\(\int \frac {1}{x (1+x)^{5/2} (1-x+x^2)^{5/2}} \, dx\) [504]

Optimal result

Integrand size = 23, antiderivative size = 96 \[ \int \frac {1}{x (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\frac {2}{3 \sqrt {1+x} \sqrt {1-x+x^2}}+\frac {2}{9 \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )}-\frac {2 \sqrt {1+x^3} \text {arctanh}\left (\sqrt {1+x^3}\right )}{3 \sqrt {1+x} \sqrt {1-x+x^2}} \] Output:

2/3/(1+x)^(1/2)/(x^2-x+1)^(1/2)+2/9/(1+x)^(1/2)/(x^2-x+1)^(1/2)/(x^3+1)-2/ 
3*(x^3+1)^(1/2)*arctanh((x^3+1)^(1/2))/(1+x)^(1/2)/(x^2-x+1)^(1/2)
 

Mathematica [A] (verified)

Time = 10.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\frac {\frac {2 \left (4+3 x^3\right )}{3 (1+x)^{3/2} \left (1-x+x^2\right )}-2 (1+x) \sqrt {\frac {1-x+x^2}{(1+x)^2}} \text {arctanh}\left (\frac {1}{(1+x)^{3/2} \sqrt {\frac {1-x+x^2}{(1+x)^2}}}\right )}{3 \sqrt {1-x+x^2}} \] Input:

Integrate[1/(x*(1 + x)^(5/2)*(1 - x + x^2)^(5/2)),x]
 

Output:

((2*(4 + 3*x^3))/(3*(1 + x)^(3/2)*(1 - x + x^2)) - 2*(1 + x)*Sqrt[(1 - x + 
 x^2)/(1 + x)^2]*ArcTanh[1/((1 + x)^(3/2)*Sqrt[(1 - x + x^2)/(1 + x)^2])]) 
/(3*Sqrt[1 - x + x^2])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.72, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {1210, 798, 61, 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.

\(\displaystyle \int \frac {1}{x (x+1)^{5/2} \left (x^2-x+1\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1210

\(\displaystyle \frac {\sqrt {x^3+1} \int \frac {1}{x \left (x^3+1\right )^{5/2}}dx}{\sqrt {x+1} \sqrt {x^2-x+1}}\)

\(\Big \downarrow \) 798

\(\displaystyle \frac {\sqrt {x^3+1} \int \frac {1}{x^3 \left (x^3+1\right )^{5/2}}dx^3}{3 \sqrt {x+1} \sqrt {x^2-x+1}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {\sqrt {x^3+1} \left (\int \frac {1}{x^3 \left (x^3+1\right )^{3/2}}dx^3+\frac {2}{3 \left (x^3+1\right )^{3/2}}\right )}{3 \sqrt {x+1} \sqrt {x^2-x+1}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {\sqrt {x^3+1} \left (\int \frac {1}{x^3 \sqrt {x^3+1}}dx^3+\frac {2}{\sqrt {x^3+1}}+\frac {2}{3 \left (x^3+1\right )^{3/2}}\right )}{3 \sqrt {x+1} \sqrt {x^2-x+1}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\sqrt {x^3+1} \left (2 \int \frac {1}{x^6-1}d\sqrt {x^3+1}+\frac {2}{\sqrt {x^3+1}}+\frac {2}{3 \left (x^3+1\right )^{3/2}}\right )}{3 \sqrt {x+1} \sqrt {x^2-x+1}}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {\sqrt {x^3+1} \left (-2 \text {arctanh}\left (\sqrt {x^3+1}\right )+\frac {2}{\sqrt {x^3+1}}+\frac {2}{3 \left (x^3+1\right )^{3/2}}\right )}{3 \sqrt {x+1} \sqrt {x^2-x+1}}\)

Input:

Int[1/(x*(1 + x)^(5/2)*(1 - x + x^2)^(5/2)),x]
 

Output:

(Sqrt[1 + x^3]*(2/(3*(1 + x^3)^(3/2)) + 2/Sqrt[1 + x^3] - 2*ArcTanh[Sqrt[1 
 + x^3]]))/(3*Sqrt[1 + x]*Sqrt[1 - x + x^2])
 

Defintions of rubi rules used

rule 61

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]
 

rule 73

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]
 

rule 220

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])
 

rule 798

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]]
 

rule 1210

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]
 

Maple [A] (verified)

Time = 2.70 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.62

method	result	size
elliptic	\(\frac {\sqrt {\left (x +1\right ) \left (x^{2}-x +1\right )}\, \left (\frac {2}{9 \left (x^{3}+1\right )^{\frac {3}{2}}}+\frac {2}{3 \sqrt {x^{3}+1}}-\frac {2 \,\operatorname {arctanh}\left (\sqrt {x^{3}+1}\right )}{3}\right )}{\sqrt {x +1}\, \sqrt {x^{2}-x +1}}\)	\(60\)
default	\(-\frac {2 \left (3 \,\operatorname {arctanh}\left (\sqrt {x^{3}+1}\right ) \sqrt {x^{3}+1}\, x^{3}-3 x^{3}+3 \,\operatorname {arctanh}\left (\sqrt {x^{3}+1}\right ) \sqrt {x^{3}+1}-4\right )}{9 \left (x^{3}+1\right ) \sqrt {x^{2}-x +1}\, \sqrt {x +1}}\)	\(69\)

Input:

int(1/x/(x+1)^(5/2)/(x^2-x+1)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

((x+1)*(x^2-x+1))^(1/2)/(x+1)^(1/2)/(x^2-x+1)^(1/2)*(2/9/(x^3+1)^(3/2)+2/3 
/(x^3+1)^(1/2)-2/3*arctanh((x^3+1)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (3 \, x^{3} + 4\right )} \sqrt {x^{2} - x + 1} \sqrt {x + 1} - 3 \, {\left (x^{6} + 2 \, x^{3} + 1\right )} \log \left (\sqrt {x^{2} - x + 1} \sqrt {x + 1} + 1\right ) + 3 \, {\left (x^{6} + 2 \, x^{3} + 1\right )} \log \left (\sqrt {x^{2} - x + 1} \sqrt {x + 1} - 1\right )}{9 \, {\left (x^{6} + 2 \, x^{3} + 1\right )}} \] Input:

integrate(1/x/(1+x)^(5/2)/(x^2-x+1)^(5/2),x, algorithm="fricas")
 

Output:

1/9*(2*(3*x^3 + 4)*sqrt(x^2 - x + 1)*sqrt(x + 1) - 3*(x^6 + 2*x^3 + 1)*log 
(sqrt(x^2 - x + 1)*sqrt(x + 1) + 1) + 3*(x^6 + 2*x^3 + 1)*log(sqrt(x^2 - x 
 + 1)*sqrt(x + 1) - 1))/(x^6 + 2*x^3 + 1)
 

Sympy [F]

\[ \int \frac {1}{x (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int \frac {1}{x \left (x + 1\right )^{\frac {5}{2}} \left (x^{2} - x + 1\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/x/(1+x)**(5/2)/(x**2-x+1)**(5/2),x)
 

Output:

Integral(1/(x*(x + 1)**(5/2)*(x**2 - x + 1)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {1}{x (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (x^{2} - x + 1\right )}^{\frac {5}{2}} {\left (x + 1\right )}^{\frac {5}{2}} x} \,d x } \] Input:

integrate(1/x/(1+x)^(5/2)/(x^2-x+1)^(5/2),x, algorithm="maxima")
 

Output:

integrate(1/((x^2 - x + 1)^(5/2)*(x + 1)^(5/2)*x), x)
 

Giac [F]

\[ \int \frac {1}{x (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (x^{2} - x + 1\right )}^{\frac {5}{2}} {\left (x + 1\right )}^{\frac {5}{2}} x} \,d x } \] Input:

integrate(1/x/(1+x)^(5/2)/(x^2-x+1)^(5/2),x, algorithm="giac")
 

Output:

integrate(1/((x^2 - x + 1)^(5/2)*(x + 1)^(5/2)*x), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int \frac {1}{x\,{\left (x+1\right )}^{5/2}\,{\left (x^2-x+1\right )}^{5/2}} \,d x \] Input:

int(1/(x*(x + 1)^(5/2)*(x^2 - x + 1)^(5/2)),x)
 

Output:

int(1/(x*(x + 1)^(5/2)*(x^2 - x + 1)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x (1+x)^{5/2} \left (1-x+x^2\right )^{5/2}} \, dx=\int \frac {1}{x \left (x +1\right )^{\frac {5}{2}} \left (x^{2}-x +1\right )^{\frac {5}{2}}}d x \] Input:

int(1/x/(1+x)^(5/2)/(x^2-x+1)^(5/2),x)
 

Output:

int(1/x/(1+x)^(5/2)/(x^2-x+1)^(5/2),x)