3.1.30 \(\int \frac {\log (x+\sqrt {1+x})}{1+x^2} \, dx\) [30]

3.1.30.1 Optimal result

Integrand size = 18, antiderivative size = 981 \[ \int \frac {\log \left (x+\sqrt {1+x}\right )}{1+x^2} \, dx =\text {Too large to display} \]

output

-1/2*I*ln((1-I)^(1/2)-(1+x)^(1/2))*ln((1-5^(1/2)+2*(1+x)^(1/2))/(1+2*(1-I) 
^(1/2)-5^(1/2)))+1/2*I*ln((1+I)^(1/2)+(1+x)^(1/2))*ln((1-5^(1/2)+2*(1+x)^( 
1/2))/(1-2*(1+I)^(1/2)-5^(1/2)))-1/2*I*ln((1-I)^(1/2)-(1+x)^(1/2))*ln((1+5 
^(1/2)+2*(1+x)^(1/2))/(1+2*(1-I)^(1/2)+5^(1/2)))-1/2*I*ln(x+(1+x)^(1/2))*l 
n((1+I)^(1/2)+(1+x)^(1/2))+1/2*I*ln(x+(1+x)^(1/2))*ln((1-I)^(1/2)+(1+x)^(1 
/2))-1/2*I*polylog(2,-2*((1-I)^(1/2)+(1+x)^(1/2))/(1-2*(1-I)^(1/2)+5^(1/2) 
))+1/2*I*ln((1-I)^(1/2)-(1+x)^(1/2))*ln(x+(1+x)^(1/2))+1/2*I*ln((1+I)^(1/2 
)-(1+x)^(1/2))*ln((1+5^(1/2)+2*(1+x)^(1/2))/(1+2*(1+I)^(1/2)+5^(1/2)))-1/2 
*I*ln((1+I)^(1/2)-(1+x)^(1/2))*ln(x+(1+x)^(1/2))-1/2*I*ln((1-I)^(1/2)+(1+x 
)^(1/2))*ln((1-5^(1/2)+2*(1+x)^(1/2))/(1-2*(1-I)^(1/2)-5^(1/2)))+1/2*I*ln( 
(1+I)^(1/2)-(1+x)^(1/2))*ln((1-5^(1/2)+2*(1+x)^(1/2))/(1+2*(1+I)^(1/2)-5^( 
1/2)))+1/2*I*ln((1+I)^(1/2)+(1+x)^(1/2))*ln((1+5^(1/2)+2*(1+x)^(1/2))/(1-2 
*(1+I)^(1/2)+5^(1/2)))-1/2*I*polylog(2,-2*((1-I)^(1/2)+(1+x)^(1/2))/(1-2*( 
1-I)^(1/2)-5^(1/2)))+1/2*I*polylog(2,2*((1+I)^(1/2)-(1+x)^(1/2))/(1+2*(1+I 
)^(1/2)+5^(1/2)))+1/2*I*polylog(2,-2*((1+I)^(1/2)+(1+x)^(1/2))/(1-2*(1+I)^ 
(1/2)-5^(1/2)))+1/2*I*polylog(2,-2*((1+I)^(1/2)+(1+x)^(1/2))/(1-2*(1+I)^(1 
/2)+5^(1/2)))-1/2*I*ln((1-I)^(1/2)+(1+x)^(1/2))*ln((1+5^(1/2)+2*(1+x)^(1/2 
))/(1-2*(1-I)^(1/2)+5^(1/2)))-1/2*I*polylog(2,2*((1-I)^(1/2)-(1+x)^(1/2))/ 
(1+2*(1-I)^(1/2)-5^(1/2)))-1/2*I*polylog(2,2*((1-I)^(1/2)-(1+x)^(1/2))/(1+ 
2*(1-I)^(1/2)+5^(1/2)))+1/2*I*polylog(2,2*((1+I)^(1/2)-(1+x)^(1/2))/(1+...
 

3.1.30.2 Mathematica [F]

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

input

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

output

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

3.1.30.3 Rubi [A] (verified)

Time = 1.37 (sec) , antiderivative size = 983, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3010, 7292, 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.

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

\(\Big \downarrow \) 3010

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 3008

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

\(\Big \downarrow \) 2009

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

input

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

output

2*((I/4)*Log[Sqrt[1 - I] - Sqrt[1 + x]]*Log[x + Sqrt[1 + x]] - (I/4)*Log[S 
qrt[1 + I] - Sqrt[1 + x]]*Log[x + Sqrt[1 + x]] + (I/4)*Log[Sqrt[1 - I] + S 
qrt[1 + x]]*Log[x + Sqrt[1 + x]] - (I/4)*Log[Sqrt[1 + I] + Sqrt[1 + x]]*Lo 
g[x + Sqrt[1 + x]] - (I/4)*Log[Sqrt[1 - I] + Sqrt[1 + x]]*Log[(1 - Sqrt[5] 
 + 2*Sqrt[1 + x])/(1 - 2*Sqrt[1 - I] - Sqrt[5])] - (I/4)*Log[Sqrt[1 - I] - 
 Sqrt[1 + x]]*Log[(1 - Sqrt[5] + 2*Sqrt[1 + x])/(1 + 2*Sqrt[1 - I] - Sqrt[ 
5])] + (I/4)*Log[Sqrt[1 + I] + Sqrt[1 + x]]*Log[(1 - Sqrt[5] + 2*Sqrt[1 + 
x])/(1 - 2*Sqrt[1 + I] - Sqrt[5])] + (I/4)*Log[Sqrt[1 + I] - Sqrt[1 + x]]* 
Log[(1 - Sqrt[5] + 2*Sqrt[1 + x])/(1 + 2*Sqrt[1 + I] - Sqrt[5])] - (I/4)*L 
og[Sqrt[1 - I] + Sqrt[1 + x]]*Log[(1 + Sqrt[5] + 2*Sqrt[1 + x])/(1 - 2*Sqr 
t[1 - I] + Sqrt[5])] - (I/4)*Log[Sqrt[1 - I] - Sqrt[1 + x]]*Log[(1 + Sqrt[ 
5] + 2*Sqrt[1 + x])/(1 + 2*Sqrt[1 - I] + Sqrt[5])] + (I/4)*Log[Sqrt[1 + I] 
 + Sqrt[1 + x]]*Log[(1 + Sqrt[5] + 2*Sqrt[1 + x])/(1 - 2*Sqrt[1 + I] + Sqr 
t[5])] + (I/4)*Log[Sqrt[1 + I] - Sqrt[1 + x]]*Log[(1 + Sqrt[5] + 2*Sqrt[1 
+ x])/(1 + 2*Sqrt[1 + I] + Sqrt[5])] - (I/4)*PolyLog[2, (2*(Sqrt[1 - I] - 
Sqrt[1 + x]))/(1 + 2*Sqrt[1 - I] - Sqrt[5])] - (I/4)*PolyLog[2, (2*(Sqrt[1 
 - I] - Sqrt[1 + x]))/(1 + 2*Sqrt[1 - I] + Sqrt[5])] + (I/4)*PolyLog[2, (2 
*(Sqrt[1 + I] - Sqrt[1 + x]))/(1 + 2*Sqrt[1 + I] - Sqrt[5])] + (I/4)*PolyL 
og[2, (2*(Sqrt[1 + I] - Sqrt[1 + x]))/(1 + 2*Sqrt[1 + I] + Sqrt[5])] - (I/ 
4)*PolyLog[2, (-2*(Sqrt[1 - I] + Sqrt[1 + x]))/(1 - 2*Sqrt[1 - I] - Sqr...
 

3.1.30.3.1 Defintions of rubi rules used

rule 2009

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

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 3010

Int[((a_.) + Log[u_]*(b_.))*(RFx_), x_Symbol] :> With[{lst = SubstForFracti 
onalPowerOfLinear[RFx*(a + b*Log[u]), x]}, Simp[lst[[2]]*lst[[4]]   Subst[I 
nt[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x] /;  !FalseQ[lst]] /; FreeQ[{ 
a, b}, x] && RationalFunctionQ[RFx, x]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

3.1.30.4 Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 722, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {i \left (\ln \left (\sqrt {1+x}-\sqrt {1-i}\right ) \ln \left (x +\sqrt {1+x}\right )-\ln \left (\sqrt {1+x}-\sqrt {1-i}\right ) \ln \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1-i}-\sqrt {5}}\right )-\ln \left (\sqrt {1+x}-\sqrt {1-i}\right ) \ln \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1-i}+\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1-i}-\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1-i}+\sqrt {5}}\right )\right )}{2}+\frac {i \left (\ln \left (\sqrt {1-i}+\sqrt {1+x}\right ) \ln \left (x +\sqrt {1+x}\right )-\ln \left (\sqrt {1-i}+\sqrt {1+x}\right ) \ln \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1-i}-\sqrt {5}}\right )-\ln \left (\sqrt {1-i}+\sqrt {1+x}\right ) \ln \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1-i}+\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1-i}-\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1-i}+\sqrt {5}}\right )\right )}{2}-\frac {i \left (\ln \left (\sqrt {1+x}-\sqrt {1+i}\right ) \ln \left (x +\sqrt {1+x}\right )-\ln \left (\sqrt {1+x}-\sqrt {1+i}\right ) \ln \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1+i}-\sqrt {5}}\right )-\ln \left (\sqrt {1+x}-\sqrt {1+i}\right ) \ln \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1+i}+\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1+i}-\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1+i}+\sqrt {5}}\right )\right )}{2}-\frac {i \left (\ln \left (\sqrt {1+i}+\sqrt {1+x}\right ) \ln \left (x +\sqrt {1+x}\right )-\ln \left (\sqrt {1+i}+\sqrt {1+x}\right ) \ln \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1+i}-\sqrt {5}}\right )-\ln \left (\sqrt {1+i}+\sqrt {1+x}\right ) \ln \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1+i}+\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1+i}-\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1+i}+\sqrt {5}}\right )\right )}{2}\)	\(722\)
default	\(\frac {i \left (\ln \left (\sqrt {1+x}-\sqrt {1-i}\right ) \ln \left (x +\sqrt {1+x}\right )-\ln \left (\sqrt {1+x}-\sqrt {1-i}\right ) \ln \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1-i}-\sqrt {5}}\right )-\ln \left (\sqrt {1+x}-\sqrt {1-i}\right ) \ln \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1-i}+\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1-i}-\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1-i}+\sqrt {5}}\right )\right )}{2}+\frac {i \left (\ln \left (\sqrt {1-i}+\sqrt {1+x}\right ) \ln \left (x +\sqrt {1+x}\right )-\ln \left (\sqrt {1-i}+\sqrt {1+x}\right ) \ln \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1-i}-\sqrt {5}}\right )-\ln \left (\sqrt {1-i}+\sqrt {1+x}\right ) \ln \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1-i}+\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1-i}-\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1-i}+\sqrt {5}}\right )\right )}{2}-\frac {i \left (\ln \left (\sqrt {1+x}-\sqrt {1+i}\right ) \ln \left (x +\sqrt {1+x}\right )-\ln \left (\sqrt {1+x}-\sqrt {1+i}\right ) \ln \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1+i}-\sqrt {5}}\right )-\ln \left (\sqrt {1+x}-\sqrt {1+i}\right ) \ln \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1+i}+\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1+i}-\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1+i}+\sqrt {5}}\right )\right )}{2}-\frac {i \left (\ln \left (\sqrt {1+i}+\sqrt {1+x}\right ) \ln \left (x +\sqrt {1+x}\right )-\ln \left (\sqrt {1+i}+\sqrt {1+x}\right ) \ln \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1+i}-\sqrt {5}}\right )-\ln \left (\sqrt {1+i}+\sqrt {1+x}\right ) \ln \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1+i}+\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1+i}-\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1+i}+\sqrt {5}}\right )\right )}{2}\)	\(722\)
parts	\(\frac {i \left (\ln \left (\sqrt {1+x}-\sqrt {1-i}\right ) \ln \left (x +\sqrt {1+x}\right )-\ln \left (\sqrt {1+x}-\sqrt {1-i}\right ) \ln \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1-i}-\sqrt {5}}\right )-\ln \left (\sqrt {1+x}-\sqrt {1-i}\right ) \ln \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1-i}+\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1-i}-\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1-i}+\sqrt {5}}\right )\right )}{2}+\frac {i \left (\ln \left (\sqrt {1-i}+\sqrt {1+x}\right ) \ln \left (x +\sqrt {1+x}\right )-\ln \left (\sqrt {1-i}+\sqrt {1+x}\right ) \ln \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1-i}-\sqrt {5}}\right )-\ln \left (\sqrt {1-i}+\sqrt {1+x}\right ) \ln \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1-i}+\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1-i}-\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1-i}+\sqrt {5}}\right )\right )}{2}-\frac {i \left (\ln \left (\sqrt {1+x}-\sqrt {1+i}\right ) \ln \left (x +\sqrt {1+x}\right )-\ln \left (\sqrt {1+x}-\sqrt {1+i}\right ) \ln \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1+i}-\sqrt {5}}\right )-\ln \left (\sqrt {1+x}-\sqrt {1+i}\right ) \ln \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1+i}+\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1+i}-\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1+2 \sqrt {1+i}+\sqrt {5}}\right )\right )}{2}-\frac {i \left (\ln \left (\sqrt {1+i}+\sqrt {1+x}\right ) \ln \left (x +\sqrt {1+x}\right )-\ln \left (\sqrt {1+i}+\sqrt {1+x}\right ) \ln \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1+i}-\sqrt {5}}\right )-\ln \left (\sqrt {1+i}+\sqrt {1+x}\right ) \ln \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1+i}+\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1+i}-\sqrt {5}}\right )-\operatorname {dilog}\left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-2 \sqrt {1+i}+\sqrt {5}}\right )\right )}{2}\)	\(722\)

input

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

output

1/2*I*(ln((1+x)^(1/2)-(1-I)^(1/2))*ln(x+(1+x)^(1/2))-ln((1+x)^(1/2)-(1-I)^ 
(1/2))*ln((1-5^(1/2)+2*(1+x)^(1/2))/(1+2*(1-I)^(1/2)-5^(1/2)))-ln((1+x)^(1 
/2)-(1-I)^(1/2))*ln((1+5^(1/2)+2*(1+x)^(1/2))/(1+2*(1-I)^(1/2)+5^(1/2)))-d 
ilog((1-5^(1/2)+2*(1+x)^(1/2))/(1+2*(1-I)^(1/2)-5^(1/2)))-dilog((1+5^(1/2) 
+2*(1+x)^(1/2))/(1+2*(1-I)^(1/2)+5^(1/2))))+1/2*I*(ln((1-I)^(1/2)+(1+x)^(1 
/2))*ln(x+(1+x)^(1/2))-ln((1-I)^(1/2)+(1+x)^(1/2))*ln((1-5^(1/2)+2*(1+x)^( 
1/2))/(1-2*(1-I)^(1/2)-5^(1/2)))-ln((1-I)^(1/2)+(1+x)^(1/2))*ln((1+5^(1/2) 
+2*(1+x)^(1/2))/(1-2*(1-I)^(1/2)+5^(1/2)))-dilog((1-5^(1/2)+2*(1+x)^(1/2)) 
/(1-2*(1-I)^(1/2)-5^(1/2)))-dilog((1+5^(1/2)+2*(1+x)^(1/2))/(1-2*(1-I)^(1/ 
2)+5^(1/2))))-1/2*I*(ln((1+x)^(1/2)-(1+I)^(1/2))*ln(x+(1+x)^(1/2))-ln((1+x 
)^(1/2)-(1+I)^(1/2))*ln((1-5^(1/2)+2*(1+x)^(1/2))/(1+2*(1+I)^(1/2)-5^(1/2) 
))-ln((1+x)^(1/2)-(1+I)^(1/2))*ln((1+5^(1/2)+2*(1+x)^(1/2))/(1+2*(1+I)^(1/ 
2)+5^(1/2)))-dilog((1-5^(1/2)+2*(1+x)^(1/2))/(1+2*(1+I)^(1/2)-5^(1/2)))-di 
log((1+5^(1/2)+2*(1+x)^(1/2))/(1+2*(1+I)^(1/2)+5^(1/2))))-1/2*I*(ln((1+I)^ 
(1/2)+(1+x)^(1/2))*ln(x+(1+x)^(1/2))-ln((1+I)^(1/2)+(1+x)^(1/2))*ln((1-5^( 
1/2)+2*(1+x)^(1/2))/(1-2*(1+I)^(1/2)-5^(1/2)))-ln((1+I)^(1/2)+(1+x)^(1/2)) 
*ln((1+5^(1/2)+2*(1+x)^(1/2))/(1-2*(1+I)^(1/2)+5^(1/2)))-dilog((1-5^(1/2)+ 
2*(1+x)^(1/2))/(1-2*(1+I)^(1/2)-5^(1/2)))-dilog((1+5^(1/2)+2*(1+x)^(1/2))/ 
(1-2*(1+I)^(1/2)+5^(1/2))))
 

3.1.30.5 Fricas [F]

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

input

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

output

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

3.1.30.6 Sympy [F]

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

input

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

output

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

3.1.30.7 Maxima [F]

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

input

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

output

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

3.1.30.8 Giac [F]

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

input

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

output

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

3.1.30.9 Mupad [F(-1)]

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

input

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

output

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

3.1.30.10 Reduce [F]

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

input

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

output

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