Integrand size = 23, antiderivative size = 66 \[ \int \frac {1}{x (1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\frac {2}{3 \sqrt {1+x} \sqrt {1-x+x^2}}-\frac {2 \sqrt {1+x^3} \text {arctanh}\left (\sqrt {1+x^3}\right )}{3 \sqrt {1+x} \sqrt {1-x+x^2}} \]
2/3/(1+x)^(1/2)/(x^2-x+1)^(1/2)-2/3*arctanh((x^3+1)^(1/2))*(x^3+1)^(1/2)/( 1+x)^(1/2)/(x^2-x+1)^(1/2)
Time = 11.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x (1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\frac {2 \left (\sqrt {1+x}-(1+x)^2 \sqrt {\frac {1-x+x^2}{(1+x)^2}} \text {arctanh}\left (\frac {1}{(1+x)^{3/2} \sqrt {\frac {1-x+x^2}{(1+x)^2}}}\right )\right )}{3 (1+x) \sqrt {1-x+x^2}} \]
(2*(Sqrt[1 + x] - (1 + x)^2*Sqrt[(1 - x + x^2)/(1 + x)^2]*ArcTanh[1/((1 + x)^(3/2)*Sqrt[(1 - x + x^2)/(1 + x)^2])]))/(3*(1 + x)*Sqrt[1 - x + x^2])
Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1210, 798, 61, 73, 220}
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.
\(\displaystyle \int \frac {1}{x (x+1)^{3/2} \left (x^2-x+1\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1210 |
\(\displaystyle \frac {\sqrt {x^3+1} \int \frac {1}{x \left (x^3+1\right )^{3/2}}dx}{\sqrt {x+1} \sqrt {x^2-x+1}}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\sqrt {x^3+1} \int \frac {1}{x^3 \left (x^3+1\right )^{3/2}}dx^3}{3 \sqrt {x+1} \sqrt {x^2-x+1}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {\sqrt {x^3+1} \left (\int \frac {1}{x^3 \sqrt {x^3+1}}dx^3+\frac {2}{\sqrt {x^3+1}}\right )}{3 \sqrt {x+1} \sqrt {x^2-x+1}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\sqrt {x^3+1} \left (2 \int \frac {1}{x^6-1}d\sqrt {x^3+1}+\frac {2}{\sqrt {x^3+1}}\right )}{3 \sqrt {x+1} \sqrt {x^2-x+1}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {\sqrt {x^3+1} \left (\frac {2}{\sqrt {x^3+1}}-2 \text {arctanh}\left (\sqrt {x^3+1}\right )\right )}{3 \sqrt {x+1} \sqrt {x^2-x+1}}\) |
(Sqrt[1 + x^3]*(2/Sqrt[1 + x^3] - 2*ArcTanh[Sqrt[1 + x^3]]))/(3*Sqrt[1 + x ]*Sqrt[1 - x + x^2])
3.6.14.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^FracPart[p]*((a + b*x + c*x^2)^FracPart[p]/(a*d + c*e*x^3)^FracPart[p]) Int[(f + g*x)^n*(a*d + c* e*x^3)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0] && EqQ[m, p]
Time = 0.67 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.65
method | result | size |
default | \(-\frac {2 \sqrt {1+x}\, \sqrt {x^{2}-x +1}\, \left (\operatorname {arctanh}\left (\sqrt {x^{3}+1}\right ) \sqrt {x^{3}+1}-1\right )}{3 \left (x^{3}+1\right )}\) | \(43\) |
elliptic | \(\frac {\sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}\, \left (\frac {2}{3 \sqrt {x^{3}+1}}-\frac {2 \,\operatorname {arctanh}\left (\sqrt {x^{3}+1}\right )}{3}\right )}{\sqrt {1+x}\, \sqrt {x^{2}-x +1}}\) | \(51\) |
risch | \(\frac {2}{3 \sqrt {1+x}\, \sqrt {x^{2}-x +1}}-\frac {2 \,\operatorname {arctanh}\left (\sqrt {x^{3}+1}\right ) \sqrt {\left (1+x \right ) \left (x^{2}-x +1\right )}}{3 \sqrt {1+x}\, \sqrt {x^{2}-x +1}}\) | \(58\) |
Time = 0.34 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x (1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=-\frac {{\left (x^{3} + 1\right )} \log \left (\sqrt {x^{2} - x + 1} \sqrt {x + 1} + 1\right ) - {\left (x^{3} + 1\right )} \log \left (\sqrt {x^{2} - x + 1} \sqrt {x + 1} - 1\right ) - 2 \, \sqrt {x^{2} - x + 1} \sqrt {x + 1}}{3 \, {\left (x^{3} + 1\right )}} \]
-1/3*((x^3 + 1)*log(sqrt(x^2 - x + 1)*sqrt(x + 1) + 1) - (x^3 + 1)*log(sqr t(x^2 - x + 1)*sqrt(x + 1) - 1) - 2*sqrt(x^2 - x + 1)*sqrt(x + 1))/(x^3 + 1)
\[ \int \frac {1}{x (1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\int \frac {1}{x \left (x + 1\right )^{\frac {3}{2}} \left (x^{2} - x + 1\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{x (1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{2} - x + 1\right )}^{\frac {3}{2}} {\left (x + 1\right )}^{\frac {3}{2}} x} \,d x } \]
\[ \int \frac {1}{x (1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{2} - x + 1\right )}^{\frac {3}{2}} {\left (x + 1\right )}^{\frac {3}{2}} x} \,d x } \]
Timed out. \[ \int \frac {1}{x (1+x)^{3/2} \left (1-x+x^2\right )^{3/2}} \, dx=\int \frac {1}{x\,{\left (x+1\right )}^{3/2}\,{\left (x^2-x+1\right )}^{3/2}} \,d x \]