3.51.47 \(\int \frac {15+10 e^2-20 x+(2+6 x-6 x^2+e^2 (-4+4 x)) \log (3)}{500+1500 x+125 x^2-1500 x^3+500 x^4+e^4 (2000-2000 x+500 x^2)+e^2 (-2000-2000 x+3500 x^2-1000 x^3)+(200 x+600 x^2+50 x^3-600 x^4+200 x^5+e^4 (800 x-800 x^2+200 x^3)+e^2 (-800 x-800 x^2+1400 x^3-400 x^4)) \log (3)+(20 x^2+60 x^3+5 x^4-60 x^5+20 x^6+e^4 (80 x^2-80 x^3+20 x^4)+e^2 (-80 x^2-80 x^3+140 x^4-40 x^5)) \log ^2(3)} \, dx\)

Optimal. Leaf size=29 \[ \frac {1}{5 (-2+x) \left (1-2 e^2+2 x\right ) (5+x \log (3))} \]

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Rubi [B]  time = 0.29, antiderivative size = 102, normalized size of antiderivative = 3.52, number of steps used = 2, number of rules used = 1, integrand size = 232, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.004, Rules used = {2074} \begin {gather*} \frac {\log ^2(3)}{5 (5+\log (9)) \left (10-\log (3)+e^2 \log (9)\right ) (x \log (3)+5)}-\frac {4}{5 \left (5-2 e^2\right ) \left (2 x-2 e^2+1\right ) \left (10-\log (3)+e^2 \log (9)\right )}-\frac {1}{5 \left (5-2 e^2\right ) (2-x) (5+\log (9))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15 + 10*E^2 - 20*x + (2 + 6*x - 6*x^2 + E^2*(-4 + 4*x))*Log[3])/(500 + 1500*x + 125*x^2 - 1500*x^3 + 500*
x^4 + E^4*(2000 - 2000*x + 500*x^2) + E^2*(-2000 - 2000*x + 3500*x^2 - 1000*x^3) + (200*x + 600*x^2 + 50*x^3 -
600*x^4 + 200*x^5 + E^4*(800*x - 800*x^2 + 200*x^3) + E^2*(-800*x - 800*x^2 + 1400*x^3 - 400*x^4))*Log[3] + (
20*x^2 + 60*x^3 + 5*x^4 - 60*x^5 + 20*x^6 + E^4*(80*x^2 - 80*x^3 + 20*x^4) + E^2*(-80*x^2 - 80*x^3 + 140*x^4 -
40*x^5))*Log[3]^2),x]

[Out]

-1/5*1/((5 - 2*E^2)*(2 - x)*(5 + Log[9])) - 4/(5*(5 - 2*E^2)*(1 - 2*E^2 + 2*x)*(10 - Log[3] + E^2*Log[9])) + L
og[3]^2/(5*(5 + x*Log[3])*(5 + Log[9])*(10 - Log[3] + E^2*Log[9]))

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1}{5 \left (-5+2 e^2\right ) (-2+x)^2 (5+\log (9))}+\frac {\log ^3(3)}{5 (5+x \log (3))^2 (5+\log (9)) \left (-10+\log (3)-e^2 \log (9)\right )}-\frac {8}{5 \left (-5+2 e^2\right ) \left (-1+2 e^2-2 x\right )^2 \left (10-\log (3)+e^2 \log (9)\right )}\right ) \, dx\\ &=-\frac {1}{5 \left (5-2 e^2\right ) (2-x) (5+\log (9))}-\frac {4}{5 \left (5-2 e^2\right ) \left (1-2 e^2+2 x\right ) \left (10-\log (3)+e^2 \log (9)\right )}+\frac {\log ^2(3)}{5 (5+x \log (3)) (5+\log (9)) \left (10-\log (3)+e^2 \log (9)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.32, size = 151, normalized size = 5.21 \begin {gather*} \frac {1}{5} \left (\frac {25-e^2 (10+\log (81))+\log (59049)}{\left (5-2 e^2\right )^2 (-2+x) (5+\log (9))^2}-\frac {4 \left (-50+e^4 \log (81)+\log (243)-e^2 (-20+\log (531441))\right )}{\left (5-2 e^2\right )^2 \left (-1+2 e^2-2 x\right ) \left (10-\log (3)+e^2 \log (9)\right )^2}+\frac {\log ^2(3) \left (50-\log (3) \log (9)+e^2 \left (4 \log ^2(3)+\log (59049)\right )+\log (14348907)\right )}{(5+x \log (3)) (5+\log (9))^2 \left (10-\log (3)+e^2 \log (9)\right )^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15 + 10*E^2 - 20*x + (2 + 6*x - 6*x^2 + E^2*(-4 + 4*x))*Log[3])/(500 + 1500*x + 125*x^2 - 1500*x^3
+ 500*x^4 + E^4*(2000 - 2000*x + 500*x^2) + E^2*(-2000 - 2000*x + 3500*x^2 - 1000*x^3) + (200*x + 600*x^2 + 50
*x^3 - 600*x^4 + 200*x^5 + E^4*(800*x - 800*x^2 + 200*x^3) + E^2*(-800*x - 800*x^2 + 1400*x^3 - 400*x^4))*Log[
3] + (20*x^2 + 60*x^3 + 5*x^4 - 60*x^5 + 20*x^6 + E^4*(80*x^2 - 80*x^3 + 20*x^4) + E^2*(-80*x^2 - 80*x^3 + 140
*x^4 - 40*x^5))*Log[3]^2),x]

[Out]

((25 - E^2*(10 + Log[81]) + Log[59049])/((5 - 2*E^2)^2*(-2 + x)*(5 + Log[9])^2) - (4*(-50 + E^4*Log[81] + Log[
243] - E^2*(-20 + Log[531441])))/((5 - 2*E^2)^2*(-1 + 2*E^2 - 2*x)*(10 - Log[3] + E^2*Log[9])^2) + (Log[3]^2*(
50 - Log[3]*Log[9] + E^2*(4*Log[3]^2 + Log[59049]) + Log[14348907]))/((5 + x*Log[3])*(5 + Log[9])^2*(10 - Log[
3] + E^2*Log[9])^2))/5

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fricas [A]  time = 0.76, size = 49, normalized size = 1.69 \begin {gather*} \frac {1}{5 \, {\left (10 \, x^{2} - 10 \, {\left (x - 2\right )} e^{2} + {\left (2 \, x^{3} - 3 \, x^{2} - 2 \, {\left (x^{2} - 2 \, x\right )} e^{2} - 2 \, x\right )} \log \relax (3) - 15 \, x - 10\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x-4)*exp(2)-6*x^2+6*x+2)*log(3)+10*exp(2)-20*x+15)/(((20*x^4-80*x^3+80*x^2)*exp(2)^2+(-40*x^5+1
40*x^4-80*x^3-80*x^2)*exp(2)+20*x^6-60*x^5+5*x^4+60*x^3+20*x^2)*log(3)^2+((200*x^3-800*x^2+800*x)*exp(2)^2+(-4
00*x^4+1400*x^3-800*x^2-800*x)*exp(2)+200*x^5-600*x^4+50*x^3+600*x^2+200*x)*log(3)+(500*x^2-2000*x+2000)*exp(2
)^2+(-1000*x^3+3500*x^2-2000*x-2000)*exp(2)+500*x^4-1500*x^3+125*x^2+1500*x+500),x, algorithm="fricas")

[Out]

1/5/(10*x^2 - 10*(x - 2)*e^2 + (2*x^3 - 3*x^2 - 2*(x^2 - 2*x)*e^2 - 2*x)*log(3) - 15*x - 10)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x-4)*exp(2)-6*x^2+6*x+2)*log(3)+10*exp(2)-20*x+15)/(((20*x^4-80*x^3+80*x^2)*exp(2)^2+(-40*x^5+1
40*x^4-80*x^3-80*x^2)*exp(2)+20*x^6-60*x^5+5*x^4+60*x^3+20*x^2)*log(3)^2+((200*x^3-800*x^2+800*x)*exp(2)^2+(-4
00*x^4+1400*x^3-800*x^2-800*x)*exp(2)+200*x^5-600*x^4+50*x^3+600*x^2+200*x)*log(3)+(500*x^2-2000*x+2000)*exp(2
)^2+(-1000*x^3+3500*x^2-2000*x-2000)*exp(2)+500*x^4-1500*x^3+125*x^2+1500*x+500),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -1/5*((-512*exp(4)^2*exp(2)*ln(3)^3+768*
exp(4)^2*ln(3)^3-1280*exp(4)^2*ln(3)^2+512*exp(4)*exp(2)^3*ln(3)^3+768*exp(4)*exp(2)^2*ln(3)^3-1280*exp(4)*exp
(2)^2*ln(3)^2-3968*ex

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maple [A]  time = 4.56, size = 27, normalized size = 0.93

method	result	size
norman	\(-\frac {1}{5 \left (x -2\right ) \left (2 \,{\mathrm e}^{2}-2 x -1\right ) \left (x \ln \relax (3)+5\right )}\)	\(27\)
risch	\(-\frac {1}{10 \left (x^{2} {\mathrm e}^{2} \ln \relax (3)-x^{3} \ln \relax (3)-2 x \,{\mathrm e}^{2} \ln \relax (3)+\frac {3 x^{2} \ln \relax (3)}{2}+x \ln \relax (3)+5 \,{\mathrm e}^{2} x -5 x^{2}-10 \,{\mathrm e}^{2}+\frac {15 x}{2}+5\right )}\)	\(57\)
gosper	\(-\frac {1}{5 \left (2 x^{2} {\mathrm e}^{2} \ln \relax (3)-2 x^{3} \ln \relax (3)-4 x \,{\mathrm e}^{2} \ln \relax (3)+3 x^{2} \ln \relax (3)+10 \,{\mathrm e}^{2} x +2 x \ln \relax (3)-10 x^{2}-20 \,{\mathrm e}^{2}+15 x +10\right )}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x-4)*exp(2)-6*x^2+6*x+2)*ln(3)+10*exp(2)-20*x+15)/(((20*x^4-80*x^3+80*x^2)*exp(2)^2+(-40*x^5+140*x^4-
80*x^3-80*x^2)*exp(2)+20*x^6-60*x^5+5*x^4+60*x^3+20*x^2)*ln(3)^2+((200*x^3-800*x^2+800*x)*exp(2)^2+(-400*x^4+1
400*x^3-800*x^2-800*x)*exp(2)+200*x^5-600*x^4+50*x^3+600*x^2+200*x)*ln(3)+(500*x^2-2000*x+2000)*exp(2)^2+(-100
0*x^3+3500*x^2-2000*x-2000)*exp(2)+500*x^4-1500*x^3+125*x^2+1500*x+500),x,method=_RETURNVERBOSE)

[Out]

-1/5/(x-2)/(2*exp(2)-2*x-1)/(x*ln(3)+5)

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maxima [A]  time = 0.35, size = 51, normalized size = 1.76 \begin {gather*} \frac {1}{5 \, {\left (2 \, x^{3} \log \relax (3) - {\left ({\left (2 \, e^{2} + 3\right )} \log \relax (3) - 10\right )} x^{2} + {\left (2 \, {\left (2 \, e^{2} - 1\right )} \log \relax (3) - 10 \, e^{2} - 15\right )} x + 20 \, e^{2} - 10\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x-4)*exp(2)-6*x^2+6*x+2)*log(3)+10*exp(2)-20*x+15)/(((20*x^4-80*x^3+80*x^2)*exp(2)^2+(-40*x^5+1
40*x^4-80*x^3-80*x^2)*exp(2)+20*x^6-60*x^5+5*x^4+60*x^3+20*x^2)*log(3)^2+((200*x^3-800*x^2+800*x)*exp(2)^2+(-4
00*x^4+1400*x^3-800*x^2-800*x)*exp(2)+200*x^5-600*x^4+50*x^3+600*x^2+200*x)*log(3)+(500*x^2-2000*x+2000)*exp(2
)^2+(-1000*x^3+3500*x^2-2000*x-2000)*exp(2)+500*x^4-1500*x^3+125*x^2+1500*x+500),x, algorithm="maxima")

[Out]

1/5/(2*x^3*log(3) - ((2*e^2 + 3)*log(3) - 10)*x^2 + (2*(2*e^2 - 1)*log(3) - 10*e^2 - 15)*x + 20*e^2 - 10)

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mupad [B]  time = 3.85, size = 95, normalized size = 3.28 \begin {gather*} \frac {4}{5\,\left (2\,{\mathrm {e}}^2-5\right )\,\left (2\,{\mathrm {e}}^2\,\ln \relax (3)-\ln \relax (3)+10\right )\,\left (2\,x-2\,{\mathrm {e}}^2+1\right )}-\frac {1}{5\,\left (2\,{\mathrm {e}}^2-5\right )\,\left (2\,\ln \relax (3)+5\right )\,\left (x-2\right )}+\frac {{\ln \relax (3)}^2}{5\,\left (2\,\ln \relax (3)+5\right )\,\left (x\,\ln \relax (3)+5\right )\,\left (2\,{\mathrm {e}}^2\,\ln \relax (3)-\ln \relax (3)+10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*exp(2) - 20*x + log(3)*(6*x - 6*x^2 + exp(2)*(4*x - 4) + 2) + 15)/(1500*x + exp(4)*(500*x^2 - 2000*x +
2000) + log(3)^2*(exp(4)*(80*x^2 - 80*x^3 + 20*x^4) + 20*x^2 + 60*x^3 + 5*x^4 - 60*x^5 + 20*x^6 - exp(2)*(80*
x^2 + 80*x^3 - 140*x^4 + 40*x^5)) - exp(2)*(2000*x - 3500*x^2 + 1000*x^3 + 2000) + 125*x^2 - 1500*x^3 + 500*x^
4 + log(3)*(200*x + exp(4)*(800*x - 800*x^2 + 200*x^3) - exp(2)*(800*x + 800*x^2 - 1400*x^3 + 400*x^4) + 600*x
^2 + 50*x^3 - 600*x^4 + 200*x^5) + 500),x)

[Out]

4/(5*(2*exp(2) - 5)*(2*exp(2)*log(3) - log(3) + 10)*(2*x - 2*exp(2) + 1)) - 1/(5*(2*exp(2) - 5)*(2*log(3) + 5)
*(x - 2)) + log(3)^2/(5*(2*log(3) + 5)*(x*log(3) + 5)*(2*exp(2)*log(3) - log(3) + 10))

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sympy [B]  time = 10.33, size = 56, normalized size = 1.93 \begin {gather*} \frac {1}{10 x^{3} \log {\relax (3 )} + x^{2} \left (- 10 e^{2} \log {\relax (3 )} - 15 \log {\relax (3 )} + 50\right ) + x \left (- 50 e^{2} - 75 - 10 \log {\relax (3 )} + 20 e^{2} \log {\relax (3 )}\right ) - 50 + 100 e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x-4)*exp(2)-6*x**2+6*x+2)*ln(3)+10*exp(2)-20*x+15)/(((20*x**4-80*x**3+80*x**2)*exp(2)**2+(-40*x
**5+140*x**4-80*x**3-80*x**2)*exp(2)+20*x**6-60*x**5+5*x**4+60*x**3+20*x**2)*ln(3)**2+((200*x**3-800*x**2+800*
x)*exp(2)**2+(-400*x**4+1400*x**3-800*x**2-800*x)*exp(2)+200*x**5-600*x**4+50*x**3+600*x**2+200*x)*ln(3)+(500*
x**2-2000*x+2000)*exp(2)**2+(-1000*x**3+3500*x**2-2000*x-2000)*exp(2)+500*x**4-1500*x**3+125*x**2+1500*x+500),
x)

[Out]

1/(10*x**3*log(3) + x**2*(-10*exp(2)*log(3) - 15*log(3) + 50) + x*(-50*exp(2) - 75 - 10*log(3) + 20*exp(2)*log
(3)) - 50 + 100*exp(2))

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