3.104.1 \(\int \frac {e^{-x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}} (-9-6 x+e^{4 x} (18-12 x-24 x^2)+e^{8 x} (-9+18 x+24 x^2)-3 x^4 \log ^2(5))}{x^4 \log ^2(5)} \, dx\)

Optimal. Leaf size=28 \[ 3 e^{-x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}} \]

________________________________________________________________________________________

Rubi [F]  time = 11.34, 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 (-x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}\right ) \left (-9-6 x+e^{4 x} \left (18-12 x-24 x^2\right )+e^{8 x} \left (-9+18 x+24 x^2\right )-3 x^4 \log ^2(5)\right )}{x^4 \log ^2(5)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-x + (1 + E^(4*x)*(-2 - 2*x) + x + E^(8*x)*(1 + x))/(x^3*Log[5]^2))*(-9 - 6*x + E^(4*x)*(18 - 12*x - 2
4*x^2) + E^(8*x)*(-9 + 18*x + 24*x^2) - 3*x^4*Log[5]^2))/(x^4*Log[5]^2),x]

[Out]

-3*Defer[Int][E^(-x + ((-1 + E^(4*x))^2*(1 + x))/(x^3*Log[5]^2)), x] - (9*Defer[Int][E^(-x + ((-1 + E^(4*x))^2
*(1 + x))/(x^3*Log[5]^2))/x^4, x])/Log[5]^2 + (18*Defer[Int][E^(3*x + ((-1 + E^(4*x))^2*(1 + x))/(x^3*Log[5]^2
))/x^4, x])/Log[5]^2 - (9*Defer[Int][E^(7*x + ((-1 + E^(4*x))^2*(1 + x))/(x^3*Log[5]^2))/x^4, x])/Log[5]^2 - (
6*Defer[Int][E^(-x + ((-1 + E^(4*x))^2*(1 + x))/(x^3*Log[5]^2))/x^3, x])/Log[5]^2 - (12*Defer[Int][E^(3*x + ((
-1 + E^(4*x))^2*(1 + x))/(x^3*Log[5]^2))/x^3, x])/Log[5]^2 + (18*Defer[Int][E^(7*x + ((-1 + E^(4*x))^2*(1 + x)
)/(x^3*Log[5]^2))/x^3, x])/Log[5]^2 - (24*Defer[Int][E^(3*x + ((-1 + E^(4*x))^2*(1 + x))/(x^3*Log[5]^2))/x^2,
x])/Log[5]^2 + (24*Defer[Int][E^(7*x + ((-1 + E^(4*x))^2*(1 + x))/(x^3*Log[5]^2))/x^2, x])/Log[5]^2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {\exp \left (-x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}\right ) \left (-9-6 x+e^{4 x} \left (18-12 x-24 x^2\right )+e^{8 x} \left (-9+18 x+24 x^2\right )-3 x^4 \log ^2(5)\right )}{x^4} \, dx}{\log ^2(5)}\\ &=\frac {\int \left (-\frac {6 \exp \left (3 x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}\right ) \left (-3+2 x+4 x^2\right )}{x^4}+\frac {3 \exp \left (7 x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}\right ) \left (-3+6 x+8 x^2\right )}{x^4}-\frac {3 \exp \left (-x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}\right ) \left (3+2 x+x^4 \log ^2(5)\right )}{x^4}\right ) \, dx}{\log ^2(5)}\\ &=\frac {3 \int \frac {\exp \left (7 x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}\right ) \left (-3+6 x+8 x^2\right )}{x^4} \, dx}{\log ^2(5)}-\frac {3 \int \frac {\exp \left (-x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}\right ) \left (3+2 x+x^4 \log ^2(5)\right )}{x^4} \, dx}{\log ^2(5)}-\frac {6 \int \frac {\exp \left (3 x+\frac {1+e^{4 x} (-2-2 x)+x+e^{8 x} (1+x)}{x^3 \log ^2(5)}\right ) \left (-3+2 x+4 x^2\right )}{x^4} \, dx}{\log ^2(5)}\\ &=\frac {3 \int \frac {\exp \left (7 x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right ) \left (-3+6 x+8 x^2\right )}{x^4} \, dx}{\log ^2(5)}-\frac {3 \int \frac {\exp \left (-x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right ) \left (3+2 x+x^4 \log ^2(5)\right )}{x^4} \, dx}{\log ^2(5)}-\frac {6 \int \frac {\exp \left (3 x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right ) \left (-3+2 x+4 x^2\right )}{x^4} \, dx}{\log ^2(5)}\\ &=\frac {3 \int \left (-\frac {3 \exp \left (7 x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^4}+\frac {6 \exp \left (7 x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^3}+\frac {8 \exp \left (7 x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^2}\right ) \, dx}{\log ^2(5)}-\frac {3 \int \left (\frac {3 \exp \left (-x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^4}+\frac {2 \exp \left (-x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^3}+\exp \left (-x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right ) \log ^2(5)\right ) \, dx}{\log ^2(5)}-\frac {6 \int \left (-\frac {3 \exp \left (3 x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^4}+\frac {2 \exp \left (3 x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^3}+\frac {4 \exp \left (3 x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^2}\right ) \, dx}{\log ^2(5)}\\ &=-\left (3 \int \exp \left (-x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right ) \, dx\right )-\frac {6 \int \frac {\exp \left (-x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^3} \, dx}{\log ^2(5)}-\frac {9 \int \frac {\exp \left (-x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^4} \, dx}{\log ^2(5)}-\frac {9 \int \frac {\exp \left (7 x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^4} \, dx}{\log ^2(5)}-\frac {12 \int \frac {\exp \left (3 x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^3} \, dx}{\log ^2(5)}+\frac {18 \int \frac {\exp \left (3 x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^4} \, dx}{\log ^2(5)}+\frac {18 \int \frac {\exp \left (7 x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^3} \, dx}{\log ^2(5)}-\frac {24 \int \frac {\exp \left (3 x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^2} \, dx}{\log ^2(5)}+\frac {24 \int \frac {\exp \left (7 x+\frac {\left (-1+e^{4 x}\right )^2 (1+x)}{x^3 \log ^2(5)}\right )}{x^2} \, dx}{\log ^2(5)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.19, size = 43, normalized size = 1.54 \begin {gather*} 3 e^{\frac {1+x-2 e^{4 x} (1+x)+e^{8 x} (1+x)-x^4 \log ^2(5)}{x^3 \log ^2(5)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-x + (1 + E^(4*x)*(-2 - 2*x) + x + E^(8*x)*(1 + x))/(x^3*Log[5]^2))*(-9 - 6*x + E^(4*x)*(18 - 12
*x - 24*x^2) + E^(8*x)*(-9 + 18*x + 24*x^2) - 3*x^4*Log[5]^2))/(x^4*Log[5]^2),x]

[Out]

3*E^((1 + x - 2*E^(4*x)*(1 + x) + E^(8*x)*(1 + x) - x^4*Log[5]^2)/(x^3*Log[5]^2))

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 43, normalized size = 1.54 \begin {gather*} 3 \, e^{\left (-\frac {x^{4} \log \relax (5)^{2} - {\left (x + 1\right )} e^{\left (8 \, x\right )} + 2 \, {\left (x + 1\right )} e^{\left (4 \, x\right )} - x - 1}{x^{3} \log \relax (5)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x^2+18*x-9)*exp(4*x)^2+(-24*x^2-12*x+18)*exp(4*x)-3*x^4*log(5)^2-6*x-9)*exp(((x+1)*exp(4*x)^2+(
-2*x-2)*exp(4*x)+x+1)/x^3/log(5)^2)/x^4/log(5)^2/exp(x),x, algorithm="fricas")

[Out]

3*e^(-(x^4*log(5)^2 - (x + 1)*e^(8*x) + 2*(x + 1)*e^(4*x) - x - 1)/(x^3*log(5)^2))

________________________________________________________________________________________

giac [B]  time = 0.57, size = 73, normalized size = 2.61 \begin {gather*} 3 \, e^{\left (-x + \frac {e^{\left (8 \, x\right )}}{x^{2} \log \relax (5)^{2}} - \frac {2 \, e^{\left (4 \, x\right )}}{x^{2} \log \relax (5)^{2}} + \frac {1}{x^{2} \log \relax (5)^{2}} + \frac {e^{\left (8 \, x\right )}}{x^{3} \log \relax (5)^{2}} - \frac {2 \, e^{\left (4 \, x\right )}}{x^{3} \log \relax (5)^{2}} + \frac {1}{x^{3} \log \relax (5)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x^2+18*x-9)*exp(4*x)^2+(-24*x^2-12*x+18)*exp(4*x)-3*x^4*log(5)^2-6*x-9)*exp(((x+1)*exp(4*x)^2+(
-2*x-2)*exp(4*x)+x+1)/x^3/log(5)^2)/x^4/log(5)^2/exp(x),x, algorithm="giac")

[Out]

3*e^(-x + e^(8*x)/(x^2*log(5)^2) - 2*e^(4*x)/(x^2*log(5)^2) + 1/(x^2*log(5)^2) + e^(8*x)/(x^3*log(5)^2) - 2*e^
(4*x)/(x^3*log(5)^2) + 1/(x^3*log(5)^2))

________________________________________________________________________________________

maple [A]  time = 0.12, size = 52, normalized size = 1.86

method	result	size
risch	\(3 \,{\mathrm e}^{-\frac {x^{4} \ln \relax (5)^{2}-x \,{\mathrm e}^{8 x}+2 x \,{\mathrm e}^{4 x}-{\mathrm e}^{8 x}+2 \,{\mathrm e}^{4 x}-x -1}{x^{3} \ln \relax (5)^{2}}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((24*x^2+18*x-9)*exp(4*x)^2+(-24*x^2-12*x+18)*exp(4*x)-3*x^4*ln(5)^2-6*x-9)*exp(((x+1)*exp(4*x)^2+(-2*x-2)
*exp(4*x)+x+1)/x^3/ln(5)^2)/x^4/ln(5)^2/exp(x),x,method=_RETURNVERBOSE)

[Out]

3*exp(-(x^4*ln(5)^2-x*exp(8*x)+2*x*exp(4*x)-exp(8*x)+2*exp(4*x)-x-1)/x^3/ln(5)^2)

________________________________________________________________________________________

maxima [B]  time = 0.62, size = 73, normalized size = 2.61 \begin {gather*} 3 \, e^{\left (-x + \frac {e^{\left (8 \, x\right )}}{x^{2} \log \relax (5)^{2}} - \frac {2 \, e^{\left (4 \, x\right )}}{x^{2} \log \relax (5)^{2}} + \frac {1}{x^{2} \log \relax (5)^{2}} + \frac {e^{\left (8 \, x\right )}}{x^{3} \log \relax (5)^{2}} - \frac {2 \, e^{\left (4 \, x\right )}}{x^{3} \log \relax (5)^{2}} + \frac {1}{x^{3} \log \relax (5)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x^2+18*x-9)*exp(4*x)^2+(-24*x^2-12*x+18)*exp(4*x)-3*x^4*log(5)^2-6*x-9)*exp(((x+1)*exp(4*x)^2+(
-2*x-2)*exp(4*x)+x+1)/x^3/log(5)^2)/x^4/log(5)^2/exp(x),x, algorithm="maxima")

[Out]

3*e^(-x + e^(8*x)/(x^2*log(5)^2) - 2*e^(4*x)/(x^2*log(5)^2) + 1/(x^2*log(5)^2) + e^(8*x)/(x^3*log(5)^2) - 2*e^
(4*x)/(x^3*log(5)^2) + 1/(x^3*log(5)^2))

________________________________________________________________________________________

mupad [B]  time = 7.54, size = 78, normalized size = 2.79 \begin {gather*} 3\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{\frac {1}{x^2\,{\ln \relax (5)}^2}}\,{\mathrm {e}}^{\frac {1}{x^3\,{\ln \relax (5)}^2}}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^{4\,x}}{x^2\,{\ln \relax (5)}^2}}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^{4\,x}}{x^3\,{\ln \relax (5)}^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{8\,x}}{x^2\,{\ln \relax (5)}^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{8\,x}}{x^3\,{\ln \relax (5)}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*exp((x + exp(8*x)*(x + 1) - exp(4*x)*(2*x + 2) + 1)/(x^3*log(5)^2))*(6*x + 3*x^4*log(5)^2 + exp(
4*x)*(12*x + 24*x^2 - 18) - exp(8*x)*(18*x + 24*x^2 - 9) + 9))/(x^4*log(5)^2),x)

[Out]

3*exp(-x)*exp(1/(x^2*log(5)^2))*exp(1/(x^3*log(5)^2))*exp(-(2*exp(4*x))/(x^2*log(5)^2))*exp(-(2*exp(4*x))/(x^3
*log(5)^2))*exp(exp(8*x)/(x^2*log(5)^2))*exp(exp(8*x)/(x^3*log(5)^2))

________________________________________________________________________________________

sympy [A]  time = 0.39, size = 37, normalized size = 1.32 \begin {gather*} 3 e^{- x} e^{\frac {x + \left (- 2 x - 2\right ) e^{4 x} + \left (x + 1\right ) e^{8 x} + 1}{x^{3} \log {\relax (5 )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((24*x**2+18*x-9)*exp(4*x)**2+(-24*x**2-12*x+18)*exp(4*x)-3*x**4*ln(5)**2-6*x-9)*exp(((x+1)*exp(4*x)
**2+(-2*x-2)*exp(4*x)+x+1)/x**3/ln(5)**2)/x**4/ln(5)**2/exp(x),x)

[Out]

3*exp(-x)*exp((x + (-2*x - 2)*exp(4*x) + (x + 1)*exp(8*x) + 1)/(x**3*log(5)**2))

________________________________________________________________________________________