\(\int \frac {-16-5 \log (5) \log (32)}{5 x^2 \log (5)} \, dx\) [568]

Optimal result

Integrand size = 19, antiderivative size = 19 \[ \int \frac {-16-5 \log (5) \log (32)}{5 x^2 \log (5)} \, dx=5+\frac {16}{5 x \log (5)}+\frac {\log (32)}{x} \]

[Out]

5+5*ln(2)/x+16/5/x/ln(5)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 30} \[ \int \frac {-16-5 \log (5) \log (32)}{5 x^2 \log (5)} \, dx=\frac {16+5 \log (5) \log (32)}{5 x \log (5)} \]

[In]

Int[(-16 - 5*Log[5]*Log[32])/(5*x^2*Log[5]),x]

[Out]

(16 + 5*Log[5]*Log[32])/(5*x*Log[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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {(-16-5 \log (5) \log (32)) \int \frac {1}{x^2} \, dx}{5 \log (5)} \\ & = \frac {16+5 \log (5) \log (32)}{5 x \log (5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-16-5 \log (5) \log (32)}{5 x^2 \log (5)} \, dx=\frac {16+5 \log (5) \log (32)}{5 x \log (5)} \]

[In]

Integrate[(-16 - 5*Log[5]*Log[32])/(5*x^2*Log[5]),x]

[Out]

(16 + 5*Log[5]*Log[32])/(5*x*Log[5])

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
gosper	\(\frac {25 \ln \left (2\right ) \ln \left (5\right )+16}{5 x \ln \left (5\right )}\)	\(18\)
default	\(-\frac {-25 \ln \left (2\right ) \ln \left (5\right )-16}{5 x \ln \left (5\right )}\)	\(18\)
norman	\(\frac {25 \ln \left (2\right ) \ln \left (5\right )+16}{5 x \ln \left (5\right )}\)	\(18\)
risch	\(\frac {5 \ln \left (2\right )}{x}+\frac {16}{5 x \ln \left (5\right )}\)	\(18\)
parallelrisch	\(-\frac {-25 \ln \left (2\right ) \ln \left (5\right )-16}{5 x \ln \left (5\right )}\)	\(18\)

[In]

int(1/5*(-25*ln(2)*ln(5)-16)/x^2/ln(5),x,method=_RETURNVERBOSE)

[Out]

1/5*(25*ln(2)*ln(5)+16)/x/ln(5)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16-5 \log (5) \log (32)}{5 x^2 \log (5)} \, dx=\frac {25 \, \log \left (5\right ) \log \left (2\right ) + 16}{5 \, x \log \left (5\right )} \]

[In]

integrate(1/5*(-25*log(2)*log(5)-16)/x^2/log(5),x, algorithm="fricas")

[Out]

1/5*(25*log(5)*log(2) + 16)/(x*log(5))

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-16-5 \log (5) \log (32)}{5 x^2 \log (5)} \, dx=- \frac {- 25 \log {\left (2 \right )} \log {\left (5 \right )} - 16}{5 x \log {\left (5 \right )}} \]

[In]

integrate(1/5*(-25*ln(2)*ln(5)-16)/x**2/ln(5),x)

[Out]

-(-25*log(2)*log(5) - 16)/(5*x*log(5))

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16-5 \log (5) \log (32)}{5 x^2 \log (5)} \, dx=\frac {25 \, \log \left (5\right ) \log \left (2\right ) + 16}{5 \, x \log \left (5\right )} \]

[In]

integrate(1/5*(-25*log(2)*log(5)-16)/x^2/log(5),x, algorithm="maxima")

[Out]

1/5*(25*log(5)*log(2) + 16)/(x*log(5))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-16-5 \log (5) \log (32)}{5 x^2 \log (5)} \, dx=\frac {25 \, \log \left (5\right ) \log \left (2\right ) + 16}{5 \, x \log \left (5\right )} \]

[In]

integrate(1/5*(-25*log(2)*log(5)-16)/x^2/log(5),x, algorithm="giac")

[Out]

1/5*(25*log(5)*log(2) + 16)/(x*log(5))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {-16-5 \log (5) \log (32)}{5 x^2 \log (5)} \, dx=\frac {5\,\ln \left (2\right )\,\ln \left (5\right )+\frac {16}{5}}{x\,\ln \left (5\right )} \]

[In]

int(-(5*log(2)*log(5) + 16/5)/(x^2*log(5)),x)

[Out]

(5*log(2)*log(5) + 16/5)/(x*log(5))