\(\int \frac {800+e^5 (20-200 x)+1000 x+e^{10} (-10+12 x)}{e^{10}} \, dx\) [8959]

Optimal result

Integrand size = 27, antiderivative size = 23 \[ \int \frac {800+e^5 (20-200 x)+1000 x+e^{10} (-10+12 x)}{e^{10}} \, dx=x^2+5 \left (1-x+\frac {10 \left (\frac {4}{5}+x\right )}{e^5}\right )^2 \]

[Out]

x^2+5*(1+10*(x+4/5)/exp(5)-x)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {12} \[ \int \frac {800+e^5 (20-200 x)+1000 x+e^{10} (-10+12 x)}{e^{10}} \, dx=\frac {500 x^2}{e^{10}}-\frac {(1-10 x)^2}{e^5}+\frac {1}{6} (5-6 x)^2+\frac {800 x}{e^{10}} \]

[In]

Int[(800 + E^5*(20 - 200*x) + 1000*x + E^10*(-10 + 12*x))/E^10,x]

[Out]

-((1 - 10*x)^2/E^5) + (5 - 6*x)^2/6 + (800*x)/E^10 + (500*x^2)/E^10

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (800+e^5 (20-200 x)+1000 x+e^{10} (-10+12 x)\right ) \, dx}{e^{10}} \\ & = -\frac {(1-10 x)^2}{e^5}+\frac {1}{6} (5-6 x)^2+\frac {800 x}{e^{10}}+\frac {500 x^2}{e^{10}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {800+e^5 (20-200 x)+1000 x+e^{10} (-10+12 x)}{e^{10}} \, dx=2 \left (-5 x+\frac {400 x}{e^{10}}+\frac {10 x}{e^5}+3 x^2+\frac {250 x^2}{e^{10}}-\frac {50 x^2}{e^5}\right ) \]

[In]

Integrate[(800 + E^5*(20 - 200*x) + 1000*x + E^10*(-10 + 12*x))/E^10,x]

[Out]

2*(-5*x + (400*x)/E^10 + (10*x)/E^5 + 3*x^2 + (250*x^2)/E^10 - (50*x^2)/E^5)

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48

method	result	size
risch	\(6 x^{2}-10 x -100 x^{2} {\mathrm e}^{-5}+20 x \,{\mathrm e}^{-5}+500 \,{\mathrm e}^{-10} x^{2}+800 \,{\mathrm e}^{-10} x\)	\(34\)
gosper	\(2 x \left (3 x \,{\mathrm e}^{10}-5 \,{\mathrm e}^{10}-50 x \,{\mathrm e}^{5}+10 \,{\mathrm e}^{5}+250 x +400\right ) {\mathrm e}^{-10}\)	\(35\)
default	\({\mathrm e}^{-10} \left (6 x^{2} {\mathrm e}^{10}-10 x \,{\mathrm e}^{10}-100 x^{2} {\mathrm e}^{5}+20 x \,{\mathrm e}^{5}+500 x^{2}+800 x \right )\)	\(43\)
parallelrisch	\({\mathrm e}^{-10} \left (6 x^{2} {\mathrm e}^{10}-10 x \,{\mathrm e}^{10}-100 x^{2} {\mathrm e}^{5}+20 x \,{\mathrm e}^{5}+500 x^{2}+800 x \right )\)	\(43\)
norman	\(\left (-10 \left ({\mathrm e}^{10}-2 \,{\mathrm e}^{5}-80\right ) {\mathrm e}^{-5} x +2 \left (3 \,{\mathrm e}^{10}-50 \,{\mathrm e}^{5}+250\right ) {\mathrm e}^{-5} x^{2}\right ) {\mathrm e}^{-5}\)	\(45\)

[In]

int(((12*x-10)*exp(5)^2+(-200*x+20)*exp(5)+1000*x+800)/exp(5)^2,x,method=_RETURNVERBOSE)

[Out]

6*x^2-10*x-100*x^2*exp(-5)+20*x*exp(-5)+500*exp(-10)*x^2+800*exp(-10)*x

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {800+e^5 (20-200 x)+1000 x+e^{10} (-10+12 x)}{e^{10}} \, dx=2 \, {\left (250 \, x^{2} + {\left (3 \, x^{2} - 5 \, x\right )} e^{10} - 10 \, {\left (5 \, x^{2} - x\right )} e^{5} + 400 \, x\right )} e^{\left (-10\right )} \]

[In]

integrate(((12*x-10)*exp(5)^2+(-200*x+20)*exp(5)+1000*x+800)/exp(5)^2,x, algorithm="fricas")

[Out]

2*(250*x^2 + (3*x^2 - 5*x)*e^10 - 10*(5*x^2 - x)*e^5 + 400*x)*e^(-10)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {800+e^5 (20-200 x)+1000 x+e^{10} (-10+12 x)}{e^{10}} \, dx=\frac {x^{2} \left (- 100 e^{5} + 500 + 6 e^{10}\right )}{e^{10}} + \frac {x \left (- 10 e^{10} + 800 + 20 e^{5}\right )}{e^{10}} \]

[In]

integrate(((12*x-10)*exp(5)**2+(-200*x+20)*exp(5)+1000*x+800)/exp(5)**2,x)

[Out]

x**2*(-100*exp(5) + 500 + 6*exp(10))*exp(-10) + x*(-10*exp(10) + 800 + 20*exp(5))*exp(-10)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {800+e^5 (20-200 x)+1000 x+e^{10} (-10+12 x)}{e^{10}} \, dx=2 \, {\left (250 \, x^{2} + {\left (3 \, x^{2} - 5 \, x\right )} e^{10} - 10 \, {\left (5 \, x^{2} - x\right )} e^{5} + 400 \, x\right )} e^{\left (-10\right )} \]

[In]

integrate(((12*x-10)*exp(5)^2+(-200*x+20)*exp(5)+1000*x+800)/exp(5)^2,x, algorithm="maxima")

[Out]

2*(250*x^2 + (3*x^2 - 5*x)*e^10 - 10*(5*x^2 - x)*e^5 + 400*x)*e^(-10)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {800+e^5 (20-200 x)+1000 x+e^{10} (-10+12 x)}{e^{10}} \, dx=2 \, {\left (250 \, x^{2} + {\left (3 \, x^{2} - 5 \, x\right )} e^{10} - 10 \, {\left (5 \, x^{2} - x\right )} e^{5} + 400 \, x\right )} e^{\left (-10\right )} \]

[In]

integrate(((12*x-10)*exp(5)^2+(-200*x+20)*exp(5)+1000*x+800)/exp(5)^2,x, algorithm="giac")

[Out]

2*(250*x^2 + (3*x^2 - 5*x)*e^10 - 10*(5*x^2 - x)*e^5 + 400*x)*e^(-10)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {800+e^5 (20-200 x)+1000 x+e^{10} (-10+12 x)}{e^{10}} \, dx=\frac {{\mathrm {e}}^{-10}\,\left (12\,{\mathrm {e}}^{10}-200\,{\mathrm {e}}^5+1000\right )\,x^2}{2}+{\mathrm {e}}^{-10}\,\left (20\,{\mathrm {e}}^5-10\,{\mathrm {e}}^{10}+800\right )\,x \]

[In]

int(exp(-10)*(1000*x + exp(10)*(12*x - 10) - exp(5)*(200*x - 20) + 800),x)

[Out]

x*exp(-10)*(20*exp(5) - 10*exp(10) + 800) + (x^2*exp(-10)*(12*exp(10) - 200*exp(5) + 1000))/2