\(\int \frac {1-6 x^2 \log (5)}{5 x} \, dx\) [777]

Optimal result

Integrand size = 16, antiderivative size = 15 \[ \int \frac {1-6 x^2 \log (5)}{5 x} \, dx=\frac {1}{5} \left (3-3 x^2 \log (5)+\log (x)\right ) \]

[Out]

3/5-3/5*x^2*ln(5)+1/5*ln(x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 14} \[ \int \frac {1-6 x^2 \log (5)}{5 x} \, dx=\frac {\log (x)}{5}-\frac {3}{5} x^2 \log (5) \]

[In]

Int[(1 - 6*x^2*Log[5])/(5*x),x]

[Out]

(-3*x^2*Log[5])/5 + Log[x]/5

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]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {1-6 x^2 \log (5)}{x} \, dx \\ & = \frac {1}{5} \int \left (\frac {1}{x}-6 x \log (5)\right ) \, dx \\ & = -\frac {3}{5} x^2 \log (5)+\frac {\log (x)}{5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {1-6 x^2 \log (5)}{5 x} \, dx=-\frac {3}{5} x^2 \log (5)+\frac {\log (x)}{5} \]

[In]

Integrate[(1 - 6*x^2*Log[5])/(5*x),x]

[Out]

(-3*x^2*Log[5])/5 + Log[x]/5

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {3 x^{2} \ln \left (5\right )}{5}+\frac {\ln \left (x \right )}{5}\)	\(13\)
norman	\(-\frac {3 x^{2} \ln \left (5\right )}{5}+\frac {\ln \left (x \right )}{5}\)	\(13\)
risch	\(-\frac {3 x^{2} \ln \left (5\right )}{5}+\frac {\ln \left (x \right )}{5}\)	\(13\)
parallelrisch	\(-\frac {3 x^{2} \ln \left (5\right )}{5}+\frac {\ln \left (x \right )}{5}\)	\(13\)

[In]

int(1/5*(-6*x^2*ln(5)+1)/x,x,method=_RETURNVERBOSE)

[Out]

-3/5*x^2*ln(5)+1/5*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {1-6 x^2 \log (5)}{5 x} \, dx=-\frac {3}{5} \, x^{2} \log \left (5\right ) + \frac {1}{5} \, \log \left (x\right ) \]

[In]

integrate(1/5*(-6*x^2*log(5)+1)/x,x, algorithm="fricas")

[Out]

-3/5*x^2*log(5) + 1/5*log(x)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {1-6 x^2 \log (5)}{5 x} \, dx=- \frac {3 x^{2} \log {\left (5 \right )}}{5} + \frac {\log {\left (x \right )}}{5} \]

[In]

integrate(1/5*(-6*x**2*ln(5)+1)/x,x)

[Out]

-3*x**2*log(5)/5 + log(x)/5

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {1-6 x^2 \log (5)}{5 x} \, dx=-\frac {3}{5} \, x^{2} \log \left (5\right ) + \frac {1}{10} \, \log \left (x^{2}\right ) \]

[In]

integrate(1/5*(-6*x^2*log(5)+1)/x,x, algorithm="maxima")

[Out]

-3/5*x^2*log(5) + 1/10*log(x^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {1-6 x^2 \log (5)}{5 x} \, dx=-\frac {3}{5} \, x^{2} \log \left (5\right ) + \frac {1}{10} \, \log \left (x^{2}\right ) \]

[In]

integrate(1/5*(-6*x^2*log(5)+1)/x,x, algorithm="giac")

[Out]

-3/5*x^2*log(5) + 1/10*log(x^2)

Mupad [B] (verification not implemented)

Time = 8.15 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {1-6 x^2 \log (5)}{5 x} \, dx=\frac {\ln \left (x\right )}{5}-\frac {3\,x^2\,\ln \left (5\right )}{5} \]

[In]

int(-((6*x^2*log(5))/5 - 1/5)/x,x)

[Out]

log(x)/5 - (3*x^2*log(5))/5