\(\int \frac {8 x+(5+48 x) \log (4)}{5 x \log (4)} \, dx\) [2673]

Optimal result

Integrand size = 23, antiderivative size = 20 \[ \int \frac {8 x+(5+48 x) \log (4)}{5 x \log (4)} \, dx=-1+8 \left (x+\frac {1}{5} \left (x+\frac {x}{\log (4)}\right )\right )+\log (x) \]

[Out]

ln(x)-1+48/5*x+4/5*x/ln(2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {12, 192, 45} \[ \int \frac {8 x+(5+48 x) \log (4)}{5 x \log (4)} \, dx=\frac {8 x (1+6 \log (4))}{5 \log (4)}+\log (x) \]

[In]

Int[(8*x + (5 + 48*x)*Log[4])/(5*x*Log[4]),x]

[Out]

(8*x*(1 + 6*Log[4]))/(5*Log[4]) + Log[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])

Rule 192

Int[(u_)^(m_.)*(v_)^(n_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^n, x] /; FreeQ[{m, n}, x] &&
LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {8 x+(5+48 x) \log (4)}{x} \, dx}{5 \log (4)} \\ & = \frac {\int \frac {5 \log (4)+8 x (1+6 \log (4))}{x} \, dx}{5 \log (4)} \\ & = \frac {\int \left (\frac {5 \log (4)}{x}+8 (1+6 \log (4))\right ) \, dx}{5 \log (4)} \\ & = \frac {8 x (1+6 \log (4))}{5 \log (4)}+\log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {8 x+(5+48 x) \log (4)}{5 x \log (4)} \, dx=\frac {8}{5} x \left (6+\frac {1}{\log (4)}\right )+\log (x) \]

[In]

Integrate[(8*x + (5 + 48*x)*Log[4])/(5*x*Log[4]),x]

[Out]

(8*x*(6 + Log[4]^(-1)))/5 + Log[x]

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70

method	result	size
risch	\(\frac {48 x}{5}+\frac {4 x}{5 \ln \left (2\right )}+\ln \left (x \right )\)	\(14\)
norman	\(\frac {4 \left (12 \ln \left (2\right )+1\right ) x}{5 \ln \left (2\right )}+\ln \left (x \right )\)	\(17\)
default	\(\frac {48 x \ln \left (2\right )+4 x +5 \ln \left (2\right ) \ln \left (x \right )}{5 \ln \left (2\right )}\)	\(22\)
parallelrisch	\(\frac {10 \ln \left (2\right ) \ln \left (x \right )+96 x \ln \left (2\right )+8 x}{10 \ln \left (2\right )}\)	\(22\)

[In]

int(1/10*(2*(48*x+5)*ln(2)+8*x)/x/ln(2),x,method=_RETURNVERBOSE)

[Out]

48/5*x+4/5*x/ln(2)+ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {8 x+(5+48 x) \log (4)}{5 x \log (4)} \, dx=\frac {48 \, x \log \left (2\right ) + 5 \, \log \left (2\right ) \log \left (x\right ) + 4 \, x}{5 \, \log \left (2\right )} \]

[In]

integrate(1/10*(2*(48*x+5)*log(2)+8*x)/x/log(2),x, algorithm="fricas")

[Out]

1/5*(48*x*log(2) + 5*log(2)*log(x) + 4*x)/log(2)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {8 x+(5+48 x) \log (4)}{5 x \log (4)} \, dx=\frac {x \left (4 + 48 \log {\left (2 \right )}\right ) + 5 \log {\left (2 \right )} \log {\left (x \right )}}{5 \log {\left (2 \right )}} \]

[In]

integrate(1/10*(2*(48*x+5)*ln(2)+8*x)/x/ln(2),x)

[Out]

(x*(4 + 48*log(2)) + 5*log(2)*log(x))/(5*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {8 x+(5+48 x) \log (4)}{5 x \log (4)} \, dx=\frac {4 \, x {\left (12 \, \log \left (2\right ) + 1\right )} + 5 \, \log \left (2\right ) \log \left (x\right )}{5 \, \log \left (2\right )} \]

[In]

integrate(1/10*(2*(48*x+5)*log(2)+8*x)/x/log(2),x, algorithm="maxima")

[Out]

1/5*(4*x*(12*log(2) + 1) + 5*log(2)*log(x))/log(2)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {8 x+(5+48 x) \log (4)}{5 x \log (4)} \, dx=\frac {48 \, x \log \left (2\right ) + 5 \, \log \left (2\right ) \log \left ({\left | x \right |}\right ) + 4 \, x}{5 \, \log \left (2\right )} \]

[In]

integrate(1/10*(2*(48*x+5)*log(2)+8*x)/x/log(2),x, algorithm="giac")

[Out]

1/5*(48*x*log(2) + 5*log(2)*log(abs(x)) + 4*x)/log(2)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {8 x+(5+48 x) \log (4)}{5 x \log (4)} \, dx=\ln \left (x\right )+\frac {x\,\left (48\,\ln \left (2\right )+4\right )}{5\,\ln \left (2\right )} \]

[In]

int(((4*x)/5 + (log(2)*(48*x + 5))/5)/(x*log(2)),x)

[Out]

log(x) + (x*(48*log(2) + 4))/(5*log(2))