\(\int \frac {3 x+8 \log (x^2)}{2 x} \, dx\) [2356]

Optimal result

Integrand size = 17, antiderivative size = 13 \[ \int \frac {3 x+8 \log \left (x^2\right )}{2 x} \, dx=1+\frac {3 x}{2}+\log ^2\left (x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {12, 14, 2338} \[ \int \frac {3 x+8 \log \left (x^2\right )}{2 x} \, dx=\log ^2\left (x^2\right )+\frac {3 x}{2} \]

[In]

Int[(3*x + 8*Log[x^2])/(2*x),x]

[Out]

(3*x)/2 + Log[x^2]^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {3 x+8 \log \left (x^2\right )}{x} \, dx \\ & = \frac {1}{2} \int \left (3+\frac {8 \log \left (x^2\right )}{x}\right ) \, dx \\ & = \frac {3 x}{2}+4 \int \frac {\log \left (x^2\right )}{x} \, dx \\ & = \frac {3 x}{2}+\log ^2\left (x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {3 x+8 \log \left (x^2\right )}{2 x} \, dx=\frac {3 x}{2}+\log ^2\left (x^2\right ) \]

[In]

Integrate[(3*x + 8*Log[x^2])/(2*x),x]

[Out]

(3*x)/2 + Log[x^2]^2

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {3 x}{2}+\ln \left (x^{2}\right )^{2}\)	\(11\)
norman	\(\frac {3 x}{2}+\ln \left (x^{2}\right )^{2}\)	\(11\)
risch	\(\frac {3 x}{2}+\ln \left (x^{2}\right )^{2}\)	\(11\)
parts	\(\frac {3 x}{2}+\ln \left (x^{2}\right )^{2}\)	\(11\)

[In]

int(1/2*(8*ln(x^2)+3*x)/x,x,method=_RETURNVERBOSE)

[Out]

3/2*x+ln(x^2)^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {3 x+8 \log \left (x^2\right )}{2 x} \, dx=\log \left (x^{2}\right )^{2} + \frac {3}{2} \, x \]

[In]

integrate(1/2*(8*log(x^2)+3*x)/x,x, algorithm="fricas")

[Out]

log(x^2)^2 + 3/2*x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {3 x+8 \log \left (x^2\right )}{2 x} \, dx=\frac {3 x}{2} + \log {\left (x^{2} \right )}^{2} \]

[In]

integrate(1/2*(8*ln(x**2)+3*x)/x,x)

[Out]

3*x/2 + log(x**2)**2

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {3 x+8 \log \left (x^2\right )}{2 x} \, dx=\log \left (x^{2}\right )^{2} + \frac {3}{2} \, x \]

[In]

integrate(1/2*(8*log(x^2)+3*x)/x,x, algorithm="maxima")

[Out]

log(x^2)^2 + 3/2*x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {3 x+8 \log \left (x^2\right )}{2 x} \, dx=\log \left (x^{2}\right )^{2} + \frac {3}{2} \, x \]

[In]

integrate(1/2*(8*log(x^2)+3*x)/x,x, algorithm="giac")

[Out]

log(x^2)^2 + 3/2*x

Mupad [B] (verification not implemented)

Time = 8.94 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {3 x+8 \log \left (x^2\right )}{2 x} \, dx={\ln \left (x^2\right )}^2+\frac {3\,x}{2} \]

[In]

int(((3*x)/2 + 4*log(x^2))/x,x)

[Out]

(3*x)/2 + log(x^2)^2