3.83.39 \(\int \frac {e^{\frac {e^{2 x}+(-x+x^2) \log (\frac {1}{13} (65+5 x \log (2)))}{\log (\frac {1}{13} (65+5 x \log (2)))}} (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log (\frac {1}{13} (65+5 x \log (2)))+(-13+26 x+(-x+2 x^2) \log (2)) \log ^2(\frac {1}{13} (65+5 x \log (2))))}{(13+x \log (2)) \log ^2(\frac {1}{13} (65+5 x \log (2)))} \, dx\)

Optimal. Leaf size=27 \[ e^{-x+x^2+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}} \]

________________________________________________________________________________________

Rubi [F]  time = 10.80, 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 {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (\frac {1}{13} (65+5 x \log (2))\right )}\right ) \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((E^(2*x) + (-x + x^2)*Log[(65 + 5*x*Log[2])/13])/Log[(65 + 5*x*Log[2])/13])*(-(E^(2*x)*Log[2]) + E^(2*
x)*(26 + 2*x*Log[2])*Log[(65 + 5*x*Log[2])/13] + (-13 + 26*x + (-x + 2*x^2)*Log[2])*Log[(65 + 5*x*Log[2])/13]^
2))/((13 + x*Log[2])*Log[(65 + 5*x*Log[2])/13]^2),x]

[Out]

-Defer[Int][E^((-1 + x)*x + E^(2*x)/Log[5 + (5*x*Log[2])/13]), x] + 2*Defer[Int][E^((-1 + x)*x + E^(2*x)/Log[5
+ (5*x*Log[2])/13])*x, x] - Log[2]*Defer[Int][E^(x + x^2 + E^(2*x)/Log[5 + (5*x*Log[2])/13])/((13 + x*Log[2])
*Log[5 + (5*x*Log[2])/13]^2), x] + 2*Defer[Int][E^(x + x^2 + E^(2*x)/Log[5 + (5*x*Log[2])/13])/Log[5 + (5*x*Lo
g[2])/13], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (5+\frac {5}{13} x \log (2)\right )}\right ) \left (-e^{2 x} \log (2)+e^{2 x} (26+2 x \log (2)) \log \left (\frac {1}{13} (65+5 x \log (2))\right )+\left (-13+26 x+\left (-x+2 x^2\right ) \log (2)\right ) \log ^2\left (\frac {1}{13} (65+5 x \log (2))\right )\right )}{(13+x \log (2)) \log ^2\left (5+\frac {5}{13} x \log (2)\right )} \, dx\\ &=\int \left (-\exp \left (\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (5+\frac {5}{13} x \log (2)\right )}\right )+2 \exp \left (\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (5+\frac {5}{13} x \log (2)\right )}\right ) x+\frac {\exp \left (2 x+\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (5+\frac {5}{13} x \log (2)\right )}\right ) \left (-\log (2)+26 \log \left (5+\frac {5}{13} x \log (2)\right )+2 x \log (2) \log \left (5+\frac {5}{13} x \log (2)\right )\right )}{(13+x \log (2)) \log ^2\left (5+\frac {5}{13} x \log (2)\right )}\right ) \, dx\\ &=2 \int \exp \left (\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (5+\frac {5}{13} x \log (2)\right )}\right ) x \, dx-\int \exp \left (\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (5+\frac {5}{13} x \log (2)\right )}\right ) \, dx+\int \frac {\exp \left (2 x+\frac {e^{2 x}+\left (-x+x^2\right ) \log \left (\frac {1}{13} (65+5 x \log (2))\right )}{\log \left (5+\frac {5}{13} x \log (2)\right )}\right ) \left (-\log (2)+26 \log \left (5+\frac {5}{13} x \log (2)\right )+2 x \log (2) \log \left (5+\frac {5}{13} x \log (2)\right )\right )}{(13+x \log (2)) \log ^2\left (5+\frac {5}{13} x \log (2)\right )} \, dx\\ &=2 \int e^{(-1+x) x+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}} x \, dx-\int e^{(-1+x) x+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}} \, dx+\int \frac {e^{x+x^2+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}} \left (-\log (2)+2 (13+x \log (2)) \log \left (5+\frac {5}{13} x \log (2)\right )\right )}{(13+x \log (2)) \log ^2\left (5+\frac {5}{13} x \log (2)\right )} \, dx\\ &=2 \int e^{(-1+x) x+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}} x \, dx-\int e^{(-1+x) x+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}} \, dx+\int \left (-\frac {e^{x+x^2+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}} \log (2)}{(13+x \log (2)) \log ^2\left (5+\frac {5}{13} x \log (2)\right )}+\frac {2 e^{x+x^2+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}}}{\log \left (5+\frac {5}{13} x \log (2)\right )}\right ) \, dx\\ &=2 \int e^{(-1+x) x+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}} x \, dx+2 \int \frac {e^{x+x^2+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}}}{\log \left (5+\frac {5}{13} x \log (2)\right )} \, dx-\log (2) \int \frac {e^{x+x^2+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}}}{(13+x \log (2)) \log ^2\left (5+\frac {5}{13} x \log (2)\right )} \, dx-\int e^{(-1+x) x+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.19, size = 27, normalized size = 1.00 \begin {gather*} e^{-x+x^2+\frac {e^{2 x}}{\log \left (5+\frac {5}{13} x \log (2)\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((E^(2*x) + (-x + x^2)*Log[(65 + 5*x*Log[2])/13])/Log[(65 + 5*x*Log[2])/13])*(-(E^(2*x)*Log[2]) +
E^(2*x)*(26 + 2*x*Log[2])*Log[(65 + 5*x*Log[2])/13] + (-13 + 26*x + (-x + 2*x^2)*Log[2])*Log[(65 + 5*x*Log[2]
)/13]^2))/((13 + x*Log[2])*Log[(65 + 5*x*Log[2])/13]^2),x]

[Out]

E^(-x + x^2 + E^(2*x)/Log[5 + (5*x*Log[2])/13])

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 33, normalized size = 1.22 \begin {gather*} e^{\left (\frac {{\left (x^{2} - x\right )} \log \left (\frac {5}{13} \, x \log \relax (2) + 5\right ) + e^{\left (2 \, x\right )}}{\log \left (\frac {5}{13} \, x \log \relax (2) + 5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-x)*log(2)+26*x-13)*log(5/13*x*log(2)+5)^2+(2*x*log(2)+26)*exp(2*x)*log(5/13*x*log(2)+5)-log
(2)*exp(2*x))*exp(((x^2-x)*log(5/13*x*log(2)+5)+exp(2*x))/log(5/13*x*log(2)+5))/(x*log(2)+13)/log(5/13*x*log(2
)+5)^2,x, algorithm="fricas")

[Out]

e^(((x^2 - x)*log(5/13*x*log(2) + 5) + e^(2*x))/log(5/13*x*log(2) + 5))

________________________________________________________________________________________

giac [A]  time = 4.32, size = 23, normalized size = 0.85 \begin {gather*} e^{\left (x^{2} - x + \frac {e^{\left (2 \, x\right )}}{\log \left (\frac {5}{13} \, x \log \relax (2) + 5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-x)*log(2)+26*x-13)*log(5/13*x*log(2)+5)^2+(2*x*log(2)+26)*exp(2*x)*log(5/13*x*log(2)+5)-log
(2)*exp(2*x))*exp(((x^2-x)*log(5/13*x*log(2)+5)+exp(2*x))/log(5/13*x*log(2)+5))/(x*log(2)+13)/log(5/13*x*log(2
)+5)^2,x, algorithm="giac")

[Out]

e^(x^2 - x + e^(2*x)/log(5/13*x*log(2) + 5))

________________________________________________________________________________________

maple [A]  time = 0.32, size = 41, normalized size = 1.52

method	result	size
risch	\({\mathrm e}^{\frac {\ln \left (\frac {5 x \ln \relax (2)}{13}+5\right ) x^{2}-\ln \left (\frac {5 x \ln \relax (2)}{13}+5\right ) x +{\mathrm e}^{2 x}}{\ln \left (\frac {5 x \ln \relax (2)}{13}+5\right )}}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2-x)*ln(2)+26*x-13)*ln(5/13*x*ln(2)+5)^2+(2*x*ln(2)+26)*exp(2*x)*ln(5/13*x*ln(2)+5)-ln(2)*exp(2*x))
*exp(((x^2-x)*ln(5/13*x*ln(2)+5)+exp(2*x))/ln(5/13*x*ln(2)+5))/(x*ln(2)+13)/ln(5/13*x*ln(2)+5)^2,x,method=_RET
URNVERBOSE)

[Out]

exp((ln(5/13*x*ln(2)+5)*x^2-ln(5/13*x*ln(2)+5)*x+exp(2*x))/ln(5/13*x*ln(2)+5))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-x)*log(2)+26*x-13)*log(5/13*x*log(2)+5)^2+(2*x*log(2)+26)*exp(2*x)*log(5/13*x*log(2)+5)-log
(2)*exp(2*x))*exp(((x^2-x)*log(5/13*x*log(2)+5)+exp(2*x))/log(5/13*x*log(2)+5))/(x*log(2)+13)/log(5/13*x*log(2
)+5)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

________________________________________________________________________________________

mupad [B]  time = 5.19, size = 25, normalized size = 0.93 \begin {gather*} {\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{2\,x}}{\ln \left (\frac {5\,x\,\ln \relax (2)}{13}+5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(2*x) - log((5*x*log(2))/13 + 5)*(x - x^2))/log((5*x*log(2))/13 + 5))*(exp(2*x)*log(2) + log((5*
x*log(2))/13 + 5)^2*(log(2)*(x - 2*x^2) - 26*x + 13) - log((5*x*log(2))/13 + 5)*exp(2*x)*(2*x*log(2) + 26)))/(
log((5*x*log(2))/13 + 5)^2*(x*log(2) + 13)),x)

[Out]

exp(-x)*exp(x^2)*exp(exp(2*x)/log((5*x*log(2))/13 + 5))

________________________________________________________________________________________

sympy [A]  time = 0.94, size = 34, normalized size = 1.26 \begin {gather*} e^{\frac {\left (x^{2} - x\right ) \log {\left (\frac {5 x \log {\relax (2 )}}{13} + 5 \right )} + e^{2 x}}{\log {\left (\frac {5 x \log {\relax (2 )}}{13} + 5 \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2-x)*ln(2)+26*x-13)*ln(5/13*x*ln(2)+5)**2+(2*x*ln(2)+26)*exp(2*x)*ln(5/13*x*ln(2)+5)-ln(2)*e
xp(2*x))*exp(((x**2-x)*ln(5/13*x*ln(2)+5)+exp(2*x))/ln(5/13*x*ln(2)+5))/(x*ln(2)+13)/ln(5/13*x*ln(2)+5)**2,x)

[Out]

exp(((x**2 - x)*log(5*x*log(2)/13 + 5) + exp(2*x))/log(5*x*log(2)/13 + 5))

________________________________________________________________________________________