\(\int \frac {\log ^2(-1+x+x^2)}{x^3} \, dx\) [100]

Optimal result

Integrand size = 13, antiderivative size = 443 \[ \int \frac {\log ^2\left (-1+x+x^2\right )}{x^3} \, dx=\log (x)-\frac {1}{2} \left (1+\sqrt {5}\right ) \log \left (1-\sqrt {5}+2 x\right )+3 \log \left (\frac {1}{2} \left (-1+\sqrt {5}\right )\right ) \log \left (1-\sqrt {5}+2 x\right )-\frac {1}{4} \left (3+\sqrt {5}\right ) \log ^2\left (1-\sqrt {5}+2 x\right )-\frac {1}{2} \left (1-\sqrt {5}\right ) \log \left (1+\sqrt {5}+2 x\right )-\frac {1}{2} \left (3-\sqrt {5}\right ) \log \left (-\frac {1-\sqrt {5}+2 x}{2 \sqrt {5}}\right ) \log \left (1+\sqrt {5}+2 x\right )-\frac {1}{4} \left (3-\sqrt {5}\right ) \log ^2\left (1+\sqrt {5}+2 x\right )-\frac {1}{2} \left (3+\sqrt {5}\right ) \log \left (1-\sqrt {5}+2 x\right ) \log \left (\frac {1+\sqrt {5}+2 x}{2 \sqrt {5}}\right )+3 \log (x) \log \left (1+\frac {2 x}{1+\sqrt {5}}\right )+\frac {\log \left (-1+x+x^2\right )}{x}-3 \log (x) \log \left (-1+x+x^2\right )+\frac {1}{2} \left (3+\sqrt {5}\right ) \log \left (1-\sqrt {5}+2 x\right ) \log \left (-1+x+x^2\right )+\frac {1}{2} \left (3-\sqrt {5}\right ) \log \left (1+\sqrt {5}+2 x\right ) \log \left (-1+x+x^2\right )-\frac {\log ^2\left (-1+x+x^2\right )}{2 x^2}+3 \operatorname {PolyLog}\left (2,-\frac {2 x}{1+\sqrt {5}}\right )-\frac {1}{2} \left (3+\sqrt {5}\right ) \operatorname {PolyLog}\left (2,-\frac {1-\sqrt {5}+2 x}{2 \sqrt {5}}\right )-\frac {1}{2} \left (3-\sqrt {5}\right ) \operatorname {PolyLog}\left (2,\frac {1+\sqrt {5}+2 x}{2 \sqrt {5}}\right )-3 \operatorname {PolyLog}\left (2,1+\frac {2 x}{1-\sqrt {5}}\right ) \] Output:

ln(x)-1/2*(5^(1/2)+1)*ln(1-5^(1/2)+2*x)+3*ln(1/2*5^(1/2)-1/2)*ln(1-5^(1/2) 
+2*x)-1/4*(3+5^(1/2))*ln(1-5^(1/2)+2*x)^2-1/2*(-5^(1/2)+1)*ln(1+5^(1/2)+2* 
x)-1/2*(3-5^(1/2))*ln(-1/10*(1-5^(1/2)+2*x)*5^(1/2))*ln(1+5^(1/2)+2*x)-1/4 
*(3-5^(1/2))*ln(1+5^(1/2)+2*x)^2-1/2*(3+5^(1/2))*ln(1-5^(1/2)+2*x)*ln(1/10 
*(1+5^(1/2)+2*x)*5^(1/2))+3*ln(x)*ln(1+2*x/(5^(1/2)+1))+ln(x^2+x-1)/x-3*ln 
(x)*ln(x^2+x-1)+1/2*(3+5^(1/2))*ln(1-5^(1/2)+2*x)*ln(x^2+x-1)+1/2*(3-5^(1/ 
2))*ln(1+5^(1/2)+2*x)*ln(x^2+x-1)-1/2*ln(x^2+x-1)^2/x^2+3*polylog(2,-2*x/( 
5^(1/2)+1))-1/2*(3+5^(1/2))*polylog(2,-1/10*(1-5^(1/2)+2*x)*5^(1/2))-1/2*( 
3-5^(1/2))*polylog(2,1/10*(1+5^(1/2)+2*x)*5^(1/2))-3*polylog(2,1+2*x/(-5^( 
1/2)+1))
 

Mathematica [A] (warning: unable to verify)

Time = 0.79 (sec) , antiderivative size = 826, normalized size of antiderivative = 1.86 \[ \int \frac {\log ^2\left (-1+x+x^2\right )}{x^3} \, dx =\text {Too large to display} \] Input:

Integrate[Log[-1 + x + x^2]^2/x^3,x]
 

Output:

(-2*Log[-1 + x + x^2]^2 + x*(4*x*Log[x] - 12*x*Log[(1 + Sqrt[5])/2]*Log[x] 
 - 6*x*Log[-1 + Sqrt[5] - 2*x]*Log[1/2 - Sqrt[5]/2 + x] - 2*Sqrt[5]*x*Log[ 
-1 + Sqrt[5] - 2*x]*Log[1/2 - Sqrt[5]/2 + x] + 12*x*Log[x]*Log[1/2 - Sqrt[ 
5]/2 + x] - 12*x*Log[(2*x)/(-1 + Sqrt[5])]*Log[1/2 - Sqrt[5]/2 + x] + 3*x* 
Log[1/2 - Sqrt[5]/2 + x]^2 + Sqrt[5]*x*Log[1/2 - Sqrt[5]/2 + x]^2 - 6*x*Lo 
g[-1 + Sqrt[5] - 2*x]*Log[(1 + Sqrt[5])/2 + x] - 2*Sqrt[5]*x*Log[-1 + Sqrt 
[5] - 2*x]*Log[(1 + Sqrt[5])/2 + x] + 12*x*Log[x]*Log[(1 + Sqrt[5])/2 + x] 
 + 3*x*Log[(1 + Sqrt[5])/2 + x]^2 - Sqrt[5]*x*Log[(1 + Sqrt[5])/2 + x]^2 - 
 2*x*Log[1 - Sqrt[5] + 2*x] - 2*Sqrt[5]*x*Log[1 - Sqrt[5] + 2*x] + 3*x*Log 
[5]*Log[1 - Sqrt[5] + 2*x] + Sqrt[5]*x*Log[5]*Log[1 - Sqrt[5] + 2*x] - 2*x 
*Log[1 + Sqrt[5] + 2*x] + 2*Sqrt[5]*x*Log[1 + Sqrt[5] + 2*x] - 6*x*Log[1/2 
 - Sqrt[5]/2 + x]*Log[1 + Sqrt[5] + 2*x] + 2*Sqrt[5]*x*Log[1/2 - Sqrt[5]/2 
 + x]*Log[1 + Sqrt[5] + 2*x] - 6*x*Log[(1 + Sqrt[5])/2 + x]*Log[1 + Sqrt[5 
] + 2*x] + 2*Sqrt[5]*x*Log[(1 + Sqrt[5])/2 + x]*Log[1 + Sqrt[5] + 2*x] + 6 
*x*Log[1/2 - Sqrt[5]/2 + x]*Log[(1 + Sqrt[5] + 2*x)/(2*Sqrt[5])] - 2*Sqrt[ 
5]*x*Log[1/2 - Sqrt[5]/2 + x]*Log[(1 + Sqrt[5] + 2*x)/(2*Sqrt[5])] + 4*Log 
[-1 + x + x^2] + 6*x*Log[-1 + Sqrt[5] - 2*x]*Log[-1 + x + x^2] + 2*Sqrt[5] 
*x*Log[-1 + Sqrt[5] - 2*x]*Log[-1 + x + x^2] - 12*x*Log[x]*Log[-1 + x + x^ 
2] + 6*x*Log[1 + Sqrt[5] + 2*x]*Log[-1 + x + x^2] - 2*Sqrt[5]*x*Log[1 + Sq 
rt[5] + 2*x]*Log[-1 + x + x^2] - 4*Sqrt[5]*x*PolyLog[2, (-1 + Sqrt[5] -...
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3005, 25, 3008, 2009}

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.

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

\(\Big \downarrow \) 3005

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3008

\(\displaystyle -\int \left (\frac {3 \log \left (x^2+x-1\right )}{x}+\frac {(-3 x-4) \log \left (x^2+x-1\right )}{x^2+x-1}+\frac {\log \left (x^2+x-1\right )}{x^2}\right )dx-\frac {\log ^2\left (x^2+x-1\right )}{2 x^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle 3 \operatorname {PolyLog}\left (2,-\frac {2 x}{1+\sqrt {5}}\right )-\frac {1}{2} \left (3+\sqrt {5}\right ) \operatorname {PolyLog}\left (2,-\frac {2 x-\sqrt {5}+1}{2 \sqrt {5}}\right )-\frac {1}{2} \left (3-\sqrt {5}\right ) \operatorname {PolyLog}\left (2,\frac {2 x+\sqrt {5}+1}{2 \sqrt {5}}\right )-3 \operatorname {PolyLog}\left (2,\frac {2 x}{1-\sqrt {5}}+1\right )-\frac {\log ^2\left (x^2+x-1\right )}{2 x^2}+\frac {1}{2} \left (3+\sqrt {5}\right ) \log \left (x^2+x-1\right ) \log \left (2 x-\sqrt {5}+1\right )-3 \log (x) \log \left (x^2+x-1\right )+\frac {1}{2} \left (3-\sqrt {5}\right ) \log \left (2 x+\sqrt {5}+1\right ) \log \left (x^2+x-1\right )+\frac {\log \left (x^2+x-1\right )}{x}-\frac {1}{4} \left (3+\sqrt {5}\right ) \log ^2\left (2 x-\sqrt {5}+1\right )-\frac {1}{4} \left (3-\sqrt {5}\right ) \log ^2\left (2 x+\sqrt {5}+1\right )-\frac {1}{2} \left (3+\sqrt {5}\right ) \log \left (\frac {2 x+\sqrt {5}+1}{2 \sqrt {5}}\right ) \log \left (2 x-\sqrt {5}+1\right )+3 \log \left (\frac {1}{2} \left (\sqrt {5}-1\right )\right ) \log \left (2 x-\sqrt {5}+1\right )-\frac {1}{2} \left (1+\sqrt {5}\right ) \log \left (2 x-\sqrt {5}+1\right )+\log (x)-\frac {1}{2} \left (3-\sqrt {5}\right ) \log \left (-\frac {2 x-\sqrt {5}+1}{2 \sqrt {5}}\right ) \log \left (2 x+\sqrt {5}+1\right )-\frac {1}{2} \left (1-\sqrt {5}\right ) \log \left (2 x+\sqrt {5}+1\right )+3 \log (x) \log \left (\frac {2 x}{1+\sqrt {5}}+1\right )\)

Input:

Int[Log[-1 + x + x^2]^2/x^3,x]
 

Output:

Log[x] - ((1 + Sqrt[5])*Log[1 - Sqrt[5] + 2*x])/2 + 3*Log[(-1 + Sqrt[5])/2 
]*Log[1 - Sqrt[5] + 2*x] - ((3 + Sqrt[5])*Log[1 - Sqrt[5] + 2*x]^2)/4 - (( 
1 - Sqrt[5])*Log[1 + Sqrt[5] + 2*x])/2 - ((3 - Sqrt[5])*Log[-1/2*(1 - Sqrt 
[5] + 2*x)/Sqrt[5]]*Log[1 + Sqrt[5] + 2*x])/2 - ((3 - Sqrt[5])*Log[1 + Sqr 
t[5] + 2*x]^2)/4 - ((3 + Sqrt[5])*Log[1 - Sqrt[5] + 2*x]*Log[(1 + Sqrt[5] 
+ 2*x)/(2*Sqrt[5])])/2 + 3*Log[x]*Log[1 + (2*x)/(1 + Sqrt[5])] + Log[-1 + 
x + x^2]/x - 3*Log[x]*Log[-1 + x + x^2] + ((3 + Sqrt[5])*Log[1 - Sqrt[5] + 
 2*x]*Log[-1 + x + x^2])/2 + ((3 - Sqrt[5])*Log[1 + Sqrt[5] + 2*x]*Log[-1 
+ x + x^2])/2 - Log[-1 + x + x^2]^2/(2*x^2) + 3*PolyLog[2, (-2*x)/(1 + Sqr 
t[5])] - ((3 + Sqrt[5])*PolyLog[2, -1/2*(1 - Sqrt[5] + 2*x)/Sqrt[5]])/2 - 
((3 - Sqrt[5])*PolyLog[2, (1 + Sqrt[5] + 2*x)/(2*Sqrt[5])])/2 - 3*PolyLog[ 
2, 1 + (2*x)/(1 - Sqrt[5])]
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3005

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_. 
), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + b*Log[c*RFx^p])^n/(e*(m + 1))) 
, x] - Simp[b*n*(p/(e*(m + 1)))   Int[SimplifyIntegrand[(d + e*x)^(m + 1)*( 
a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, 
d, e, m, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || 
 IntegerQ[m]) && NeQ[m, -1]
 

rule 3008

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With 
[{u = ExpandIntegrand[(a + b*Log[c*RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u 
]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalFuncti 
onQ[RGx, x] && IGtQ[n, 0]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.21 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.49

method	result	size
parts	\(-\frac {\ln \left (x^{2}+x -1\right )^{2}}{2 x^{2}}+\frac {\ln \left (x^{2}+x -1\right )}{x}-\frac {\ln \left (x^{2}+x -1\right )}{2}+\sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 x +1\right ) \sqrt {5}}{5}\right )+\ln \left (x \right )+\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} -1\right )}{\sum }\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{2}+x -1\right )-\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )^{2}}{2}-\operatorname {dilog}\left (\frac {\underline {\hspace {1.25 ex}}\alpha +x +1}{2 \underline {\hspace {1.25 ex}}\alpha +1}\right )-\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {\underline {\hspace {1.25 ex}}\alpha +x +1}{2 \underline {\hspace {1.25 ex}}\alpha +1}\right )\right ) \left (\underline {\hspace {1.25 ex}}\alpha +2\right )\right )-3 \ln \left (x \right ) \ln \left (x^{2}+x -1\right )+3 \ln \left (x \right ) \ln \left (\frac {-2 x -1+\sqrt {5}}{\sqrt {5}-1}\right )+3 \ln \left (x \right ) \ln \left (\frac {1+\sqrt {5}+2 x}{\sqrt {5}+1}\right )+3 \operatorname {dilog}\left (\frac {-2 x -1+\sqrt {5}}{\sqrt {5}-1}\right )+3 \operatorname {dilog}\left (\frac {1+\sqrt {5}+2 x}{\sqrt {5}+1}\right )\)	\(219\)

Input:

int(ln(x^2+x-1)^2/x^3,x,method=_RETURNVERBOSE)
 

Output:

-1/2*ln(x^2+x-1)^2/x^2+ln(x^2+x-1)/x-1/2*ln(x^2+x-1)+5^(1/2)*arctanh(1/5*( 
2*x+1)*5^(1/2))+ln(x)+Sum((ln(x-_alpha)*ln(x^2+x-1)-1/2*ln(x-_alpha)^2-dil 
og((_alpha+x+1)/(2*_alpha+1))-ln(x-_alpha)*ln((_alpha+x+1)/(2*_alpha+1)))* 
(_alpha+2),_alpha=RootOf(_Z^2+_Z-1))-3*ln(x)*ln(x^2+x-1)+3*ln(x)*ln((-2*x- 
1+5^(1/2))/(5^(1/2)-1))+3*ln(x)*ln((1+5^(1/2)+2*x)/(5^(1/2)+1))+3*dilog((- 
2*x-1+5^(1/2))/(5^(1/2)-1))+3*dilog((1+5^(1/2)+2*x)/(5^(1/2)+1))
 

Fricas [F]

\[ \int \frac {\log ^2\left (-1+x+x^2\right )}{x^3} \, dx=\int { \frac {\log \left (x^{2} + x - 1\right )^{2}}{x^{3}} \,d x } \] Input:

integrate(log(x^2+x-1)^2/x^3,x, algorithm="fricas")
 

Output:

integral(log(x^2 + x - 1)^2/x^3, x)
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {\log ^2\left (-1+x+x^2\right )}{x^3} \, dx=\text {Exception raised: RecursionError} \] Input:

integrate(ln(x**2+x-1)**2/x**3,x)
 

Output:

Exception raised: RecursionError >> maximum recursion depth exceeded while 
 calling a Python object
 

Maxima [F]

\[ \int \frac {\log ^2\left (-1+x+x^2\right )}{x^3} \, dx=\int { \frac {\log \left (x^{2} + x - 1\right )^{2}}{x^{3}} \,d x } \] Input:

integrate(log(x^2+x-1)^2/x^3,x, algorithm="maxima")
 

Output:

-1/2*log(x^2 + x - 1)^2/x^2 + integrate((2*x + 1)*log(x^2 + x - 1)/(x^4 + 
x^3 - x^2), x)
 

Giac [F]

\[ \int \frac {\log ^2\left (-1+x+x^2\right )}{x^3} \, dx=\int { \frac {\log \left (x^{2} + x - 1\right )^{2}}{x^{3}} \,d x } \] Input:

integrate(log(x^2+x-1)^2/x^3,x, algorithm="giac")
 

Output:

integrate(log(x^2 + x - 1)^2/x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\log ^2\left (-1+x+x^2\right )}{x^3} \, dx=\int \frac {{\ln \left (x^2+x-1\right )}^2}{x^3} \,d x \] Input:

int(log(x + x^2 - 1)^2/x^3,x)
 

Output:

int(log(x + x^2 - 1)^2/x^3, x)
 

Reduce [F]

\[ \int \frac {\log ^2\left (-1+x+x^2\right )}{x^3} \, dx=\frac {2 \left (\int \frac {\mathrm {log}\left (x^{2}+x -1\right )}{x^{4}+x^{3}-x^{2}}d x \right ) x^{2}+4 \left (\int \frac {\mathrm {log}\left (x^{2}+x -1\right )}{x^{3}+x^{2}-x}d x \right ) x^{2}-\mathrm {log}\left (x^{2}+x -1\right )^{2}}{2 x^{2}} \] Input:

int(log(x^2+x-1)^2/x^3,x)
 

Output:

(2*int(log(x**2 + x - 1)/(x**4 + x**3 - x**2),x)*x**2 + 4*int(log(x**2 + x 
 - 1)/(x**3 + x**2 - x),x)*x**2 - log(x**2 + x - 1)**2)/(2*x**2)