\(\int \frac {3+2 x+3 x \log (4)}{4 x} \, dx\) [5508]

Optimal result

Integrand size = 17, antiderivative size = 19 \[ \int \frac {3+2 x+3 x \log (4)}{4 x} \, dx=-1-x+\frac {3}{4} (2 x+x \log (4)+\log (x)) \]

[Out]

1/2*x-1+3/4*ln(x)+3/2*x*ln(2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {6, 12, 45} \[ \int \frac {3+2 x+3 x \log (4)}{4 x} \, dx=\frac {1}{4} x (2+\log (64))+\frac {3 \log (x)}{4} \]

[In]

Int[(3 + 2*x + 3*x*Log[4])/(4*x),x]

[Out]

(x*(2 + Log[64]))/4 + (3*Log[x])/4

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \frac {3+x (2+3 \log (4))}{4 x} \, dx \\ & = \frac {1}{4} \int \frac {3+x (2+3 \log (4))}{x} \, dx \\ & = \frac {1}{4} \int \left (2+\frac {3}{x}+\log (64)\right ) \, dx \\ & = \frac {1}{4} x (2+\log (64))+\frac {3 \log (x)}{4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {3+2 x+3 x \log (4)}{4 x} \, dx=\frac {1}{4} (x (2+\log (64))+3 \log (x)) \]

[In]

Integrate[(3 + 2*x + 3*x*Log[4])/(4*x),x]

[Out]

(x*(2 + Log[64]) + 3*Log[x])/4

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {3 x \ln \left (2\right )}{2}+\frac {x}{2}+\frac {3 \ln \left (x \right )}{4}\)	\(14\)
norman	\(\left (\frac {3 \ln \left (2\right )}{2}+\frac {1}{2}\right ) x +\frac {3 \ln \left (x \right )}{4}\)	\(14\)
risch	\(\frac {3 x \ln \left (2\right )}{2}+\frac {x}{2}+\frac {3 \ln \left (x \right )}{4}\)	\(14\)
parallelrisch	\(\frac {3 x \ln \left (2\right )}{2}+\frac {x}{2}+\frac {3 \ln \left (x \right )}{4}\)	\(14\)

[In]

int(1/4*(6*x*ln(2)+2*x+3)/x,x,method=_RETURNVERBOSE)

[Out]

3/2*x*ln(2)+1/2*x+3/4*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {3+2 x+3 x \log (4)}{4 x} \, dx=\frac {3}{2} \, x \log \left (2\right ) + \frac {1}{2} \, x + \frac {3}{4} \, \log \left (x\right ) \]

[In]

integrate(1/4*(6*x*log(2)+2*x+3)/x,x, algorithm="fricas")

[Out]

3/2*x*log(2) + 1/2*x + 3/4*log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {3+2 x+3 x \log (4)}{4 x} \, dx=\frac {x \left (2 + 6 \log {\left (2 \right )}\right )}{4} + \frac {3 \log {\left (x \right )}}{4} \]

[In]

integrate(1/4*(6*x*ln(2)+2*x+3)/x,x)

[Out]

x*(2 + 6*log(2))/4 + 3*log(x)/4

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {3+2 x+3 x \log (4)}{4 x} \, dx=\frac {1}{2} \, x {\left (3 \, \log \left (2\right ) + 1\right )} + \frac {3}{4} \, \log \left (x\right ) \]

[In]

integrate(1/4*(6*x*log(2)+2*x+3)/x,x, algorithm="maxima")

[Out]

1/2*x*(3*log(2) + 1) + 3/4*log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {3+2 x+3 x \log (4)}{4 x} \, dx=\frac {3}{2} \, x \log \left (2\right ) + \frac {1}{2} \, x + \frac {3}{4} \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/4*(6*x*log(2)+2*x+3)/x,x, algorithm="giac")

[Out]

3/2*x*log(2) + 1/2*x + 3/4*log(abs(x))

Mupad [B] (verification not implemented)

Time = 10.74 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {3+2 x+3 x \log (4)}{4 x} \, dx=\frac {3\,\ln \left (x\right )}{4}+x\,\left (\frac {\ln \left (64\right )}{4}+\frac {1}{2}\right ) \]

[In]

int((x/2 + (3*x*log(2))/2 + 3/4)/x,x)

[Out]

(3*log(x))/4 + x*(log(64)/4 + 1/2)