\(\int \frac {64+16 x+16 x \log (x)}{x} \, dx\) [6686]

Optimal result

Integrand size = 14, antiderivative size = 7 \[ \int \frac {64+16 x+16 x \log (x)}{x} \, dx=16 (4+x) \log (x) \]

[Out]

16*(4+x)*ln(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.43, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {14, 45, 2332} \[ \int \frac {64+16 x+16 x \log (x)}{x} \, dx=16 x \log (x)+64 \log (x) \]

[In]

Int[(64 + 16*x + 16*x*Log[x])/x,x]

[Out]

64*Log[x] + 16*x*Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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}& = \int \left (\frac {16 (4+x)}{x}+16 \log (x)\right ) \, dx \\ & = 16 \int \frac {4+x}{x} \, dx+16 \int \log (x) \, dx \\ & = -16 x+16 x \log (x)+16 \int \left (1+\frac {4}{x}\right ) \, dx \\ & = 64 \log (x)+16 x \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.43 \[ \int \frac {64+16 x+16 x \log (x)}{x} \, dx=64 \log (x)+16 x \log (x) \]

[In]

Integrate[(64 + 16*x + 16*x*Log[x])/x,x]

[Out]

64*Log[x] + 16*x*Log[x]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.57

method	result	size
default	\(16 x \ln \left (x \right )+64 \ln \left (x \right )\)	\(11\)
norman	\(16 x \ln \left (x \right )+64 \ln \left (x \right )\)	\(11\)
risch	\(16 x \ln \left (x \right )+64 \ln \left (x \right )\)	\(11\)
parallelrisch	\(16 x \ln \left (x \right )+64 \ln \left (x \right )\)	\(11\)
parts	\(16 x \ln \left (x \right )+64 \ln \left (x \right )\)	\(11\)

[In]

int((16*x*ln(x)+16*x+64)/x,x,method=_RETURNVERBOSE)

[Out]

16*x*ln(x)+64*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00 \[ \int \frac {64+16 x+16 x \log (x)}{x} \, dx=16 \, {\left (x + 4\right )} \log \left (x\right ) \]

[In]

integrate((16*x*log(x)+16*x+64)/x,x, algorithm="fricas")

[Out]

16*(x + 4)*log(x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.43 \[ \int \frac {64+16 x+16 x \log (x)}{x} \, dx=16 x \log {\left (x \right )} + 64 \log {\left (x \right )} \]

[In]

integrate((16*x*ln(x)+16*x+64)/x,x)

[Out]

16*x*log(x) + 64*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.43 \[ \int \frac {64+16 x+16 x \log (x)}{x} \, dx=16 \, x \log \left (x\right ) + 64 \, \log \left (x\right ) \]

[In]

integrate((16*x*log(x)+16*x+64)/x,x, algorithm="maxima")

[Out]

16*x*log(x) + 64*log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.43 \[ \int \frac {64+16 x+16 x \log (x)}{x} \, dx=16 \, x \log \left (x\right ) + 64 \, \log \left (x\right ) \]

[In]

integrate((16*x*log(x)+16*x+64)/x,x, algorithm="giac")

[Out]

16*x*log(x) + 64*log(x)

Mupad [B] (verification not implemented)

Time = 12.93 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00 \[ \int \frac {64+16 x+16 x \log (x)}{x} \, dx=16\,\ln \left (x\right )\,\left (x+4\right ) \]

[In]

int((16*x + 16*x*log(x) + 64)/x,x)

[Out]

16*log(x)*(x + 4)