\(\int \frac {-10000 e^{2 x} x+\log (2 x)}{5000 x} \, dx\) [6516]

Optimal result

Integrand size = 20, antiderivative size = 19 \[ \int \frac {-10000 e^{2 x} x+\log (2 x)}{5000 x} \, dx=2-e^{2 x}+\frac {\log ^2(2 x)}{10000} \]

[Out]

2-exp(x)^2+1/10000*ln(2*x)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 14, 2225, 2338} \[ \int \frac {-10000 e^{2 x} x+\log (2 x)}{5000 x} \, dx=\frac {\log ^2(2 x)}{10000}-e^{2 x} \]

[In]

Int[(-10000*E^(2*x)*x + Log[2*x])/(5000*x),x]

[Out]

-E^(2*x) + Log[2*x]^2/10000

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 2225

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

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 {\int \frac {-10000 e^{2 x} x+\log (2 x)}{x} \, dx}{5000} \\ & = \frac {\int \left (-10000 e^{2 x}+\frac {\log (2 x)}{x}\right ) \, dx}{5000} \\ & = \frac {\int \frac {\log (2 x)}{x} \, dx}{5000}-2 \int e^{2 x} \, dx \\ & = -e^{2 x}+\frac {\log ^2(2 x)}{10000} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-10000 e^{2 x} x+\log (2 x)}{5000 x} \, dx=-e^{2 x}+\frac {\log ^2(2 x)}{10000} \]

[In]

Integrate[(-10000*E^(2*x)*x + Log[2*x])/(5000*x),x]

[Out]

-E^(2*x) + Log[2*x]^2/10000

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {\ln \left (2 x \right )^{2}}{10000}-{\mathrm e}^{2 x}\)	\(16\)
norman	\(\frac {\ln \left (2 x \right )^{2}}{10000}-{\mathrm e}^{2 x}\)	\(16\)
risch	\(\frac {\ln \left (2 x \right )^{2}}{10000}-{\mathrm e}^{2 x}\)	\(16\)
parallelrisch	\(\frac {\ln \left (2 x \right )^{2}}{10000}-{\mathrm e}^{2 x}\)	\(16\)
parts	\(\frac {\ln \left (2 x \right )^{2}}{10000}-{\mathrm e}^{2 x}\)	\(16\)

[In]

int(1/5000*(ln(2*x)-10000*x*exp(x)^2)/x,x,method=_RETURNVERBOSE)

[Out]

1/10000*ln(2*x)^2-exp(x)^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-10000 e^{2 x} x+\log (2 x)}{5000 x} \, dx=\frac {1}{10000} \, \log \left (2 \, x\right )^{2} - e^{\left (2 \, x\right )} \]

[In]

integrate(1/5000*(log(2*x)-10000*x*exp(x)^2)/x,x, algorithm="fricas")

[Out]

1/10000*log(2*x)^2 - e^(2*x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {-10000 e^{2 x} x+\log (2 x)}{5000 x} \, dx=- e^{2 x} + \frac {\log {\left (2 x \right )}^{2}}{10000} \]

[In]

integrate(1/5000*(ln(2*x)-10000*x*exp(x)**2)/x,x)

[Out]

-exp(2*x) + log(2*x)**2/10000

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-10000 e^{2 x} x+\log (2 x)}{5000 x} \, dx=\frac {1}{10000} \, \log \left (2 \, x\right )^{2} - e^{\left (2 \, x\right )} \]

[In]

integrate(1/5000*(log(2*x)-10000*x*exp(x)^2)/x,x, algorithm="maxima")

[Out]

1/10000*log(2*x)^2 - e^(2*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {-10000 e^{2 x} x+\log (2 x)}{5000 x} \, dx=\frac {1}{10000} \, \log \left (2 \, x\right )^{2} - e^{\left (2 \, x\right )} \]

[In]

integrate(1/5000*(log(2*x)-10000*x*exp(x)^2)/x,x, algorithm="giac")

[Out]

1/10000*log(2*x)^2 - e^(2*x)

Mupad [B] (verification not implemented)

Time = 12.42 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-10000 e^{2 x} x+\log (2 x)}{5000 x} \, dx=\frac {{\ln \left (x\right )}^2}{10000}+\frac {\ln \left (2\right )\,\ln \left (x\right )}{5000}-{\mathrm {e}}^{2\,x} \]

[In]

int((log(2*x)/5000 - 2*x*exp(2*x))/x,x)

[Out]

log(x)^2/10000 - exp(2*x) + (log(2)*log(x))/5000