3.10.79 \(\int \frac {e^{-x+\frac {e^{-x} (1+128 e^x x^6 \log ^2(5))}{64 x^6 \log ^2(5)}} (-6-x)}{64 x^7 \log ^2(5)} \, dx\)

Optimal. Leaf size=20 \[ e^{2+\frac {e^{-x}}{64 x^6 \log ^2(5)}} \]

________________________________________________________________________________________

Rubi [F]  time = 0.83, 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 {e^{-x} \left (1+128 e^x x^6 \log ^2(5)\right )}{64 x^6 \log ^2(5)}\right ) (-6-x)}{64 x^7 \log ^2(5)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-x + (1 + 128*E^x*x^6*Log[5]^2)/(64*E^x*x^6*Log[5]^2))*(-6 - x))/(64*x^7*Log[5]^2),x]

[Out]

(-3*Defer[Int][E^(2 - x + 1/(64*E^x*x^6*Log[5]^2))/x^7, x])/(32*Log[5]^2) - Defer[Int][E^(2 - x + 1/(64*E^x*x^
6*Log[5]^2))/x^6, x]/(64*Log[5]^2)

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {\exp \left (-x+\frac {e^{-x} \left (1+128 e^x x^6 \log ^2(5)\right )}{64 x^6 \log ^2(5)}\right ) (-6-x)}{x^7} \, dx}{64 \log ^2(5)}\\ &=\frac {\int \frac {e^{2-x+\frac {e^{-x}}{64 x^6 \log ^2(5)}} (-6-x)}{x^7} \, dx}{64 \log ^2(5)}\\ &=\frac {\int \left (-\frac {6 e^{2-x+\frac {e^{-x}}{64 x^6 \log ^2(5)}}}{x^7}-\frac {e^{2-x+\frac {e^{-x}}{64 x^6 \log ^2(5)}}}{x^6}\right ) \, dx}{64 \log ^2(5)}\\ &=-\frac {\int \frac {e^{2-x+\frac {e^{-x}}{64 x^6 \log ^2(5)}}}{x^6} \, dx}{64 \log ^2(5)}-\frac {3 \int \frac {e^{2-x+\frac {e^{-x}}{64 x^6 \log ^2(5)}}}{x^7} \, dx}{32 \log ^2(5)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.17, size = 20, normalized size = 1.00 \begin {gather*} e^{2+\frac {e^{-x}}{64 x^6 \log ^2(5)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-x + (1 + 128*E^x*x^6*Log[5]^2)/(64*E^x*x^6*Log[5]^2))*(-6 - x))/(64*x^7*Log[5]^2),x]

[Out]

E^(2 + 1/(64*E^x*x^6*Log[5]^2))

________________________________________________________________________________________

fricas [B]  time = 0.71, size = 35, normalized size = 1.75 \begin {gather*} e^{\left (x - \frac {{\left (64 \, {\left (x^{7} - 2 \, x^{6}\right )} e^{x} \log \relax (5)^{2} - 1\right )} e^{\left (-x\right )}}{64 \, x^{6} \log \relax (5)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(-x-6)*exp(1/64*(128*x^6*log(5)^2*exp(x)+1)/x^6/log(5)^2/exp(x))/x^7/log(5)^2/exp(x),x, algorit
hm="fricas")

[Out]

e^(x - 1/64*(64*(x^7 - 2*x^6)*e^x*log(5)^2 - 1)*e^(-x)/(x^6*log(5)^2))

________________________________________________________________________________________

giac [A]  time = 0.28, size = 16, normalized size = 0.80 \begin {gather*} e^{\left (\frac {e^{\left (-x\right )}}{64 \, x^{6} \log \relax (5)^{2}} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(-x-6)*exp(1/64*(128*x^6*log(5)^2*exp(x)+1)/x^6/log(5)^2/exp(x))/x^7/log(5)^2/exp(x),x, algorit
hm="giac")

[Out]

e^(1/64*e^(-x)/(x^6*log(5)^2) + 2)

________________________________________________________________________________________

maple [A]  time = 0.14, size = 28, normalized size = 1.40

method	result	size
norman	\({\mathrm e}^{\frac {\left (128 x^{6} \ln \relax (5)^{2} {\mathrm e}^{x}+1\right ) {\mathrm e}^{-x}}{64 x^{6} \ln \relax (5)^{2}}}\)	\(28\)
risch	\({\mathrm e}^{\frac {\left (128 x^{6} \ln \relax (5)^{2} {\mathrm e}^{x}+1\right ) {\mathrm e}^{-x}}{64 x^{6} \ln \relax (5)^{2}}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/64*(-x-6)*exp(1/64*(128*x^6*ln(5)^2*exp(x)+1)/x^6/ln(5)^2/exp(x))/x^7/ln(5)^2/exp(x),x,method=_RETURNVER
BOSE)

[Out]

exp(1/64*(128*x^6*ln(5)^2*exp(x)+1)/x^6/ln(5)^2/exp(x))

________________________________________________________________________________________

maxima [A]  time = 0.84, size = 16, normalized size = 0.80 \begin {gather*} e^{\left (\frac {e^{\left (-x\right )}}{64 \, x^{6} \log \relax (5)^{2}} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(-x-6)*exp(1/64*(128*x^6*log(5)^2*exp(x)+1)/x^6/log(5)^2/exp(x))/x^7/log(5)^2/exp(x),x, algorit
hm="maxima")

[Out]

e^(1/64*e^(-x)/(x^6*log(5)^2) + 2)

________________________________________________________________________________________

mupad [B]  time = 0.93, size = 17, normalized size = 0.85 \begin {gather*} {\mathrm {e}}^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-x}}{64\,x^6\,{\ln \relax (5)}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*exp((exp(-x)*(2*x^6*exp(x)*log(5)^2 + 1/64))/(x^6*log(5)^2))*(x + 6))/(64*x^7*log(5)^2),x)

[Out]

exp(2)*exp(exp(-x)/(64*x^6*log(5)^2))

________________________________________________________________________________________

sympy [A]  time = 0.24, size = 29, normalized size = 1.45 \begin {gather*} e^{\frac {\left (2 x^{6} e^{x} \log {\relax (5 )}^{2} + \frac {1}{64}\right ) e^{- x}}{x^{6} \log {\relax (5 )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(-x-6)*exp(1/64*(128*x**6*ln(5)**2*exp(x)+1)/x**6/ln(5)**2/exp(x))/x**7/ln(5)**2/exp(x),x)

[Out]

exp((2*x**6*exp(x)*log(5)**2 + 1/64)*exp(-x)/(x**6*log(5)**2))

________________________________________________________________________________________