\(\int \frac {\log ^2(x+\sqrt {1+x})}{(1+x)^2} \, dx\) [31]

Optimal result

Integrand size = 18, antiderivative size = 555 \[ \int \frac {\log ^2\left (x+\sqrt {1+x}\right )}{(1+x)^2} \, dx=\log (1+x)+\frac {2 \log \left (x+\sqrt {1+x}\right )}{\sqrt {1+x}}-6 \log \left (\sqrt {1+x}\right ) \log \left (x+\sqrt {1+x}\right )-\frac {\log ^2\left (x+\sqrt {1+x}\right )}{1+x}-\left (1+\sqrt {5}\right ) \log \left (1-\sqrt {5}+2 \sqrt {1+x}\right )+6 \log \left (\frac {1}{2} \left (-1+\sqrt {5}\right )\right ) \log \left (1-\sqrt {5}+2 \sqrt {1+x}\right )+\left (3+\sqrt {5}\right ) \log \left (x+\sqrt {1+x}\right ) \log \left (1-\sqrt {5}+2 \sqrt {1+x}\right )-\frac {1}{2} \left (3+\sqrt {5}\right ) \log ^2\left (1-\sqrt {5}+2 \sqrt {1+x}\right )-\left (1-\sqrt {5}\right ) \log \left (1+\sqrt {5}+2 \sqrt {1+x}\right )+\left (3-\sqrt {5}\right ) \log \left (x+\sqrt {1+x}\right ) \log \left (1+\sqrt {5}+2 \sqrt {1+x}\right )-\left (3-\sqrt {5}\right ) \log \left (-\frac {1-\sqrt {5}+2 \sqrt {1+x}}{2 \sqrt {5}}\right ) \log \left (1+\sqrt {5}+2 \sqrt {1+x}\right )-\frac {1}{2} \left (3-\sqrt {5}\right ) \log ^2\left (1+\sqrt {5}+2 \sqrt {1+x}\right )-\left (3+\sqrt {5}\right ) \log \left (1-\sqrt {5}+2 \sqrt {1+x}\right ) \log \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{2 \sqrt {5}}\right )+6 \log \left (\sqrt {1+x}\right ) \log \left (1+\frac {2 \sqrt {1+x}}{1+\sqrt {5}}\right )+6 \operatorname {PolyLog}\left (2,-\frac {2 \sqrt {1+x}}{1+\sqrt {5}}\right )-\left (3+\sqrt {5}\right ) \operatorname {PolyLog}\left (2,-\frac {1-\sqrt {5}+2 \sqrt {1+x}}{2 \sqrt {5}}\right )-\left (3-\sqrt {5}\right ) \operatorname {PolyLog}\left (2,\frac {1+\sqrt {5}+2 \sqrt {1+x}}{2 \sqrt {5}}\right )-6 \operatorname {PolyLog}\left (2,1+\frac {2 \sqrt {1+x}}{1-\sqrt {5}}\right ) \] Output:

ln(1+x)-3*ln(1+x)*ln(x+(1+x)^(1/2))-ln(x+(1+x)^(1/2))^2/(1+x)+6*ln(1/2*5^( 
1/2)-1/2)*ln(1-5^(1/2)+2*(1+x)^(1/2))+3*ln(1+x)*ln(1+2*(1+x)^(1/2)/(5^(1/2 
)+1))+6*polylog(2,-2*(1+x)^(1/2)/(5^(1/2)+1))-6*polylog(2,1+2*(1+x)^(1/2)/ 
(-5^(1/2)+1))-ln(1+5^(1/2)+2*(1+x)^(1/2))*(-5^(1/2)+1)+ln(x+(1+x)^(1/2))*l 
n(1+5^(1/2)+2*(1+x)^(1/2))*(3-5^(1/2))-ln(1/10*(-1+5^(1/2)-2*(1+x)^(1/2))* 
5^(1/2))*ln(1+5^(1/2)+2*(1+x)^(1/2))*(3-5^(1/2))-1/2*ln(1+5^(1/2)+2*(1+x)^ 
(1/2))^2*(3-5^(1/2))-polylog(2,1/10*(1+5^(1/2)+2*(1+x)^(1/2))*5^(1/2))*(3- 
5^(1/2))-ln(1-5^(1/2)+2*(1+x)^(1/2))*(5^(1/2)+1)+ln(x+(1+x)^(1/2))*ln(1-5^ 
(1/2)+2*(1+x)^(1/2))*(3+5^(1/2))-1/2*ln(1-5^(1/2)+2*(1+x)^(1/2))^2*(3+5^(1 
/2))-ln(1-5^(1/2)+2*(1+x)^(1/2))*ln(1/10*(1+5^(1/2)+2*(1+x)^(1/2))*5^(1/2) 
)*(3+5^(1/2))-polylog(2,1/10*(-1+5^(1/2)-2*(1+x)^(1/2))*5^(1/2))*(3+5^(1/2 
))+2*ln(x+(1+x)^(1/2))/(1+x)^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1280\) vs. \(2(555)=1110\).

Time = 6.92 (sec) , antiderivative size = 1280, normalized size of antiderivative = 2.31 \[ \int \frac {\log ^2\left (x+\sqrt {1+x}\right )}{(1+x)^2} \, dx =\text {Too large to display} \] Input:

Integrate[Log[x + Sqrt[1 + x]]^2/(1 + x)^2,x]
 

Output:

(2*Log[1 + x])/(-1 + Sqrt[5]) - (2*Log[1 + x])/(1 + Sqrt[5]) - (4*Log[-1 + 
 Sqrt[5] - 2*Sqrt[1 + x]])/(-1 + Sqrt[5]) + (Log[100]*Log[1/2 - Sqrt[5]/2 
+ Sqrt[1 + x]])/Sqrt[5] - 6*Log[(2*Sqrt[1 + x])/(-1 + Sqrt[5])]*Log[1/2 - 
Sqrt[5]/2 + Sqrt[1 + x]] + 3*Log[1 + x]*Log[1/2 - Sqrt[5]/2 + Sqrt[1 + x]] 
 - 3*Log[-1 + Sqrt[5] - 2*Sqrt[1 + x]]*Log[1/2 - Sqrt[5]/2 + Sqrt[1 + x]] 
- Sqrt[5]*Log[-1 + Sqrt[5] - 2*Sqrt[1 + x]]*Log[1/2 - Sqrt[5]/2 + Sqrt[1 + 
 x]] + (3*Log[1/2 - Sqrt[5]/2 + Sqrt[1 + x]]^2)/2 + (Sqrt[5]*Log[1/2 - Sqr 
t[5]/2 + Sqrt[1 + x]]^2)/2 + (Log[8]*Log[(1 + Sqrt[5])/2 + Sqrt[1 + x]])/( 
2*Sqrt[5]) - 3*Log[-1 + Sqrt[5] - 2*Sqrt[1 + x]]*Log[(1 + Sqrt[5])/2 + Sqr 
t[1 + x]] - Sqrt[5]*Log[-1 + Sqrt[5] - 2*Sqrt[1 + x]]*Log[(1 + Sqrt[5])/2 
+ Sqrt[1 + x]] + (3*Log[(1 + Sqrt[5])/2 + Sqrt[1 + x]]^2)/2 - Log[(1 + Sqr 
t[5])/2 + Sqrt[1 + x]]^2/Sqrt[5] + (2*Log[x + Sqrt[1 + x]])/Sqrt[1 + x] - 
3*Log[1 + x]*Log[x + Sqrt[1 + x]] + 3*Log[-1 + Sqrt[5] - 2*Sqrt[1 + x]]*Lo 
g[x + Sqrt[1 + x]] + Sqrt[5]*Log[-1 + Sqrt[5] - 2*Sqrt[1 + x]]*Log[x + Sqr 
t[1 + x]] - Log[x + Sqrt[1 + x]]^2/(1 + x) + (4*Log[1 + Sqrt[5] + 2*Sqrt[1 
 + x]])/(1 + Sqrt[5]) - 3*Log[1/2 - Sqrt[5]/2 + Sqrt[1 + x]]*Log[1 + Sqrt[ 
5] + 2*Sqrt[1 + x]] + Sqrt[5]*Log[1/2 - Sqrt[5]/2 + Sqrt[1 + x]]*Log[1 + S 
qrt[5] + 2*Sqrt[1 + x]] - 3*Log[(1 + Sqrt[5])/2 + Sqrt[1 + x]]*Log[1 + Sqr 
t[5] + 2*Sqrt[1 + x]] + (7*Log[(1 + Sqrt[5])/2 + Sqrt[1 + x]]*Log[1 + Sqrt 
[5] + 2*Sqrt[1 + x]])/(2*Sqrt[5]) + 3*Log[x + Sqrt[1 + x]]*Log[1 + Sqrt...
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 580, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {7267, 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.

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

\(\Big \downarrow \) 7267

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

\(\Big \downarrow \) 3005

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3008

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

\(\Big \downarrow \) 2009

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

Input:

Int[Log[x + Sqrt[1 + x]]^2/(1 + x)^2,x]
 

Output:

2*(Log[Sqrt[1 + x]] + Log[x + Sqrt[1 + x]]/Sqrt[1 + x] - 3*Log[Sqrt[1 + x] 
]*Log[x + Sqrt[1 + x]] - Log[x + Sqrt[1 + x]]^2/(2*(1 + x)) - ((1 + Sqrt[5 
])*Log[1 - Sqrt[5] + 2*Sqrt[1 + x]])/2 + 3*Log[(-1 + Sqrt[5])/2]*Log[1 - S 
qrt[5] + 2*Sqrt[1 + x]] + ((3 + Sqrt[5])*Log[x + Sqrt[1 + x]]*Log[1 - Sqrt 
[5] + 2*Sqrt[1 + x]])/2 - ((3 + Sqrt[5])*Log[1 - Sqrt[5] + 2*Sqrt[1 + x]]^ 
2)/4 - ((1 - Sqrt[5])*Log[1 + Sqrt[5] + 2*Sqrt[1 + x]])/2 + ((3 - Sqrt[5]) 
*Log[x + Sqrt[1 + x]]*Log[1 + Sqrt[5] + 2*Sqrt[1 + x]])/2 - ((3 - Sqrt[5]) 
*Log[-1/2*(1 - Sqrt[5] + 2*Sqrt[1 + x])/Sqrt[5]]*Log[1 + Sqrt[5] + 2*Sqrt[ 
1 + x]])/2 - ((3 - Sqrt[5])*Log[1 + Sqrt[5] + 2*Sqrt[1 + x]]^2)/4 - ((3 + 
Sqrt[5])*Log[1 - Sqrt[5] + 2*Sqrt[1 + x]]*Log[(1 + Sqrt[5] + 2*Sqrt[1 + x] 
)/(2*Sqrt[5])])/2 + 3*Log[Sqrt[1 + x]]*Log[1 + (2*Sqrt[1 + x])/(1 + Sqrt[5 
])] + 3*PolyLog[2, (-2*Sqrt[1 + x])/(1 + Sqrt[5])] - ((3 + Sqrt[5])*PolyLo 
g[2, -1/2*(1 - Sqrt[5] + 2*Sqrt[1 + x])/Sqrt[5]])/2 - ((3 - Sqrt[5])*PolyL 
og[2, (1 + Sqrt[5] + 2*Sqrt[1 + x])/(2*Sqrt[5])])/2 - 3*PolyLog[2, 1 + (2* 
Sqrt[1 + 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]
 

rule 7267

Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si 
mp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x 
] /;  !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
 

Maple [F]

\[\int \frac {\ln \left (x +\sqrt {1+x}\right )^{2}}{\left (1+x \right )^{2}}d x\]

Input:

int(ln(x+(1+x)^(1/2))^2/(1+x)^2,x)
 

Output:

int(ln(x+(1+x)^(1/2))^2/(1+x)^2,x)
 

Fricas [F]

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

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

Output:

integral(log(x + sqrt(x + 1))^2/(x^2 + 2*x + 1), x)
 

Sympy [F]

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

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

Output:

Integral(log(x + sqrt(x + 1))**2/(x + 1)**2, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

integrate(log(x + sqrt(x + 1))^2/(x + 1)^2, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(log(sqrt(x + 1) + x)**2/(x**2 + 2*x + 1),x)