\(\int \frac {\frac {150 x}{e^2}-\frac {250 x^2}{e^4}}{x} \, dx\) [9792]

Optimal result

Integrand size = 19, antiderivative size = 16 \[ \int \frac {\frac {150 x}{e^2}-\frac {250 x^2}{e^4}}{x} \, dx=5 \left (9-\left (-3+\frac {5 x}{e^2}\right )^2\right ) \]

[Out]

45-5*(exp(ln(x)+ln(5)-2)-3)^2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {14} \[ \int \frac {\frac {150 x}{e^2}-\frac {250 x^2}{e^4}}{x} \, dx=\frac {150 x}{e^2}-\frac {125 x^2}{e^4} \]

[In]

Int[((150*x)/E^2 - (250*x^2)/E^4)/x,x]

[Out]

(150*x)/E^2 - (125*x^2)/E^4

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}& = \int \left (\frac {150}{e^2}-\frac {250 x}{e^4}\right ) \, dx \\ & = \frac {150 x}{e^2}-\frac {125 x^2}{e^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {\frac {150 x}{e^2}-\frac {250 x^2}{e^4}}{x} \, dx=\frac {50 \left (3 e^2 x-\frac {5 x^2}{2}\right )}{e^4} \]

[In]

Integrate[((150*x)/E^2 - (250*x^2)/E^4)/x,x]

[Out]

(50*(3*E^2*x - (5*x^2)/2))/E^4

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00

method	result	size
risch	\(-125 \,{\mathrm e}^{-4} x^{2}+150 x \,{\mathrm e}^{-2}\)	\(16\)
norman	\(\left (150 x -125 \,{\mathrm e}^{-2} x^{2}\right ) {\mathrm e}^{-2}\)	\(19\)
derivativedivides	\(-125 \,{\mathrm e}^{-4} x^{2}+30 \,{\mathrm e}^{\ln \left (x \right )+\ln \left (5\right )-2}\)	\(22\)
default	\(-125 \,{\mathrm e}^{-4} x^{2}+30 \,{\mathrm e}^{\ln \left (x \right )+\ln \left (5\right )-2}\)	\(22\)
parts	\(-125 \,{\mathrm e}^{-4} x^{2}+30 \,{\mathrm e}^{\ln \left (x \right )+\ln \left (5\right )-2}\)	\(22\)
parallelrisch	\(\frac {-125 \,{\mathrm e}^{-4} x^{4}+30 \,{\mathrm e}^{\ln \left (x \right )+\ln \left (5\right )-2} x^{2}}{x^{2}}\)	\(32\)

[In]

int((-10*exp(ln(x)+ln(5)-2)^2+30*exp(ln(x)+ln(5)-2))/x,x,method=_RETURNVERBOSE)

[Out]

-125*exp(-2)^2*x^2+150*x*exp(-2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {\frac {150 x}{e^2}-\frac {250 x^2}{e^4}}{x} \, dx=-5 \, x^{2} e^{\left (2 \, \log \left (5\right ) - 4\right )} + 30 \, x e^{\left (\log \left (5\right ) - 2\right )} \]

[In]

integrate((-10*exp(log(x)+log(5)-2)^2+30*exp(log(x)+log(5)-2))/x,x, algorithm="fricas")

[Out]

-5*x^2*e^(2*log(5) - 4) + 30*x*e^(log(5) - 2)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {\frac {150 x}{e^2}-\frac {250 x^2}{e^4}}{x} \, dx=- \frac {125 x^{2}}{e^{4}} + \frac {150 x}{e^{2}} \]

[In]

integrate((-10*exp(ln(x)+ln(5)-2)**2+30*exp(ln(x)+ln(5)-2))/x,x)

[Out]

-125*x**2*exp(-4) + 150*x*exp(-2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {\frac {150 x}{e^2}-\frac {250 x^2}{e^4}}{x} \, dx=-25 \, {\left (5 \, x^{2} - 6 \, x e^{2}\right )} e^{\left (-4\right )} \]

[In]

integrate((-10*exp(log(x)+log(5)-2)^2+30*exp(log(x)+log(5)-2))/x,x, algorithm="maxima")

[Out]

-25*(5*x^2 - 6*x*e^2)*e^(-4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {\frac {150 x}{e^2}-\frac {250 x^2}{e^4}}{x} \, dx=-25 \, {\left (5 \, x^{2} - 6 \, x e^{2}\right )} e^{\left (-4\right )} \]

[In]

integrate((-10*exp(log(x)+log(5)-2)^2+30*exp(log(x)+log(5)-2))/x,x, algorithm="giac")

[Out]

-25*(5*x^2 - 6*x*e^2)*e^(-4)

Mupad [B] (verification not implemented)

Time = 16.44 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {\frac {150 x}{e^2}-\frac {250 x^2}{e^4}}{x} \, dx=-25\,x\,{\mathrm {e}}^{-4}\,\left (5\,x-6\,{\mathrm {e}}^2\right ) \]

[In]

int((30*exp(log(5) + log(x) - 2) - 10*exp(2*log(5) + 2*log(x) - 4))/x,x)

[Out]

-25*x*exp(-4)*(5*x - 6*exp(2))