\(\int \frac {\sqrt {1+\sqrt {1+x}} \log (1+x)}{x} \, dx\) [8]

Optimal result

Integrand size = 21, antiderivative size = 308 \[ \int \frac {\sqrt {1+\sqrt {1+x}} \log (1+x)}{x} \, dx=-16 \sqrt {1+\sqrt {1+x}}+16 \text {arctanh}\left (\sqrt {1+\sqrt {1+x}}\right )+4 \sqrt {1+\sqrt {1+x}} \log (1+x)-2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sqrt {1+x}}}{\sqrt {2}}\right ) \log (1+x)+4 \sqrt {2} \text {arctanh}\left (\frac {1}{\sqrt {2}}\right ) \log \left (1-\sqrt {1+\sqrt {1+x}}\right )-4 \sqrt {2} \text {arctanh}\left (\frac {1}{\sqrt {2}}\right ) \log \left (1+\sqrt {1+\sqrt {1+x}}\right )+2 \sqrt {2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {2} \left (1-\sqrt {1+\sqrt {1+x}}\right )}{2-\sqrt {2}}\right )-2 \sqrt {2} \operatorname {PolyLog}\left (2,\frac {\sqrt {2} \left (1-\sqrt {1+\sqrt {1+x}}\right )}{2+\sqrt {2}}\right )-2 \sqrt {2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {2} \left (1+\sqrt {1+\sqrt {1+x}}\right )}{2-\sqrt {2}}\right )+2 \sqrt {2} \operatorname {PolyLog}\left (2,\frac {\sqrt {2} \left (1+\sqrt {1+\sqrt {1+x}}\right )}{2+\sqrt {2}}\right ) \]

[Out]

16*arctanh((1+(1+x)^(1/2))^(1/2))-2*arctanh(1/2*(1+(1+x)^(1/2))^(1/2)*2^(1/2))*ln(1+x)*2^(1/2)+4*arctanh(1/2*2
^(1/2))*ln(1-(1+(1+x)^(1/2))^(1/2))*2^(1/2)-4*arctanh(1/2*2^(1/2))*ln(1+(1+(1+x)^(1/2))^(1/2))*2^(1/2)+2*polyl
og(2,-2^(1/2)*(1-(1+(1+x)^(1/2))^(1/2))/(2-2^(1/2)))*2^(1/2)-2*polylog(2,2^(1/2)*(1-(1+(1+x)^(1/2))^(1/2))/(2+
2^(1/2)))*2^(1/2)-2*polylog(2,-2^(1/2)*(1+(1+(1+x)^(1/2))^(1/2))/(2-2^(1/2)))*2^(1/2)+2*polylog(2,2^(1/2)*(1+(
1+(1+x)^(1/2))^(1/2))/(2+2^(1/2)))*2^(1/2)-16*(1+(1+x)^(1/2))^(1/2)+4*ln(1+x)*(1+(1+x)^(1/2))^(1/2)

Rubi [F]

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

[In]

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

[Out]

Defer[Int][(Sqrt[1 + Sqrt[1 + x]]*Log[1 + x])/x, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sqrt {1+\sqrt {1+x}} \log (1+x)}{x} \, dx \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 0.13 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt {1+\sqrt {1+x}} \log (1+x)}{x} \, dx=-16 \sqrt {1+\sqrt {1+x}}+16 \text {arctanh}\left (\sqrt {1+\sqrt {1+x}}\right )+4 \sqrt {1+\sqrt {1+x}} \log (1+x)+\sqrt {2} \log (1+x) \left (\log \left (\sqrt {2}-\sqrt {1+\sqrt {1+x}}\right )-\log \left (\sqrt {2}+\sqrt {1+\sqrt {1+x}}\right )\right )-2 \sqrt {2} \left (\log \left (-1+\sqrt {2}\right ) \log \left (1-\sqrt {1+\sqrt {1+x}}\right )-\log \left (1+\sqrt {2}\right ) \log \left (1-\sqrt {1+\sqrt {1+x}}\right )-\log \left (-1+\sqrt {2}\right ) \log \left (1+\sqrt {1+\sqrt {1+x}}\right )+\log \left (1+\sqrt {2}\right ) \log \left (1+\sqrt {1+\sqrt {1+x}}\right )+\operatorname {PolyLog}\left (2,-\left (\left (-1+\sqrt {2}\right ) \left (-1+\sqrt {1+\sqrt {1+x}}\right )\right )\right )-\operatorname {PolyLog}\left (2,\left (1+\sqrt {2}\right ) \left (-1+\sqrt {1+\sqrt {1+x}}\right )\right )-\operatorname {PolyLog}\left (2,\left (-1+\sqrt {2}\right ) \left (1+\sqrt {1+\sqrt {1+x}}\right )\right )+\operatorname {PolyLog}\left (2,-\left (\left (1+\sqrt {2}\right ) \left (1+\sqrt {1+\sqrt {1+x}}\right )\right )\right )\right ) \]

[In]

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

[Out]

-16*Sqrt[1 + Sqrt[1 + x]] + 16*ArcTanh[Sqrt[1 + Sqrt[1 + x]]] + 4*Sqrt[1 + Sqrt[1 + x]]*Log[1 + x] + Sqrt[2]*L
og[1 + x]*(Log[Sqrt[2] - Sqrt[1 + Sqrt[1 + x]]] - Log[Sqrt[2] + Sqrt[1 + Sqrt[1 + x]]]) - 2*Sqrt[2]*(Log[-1 +
Sqrt[2]]*Log[1 - Sqrt[1 + Sqrt[1 + x]]] - Log[1 + Sqrt[2]]*Log[1 - Sqrt[1 + Sqrt[1 + x]]] - Log[-1 + Sqrt[2]]*
Log[1 + Sqrt[1 + Sqrt[1 + x]]] + Log[1 + Sqrt[2]]*Log[1 + Sqrt[1 + Sqrt[1 + x]]] + PolyLog[2, -((-1 + Sqrt[2])
*(-1 + Sqrt[1 + Sqrt[1 + x]]))] - PolyLog[2, (1 + Sqrt[2])*(-1 + Sqrt[1 + Sqrt[1 + x]])] - PolyLog[2, (-1 + Sq
rt[2])*(1 + Sqrt[1 + Sqrt[1 + x]])] + PolyLog[2, -((1 + Sqrt[2])*(1 + Sqrt[1 + Sqrt[1 + x]]))])

Maple [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(4 \ln \left (1+x \right ) \sqrt {1+\sqrt {1+x}}-16 \sqrt {1+\sqrt {1+x}}-8 \ln \left (\sqrt {1+\sqrt {1+x}}-1\right )+8 \ln \left (1+\sqrt {1+\sqrt {1+x}}\right )+8 \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )}{\sum }\frac {\left (\frac {\ln \left (\sqrt {1+\sqrt {1+x}}-\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (1+x \right )}{2}-\operatorname {dilog}\left (\frac {\sqrt {1+\sqrt {1+x}}-1}{-1+\underline {\hspace {1.25 ex}}\alpha }\right )-\ln \left (\sqrt {1+\sqrt {1+x}}-\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {\sqrt {1+\sqrt {1+x}}-1}{-1+\underline {\hspace {1.25 ex}}\alpha }\right )-\operatorname {dilog}\left (\frac {1+\sqrt {1+\sqrt {1+x}}}{1+\underline {\hspace {1.25 ex}}\alpha }\right )-\ln \left (\sqrt {1+\sqrt {1+x}}-\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {1+\sqrt {1+\sqrt {1+x}}}{1+\underline {\hspace {1.25 ex}}\alpha }\right )\right ) \underline {\hspace {1.25 ex}}\alpha }{4}\right )\)	\(199\)
default	\(4 \ln \left (1+x \right ) \sqrt {1+\sqrt {1+x}}-16 \sqrt {1+\sqrt {1+x}}-8 \ln \left (\sqrt {1+\sqrt {1+x}}-1\right )+8 \ln \left (1+\sqrt {1+\sqrt {1+x}}\right )+8 \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )}{\sum }\frac {\left (\frac {\ln \left (\sqrt {1+\sqrt {1+x}}-\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (1+x \right )}{2}-\operatorname {dilog}\left (\frac {\sqrt {1+\sqrt {1+x}}-1}{-1+\underline {\hspace {1.25 ex}}\alpha }\right )-\ln \left (\sqrt {1+\sqrt {1+x}}-\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {\sqrt {1+\sqrt {1+x}}-1}{-1+\underline {\hspace {1.25 ex}}\alpha }\right )-\operatorname {dilog}\left (\frac {1+\sqrt {1+\sqrt {1+x}}}{1+\underline {\hspace {1.25 ex}}\alpha }\right )-\ln \left (\sqrt {1+\sqrt {1+x}}-\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {1+\sqrt {1+\sqrt {1+x}}}{1+\underline {\hspace {1.25 ex}}\alpha }\right )\right ) \underline {\hspace {1.25 ex}}\alpha }{4}\right )\)	\(199\)

[In]

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

[Out]

4*ln(1+x)*(1+(1+x)^(1/2))^(1/2)-16*(1+(1+x)^(1/2))^(1/2)-8*ln((1+(1+x)^(1/2))^(1/2)-1)+8*ln(1+(1+(1+x)^(1/2))^
(1/2))+8*Sum(1/4*(1/2*ln((1+(1+x)^(1/2))^(1/2)-_alpha)*ln(1+x)-dilog(((1+(1+x)^(1/2))^(1/2)-1)/(-1+_alpha))-ln
((1+(1+x)^(1/2))^(1/2)-_alpha)*ln(((1+(1+x)^(1/2))^(1/2)-1)/(-1+_alpha))-dilog((1+(1+(1+x)^(1/2))^(1/2))/(1+_a
lpha))-ln((1+(1+x)^(1/2))^(1/2)-_alpha)*ln((1+(1+(1+x)^(1/2))^(1/2))/(1+_alpha)))*_alpha,_alpha=RootOf(_Z^2-2)
)

Fricas [F(-2)]

Exception generated. \[ \int \frac {\sqrt {1+\sqrt {1+x}} \log (1+x)}{x} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {1+\sqrt {1+x}} \log (1+x)}{x} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.23 \[ \int \frac {\sqrt {1+\sqrt {1+x}} \log (1+x)}{x} \, dx={\left (\sqrt {2} \log \left (-\frac {\sqrt {2} - \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}}\right ) + 4 \, \sqrt {\sqrt {x + 1} + 1}\right )} \log \left (x + 1\right ) + 2 \, \sqrt {2} {\left (\log \left (\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}\right ) \log \left (-\frac {\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} + 1} + 1\right ) + {\rm Li}_2\left (\frac {\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} + 1}\right )\right )} - 2 \, \sqrt {2} {\left (\log \left (-\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}\right ) \log \left (-\frac {\sqrt {2} - \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} + 1} + 1\right ) + {\rm Li}_2\left (\frac {\sqrt {2} - \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} + 1}\right )\right )} + 2 \, \sqrt {2} {\left (\log \left (\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}\right ) \log \left (-\frac {\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} - 1} + 1\right ) + {\rm Li}_2\left (\frac {\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} - 1}\right )\right )} - 2 \, \sqrt {2} {\left (\log \left (-\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}\right ) \log \left (-\frac {\sqrt {2} - \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} - 1} + 1\right ) + {\rm Li}_2\left (\frac {\sqrt {2} - \sqrt {\sqrt {x + 1} + 1}}{\sqrt {2} - 1}\right )\right )} - 16 \, \sqrt {\sqrt {x + 1} + 1} + 8 \, \log \left (\sqrt {\sqrt {x + 1} + 1} + 1\right ) - 8 \, \log \left (\sqrt {\sqrt {x + 1} + 1} - 1\right ) \]

[In]

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

[Out]

(sqrt(2)*log(-(sqrt(2) - sqrt(sqrt(x + 1) + 1))/(sqrt(2) + sqrt(sqrt(x + 1) + 1))) + 4*sqrt(sqrt(x + 1) + 1))*
log(x + 1) + 2*sqrt(2)*(log(sqrt(2) + sqrt(sqrt(x + 1) + 1))*log(-(sqrt(2) + sqrt(sqrt(x + 1) + 1))/(sqrt(2) +
1) + 1) + dilog((sqrt(2) + sqrt(sqrt(x + 1) + 1))/(sqrt(2) + 1))) - 2*sqrt(2)*(log(-sqrt(2) + sqrt(sqrt(x + 1
) + 1))*log(-(sqrt(2) - sqrt(sqrt(x + 1) + 1))/(sqrt(2) + 1) + 1) + dilog((sqrt(2) - sqrt(sqrt(x + 1) + 1))/(s
qrt(2) + 1))) + 2*sqrt(2)*(log(sqrt(2) + sqrt(sqrt(x + 1) + 1))*log(-(sqrt(2) + sqrt(sqrt(x + 1) + 1))/(sqrt(2
) - 1) + 1) + dilog((sqrt(2) + sqrt(sqrt(x + 1) + 1))/(sqrt(2) - 1))) - 2*sqrt(2)*(log(-sqrt(2) + sqrt(sqrt(x
+ 1) + 1))*log(-(sqrt(2) - sqrt(sqrt(x + 1) + 1))/(sqrt(2) - 1) + 1) + dilog((sqrt(2) - sqrt(sqrt(x + 1) + 1))
/(sqrt(2) - 1))) - 16*sqrt(sqrt(x + 1) + 1) + 8*log(sqrt(sqrt(x + 1) + 1) + 1) - 8*log(sqrt(sqrt(x + 1) + 1) -
1)

Giac [F]

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

[In]

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

[Out]

integrate(sqrt(sqrt(x + 1) + 1)*log(x + 1)/x, x)

Mupad [F(-1)]

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

[In]

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

[Out]

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