3.58.53 \(\int \frac {1}{125} (-6+140 x-450 x^2-500 x^3+e^{1+2 x+x^2+(-2-2 x) \log (3)+\log ^2(3)} (250+250 x-250 \log (3))) \, dx\)

Optimal. Leaf size=30 \[ e^{(-1-x+\log (3))^2}+x^2-\left (\left (\frac {1}{5}-x\right )^2+x\right )^2 \]

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Rubi [A]  time = 0.08, antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 4, number of rules used = 3, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 2227, 2209} \begin {gather*} -x^4-\frac {6 x^3}{5}+\frac {14 x^2}{25}-\frac {6 x}{125}+e^{(x+1-\log (3))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6 + 140*x - 450*x^2 - 500*x^3 + E^(1 + 2*x + x^2 + (-2 - 2*x)*Log[3] + Log[3]^2)*(250 + 250*x - 250*Log[
3]))/125,x]

[Out]

E^(1 + x - Log[3])^2 - (6*x)/125 + (14*x^2)/25 - (6*x^3)/5 - x^4

Rule 12

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

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2227

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{125} \int \left (-6+140 x-450 x^2-500 x^3+e^{1+2 x+x^2+(-2-2 x) \log (3)+\log ^2(3)} (250+250 x-250 \log (3))\right ) \, dx\\ &=-\frac {6 x}{125}+\frac {14 x^2}{25}-\frac {6 x^3}{5}-x^4+\frac {1}{125} \int e^{1+2 x+x^2+(-2-2 x) \log (3)+\log ^2(3)} (250+250 x-250 \log (3)) \, dx\\ &=-\frac {6 x}{125}+\frac {14 x^2}{25}-\frac {6 x^3}{5}-x^4+\frac {1}{125} \int e^{(1+x-\log (3))^2} (250+250 x-250 \log (3)) \, dx\\ &=e^{(1+x-\log (3))^2}-\frac {6 x}{125}+\frac {14 x^2}{25}-\frac {6 x^3}{5}-x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.36, size = 47, normalized size = 1.57 \begin {gather*} 3^{-2-2 x} e^{1+2 x+x^2+\log ^2(3)}-\frac {6 x}{125}+\frac {14 x^2}{25}-\frac {6 x^3}{5}-x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6 + 140*x - 450*x^2 - 500*x^3 + E^(1 + 2*x + x^2 + (-2 - 2*x)*Log[3] + Log[3]^2)*(250 + 250*x - 25
0*Log[3]))/125,x]

[Out]

3^(-2 - 2*x)*E^(1 + 2*x + x^2 + Log[3]^2) - (6*x)/125 + (14*x^2)/25 - (6*x^3)/5 - x^4

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fricas [A]  time = 0.49, size = 39, normalized size = 1.30 \begin {gather*} -x^{4} - \frac {6}{5} \, x^{3} + \frac {14}{25} \, x^{2} - \frac {6}{125} \, x + e^{\left (x^{2} - 2 \, {\left (x + 1\right )} \log \relax (3) + \log \relax (3)^{2} + 2 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(-250*log(3)+250*x+250)*exp(log(3)^2+(-2*x-2)*log(3)+x^2+2*x+1)-4*x^3-18/5*x^2+28/25*x-6/125,x
, algorithm="fricas")

[Out]

-x^4 - 6/5*x^3 + 14/25*x^2 - 6/125*x + e^(x^2 - 2*(x + 1)*log(3) + log(3)^2 + 2*x + 1)

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giac [A]  time = 0.19, size = 41, normalized size = 1.37 \begin {gather*} -x^{4} - \frac {6}{5} \, x^{3} + \frac {14}{25} \, x^{2} - \frac {6}{125} \, x + e^{\left (x^{2} - 2 \, x \log \relax (3) + \log \relax (3)^{2} + 2 \, x - 2 \, \log \relax (3) + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(-250*log(3)+250*x+250)*exp(log(3)^2+(-2*x-2)*log(3)+x^2+2*x+1)-4*x^3-18/5*x^2+28/25*x-6/125,x
, algorithm="giac")

[Out]

-x^4 - 6/5*x^3 + 14/25*x^2 - 6/125*x + e^(x^2 - 2*x*log(3) + log(3)^2 + 2*x - 2*log(3) + 1)

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maple [A]  time = 0.13, size = 41, normalized size = 1.37

method	result	size
norman	\(-\frac {6 x}{125}+\frac {14 x^{2}}{25}-\frac {6 x^{3}}{5}-x^{4}+{\mathrm e}^{\ln \relax (3)^{2}+\left (-2 x -2\right ) \ln \relax (3)+x^{2}+2 x +1}\)	\(41\)
risch	\(-\frac {6 x}{125}+\frac {14 x^{2}}{25}-\frac {6 x^{3}}{5}-x^{4}+3^{-2 x -2} {\mathrm e}^{\ln \relax (3)^{2}+1+x^{2}+2 x}\)	\(41\)
default	\(-\frac {6 x}{125}+\frac {14 x^{2}}{25}-\frac {6 x^{3}}{5}-x^{4}+{\mathrm e}^{x^{2}+\left (-2 \ln \relax (3)+2\right ) x +\ln \relax (3)^{2}-2 \ln \relax (3)+1}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/125*(-250*ln(3)+250*x+250)*exp(ln(3)^2+(-2*x-2)*ln(3)+x^2+2*x+1)-4*x^3-18/5*x^2+28/25*x-6/125,x,method=_
RETURNVERBOSE)

[Out]

-6/125*x+14/25*x^2-6/5*x^3-x^4+exp(ln(3)^2+(-2*x-2)*ln(3)+x^2+2*x+1)

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maxima [A]  time = 0.35, size = 39, normalized size = 1.30 \begin {gather*} -x^{4} - \frac {6}{5} \, x^{3} + \frac {14}{25} \, x^{2} - \frac {6}{125} \, x + e^{\left (x^{2} - 2 \, {\left (x + 1\right )} \log \relax (3) + \log \relax (3)^{2} + 2 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(-250*log(3)+250*x+250)*exp(log(3)^2+(-2*x-2)*log(3)+x^2+2*x+1)-4*x^3-18/5*x^2+28/25*x-6/125,x
, algorithm="maxima")

[Out]

-x^4 - 6/5*x^3 + 14/25*x^2 - 6/125*x + e^(x^2 - 2*(x + 1)*log(3) + log(3)^2 + 2*x + 1)

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mupad [B]  time = 3.59, size = 41, normalized size = 1.37 \begin {gather*} \frac {{\mathrm {e}}^{x^2+2\,x+{\ln \relax (3)}^2+1}}{9\,3^{2\,x}}-\frac {6\,x}{125}+\frac {14\,x^2}{25}-\frac {6\,x^3}{5}-x^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((28*x)/25 + (exp(2*x - log(3)*(2*x + 2) + log(3)^2 + x^2 + 1)*(250*x - 250*log(3) + 250))/125 - (18*x^2)/5
- 4*x^3 - 6/125,x)

[Out]

exp(2*x + log(3)^2 + x^2 + 1)/(9*3^(2*x)) - (6*x)/125 + (14*x^2)/25 - (6*x^3)/5 - x^4

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sympy [B]  time = 0.14, size = 46, normalized size = 1.53 \begin {gather*} - x^{4} - \frac {6 x^{3}}{5} + \frac {14 x^{2}}{25} - \frac {6 x}{125} + e^{x^{2} + 2 x + \left (- 2 x - 2\right ) \log {\relax (3 )} + 1 + \log {\relax (3 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(-250*ln(3)+250*x+250)*exp(ln(3)**2+(-2*x-2)*ln(3)+x**2+2*x+1)-4*x**3-18/5*x**2+28/25*x-6/125,
x)

[Out]

-x**4 - 6*x**3/5 + 14*x**2/25 - 6*x/125 + exp(x**2 + 2*x + (-2*x - 2)*log(3) + 1 + log(3)**2)

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