\(\int \log (1+x+\sqrt {1+x}) \, dx\) [238]

Optimal result

Integrand size = 11, antiderivative size = 32 \[ \int \log \left (1+x+\sqrt {1+x}\right ) \, dx=-x+\sqrt {1+x}+\frac {1}{2} \log (1+x)+x \log \left (1+x+\sqrt {1+x}\right ) \]

[Out]

-x+1/2*ln(1+x)+x*ln(1+x+(1+x)^(1/2))+(1+x)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2628} \[ \int \log \left (1+x+\sqrt {1+x}\right ) \, dx=-x+\sqrt {x+1}+x \log \left (x+\sqrt {x+1}+1\right )+\frac {1}{2} \log (x+1) \]

[In]

Int[Log[1 + x + Sqrt[1 + x]],x]

[Out]

-x + Sqrt[1 + x] + Log[1 + x]/2 + x*Log[1 + x + Sqrt[1 + x]]

Rule 2628

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*(D[u, x]/u), x], x] /; InverseFunctionFr
eeQ[u, x]

Rubi steps \begin{align*} \text {integral}& = x \log \left (1+x+\sqrt {1+x}\right )-\int \frac {x \left (1+\frac {1}{2 \sqrt {1+x}}\right )}{1+x+\sqrt {1+x}} \, dx \\ & = x \log \left (1+x+\sqrt {1+x}\right )-2 \text {Subst}\left (\int \left (-\frac {1}{2}-\frac {1}{2 x}+x\right ) \, dx,x,\sqrt {1+x}\right ) \\ & = -x+\sqrt {1+x}+\frac {1}{2} \log (1+x)+x \log \left (1+x+\sqrt {1+x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \log \left (1+x+\sqrt {1+x}\right ) \, dx=-x+\sqrt {1+x}-\log \left (1+\sqrt {1+x}\right )+(1+x) \log \left (1+x+\sqrt {1+x}\right ) \]

[In]

Integrate[Log[1 + x + Sqrt[1 + x]],x]

[Out]

-x + Sqrt[1 + x] - Log[1 + Sqrt[1 + x]] + (1 + x)*Log[1 + x + Sqrt[1 + x]]

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88

method	result	size
parts	\(x \ln \left (1+x +\sqrt {x +1}\right )+\sqrt {x +1}+\frac {\ln \left (x +1\right )}{2}-x -1\)	\(28\)
derivativedivides	\(\left (x +1\right ) \ln \left (1+x +\sqrt {x +1}\right )-x -1+\sqrt {x +1}-\ln \left (\sqrt {x +1}+1\right )\)	\(34\)
default	\(\left (x +1\right ) \ln \left (1+x +\sqrt {x +1}\right )-x -1+\sqrt {x +1}-\ln \left (\sqrt {x +1}+1\right )\)	\(34\)

[In]

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

[Out]

x*ln(1+x+(x+1)^(1/2))+(x+1)^(1/2)+1/2*ln(x+1)-x-1

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \log \left (1+x+\sqrt {1+x}\right ) \, dx={\left (x - 1\right )} \log \left (x + \sqrt {x + 1} + 1\right ) - x + \sqrt {x + 1} + \log \left (\sqrt {x + 1} + 1\right ) + 2 \, \log \left (\sqrt {x + 1}\right ) \]

[In]

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

[Out]

(x - 1)*log(x + sqrt(x + 1) + 1) - x + sqrt(x + 1) + log(sqrt(x + 1) + 1) + 2*log(sqrt(x + 1))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (27) = 54\).

Time = 0.41 (sec) , antiderivative size = 184, normalized size of antiderivative = 5.75 \[ \int \log \left (1+x+\sqrt {1+x}\right ) \, dx=\frac {x \sqrt {x + 1} \log {\left (x + \sqrt {x + 1} + 1 \right )}}{\sqrt {x + 1} + 1} - \frac {x \sqrt {x + 1}}{\sqrt {x + 1} + 1} + \frac {x \log {\left (x + \sqrt {x + 1} + 1 \right )}}{\sqrt {x + 1} + 1} - \frac {\sqrt {x + 1} \log {\left (\sqrt {x + 1} + 1 \right )}}{\sqrt {x + 1} + 1} + \frac {\sqrt {x + 1} \log {\left (x + \sqrt {x + 1} + 1 \right )}}{\sqrt {x + 1} + 1} + \frac {\sqrt {x + 1}}{\sqrt {x + 1} + 1} - \frac {\log {\left (\sqrt {x + 1} + 1 \right )}}{\sqrt {x + 1} + 1} + \frac {\log {\left (x + \sqrt {x + 1} + 1 \right )}}{\sqrt {x + 1} + 1} + \frac {1}{\sqrt {x + 1} + 1} \]

[In]

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

[Out]

x*sqrt(x + 1)*log(x + sqrt(x + 1) + 1)/(sqrt(x + 1) + 1) - x*sqrt(x + 1)/(sqrt(x + 1) + 1) + x*log(x + sqrt(x
+ 1) + 1)/(sqrt(x + 1) + 1) - sqrt(x + 1)*log(sqrt(x + 1) + 1)/(sqrt(x + 1) + 1) + sqrt(x + 1)*log(x + sqrt(x
+ 1) + 1)/(sqrt(x + 1) + 1) + sqrt(x + 1)/(sqrt(x + 1) + 1) - log(sqrt(x + 1) + 1)/(sqrt(x + 1) + 1) + log(x +
sqrt(x + 1) + 1)/(sqrt(x + 1) + 1) + 1/(sqrt(x + 1) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \log \left (1+x+\sqrt {1+x}\right ) \, dx={\left (x + 1\right )} \log \left (x + \sqrt {x + 1} + 1\right ) - x + \sqrt {x + 1} - \log \left (\sqrt {x + 1} + 1\right ) - 1 \]

[In]

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

[Out]

(x + 1)*log(x + sqrt(x + 1) + 1) - x + sqrt(x + 1) - log(sqrt(x + 1) + 1) - 1

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \log \left (1+x+\sqrt {1+x}\right ) \, dx={\left (x + 1\right )} \log \left (x + \sqrt {x + 1} + 1\right ) - x + \sqrt {x + 1} - \log \left (\sqrt {x + 1} + 1\right ) - 1 \]

[In]

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

[Out]

(x + 1)*log(x + sqrt(x + 1) + 1) - x + sqrt(x + 1) - log(sqrt(x + 1) + 1) - 1

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \log \left (1+x+\sqrt {1+x}\right ) \, dx=\ln \left (\sqrt {x+1}\right )-x+\sqrt {x+1}+x\,\ln \left (x+\sqrt {x+1}+1\right ) \]

[In]

int(log(x + (x + 1)^(1/2) + 1),x)

[Out]

log((x + 1)^(1/2)) - x + (x + 1)^(1/2) + x*log(x + (x + 1)^(1/2) + 1)