3.11.40 \(\int \frac {e^{\frac {12+e^{e^{2 x}-2 e^x \log (\frac {13}{4})+\log ^2(\frac {13}{4})} (3-3 x)-11 x}{-4+e^{e^{2 x}-2 e^x \log (\frac {13}{4})+\log ^2(\frac {13}{4})} (-1+x)+4 x}} (-4+e^{e^{2 x}-2 e^x \log (\frac {13}{4})+\log ^2(\frac {13}{4})} (-1+e^{2 x} (2 x-2 x^2)+e^x (-2 x+2 x^2) \log (\frac {13}{4})))}{16-32 x+16 x^2+e^{2 e^{2 x}-4 e^x \log (\frac {13}{4})+2 \log ^2(\frac {13}{4})} (1-2 x+x^2)+e^{e^{2 x}-2 e^x \log (\frac {13}{4})+\log ^2(\frac {13}{4})} (8-16 x+8 x^2)} \, dx\)

Optimal. Leaf size=29 \[ e^{-3+\frac {x}{\left (4+e^{\left (e^x-\log \left (\frac {13}{4}\right )\right )^2}\right ) (-1+x)}} \]

________________________________________________________________________________________

Rubi [F]  time = 180.00, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((12 + E^(E^(2*x) - 2*E^x*Log[13/4] + Log[13/4]^2)*(3 - 3*x) - 11*x)/(-4 + E^(E^(2*x) - 2*E^x*Log[13/4]
+ Log[13/4]^2)*(-1 + x) + 4*x))*(-4 + E^(E^(2*x) - 2*E^x*Log[13/4] + Log[13/4]^2)*(-1 + E^(2*x)*(2*x - 2*x^2)
+ E^x*(-2*x + 2*x^2)*Log[13/4])))/(16 - 32*x + 16*x^2 + E^(2*E^(2*x) - 4*E^x*Log[13/4] + 2*Log[13/4]^2)*(1 -
2*x + x^2) + E^(E^(2*x) - 2*E^x*Log[13/4] + Log[13/4]^2)*(8 - 16*x + 8*x^2)),x]

[Out]

$Aborted

Rubi steps

Aborted

________________________________________________________________________________________

Mathematica [F]  time = 81.70, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {12+e^{e^{2 x}-2 e^x \log \left (\frac {13}{4}\right )+\log ^2\left (\frac {13}{4}\right )} (3-3 x)-11 x}{-4+e^{e^{2 x}-2 e^x \log \left (\frac {13}{4}\right )+\log ^2\left (\frac {13}{4}\right )} (-1+x)+4 x}} \left (-4+e^{e^{2 x}-2 e^x \log \left (\frac {13}{4}\right )+\log ^2\left (\frac {13}{4}\right )} \left (-1+e^{2 x} \left (2 x-2 x^2\right )+e^x \left (-2 x+2 x^2\right ) \log \left (\frac {13}{4}\right )\right )\right )}{16-32 x+16 x^2+e^{2 e^{2 x}-4 e^x \log \left (\frac {13}{4}\right )+2 \log ^2\left (\frac {13}{4}\right )} \left (1-2 x+x^2\right )+e^{e^{2 x}-2 e^x \log \left (\frac {13}{4}\right )+\log ^2\left (\frac {13}{4}\right )} \left (8-16 x+8 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((12 + E^(E^(2*x) - 2*E^x*Log[13/4] + Log[13/4]^2)*(3 - 3*x) - 11*x)/(-4 + E^(E^(2*x) - 2*E^x*Log
[13/4] + Log[13/4]^2)*(-1 + x) + 4*x))*(-4 + E^(E^(2*x) - 2*E^x*Log[13/4] + Log[13/4]^2)*(-1 + E^(2*x)*(2*x -
2*x^2) + E^x*(-2*x + 2*x^2)*Log[13/4])))/(16 - 32*x + 16*x^2 + E^(2*E^(2*x) - 4*E^x*Log[13/4] + 2*Log[13/4]^2)
*(1 - 2*x + x^2) + E^(E^(2*x) - 2*E^x*Log[13/4] + Log[13/4]^2)*(8 - 16*x + 8*x^2)),x]

[Out]

Integrate[(E^((12 + E^(E^(2*x) - 2*E^x*Log[13/4] + Log[13/4]^2)*(3 - 3*x) - 11*x)/(-4 + E^(E^(2*x) - 2*E^x*Log
[13/4] + Log[13/4]^2)*(-1 + x) + 4*x))*(-4 + E^(E^(2*x) - 2*E^x*Log[13/4] + Log[13/4]^2)*(-1 + E^(2*x)*(2*x -
2*x^2) + E^x*(-2*x + 2*x^2)*Log[13/4])))/(16 - 32*x + 16*x^2 + E^(2*E^(2*x) - 4*E^x*Log[13/4] + 2*Log[13/4]^2)
*(1 - 2*x + x^2) + E^(E^(2*x) - 2*E^x*Log[13/4] + Log[13/4]^2)*(8 - 16*x + 8*x^2)), x]

________________________________________________________________________________________

fricas [B]  time = 0.53, size = 56, normalized size = 1.93 \begin {gather*} e^{\left (-\frac {3 \, {\left (x - 1\right )} e^{\left (-2 \, e^{x} \log \left (\frac {13}{4}\right ) + \log \left (\frac {13}{4}\right )^{2} + e^{\left (2 \, x\right )}\right )} + 11 \, x - 12}{{\left (x - 1\right )} e^{\left (-2 \, e^{x} \log \left (\frac {13}{4}\right ) + \log \left (\frac {13}{4}\right )^{2} + e^{\left (2 \, x\right )}\right )} + 4 \, x - 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+2*x)*exp(x)^2+(2*x^2-2*x)*log(13/4)*exp(x)-1)*exp(exp(x)^2-2*log(13/4)*exp(x)+log(13/4)^2)
-4)*exp(((-3*x+3)*exp(exp(x)^2-2*log(13/4)*exp(x)+log(13/4)^2)-11*x+12)/((x-1)*exp(exp(x)^2-2*log(13/4)*exp(x)
+log(13/4)^2)+4*x-4))/((x^2-2*x+1)*exp(exp(x)^2-2*log(13/4)*exp(x)+log(13/4)^2)^2+(8*x^2-16*x+8)*exp(exp(x)^2-
2*log(13/4)*exp(x)+log(13/4)^2)+16*x^2-32*x+16),x, algorithm="fricas")

[Out]

e^(-(3*(x - 1)*e^(-2*e^x*log(13/4) + log(13/4)^2 + e^(2*x)) + 11*x - 12)/((x - 1)*e^(-2*e^x*log(13/4) + log(13
/4)^2 + e^(2*x)) + 4*x - 4))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+2*x)*exp(x)^2+(2*x^2-2*x)*log(13/4)*exp(x)-1)*exp(exp(x)^2-2*log(13/4)*exp(x)+log(13/4)^2)
-4)*exp(((-3*x+3)*exp(exp(x)^2-2*log(13/4)*exp(x)+log(13/4)^2)-11*x+12)/((x-1)*exp(exp(x)^2-2*log(13/4)*exp(x)
+log(13/4)^2)+4*x-4))/((x^2-2*x+1)*exp(exp(x)^2-2*log(13/4)*exp(x)+log(13/4)^2)^2+(8*x^2-16*x+8)*exp(exp(x)^2-
2*log(13/4)*exp(x)+log(13/4)^2)+16*x^2-32*x+16),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Evaluation time: 6.52Unable to divide, perhaps due to rounding error%%%{524288,[0,8,12,22,1,1,8]%%%}+%%%{-7
864320,[0,8

________________________________________________________________________________________

maple [B]  time = 0.38, size = 108, normalized size = 3.72

method	result	size
risch	\({\mathrm e}^{-\frac {3 \,16^{{\mathrm e}^{x}} \left (\frac {1}{169}\right )^{{\mathrm e}^{x}} \left (\frac {1}{28561}\right )^{\ln \relax (2)} {\mathrm e}^{4 \ln \relax (2)^{2}+\ln \left (13\right )^{2}+{\mathrm e}^{2 x}} x -3 \,16^{{\mathrm e}^{x}} \left (\frac {1}{169}\right )^{{\mathrm e}^{x}} \left (\frac {1}{28561}\right )^{\ln \relax (2)} {\mathrm e}^{4 \ln \relax (2)^{2}+\ln \left (13\right )^{2}+{\mathrm e}^{2 x}}+11 x -12}{\left (x -1\right ) \left (16^{{\mathrm e}^{x}} \left (\frac {1}{169}\right )^{{\mathrm e}^{x}} \left (\frac {1}{28561}\right )^{\ln \relax (2)} {\mathrm e}^{4 \ln \relax (2)^{2}+\ln \left (13\right )^{2}+{\mathrm e}^{2 x}}+4\right )}}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^2+2*x)*exp(x)^2+(2*x^2-2*x)*ln(13/4)*exp(x)-1)*exp(exp(x)^2-2*ln(13/4)*exp(x)+ln(13/4)^2)-4)*exp((
(-3*x+3)*exp(exp(x)^2-2*ln(13/4)*exp(x)+ln(13/4)^2)-11*x+12)/((x-1)*exp(exp(x)^2-2*ln(13/4)*exp(x)+ln(13/4)^2)
+4*x-4))/((x^2-2*x+1)*exp(exp(x)^2-2*ln(13/4)*exp(x)+ln(13/4)^2)^2+(8*x^2-16*x+8)*exp(exp(x)^2-2*ln(13/4)*exp(
x)+ln(13/4)^2)+16*x^2-32*x+16),x,method=_RETURNVERBOSE)

[Out]

exp(-(3*16^exp(x)*(1/169)^exp(x)*(1/28561)^ln(2)*exp(4*ln(2)^2+ln(13)^2+exp(2*x))*x-3*16^exp(x)*(1/169)^exp(x)
*(1/28561)^ln(2)*exp(4*ln(2)^2+ln(13)^2+exp(2*x))+11*x-12)/(x-1)/(16^exp(x)*(1/169)^exp(x)*(1/28561)^ln(2)*exp
(4*ln(2)^2+ln(13)^2+exp(2*x))+4))

________________________________________________________________________________________

maxima [B]  time = 2.60, size = 206, normalized size = 7.10 \begin {gather*} e^{\left (\frac {13^{2 \, e^{x}} 2^{4 \, \log \left (13\right )}}{{\left (2^{4 \, \log \left (13\right ) + 2} x - 2^{4 \, \log \left (13\right ) + 2}\right )} 13^{2 \, e^{x}} + {\left (x e^{\left (\log \left (13\right )^{2} + 4 \, \log \relax (2)^{2}\right )} - e^{\left (\log \left (13\right )^{2} + 4 \, \log \relax (2)^{2}\right )}\right )} e^{\left (4 \, e^{x} \log \relax (2) + e^{\left (2 \, x\right )}\right )}} - \frac {11 \cdot 13^{2 \, e^{x}} 2^{4 \, \log \left (13\right )}}{13^{2 \, e^{x}} 2^{4 \, \log \left (13\right ) + 2} + e^{\left (\log \left (13\right )^{2} + 4 \, e^{x} \log \relax (2) + 4 \, \log \relax (2)^{2} + e^{\left (2 \, x\right )}\right )}} - \frac {3 \, e^{\left (\log \left (13\right )^{2} + 4 \, e^{x} \log \relax (2) + 4 \, \log \relax (2)^{2} + e^{\left (2 \, x\right )}\right )}}{13^{2 \, e^{x}} 2^{4 \, \log \left (13\right ) + 2} + e^{\left (\log \left (13\right )^{2} + 4 \, e^{x} \log \relax (2) + 4 \, \log \relax (2)^{2} + e^{\left (2 \, x\right )}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+2*x)*exp(x)^2+(2*x^2-2*x)*log(13/4)*exp(x)-1)*exp(exp(x)^2-2*log(13/4)*exp(x)+log(13/4)^2)
-4)*exp(((-3*x+3)*exp(exp(x)^2-2*log(13/4)*exp(x)+log(13/4)^2)-11*x+12)/((x-1)*exp(exp(x)^2-2*log(13/4)*exp(x)
+log(13/4)^2)+4*x-4))/((x^2-2*x+1)*exp(exp(x)^2-2*log(13/4)*exp(x)+log(13/4)^2)^2+(8*x^2-16*x+8)*exp(exp(x)^2-
2*log(13/4)*exp(x)+log(13/4)^2)+16*x^2-32*x+16),x, algorithm="maxima")

[Out]

e^(13^(2*e^x)*2^(4*log(13))/((2^(4*log(13) + 2)*x - 2^(4*log(13) + 2))*13^(2*e^x) + (x*e^(log(13)^2 + 4*log(2)
^2) - e^(log(13)^2 + 4*log(2)^2))*e^(4*e^x*log(2) + e^(2*x))) - 11*13^(2*e^x)*2^(4*log(13))/(13^(2*e^x)*2^(4*l
og(13) + 2) + e^(log(13)^2 + 4*e^x*log(2) + 4*log(2)^2 + e^(2*x))) - 3*e^(log(13)^2 + 4*e^x*log(2) + 4*log(2)^
2 + e^(2*x))/(13^(2*e^x)*2^(4*log(13) + 2) + e^(log(13)^2 + 4*e^x*log(2) + 4*log(2)^2 + e^(2*x))))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{-\frac {11\,x+{\mathrm {e}}^{{\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x\,\ln \left (\frac {13}{4}\right )+{\ln \left (\frac {13}{4}\right )}^2}\,\left (3\,x-3\right )-12}{4\,x+{\mathrm {e}}^{{\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x\,\ln \left (\frac {13}{4}\right )+{\ln \left (\frac {13}{4}\right )}^2}\,\left (x-1\right )-4}}\,\left ({\mathrm {e}}^{{\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x\,\ln \left (\frac {13}{4}\right )+{\ln \left (\frac {13}{4}\right )}^2}\,\left ({\mathrm {e}}^x\,\ln \left (\frac {13}{4}\right )\,\left (2\,x-2\,x^2\right )-{\mathrm {e}}^{2\,x}\,\left (2\,x-2\,x^2\right )+1\right )+4\right )}{{\mathrm {e}}^{{\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x\,\ln \left (\frac {13}{4}\right )+{\ln \left (\frac {13}{4}\right )}^2}\,\left (8\,x^2-16\,x+8\right )-32\,x+{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}-4\,{\mathrm {e}}^x\,\ln \left (\frac {13}{4}\right )+2\,{\ln \left (\frac {13}{4}\right )}^2}\,\left (x^2-2\,x+1\right )+16\,x^2+16} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(11*x + exp(exp(2*x) - 2*exp(x)*log(13/4) + log(13/4)^2)*(3*x - 3) - 12)/(4*x + exp(exp(2*x) - 2*ex
p(x)*log(13/4) + log(13/4)^2)*(x - 1) - 4))*(exp(exp(2*x) - 2*exp(x)*log(13/4) + log(13/4)^2)*(exp(x)*log(13/4
)*(2*x - 2*x^2) - exp(2*x)*(2*x - 2*x^2) + 1) + 4))/(exp(exp(2*x) - 2*exp(x)*log(13/4) + log(13/4)^2)*(8*x^2 -
16*x + 8) - 32*x + exp(2*exp(2*x) - 4*exp(x)*log(13/4) + 2*log(13/4)^2)*(x^2 - 2*x + 1) + 16*x^2 + 16),x)

[Out]

int(-(exp(-(11*x + exp(exp(2*x) - 2*exp(x)*log(13/4) + log(13/4)^2)*(3*x - 3) - 12)/(4*x + exp(exp(2*x) - 2*ex
p(x)*log(13/4) + log(13/4)^2)*(x - 1) - 4))*(exp(exp(2*x) - 2*exp(x)*log(13/4) + log(13/4)^2)*(exp(x)*log(13/4
)*(2*x - 2*x^2) - exp(2*x)*(2*x - 2*x^2) + 1) + 4))/(exp(exp(2*x) - 2*exp(x)*log(13/4) + log(13/4)^2)*(8*x^2 -
16*x + 8) - 32*x + exp(2*exp(2*x) - 4*exp(x)*log(13/4) + 2*log(13/4)^2)*(x^2 - 2*x + 1) + 16*x^2 + 16), x)

________________________________________________________________________________________

sympy [B]  time = 8.04, size = 66, normalized size = 2.28 \begin {gather*} e^{\frac {- 11 x + \left (3 - 3 x\right ) e^{e^{2 x} - 2 e^{x} \log {\left (\frac {13}{4} \right )} + \log {\left (\frac {13}{4} \right )}^{2}} + 12}{4 x + \left (x - 1\right ) e^{e^{2 x} - 2 e^{x} \log {\left (\frac {13}{4} \right )} + \log {\left (\frac {13}{4} \right )}^{2}} - 4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**2+2*x)*exp(x)**2+(2*x**2-2*x)*ln(13/4)*exp(x)-1)*exp(exp(x)**2-2*ln(13/4)*exp(x)+ln(13/4)**
2)-4)*exp(((-3*x+3)*exp(exp(x)**2-2*ln(13/4)*exp(x)+ln(13/4)**2)-11*x+12)/((x-1)*exp(exp(x)**2-2*ln(13/4)*exp(
x)+ln(13/4)**2)+4*x-4))/((x**2-2*x+1)*exp(exp(x)**2-2*ln(13/4)*exp(x)+ln(13/4)**2)**2+(8*x**2-16*x+8)*exp(exp(
x)**2-2*ln(13/4)*exp(x)+ln(13/4)**2)+16*x**2-32*x+16),x)

[Out]

exp((-11*x + (3 - 3*x)*exp(exp(2*x) - 2*exp(x)*log(13/4) + log(13/4)**2) + 12)/(4*x + (x - 1)*exp(exp(2*x) - 2
*exp(x)*log(13/4) + log(13/4)**2) - 4))

________________________________________________________________________________________