\(\int (50-10 e^{1+2 x}) \, dx\) [6769]

Optimal result

Integrand size = 11, antiderivative size = 20 \[ \int \left (50-10 e^{1+2 x}\right ) \, dx=5 \left (1-e^{1+2 x}+5 (-24+2 x)\right ) \]

[Out]

50*x-595-5*exp(1+2*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2225} \[ \int \left (50-10 e^{1+2 x}\right ) \, dx=50 x-5 e^{2 x+1} \]

[In]

Int[50 - 10*E^(1 + 2*x),x]

[Out]

-5*E^(1 + 2*x) + 50*x

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]

Rubi steps \begin{align*} \text {integral}& = 50 x-10 \int e^{1+2 x} \, dx \\ & = -5 e^{1+2 x}+50 x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \left (50-10 e^{1+2 x}\right ) \, dx=-10 \left (\frac {1}{2} e^{1+2 x}-5 x\right ) \]

[In]

Integrate[50 - 10*E^(1 + 2*x),x]

[Out]

-10*(E^(1 + 2*x)/2 - 5*x)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65

method	result	size
default	\(50 x -5 \,{\mathrm e}^{1+2 x}\)	\(13\)
norman	\(50 x -5 \,{\mathrm e}^{1+2 x}\)	\(13\)
risch	\(50 x -5 \,{\mathrm e}^{1+2 x}\)	\(13\)
parallelrisch	\(50 x -5 \,{\mathrm e}^{1+2 x}\)	\(13\)
parts	\(50 x -5 \,{\mathrm e}^{1+2 x}\)	\(13\)
derivativedivides	\(-5 \,{\mathrm e}^{1+2 x}+25 \ln \left ({\mathrm e}^{1+2 x}\right )\)	\(19\)

[In]

int(-10*exp(1+2*x)+50,x,method=_RETURNVERBOSE)

[Out]

50*x-5*exp(1+2*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \left (50-10 e^{1+2 x}\right ) \, dx=50 \, x - 5 \, e^{\left (2 \, x + 1\right )} \]

[In]

integrate(-10*exp(1+2*x)+50,x, algorithm="fricas")

[Out]

50*x - 5*e^(2*x + 1)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.50 \[ \int \left (50-10 e^{1+2 x}\right ) \, dx=50 x - 5 e^{2 x + 1} \]

[In]

integrate(-10*exp(1+2*x)+50,x)

[Out]

50*x - 5*exp(2*x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \left (50-10 e^{1+2 x}\right ) \, dx=50 \, x - 5 \, e^{\left (2 \, x + 1\right )} \]

[In]

integrate(-10*exp(1+2*x)+50,x, algorithm="maxima")

[Out]

50*x - 5*e^(2*x + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \left (50-10 e^{1+2 x}\right ) \, dx=50 \, x - 5 \, e^{\left (2 \, x + 1\right )} \]

[In]

integrate(-10*exp(1+2*x)+50,x, algorithm="giac")

[Out]

50*x - 5*e^(2*x + 1)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \left (50-10 e^{1+2 x}\right ) \, dx=50\,x-5\,{\mathrm {e}}^{2\,x}\,\mathrm {e} \]

[In]

int(50 - 10*exp(2*x + 1),x)

[Out]

50*x - 5*exp(2*x)*exp(1)