x^2+x+1}}{2 x^2}\right )+\frac {2 (1-x)}{3 x^2 \sqrt {x^2+x+1}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{4} \left (\frac {37 \sqrt {x^2+x+1}}{x}-\frac {9}{2} \text {arctanh}\left (\frac {x+2}{2 \sqrt {x^2+x+1}}\right )\right )-\frac {7 \sqrt {x^2+x+1}}{2 x^2}\right )+\frac {2 (1-x)}{3 x^2 \sqrt {x^2+x+1}}\) |
(2*(1 - x))/(3*x^2*Sqrt[1 + x + x^2]) + ((-7*Sqrt[1 + x + x^2])/(2*x^2) + ((37*Sqrt[1 + x + x^2])/x - (9*ArcTanh[(2 + x)/(2*Sqrt[1 + x + x^2])])/2)/ 4)/3
3.3.77.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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/(((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)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e) *x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^ 2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m + 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p, -1] && 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[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e *f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^ (m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x ] && EqQ[Simplify[m + 2*p + 3], 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + b* x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1) *(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[ (c*d*f - f*b*e + a*e*g)*(m + 1) + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && LtQ[m, -1 ] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Time = 0.52 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58
method | result | size |
risch | \(\frac {37 x^{3}+23 x^{2}+15 x -6}{12 \sqrt {x^{2}+x +1}\, x^{2}}-\frac {3 \,\operatorname {arctanh}\left (\frac {2+x}{2 \sqrt {x^{2}+x +1}}\right )}{8}\) | \(46\) |
trager | \(\frac {37 x^{3}+23 x^{2}+15 x -6}{12 \sqrt {x^{2}+x +1}\, x^{2}}-\frac {3 \ln \left (\frac {2 \sqrt {x^{2}+x +1}+2+x}{x}\right )}{8}\) | \(50\) |
default | \(-\frac {1}{2 x^{2} \sqrt {x^{2}+x +1}}+\frac {5}{4 x \sqrt {x^{2}+x +1}}+\frac {3}{8 \sqrt {x^{2}+x +1}}+\frac {\frac {37}{24}+\frac {37 x}{12}}{\sqrt {x^{2}+x +1}}-\frac {3 \,\operatorname {arctanh}\left (\frac {2+x}{2 \sqrt {x^{2}+x +1}}\right )}{8}\) | \(69\) |
Time = 0.24 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.35 \[ \int \frac {1}{x^3 \left (1+x+x^2\right )^{3/2}} \, dx=\frac {74 \, x^{4} + 74 \, x^{3} + 74 \, x^{2} - 9 \, {\left (x^{4} + x^{3} + x^{2}\right )} \log \left (-x + \sqrt {x^{2} + x + 1} + 1\right ) + 9 \, {\left (x^{4} + x^{3} + x^{2}\right )} \log \left (-x + \sqrt {x^{2} + x + 1} - 1\right ) + 2 \, {\left (37 \, x^{3} + 23 \, x^{2} + 15 \, x - 6\right )} \sqrt {x^{2} + x + 1}}{24 \, {\left (x^{4} + x^{3} + x^{2}\right )}} \]
1/24*(74*x^4 + 74*x^3 + 74*x^2 - 9*(x^4 + x^3 + x^2)*log(-x + sqrt(x^2 + x + 1) + 1) + 9*(x^4 + x^3 + x^2)*log(-x + sqrt(x^2 + x + 1) - 1) + 2*(37*x ^3 + 23*x^2 + 15*x - 6)*sqrt(x^2 + x + 1))/(x^4 + x^3 + x^2)
\[ \int \frac {1}{x^3 \left (1+x+x^2\right )^{3/2}} \, dx=\int \frac {1}{x^{3} \left (x^{2} + x + 1\right )^{\frac {3}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^3 \left (1+x+x^2\right )^{3/2}} \, dx=\frac {37 \, x}{12 \, \sqrt {x^{2} + x + 1}} + \frac {23}{12 \, \sqrt {x^{2} + x + 1}} + \frac {5}{4 \, \sqrt {x^{2} + x + 1} x} - \frac {1}{2 \, \sqrt {x^{2} + x + 1} x^{2}} - \frac {3}{8} \, \operatorname {arsinh}\left (\frac {\sqrt {3} x}{3 \, {\left | x \right |}} + \frac {2 \, \sqrt {3}}{3 \, {\left | x \right |}}\right ) \]
37/12*x/sqrt(x^2 + x + 1) + 23/12/sqrt(x^2 + x + 1) + 5/4/(sqrt(x^2 + x + 1)*x) - 1/2/(sqrt(x^2 + x + 1)*x^2) - 3/8*arcsinh(1/3*sqrt(3)*x/abs(x) + 2 /3*sqrt(3)/abs(x))
Time = 0.29 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.48 \[ \int \frac {1}{x^3 \left (1+x+x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (2 \, x + 1\right )}}{3 \, \sqrt {x^{2} + x + 1}} - \frac {3 \, {\left (x - \sqrt {x^{2} + x + 1}\right )}^{3} + 8 \, {\left (x - \sqrt {x^{2} + x + 1}\right )}^{2} - 13 \, x + 13 \, \sqrt {x^{2} + x + 1} - 16}{4 \, {\left ({\left (x - \sqrt {x^{2} + x + 1}\right )}^{2} - 1\right )}^{2}} - \frac {3}{8} \, \log \left ({\left | -x + \sqrt {x^{2} + x + 1} + 1 \right |}\right ) + \frac {3}{8} \, \log \left ({\left | -x + \sqrt {x^{2} + x + 1} - 1 \right |}\right ) \]
2/3*(2*x + 1)/sqrt(x^2 + x + 1) - 1/4*(3*(x - sqrt(x^2 + x + 1))^3 + 8*(x - sqrt(x^2 + x + 1))^2 - 13*x + 13*sqrt(x^2 + x + 1) - 16)/((x - sqrt(x^2 + x + 1))^2 - 1)^2 - 3/8*log(abs(-x + sqrt(x^2 + x + 1) + 1)) + 3/8*log(ab s(-x + sqrt(x^2 + x + 1) - 1))
Timed out. \[ \int \frac {1}{x^3 \left (1+x+x^2\right )^{3/2}} \, dx=\int \frac {1}{x^3\,{\left (x^2+x+1\right )}^{3/2}} \,d x \]