3.32 \(\int \frac {\log (x+\sqrt {1+x})}{x} \, dx\)

Optimal. Leaf size=313 \[ -\operatorname {PolyLog}\left (2,\frac {2 \left (1-\sqrt {x+1}\right )}{3-\sqrt {5}}\right )-\operatorname {PolyLog}\left (2,\frac {2 \left (1-\sqrt {x+1}\right )}{3+\sqrt {5}}\right )-\operatorname {PolyLog}\left (2,\frac {2 \left (\sqrt {x+1}+1\right )}{1-\sqrt {5}}\right )-\operatorname {PolyLog}\left (2,\frac {2 \left (\sqrt {x+1}+1\right )}{1+\sqrt {5}}\right )+\log \left (\sqrt {x+1}-1\right ) \log \left (x+\sqrt {x+1}\right )+\log \left (\sqrt {x+1}+1\right ) \log \left (x+\sqrt {x+1}\right )-\log \left (\sqrt {x+1}-1\right ) \log \left (\frac {2 \sqrt {x+1}-\sqrt {5}+1}{3-\sqrt {5}}\right )-\log \left (\sqrt {x+1}+1\right ) \log \left (-\frac {2 \sqrt {x+1}-\sqrt {5}+1}{1+\sqrt {5}}\right )-\log \left (\sqrt {x+1}+1\right ) \log \left (-\frac {2 \sqrt {x+1}+\sqrt {5}+1}{1-\sqrt {5}}\right )-\log \left (\sqrt {x+1}-1\right ) \log \left (\frac {2 \sqrt {x+1}+\sqrt {5}+1}{3+\sqrt {5}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2530, 2528, 2524, 2418, 2394, 2393, 2391} \[ -\text {PolyLog}\left (2,\frac {2 \left (1-\sqrt {x+1}\right )}{3-\sqrt {5}}\right )-\text {PolyLog}\left (2,\frac {2 \left (1-\sqrt {x+1}\right )}{3+\sqrt {5}}\right )-\text {PolyLog}\left (2,\frac {2 \left (\sqrt {x+1}+1\right )}{1-\sqrt {5}}\right )-\text {PolyLog}\left (2,\frac {2 \left (\sqrt {x+1}+1\right )}{1+\sqrt {5}}\right )+\log \left (\sqrt {x+1}-1\right ) \log \left (x+\sqrt {x+1}\right )+\log \left (\sqrt {x+1}+1\right ) \log \left (x+\sqrt {x+1}\right )-\log \left (\sqrt {x+1}-1\right ) \log \left (\frac {2 \sqrt {x+1}-\sqrt {5}+1}{3-\sqrt {5}}\right )-\log \left (\sqrt {x+1}+1\right ) \log \left (-\frac {2 \sqrt {x+1}-\sqrt {5}+1}{1+\sqrt {5}}\right )-\log \left (\sqrt {x+1}+1\right ) \log \left (-\frac {2 \sqrt {x+1}+\sqrt {5}+1}{1-\sqrt {5}}\right )-\log \left (\sqrt {x+1}-1\right ) \log \left (\frac {2 \sqrt {x+1}+\sqrt {5}+1}{3+\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[x + Sqrt[1 + x]]/x,x]

[Out]

Log[-1 + Sqrt[1 + x]]*Log[x + Sqrt[1 + x]] + Log[1 + Sqrt[1 + x]]*Log[x + Sqrt[1 + x]] - Log[-1 + Sqrt[1 + x]]
*Log[(1 - Sqrt[5] + 2*Sqrt[1 + x])/(3 - Sqrt[5])] - Log[1 + Sqrt[1 + x]]*Log[-((1 - Sqrt[5] + 2*Sqrt[1 + x])/(
1 + Sqrt[5]))] - Log[1 + Sqrt[1 + x]]*Log[-((1 + Sqrt[5] + 2*Sqrt[1 + x])/(1 - Sqrt[5]))] - Log[-1 + Sqrt[1 +
x]]*Log[(1 + Sqrt[5] + 2*Sqrt[1 + x])/(3 + Sqrt[5])] - PolyLog[2, (2*(1 - Sqrt[1 + x]))/(3 - Sqrt[5])] - PolyL
og[2, (2*(1 - Sqrt[1 + x]))/(3 + Sqrt[5])] - PolyLog[2, (2*(1 + Sqrt[1 + x]))/(1 - Sqrt[5])] - PolyLog[2, (2*(
1 + Sqrt[1 + x]))/(1 + Sqrt[5])]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 2524

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

Rule 2528

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] && RationalF
unctionQ[RGx, x] && IGtQ[n, 0]

Rule 2530

Int[((a_.) + Log[u_]*(b_.))*(RFx_), x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[RFx*(a + b*Log[u]
), x]}, Dist[lst[[2]]*lst[[4]], Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x] /;  !FalseQ[lst]] /; Fre
eQ[{a, b}, x] && RationalFunctionQ[RFx, x]

Rubi steps

\begin {align*} \int \frac {\log \left (x+\sqrt {1+x}\right )}{x} \, dx &=2 \operatorname {Subst}\left (\int \frac {x \log \left (-1+x+x^2\right )}{-1+x^2} \, dx,x,\sqrt {1+x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (\frac {\log \left (-1+x+x^2\right )}{2 (-1+x)}+\frac {\log \left (-1+x+x^2\right )}{2 (1+x)}\right ) \, dx,x,\sqrt {1+x}\right )\\ &=\operatorname {Subst}\left (\int \frac {\log \left (-1+x+x^2\right )}{-1+x} \, dx,x,\sqrt {1+x}\right )+\operatorname {Subst}\left (\int \frac {\log \left (-1+x+x^2\right )}{1+x} \, dx,x,\sqrt {1+x}\right )\\ &=\log \left (-1+\sqrt {1+x}\right ) \log \left (x+\sqrt {1+x}\right )+\log \left (1+\sqrt {1+x}\right ) \log \left (x+\sqrt {1+x}\right )-\operatorname {Subst}\left (\int \frac {(1+2 x) \log (-1+x)}{-1+x+x^2} \, dx,x,\sqrt {1+x}\right )-\operatorname {Subst}\left (\int \frac {(1+2 x) \log (1+x)}{-1+x+x^2} \, dx,x,\sqrt {1+x}\right )\\ &=\log \left (-1+\sqrt {1+x}\right ) \log \left (x+\sqrt {1+x}\right )+\log \left (1+\sqrt {1+x}\right ) \log \left (x+\sqrt {1+x}\right )-\operatorname {Subst}\left (\int \left (\frac {2 \log (-1+x)}{1-\sqrt {5}+2 x}+\frac {2 \log (-1+x)}{1+\sqrt {5}+2 x}\right ) \, dx,x,\sqrt {1+x}\right )-\operatorname {Subst}\left (\int \left (\frac {2 \log (1+x)}{1-\sqrt {5}+2 x}+\frac {2 \log (1+x)}{1+\sqrt {5}+2 x}\right ) \, dx,x,\sqrt {1+x}\right )\\ &=\log \left (-1+\sqrt {1+x}\right ) \log \left (x+\sqrt {1+x}\right )+\log \left (1+\sqrt {1+x}\right ) \log \left (x+\sqrt {1+x}\right )-2 \operatorname {Subst}\left (\int \frac {\log (-1+x)}{1-\sqrt {5}+2 x} \, dx,x,\sqrt {1+x}\right )-2 \operatorname {Subst}\left (\int \frac {\log (-1+x)}{1+\sqrt {5}+2 x} \, dx,x,\sqrt {1+x}\right )-2 \operatorname {Subst}\left (\int \frac {\log (1+x)}{1-\sqrt {5}+2 x} \, dx,x,\sqrt {1+x}\right )-2 \operatorname {Subst}\left (\int \frac {\log (1+x)}{1+\sqrt {5}+2 x} \, dx,x,\sqrt {1+x}\right )\\ &=\log \left (-1+\sqrt {1+x}\right ) \log \left (x+\sqrt {1+x}\right )+\log \left (1+\sqrt {1+x}\right ) \log \left (x+\sqrt {1+x}\right )-\log \left (-1+\sqrt {1+x}\right ) \log \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{3-\sqrt {5}}\right )-\log \left (1+\sqrt {1+x}\right ) \log \left (-\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+\sqrt {5}}\right )-\log \left (1+\sqrt {1+x}\right ) \log \left (-\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-\sqrt {5}}\right )-\log \left (-1+\sqrt {1+x}\right ) \log \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{3+\sqrt {5}}\right )+\operatorname {Subst}\left (\int \frac {\log \left (\frac {1-\sqrt {5}+2 x}{-1-\sqrt {5}}\right )}{1+x} \, dx,x,\sqrt {1+x}\right )+\operatorname {Subst}\left (\int \frac {\log \left (\frac {1-\sqrt {5}+2 x}{3-\sqrt {5}}\right )}{-1+x} \, dx,x,\sqrt {1+x}\right )+\operatorname {Subst}\left (\int \frac {\log \left (\frac {1+\sqrt {5}+2 x}{-1+\sqrt {5}}\right )}{1+x} \, dx,x,\sqrt {1+x}\right )+\operatorname {Subst}\left (\int \frac {\log \left (\frac {1+\sqrt {5}+2 x}{3+\sqrt {5}}\right )}{-1+x} \, dx,x,\sqrt {1+x}\right )\\ &=\log \left (-1+\sqrt {1+x}\right ) \log \left (x+\sqrt {1+x}\right )+\log \left (1+\sqrt {1+x}\right ) \log \left (x+\sqrt {1+x}\right )-\log \left (-1+\sqrt {1+x}\right ) \log \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{3-\sqrt {5}}\right )-\log \left (1+\sqrt {1+x}\right ) \log \left (-\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+\sqrt {5}}\right )-\log \left (1+\sqrt {1+x}\right ) \log \left (-\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-\sqrt {5}}\right )-\log \left (-1+\sqrt {1+x}\right ) \log \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{3+\sqrt {5}}\right )+\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{-1-\sqrt {5}}\right )}{x} \, dx,x,1+\sqrt {1+x}\right )+\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{3-\sqrt {5}}\right )}{x} \, dx,x,-1+\sqrt {1+x}\right )+\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{-1+\sqrt {5}}\right )}{x} \, dx,x,1+\sqrt {1+x}\right )+\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 x}{3+\sqrt {5}}\right )}{x} \, dx,x,-1+\sqrt {1+x}\right )\\ &=\log \left (-1+\sqrt {1+x}\right ) \log \left (x+\sqrt {1+x}\right )+\log \left (1+\sqrt {1+x}\right ) \log \left (x+\sqrt {1+x}\right )-\log \left (-1+\sqrt {1+x}\right ) \log \left (\frac {1-\sqrt {5}+2 \sqrt {1+x}}{3-\sqrt {5}}\right )-\log \left (1+\sqrt {1+x}\right ) \log \left (-\frac {1-\sqrt {5}+2 \sqrt {1+x}}{1+\sqrt {5}}\right )-\log \left (1+\sqrt {1+x}\right ) \log \left (-\frac {1+\sqrt {5}+2 \sqrt {1+x}}{1-\sqrt {5}}\right )-\log \left (-1+\sqrt {1+x}\right ) \log \left (\frac {1+\sqrt {5}+2 \sqrt {1+x}}{3+\sqrt {5}}\right )-\text {Li}_2\left (-\frac {2 \left (-1+\sqrt {1+x}\right )}{3-\sqrt {5}}\right )-\text {Li}_2\left (-\frac {2 \left (-1+\sqrt {1+x}\right )}{3+\sqrt {5}}\right )-\text {Li}_2\left (\frac {2 \left (1+\sqrt {1+x}\right )}{1-\sqrt {5}}\right )-\text {Li}_2\left (\frac {2 \left (1+\sqrt {1+x}\right )}{1+\sqrt {5}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 303, normalized size = 0.97 \[ -\operatorname {PolyLog}\left (2,\frac {2 \left (\sqrt {x+1}+1\right )}{1-\sqrt {5}}\right )+\operatorname {PolyLog}\left (2,\frac {2 \sqrt {x+1}-\sqrt {5}+1}{3-\sqrt {5}}\right )+\operatorname {PolyLog}\left (2,-\frac {2 \sqrt {x+1}-\sqrt {5}+1}{1+\sqrt {5}}\right )+\operatorname {PolyLog}\left (2,\frac {2 \sqrt {x+1}+\sqrt {5}+1}{3+\sqrt {5}}\right )+\log \left (1-\sqrt {x+1}\right ) \log \left (x+\sqrt {x+1}\right )+\log \left (\sqrt {x+1}+1\right ) \log \left (x+\sqrt {x+1}\right )-\log \left (\frac {1}{2} \left (1+\sqrt {5}\right )\right ) \log \left (2 \sqrt {x+1}-\sqrt {5}+1\right )-\log \left (\frac {1}{2} \left (3-\sqrt {5}\right )\right ) \log \left (2 \sqrt {x+1}-\sqrt {5}+1\right )-\log \left (\frac {1}{2} \left (3+\sqrt {5}\right )\right ) \log \left (2 \sqrt {x+1}+\sqrt {5}+1\right )-\log \left (\sqrt {x+1}+1\right ) \log \left (-\frac {2 \sqrt {x+1}+\sqrt {5}+1}{1-\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[x + Sqrt[1 + x]]/x,x]

[Out]

Log[1 - Sqrt[1 + x]]*Log[x + Sqrt[1 + x]] + Log[1 + Sqrt[1 + x]]*Log[x + Sqrt[1 + x]] - Log[(3 - Sqrt[5])/2]*L
og[1 - Sqrt[5] + 2*Sqrt[1 + x]] - Log[(1 + Sqrt[5])/2]*Log[1 - Sqrt[5] + 2*Sqrt[1 + x]] - Log[(3 + Sqrt[5])/2]
*Log[1 + Sqrt[5] + 2*Sqrt[1 + x]] - Log[1 + Sqrt[1 + x]]*Log[-((1 + Sqrt[5] + 2*Sqrt[1 + x])/(1 - Sqrt[5]))] -
PolyLog[2, (2*(1 + Sqrt[1 + x]))/(1 - Sqrt[5])] + PolyLog[2, (1 - Sqrt[5] + 2*Sqrt[1 + x])/(3 - Sqrt[5])] + P
olyLog[2, -((1 - Sqrt[5] + 2*Sqrt[1 + x])/(1 + Sqrt[5]))] + PolyLog[2, (1 + Sqrt[5] + 2*Sqrt[1 + x])/(3 + Sqrt
[5])]

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (x + \sqrt {x + 1}\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(x + sqrt(x + 1))/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left (x + \sqrt {x + 1}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(x + sqrt(x + 1))/x, x)

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maple [A]  time = 0.02, size = 252, normalized size = 0.81 \[ -\ln \left (\frac {-1+\sqrt {5}-2 \sqrt {x +1}}{\sqrt {5}-3}\right ) \ln \left (-1+\sqrt {x +1}\right )-\ln \left (\frac {-1+\sqrt {5}-2 \sqrt {x +1}}{\sqrt {5}+1}\right ) \ln \left (1+\sqrt {x +1}\right )-\ln \left (\frac {1+\sqrt {5}+2 \sqrt {x +1}}{3+\sqrt {5}}\right ) \ln \left (-1+\sqrt {x +1}\right )-\ln \left (\frac {1+\sqrt {5}+2 \sqrt {x +1}}{\sqrt {5}-1}\right ) \ln \left (1+\sqrt {x +1}\right )+\ln \left (-1+\sqrt {x +1}\right ) \ln \left (x +\sqrt {x +1}\right )+\ln \left (1+\sqrt {x +1}\right ) \ln \left (x +\sqrt {x +1}\right )-\dilog \left (\frac {-1+\sqrt {5}-2 \sqrt {x +1}}{\sqrt {5}-3}\right )-\dilog \left (\frac {-1+\sqrt {5}-2 \sqrt {x +1}}{\sqrt {5}+1}\right )-\dilog \left (\frac {1+\sqrt {5}+2 \sqrt {x +1}}{3+\sqrt {5}}\right )-\dilog \left (\frac {1+\sqrt {5}+2 \sqrt {x +1}}{\sqrt {5}-1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(1+(x+1)^(1/2))*ln(x+(x+1)^(1/2))-ln(1+(x+1)^(1/2))*ln((-1+5^(1/2)-2*(x+1)^(1/2))/(5^(1/2)+1))-ln(1+(x+1)^(1
/2))*ln((1+5^(1/2)+2*(x+1)^(1/2))/(5^(1/2)-1))-dilog((-1+5^(1/2)-2*(x+1)^(1/2))/(5^(1/2)+1))-dilog((1+5^(1/2)+
2*(x+1)^(1/2))/(5^(1/2)-1))+ln(-1+(x+1)^(1/2))*ln(x+(x+1)^(1/2))-ln(-1+(x+1)^(1/2))*ln((-1+5^(1/2)-2*(x+1)^(1/
2))/(5^(1/2)-3))-ln(-1+(x+1)^(1/2))*ln((1+5^(1/2)+2*(x+1)^(1/2))/(3+5^(1/2)))-dilog((-1+5^(1/2)-2*(x+1)^(1/2))
/(5^(1/2)-3))-dilog((1+5^(1/2)+2*(x+1)^(1/2))/(3+5^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left (x + \sqrt {x + 1}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(x + sqrt(x + 1))/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\ln \left (x+\sqrt {x+1}\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(x + (x + 1)^(1/2))/x,x)

[Out]

int(log(x + (x + 1)^(1/2))/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log {\left (x + \sqrt {x + 1} \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(x + sqrt(x + 1))/x, x)

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