\(\int (3+2 \log (x)) \, dx\) [8744]

Optimal result

Integrand size = 6, antiderivative size = 13 \[ \int (3+2 \log (x)) \, dx=5+e^3+x+\log (4)+2 x \log (x) \]

[Out]

5+x+exp(3)+2*ln(2)+2*x*ln(x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.54, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2332} \[ \int (3+2 \log (x)) \, dx=x+2 x \log (x) \]

[In]

Int[3 + 2*Log[x],x]

[Out]

x + 2*x*Log[x]

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps \begin{align*} \text {integral}& = 3 x+2 \int \log (x) \, dx \\ & = x+2 x \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.54 \[ \int (3+2 \log (x)) \, dx=x+2 x \log (x) \]

[In]

Integrate[3 + 2*Log[x],x]

[Out]

x + 2*x*Log[x]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.62

method	result	size
default	\(2 x \ln \left (x \right )+x\)	\(8\)
norman	\(2 x \ln \left (x \right )+x\)	\(8\)
risch	\(2 x \ln \left (x \right )+x\)	\(8\)
parallelrisch	\(2 x \ln \left (x \right )+x\)	\(8\)
parts	\(2 x \ln \left (x \right )+x\)	\(8\)

[In]

int(2*ln(x)+3,x,method=_RETURNVERBOSE)

[Out]

2*x*ln(x)+x

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.54 \[ \int (3+2 \log (x)) \, dx=2 \, x \log \left (x\right ) + x \]

[In]

integrate(2*log(x)+3,x, algorithm="fricas")

[Out]

2*x*log(x) + x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.54 \[ \int (3+2 \log (x)) \, dx=2 x \log {\left (x \right )} + x \]

[In]

integrate(2*ln(x)+3,x)

[Out]

2*x*log(x) + x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.54 \[ \int (3+2 \log (x)) \, dx=2 \, x \log \left (x\right ) + x \]

[In]

integrate(2*log(x)+3,x, algorithm="maxima")

[Out]

2*x*log(x) + x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.54 \[ \int (3+2 \log (x)) \, dx=2 \, x \log \left (x\right ) + x \]

[In]

integrate(2*log(x)+3,x, algorithm="giac")

[Out]

2*x*log(x) + x

Mupad [B] (verification not implemented)

Time = 13.73 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.62 \[ \int (3+2 \log (x)) \, dx=x\,\left (2\,\ln \left (x\right )+1\right ) \]

[In]

int(2*log(x) + 3,x)

[Out]

x*(2*log(x) + 1)