3.83.31 \(\int \frac {e^{\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}} (-1+e^2-e^4-\log (3)+e^{4 x} (4-4 e^2+4 e^4+4 \log (3)))}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx\)

Optimal. Leaf size=31 \[ e^{-2+\frac {-1+e^2-e^4-\log (3)}{5+e^{4 x}-x}} \]

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Rubi [F]  time = 10.10, 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*} \int \frac {\exp \left (\frac {-11+e^2-e^4-2 e^{4 x}+2 x-\log (3)}{5+e^{4 x}-x}\right ) \left (-1+e^2-e^4-\log (3)+e^{4 x} \left (4-4 e^2+4 e^4+4 \log (3)\right )\right )}{25+e^{8 x}+e^{4 x} (10-2 x)-10 x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-11 + E^2 - E^4 - 2*E^(4*x) + 2*x - Log[3])/(5 + E^(4*x) - x))*(-1 + E^2 - E^4 - Log[3] + E^(4*x)*(4
- 4*E^2 + 4*E^4 + 4*Log[3])))/(25 + E^(8*x) + E^(4*x)*(10 - 2*x) - 10*x + x^2),x]

[Out]

-21*(1 - E^2 + E^4 + Log[3])*Defer[Int][E^((-2*E^(4*x) + 2*x - 11*(1 + (-E^2 + E^4 + Log[3])/11))/(5 + E^(4*x)
- x))/(5 + E^(4*x) - x)^2, x] + 4*(1 - E^2 + E^4 + Log[3])*Defer[Int][E^((-2*E^(4*x) + 2*x - 11*(1 + (-E^2 +
E^4 + Log[3])/11))/(5 + E^(4*x) - x))/(5 + E^(4*x) - x), x] + 4*(1 - E^2 + E^4 + Log[3])*Defer[Int][(E^((-2*E^
(4*x) + 2*x - 11*(1 + (-E^2 + E^4 + Log[3])/11))/(5 + E^(4*x) - x))*x)/(5 + E^(4*x) - x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right ) \left (1-4 e^{4 x}\right ) \left (-1+e^2-e^4-\log (3)\right )}{\left (5+e^{4 x}-x\right )^2} \, dx\\ &=\left (-1+e^2-e^4-\log (3)\right ) \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right ) \left (1-4 e^{4 x}\right )}{\left (5+e^{4 x}-x\right )^2} \, dx\\ &=\left (-1+e^2-e^4-\log (3)\right ) \int \left (-\frac {4 \exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right )}{5+e^{4 x}-x}-\frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right ) (-21+4 x)}{\left (-5-e^{4 x}+x\right )^2}\right ) \, dx\\ &=\left (1-e^2+e^4+\log (3)\right ) \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right ) (-21+4 x)}{\left (-5-e^{4 x}+x\right )^2} \, dx+\left (4 \left (1-e^2+e^4+\log (3)\right )\right ) \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right )}{5+e^{4 x}-x} \, dx\\ &=\left (1-e^2+e^4+\log (3)\right ) \int \left (-\frac {21 \exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right )}{\left (5+e^{4 x}-x\right )^2}+\frac {4 \exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right ) x}{\left (5+e^{4 x}-x\right )^2}\right ) \, dx+\left (4 \left (1-e^2+e^4+\log (3)\right )\right ) \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right )}{5+e^{4 x}-x} \, dx\\ &=\left (4 \left (1-e^2+e^4+\log (3)\right )\right ) \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right )}{5+e^{4 x}-x} \, dx+\left (4 \left (1-e^2+e^4+\log (3)\right )\right ) \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right ) x}{\left (5+e^{4 x}-x\right )^2} \, dx-\left (21 \left (1-e^2+e^4+\log (3)\right )\right ) \int \frac {\exp \left (\frac {-2 e^{4 x}+2 x-11 \left (1+\frac {1}{11} \left (-e^2+e^4+\log (3)\right )\right )}{5+e^{4 x}-x}\right )}{\left (5+e^{4 x}-x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.93, size = 44, normalized size = 1.42 \begin {gather*} 3^{-\frac {1}{5+e^{4 x}-x}} e^{-2+\frac {-1+e^2-e^4}{5+e^{4 x}-x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-11 + E^2 - E^4 - 2*E^(4*x) + 2*x - Log[3])/(5 + E^(4*x) - x))*(-1 + E^2 - E^4 - Log[3] + E^(4*
x)*(4 - 4*E^2 + 4*E^4 + 4*Log[3])))/(25 + E^(8*x) + E^(4*x)*(10 - 2*x) - 10*x + x^2),x]

[Out]

E^(-2 + (-1 + E^2 - E^4)/(5 + E^(4*x) - x))/3^(5 + E^(4*x) - x)^(-1)

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fricas [A]  time = 0.77, size = 35, normalized size = 1.13 \begin {gather*} e^{\left (-\frac {2 \, x - e^{4} + e^{2} - 2 \, e^{\left (4 \, x\right )} - \log \relax (3) - 11}{x - e^{\left (4 \, x\right )} - 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(3)+4*exp(4)-4*exp(2)+4)*exp(4*x)-log(3)-exp(4)+exp(2)-1)*exp((-2*exp(4*x)-log(3)-exp(4)+exp(
2)+2*x-11)/(exp(4*x)+5-x))/(exp(4*x)^2+(-2*x+10)*exp(4*x)+x^2-10*x+25),x, algorithm="fricas")

[Out]

e^(-(2*x - e^4 + e^2 - 2*e^(4*x) - log(3) - 11)/(x - e^(4*x) - 5))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (4 \, {\left (e^{4} - e^{2} + \log \relax (3) + 1\right )} e^{\left (4 \, x\right )} - e^{4} + e^{2} - \log \relax (3) - 1\right )} e^{\left (-\frac {2 \, x - e^{4} + e^{2} - 2 \, e^{\left (4 \, x\right )} - \log \relax (3) - 11}{x - e^{\left (4 \, x\right )} - 5}\right )}}{x^{2} - 2 \, {\left (x - 5\right )} e^{\left (4 \, x\right )} - 10 \, x + e^{\left (8 \, x\right )} + 25}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(3)+4*exp(4)-4*exp(2)+4)*exp(4*x)-log(3)-exp(4)+exp(2)-1)*exp((-2*exp(4*x)-log(3)-exp(4)+exp(
2)+2*x-11)/(exp(4*x)+5-x))/(exp(4*x)^2+(-2*x+10)*exp(4*x)+x^2-10*x+25),x, algorithm="giac")

[Out]

integrate((4*(e^4 - e^2 + log(3) + 1)*e^(4*x) - e^4 + e^2 - log(3) - 1)*e^(-(2*x - e^4 + e^2 - 2*e^(4*x) - log
(3) - 11)/(x - e^(4*x) - 5))/(x^2 - 2*(x - 5)*e^(4*x) - 10*x + e^(8*x) + 25), x)

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maple [A]  time = 0.53, size = 36, normalized size = 1.16

method	result	size
risch	\({\mathrm e}^{-\frac {-2 \,{\mathrm e}^{4 x}-\ln \relax (3)-{\mathrm e}^{4}+{\mathrm e}^{2}+2 x -11}{x -{\mathrm e}^{4 x}-5}}\)	\(36\)
norman	\(\frac {x \,{\mathrm e}^{\frac {-2 \,{\mathrm e}^{4 x}-\ln \relax (3)-{\mathrm e}^{4}+{\mathrm e}^{2}+2 x -11}{{\mathrm e}^{4 x}+5-x}}-{\mathrm e}^{4 x} {\mathrm e}^{\frac {-2 \,{\mathrm e}^{4 x}-\ln \relax (3)-{\mathrm e}^{4}+{\mathrm e}^{2}+2 x -11}{{\mathrm e}^{4 x}+5-x}}-5 \,{\mathrm e}^{\frac {-2 \,{\mathrm e}^{4 x}-\ln \relax (3)-{\mathrm e}^{4}+{\mathrm e}^{2}+2 x -11}{{\mathrm e}^{4 x}+5-x}}}{x -{\mathrm e}^{4 x}-5}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*ln(3)+4*exp(4)-4*exp(2)+4)*exp(4*x)-ln(3)-exp(4)+exp(2)-1)*exp((-2*exp(4*x)-ln(3)-exp(4)+exp(2)+2*x-11
)/(exp(4*x)+5-x))/(exp(4*x)^2+(-2*x+10)*exp(4*x)+x^2-10*x+25),x,method=_RETURNVERBOSE)

[Out]

exp(-(-2*exp(4*x)-ln(3)-exp(4)+exp(2)+2*x-11)/(x-exp(4*x)-5))

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maxima [B]  time = 0.66, size = 57, normalized size = 1.84 \begin {gather*} e^{\left (\frac {e^{4}}{x - e^{\left (4 \, x\right )} - 5} - \frac {e^{2}}{x - e^{\left (4 \, x\right )} - 5} + \frac {\log \relax (3)}{x - e^{\left (4 \, x\right )} - 5} + \frac {1}{x - e^{\left (4 \, x\right )} - 5} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(3)+4*exp(4)-4*exp(2)+4)*exp(4*x)-log(3)-exp(4)+exp(2)-1)*exp((-2*exp(4*x)-log(3)-exp(4)+exp(
2)+2*x-11)/(exp(4*x)+5-x))/(exp(4*x)^2+(-2*x+10)*exp(4*x)+x^2-10*x+25),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.59, size = 94, normalized size = 3.03 \begin {gather*} \frac {{\mathrm {e}}^{\frac {2\,x}{{\mathrm {e}}^{4\,x}-x+5}}\,{\mathrm {e}}^{-\frac {11}{{\mathrm {e}}^{4\,x}-x+5}}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^{4\,x}}{{\mathrm {e}}^{4\,x}-x+5}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{{\mathrm {e}}^{4\,x}-x+5}}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^4}{{\mathrm {e}}^{4\,x}-x+5}}}{3^{\frac {1}{{\mathrm {e}}^{4\,x}-x+5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(2*exp(4*x) - 2*x - exp(2) + exp(4) + log(3) + 11)/(exp(4*x) - x + 5))*(exp(4) - exp(2) + log(3) -
exp(4*x)*(4*exp(4) - 4*exp(2) + 4*log(3) + 4) + 1))/(exp(8*x) - 10*x - exp(4*x)*(2*x - 10) + x^2 + 25),x)

[Out]

(exp((2*x)/(exp(4*x) - x + 5))*exp(-11/(exp(4*x) - x + 5))*exp(-(2*exp(4*x))/(exp(4*x) - x + 5))*exp(exp(2)/(e
xp(4*x) - x + 5))*exp(-exp(4)/(exp(4*x) - x + 5)))/3^(1/(exp(4*x) - x + 5))

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sympy [A]  time = 0.54, size = 31, normalized size = 1.00 \begin {gather*} e^{\frac {2 x - 2 e^{4 x} - e^{4} - 11 - \log {\relax (3 )} + e^{2}}{- x + e^{4 x} + 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*ln(3)+4*exp(4)-4*exp(2)+4)*exp(4*x)-ln(3)-exp(4)+exp(2)-1)*exp((-2*exp(4*x)-ln(3)-exp(4)+exp(2)+
2*x-11)/(exp(4*x)+5-x))/(exp(4*x)**2+(-2*x+10)*exp(4*x)+x**2-10*x+25),x)

[Out]

exp((2*x - 2*exp(4*x) - exp(4) - 11 - log(3) + exp(2))/(-x + exp(4*x) + 5))

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