\(\int \frac {4-2 e^{50} x-e^{50} \log (48)}{4 e^{50}} \, dx\) [7995]

Optimal result

Integrand size = 22, antiderivative size = 15 \[ \int \frac {4-2 e^{50} x-e^{50} \log (48)}{4 e^{50}} \, dx=\frac {x}{e^{50}}-\frac {1}{4} x (x+\log (48)) \]

[Out]

x/exp(25)^2-1/4*(ln(48)+x)*x

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.60, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {9} \[ \int \frac {4-2 e^{50} x-e^{50} \log (48)}{4 e^{50}} \, dx=-\frac {\left (-2 e^{50} x+4-e^{50} \log (48)\right )^2}{16 e^{100}} \]

[In]

Int[(4 - 2*E^50*x - E^50*Log[48])/(4*E^50),x]

[Out]

-1/16*(4 - 2*E^50*x - E^50*Log[48])^2/E^100

Rule 9

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (4-2 e^{50} x-e^{50} \log (48)\right )^2}{16 e^{100}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int \frac {4-2 e^{50} x-e^{50} \log (48)}{4 e^{50}} \, dx=\frac {x}{e^{50}}-\frac {x^2}{4}-\frac {1}{4} x \log (48) \]

[In]

Integrate[(4 - 2*E^50*x - E^50*Log[48])/(4*E^50),x]

[Out]

x/E^50 - x^2/4 - (x*Log[48])/4

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.40

method	result	size
risch	\(-\frac {x \ln \left (3\right )}{4}-x \ln \left (2\right )-\frac {x^{2}}{4}+x \,{\mathrm e}^{-50}\)	\(21\)
gosper	\(-\frac {x \left ({\mathrm e}^{50} \ln \left (48\right )+x \,{\mathrm e}^{50}-4\right ) {\mathrm e}^{-50}}{4}\)	\(23\)
default	\(\frac {{\mathrm e}^{-50} \left (-x \,{\mathrm e}^{50} \ln \left (48\right )-x^{2} {\mathrm e}^{50}+4 x \right )}{4}\)	\(29\)
parallelrisch	\(\frac {{\mathrm e}^{-50} \left (-x^{2} {\mathrm e}^{50}+\left (-{\mathrm e}^{50} \ln \left (48\right )+4\right ) x \right )}{4}\)	\(29\)
norman	\(\left (-\frac {x^{2} {\mathrm e}^{25}}{4}-\frac {{\mathrm e}^{-25} \left ({\mathrm e}^{50} \ln \left (48\right )-4\right ) x}{4}\right ) {\mathrm e}^{-25}\)	\(30\)

[In]

int(1/4*(-exp(25)^2*ln(48)-2*x*exp(25)^2+4)/exp(25)^2,x,method=_RETURNVERBOSE)

[Out]

-1/4*x*ln(3)-x*ln(2)-1/4*x^2+x*exp(-50)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int \frac {4-2 e^{50} x-e^{50} \log (48)}{4 e^{50}} \, dx=-\frac {1}{4} \, {\left (x^{2} e^{50} + x e^{50} \log \left (48\right ) - 4 \, x\right )} e^{\left (-50\right )} \]

[In]

integrate(1/4*(-exp(25)^2*log(48)-2*x*exp(25)^2+4)/exp(25)^2,x, algorithm="fricas")

[Out]

-1/4*(x^2*e^50 + x*e^50*log(48) - 4*x)*e^(-50)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {4-2 e^{50} x-e^{50} \log (48)}{4 e^{50}} \, dx=- \frac {x^{2}}{4} + \frac {x \left (- e^{50} \log {\left (48 \right )} + 4\right )}{4 e^{50}} \]

[In]

integrate(1/4*(-exp(25)**2*ln(48)-2*x*exp(25)**2+4)/exp(25)**2,x)

[Out]

-x**2/4 + x*(-exp(50)*log(48) + 4)*exp(-50)/4

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int \frac {4-2 e^{50} x-e^{50} \log (48)}{4 e^{50}} \, dx=-\frac {1}{4} \, {\left (x^{2} e^{50} + x e^{50} \log \left (48\right ) - 4 \, x\right )} e^{\left (-50\right )} \]

[In]

integrate(1/4*(-exp(25)^2*log(48)-2*x*exp(25)^2+4)/exp(25)^2,x, algorithm="maxima")

[Out]

-1/4*(x^2*e^50 + x*e^50*log(48) - 4*x)*e^(-50)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int \frac {4-2 e^{50} x-e^{50} \log (48)}{4 e^{50}} \, dx=-\frac {1}{4} \, {\left (x^{2} e^{50} + x e^{50} \log \left (48\right ) - 4 \, x\right )} e^{\left (-50\right )} \]

[In]

integrate(1/4*(-exp(25)^2*log(48)-2*x*exp(25)^2+4)/exp(25)^2,x, algorithm="giac")

[Out]

-1/4*(x^2*e^50 + x*e^50*log(48) - 4*x)*e^(-50)

Mupad [B] (verification not implemented)

Time = 13.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int \frac {4-2 e^{50} x-e^{50} \log (48)}{4 e^{50}} \, dx=-{\mathrm {e}}^{-100}\,{\left (\frac {{\mathrm {e}}^{50}\,\ln \left (48\right )}{4}+\frac {x\,{\mathrm {e}}^{50}}{2}-1\right )}^2 \]

[In]

int(-exp(-50)*((exp(50)*log(48))/4 + (x*exp(50))/2 - 1),x)

[Out]

-exp(-100)*((exp(50)*log(48))/4 + (x*exp(50))/2 - 1)^2