\(\int \frac {5+e^2 (5-100 x-90 x^2-20 x^3)+5 e^2 \log (4)}{e^2} \, dx\) [9863]

Optimal result

Integrand size = 32, antiderivative size = 29 \[ \int \frac {5+e^2 \left (5-100 x-90 x^2-20 x^3\right )+5 e^2 \log (4)}{e^2} \, dx=5 \left (6+x+\frac {x}{e^2}-x^2 (3+x)^2-x (x-\log (4))\right ) \]

[Out]

30+5*x-5*x*(x-2*ln(2))-5*x^2*(3+x)^2+5*x/exp(2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {12} \[ \int \frac {5+e^2 \left (5-100 x-90 x^2-20 x^3\right )+5 e^2 \log (4)}{e^2} \, dx=-5 x^4-30 x^3-50 x^2+5 x+5 x \left (\frac {1}{e^2}+\log (4)\right ) \]

[In]

Int[(5 + E^2*(5 - 100*x - 90*x^2 - 20*x^3) + 5*E^2*Log[4])/E^2,x]

[Out]

5*x - 50*x^2 - 30*x^3 - 5*x^4 + 5*x*(E^(-2) + Log[4])

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 (5+e^2 \left (5-100 x-90 x^2-20 x^3\right )+5 e^2 \log (4)\right ) \, dx}{e^2} \\ & = 5 x \left (\frac {1}{e^2}+\log (4)\right )+\int \left (5-100 x-90 x^2-20 x^3\right ) \, dx \\ & = 5 x-50 x^2-30 x^3-5 x^4+5 x \left (\frac {1}{e^2}+\log (4)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {5+e^2 \left (5-100 x-90 x^2-20 x^3\right )+5 e^2 \log (4)}{e^2} \, dx=5 \left (x+\frac {x}{e^2}-10 x^2-6 x^3-x^4+x \log (4)\right ) \]

[In]

Integrate[(5 + E^2*(5 - 100*x - 90*x^2 - 20*x^3) + 5*E^2*Log[4])/E^2,x]

[Out]

5*(x + x/E^2 - 10*x^2 - 6*x^3 - x^4 + x*Log[4])

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03

method	result	size
risch	\(-5 x^{4}-30 x^{3}+10 x \ln \left (2\right )-50 x^{2}+5 x +5 x \,{\mathrm e}^{-2}\)	\(30\)
norman	\(-50 x^{2}-30 x^{3}-5 x^{4}+5 \left (2 \,{\mathrm e}^{2} \ln \left (2\right )+{\mathrm e}^{2}+1\right ) {\mathrm e}^{-2} x\)	\(34\)
gosper	\(5 x \left (-x^{3} {\mathrm e}^{2}-6 x^{2} {\mathrm e}^{2}+2 \,{\mathrm e}^{2} \ln \left (2\right )-10 \,{\mathrm e}^{2} x +{\mathrm e}^{2}+1\right ) {\mathrm e}^{-2}\)	\(37\)
parallelrisch	\({\mathrm e}^{-2} \left ({\mathrm e}^{2} \left (-5 x^{4}-30 x^{3}-50 x^{2}+5 x \right )+\left (10 \,{\mathrm e}^{2} \ln \left (2\right )+5\right ) x \right )\)	\(39\)
default	\({\mathrm e}^{-2} \left (-5 x^{4} {\mathrm e}^{2}-30 x^{3} {\mathrm e}^{2}+10 x \,{\mathrm e}^{2} \ln \left (2\right )-50 x^{2} {\mathrm e}^{2}+5 \,{\mathrm e}^{2} x +5 x \right )\)	\(43\)

[In]

int((10*exp(2)*ln(2)+(-20*x^3-90*x^2-100*x+5)*exp(2)+5)/exp(2),x,method=_RETURNVERBOSE)

[Out]

-5*x^4-30*x^3+10*x*ln(2)-50*x^2+5*x+5*x*exp(-2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {5+e^2 \left (5-100 x-90 x^2-20 x^3\right )+5 e^2 \log (4)}{e^2} \, dx=5 \, {\left (2 \, x e^{2} \log \left (2\right ) - {\left (x^{4} + 6 \, x^{3} + 10 \, x^{2} - x\right )} e^{2} + x\right )} e^{\left (-2\right )} \]

[In]

integrate((10*exp(2)*log(2)+(-20*x^3-90*x^2-100*x+5)*exp(2)+5)/exp(2),x, algorithm="fricas")

[Out]

5*(2*x*e^2*log(2) - (x^4 + 6*x^3 + 10*x^2 - x)*e^2 + x)*e^(-2)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {5+e^2 \left (5-100 x-90 x^2-20 x^3\right )+5 e^2 \log (4)}{e^2} \, dx=- 5 x^{4} - 30 x^{3} - 50 x^{2} + \frac {x \left (5 + 5 e^{2} + 10 e^{2} \log {\left (2 \right )}\right )}{e^{2}} \]

[In]

integrate((10*exp(2)*ln(2)+(-20*x**3-90*x**2-100*x+5)*exp(2)+5)/exp(2),x)

[Out]

-5*x**4 - 30*x**3 - 50*x**2 + x*(5 + 5*exp(2) + 10*exp(2)*log(2))*exp(-2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {5+e^2 \left (5-100 x-90 x^2-20 x^3\right )+5 e^2 \log (4)}{e^2} \, dx=5 \, {\left (2 \, x e^{2} \log \left (2\right ) - {\left (x^{4} + 6 \, x^{3} + 10 \, x^{2} - x\right )} e^{2} + x\right )} e^{\left (-2\right )} \]

[In]

integrate((10*exp(2)*log(2)+(-20*x^3-90*x^2-100*x+5)*exp(2)+5)/exp(2),x, algorithm="maxima")

[Out]

5*(2*x*e^2*log(2) - (x^4 + 6*x^3 + 10*x^2 - x)*e^2 + x)*e^(-2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {5+e^2 \left (5-100 x-90 x^2-20 x^3\right )+5 e^2 \log (4)}{e^2} \, dx=5 \, {\left (2 \, x e^{2} \log \left (2\right ) - {\left (x^{4} + 6 \, x^{3} + 10 \, x^{2} - x\right )} e^{2} + x\right )} e^{\left (-2\right )} \]

[In]

integrate((10*exp(2)*log(2)+(-20*x^3-90*x^2-100*x+5)*exp(2)+5)/exp(2),x, algorithm="giac")

[Out]

5*(2*x*e^2*log(2) - (x^4 + 6*x^3 + 10*x^2 - x)*e^2 + x)*e^(-2)

Mupad [B] (verification not implemented)

Time = 16.73 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {5+e^2 \left (5-100 x-90 x^2-20 x^3\right )+5 e^2 \log (4)}{e^2} \, dx=-5\,x^4-30\,x^3-50\,x^2+\left (5\,{\mathrm {e}}^{-2}+10\,\ln \left (2\right )+5\right )\,x \]

[In]

int(exp(-2)*(10*exp(2)*log(2) - exp(2)*(100*x + 90*x^2 + 20*x^3 - 5) + 5),x)

[Out]

x*(5*exp(-2) + 10*log(2) + 5) - 50*x^2 - 30*x^3 - 5*x^4