3.38.88 \(\int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} (16-10 e^{10}-5 x+10 \log (25 x))}{20-20 x+5 x^2} \, dx\)

Optimal. Leaf size=26 \[ e^{\frac {x \left (e^{10}-\frac {6}{5 x}-\log (25 x)\right )}{-2+x}} \]

________________________________________________________________________________________

Rubi [F]  time = 1.93, 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 {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-6 + 5*E^10*x - 5*x*Log[25*x])/(-10 + 5*x))*(16 - 10*E^10 - 5*x + 10*Log[25*x]))/(20 - 20*x + 5*x^2),
x]

[Out]

(2*(3 - 5*E^10)*Defer[Int][E^((-6 + 5*E^10*x - 5*x*Log[25*x])/(-10 + 5*x))/(-2 + x)^2, x])/5 - Defer[Int][E^((
-6 + 5*E^10*x - 5*x*Log[25*x])/(-10 + 5*x))/(-2 + x), x] + 2*Defer[Int][(E^((-6 + 5*E^10*x - 5*x*Log[25*x])/(-
10 + 5*x))*Log[25*x])/(-2 + x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{5 (-2+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{(-2+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16 \left (1-\frac {5 e^{10}}{8}\right )-5 x+10 \log (25 x)\right )}{(2-x)^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x\right )}{(-2+x)^2}+\frac {10 e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \log (25 x)}{(-2+x)^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x\right )}{(-2+x)^2} \, dx+2 \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \log (25 x)}{(-2+x)^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {2 e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (-3+5 e^{10}\right )}{(-2+x)^2}-\frac {5 e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}}}{-2+x}\right ) \, dx+2 \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \log (25 x)}{(-2+x)^2} \, dx\\ &=2 \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \log (25 x)}{(-2+x)^2} \, dx+\frac {1}{5} \left (2 \left (3-5 e^{10}\right )\right ) \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}}}{(-2+x)^2} \, dx-\int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}}}{-2+x} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.37, size = 44, normalized size = 1.69 \begin {gather*} 5^{-1+\frac {2+x}{2-x}} e^{\frac {6-5 e^{10} x}{10-5 x}} x^{-\frac {x}{-2+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-6 + 5*E^10*x - 5*x*Log[25*x])/(-10 + 5*x))*(16 - 10*E^10 - 5*x + 10*Log[25*x]))/(20 - 20*x + 5
*x^2),x]

[Out]

(5^(-1 + (2 + x)/(2 - x))*E^((6 - 5*E^10*x)/(10 - 5*x)))/x^(x/(-2 + x))

________________________________________________________________________________________

fricas [A]  time = 0.63, size = 22, normalized size = 0.85 \begin {gather*} e^{\left (\frac {5 \, x e^{10} - 5 \, x \log \left (25 \, x\right ) - 6}{5 \, {\left (x - 2\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*log(25*x)+5*x*exp(5)^2-6)/(5*x-10))/(5*x^2-20*x+20),x, a
lgorithm="fricas")

[Out]

e^(1/5*(5*x*e^10 - 5*x*log(25*x) - 6)/(x - 2))

________________________________________________________________________________________

giac [A]  time = 0.25, size = 30, normalized size = 1.15 \begin {gather*} e^{\left (\frac {x e^{10}}{x - 2} - \frac {x \log \left (25 \, x\right )}{x - 2} - \frac {6}{5 \, {\left (x - 2\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*log(25*x)+5*x*exp(5)^2-6)/(5*x-10))/(5*x^2-20*x+20),x, a
lgorithm="giac")

[Out]

e^(x*e^10/(x - 2) - x*log(25*x)/(x - 2) - 6/5/(x - 2))

________________________________________________________________________________________

maple [A]  time = 0.14, size = 23, normalized size = 0.88

method	result	size
risch	\({\mathrm e}^{\frac {-5 x \ln \left (25 x \right )+5 x \,{\mathrm e}^{10}-6}{5 x -10}}\)	\(23\)
norman	\(\frac {x \,{\mathrm e}^{\frac {-5 x \ln \left (25 x \right )+5 x \,{\mathrm e}^{10}-6}{5 x -10}}-2 \,{\mathrm e}^{\frac {-5 x \ln \left (25 x \right )+5 x \,{\mathrm e}^{10}-6}{5 x -10}}}{x -2}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*ln(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*ln(25*x)+5*x*exp(5)^2-6)/(5*x-10))/(5*x^2-20*x+20),x,method=_RE
TURNVERBOSE)

[Out]

exp(1/5*(-5*x*ln(25*x)+5*x*exp(10)-6)/(x-2))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{5} \, \int \frac {{\left (5 \, x + 10 \, e^{10} - 10 \, \log \left (25 \, x\right ) - 16\right )} e^{\left (\frac {5 \, x e^{10} - 5 \, x \log \left (25 \, x\right ) - 6}{5 \, {\left (x - 2\right )}}\right )}}{x^{2} - 4 \, x + 4}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*log(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*log(25*x)+5*x*exp(5)^2-6)/(5*x-10))/(5*x^2-20*x+20),x, a
lgorithm="maxima")

[Out]

-1/5*integrate((5*x + 10*e^10 - 10*log(25*x) - 16)*e^(1/5*(5*x*e^10 - 5*x*log(25*x) - 6)/(x - 2))/(x^2 - 4*x +
4), x)

________________________________________________________________________________________

mupad [B]  time = 2.50, size = 52, normalized size = 2.00 \begin {gather*} \frac {{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^{10}}{5\,x-10}}\,{\mathrm {e}}^{-\frac {6}{5\,x-10}}}{5^{\frac {10\,x}{5\,x-10}}\,x^{\frac {5\,x}{5\,x-10}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(5*x*log(25*x) - 5*x*exp(10) + 6)/(5*x - 10))*(5*x - 10*log(25*x) + 10*exp(10) - 16))/(5*x^2 - 20*x
+ 20),x)

[Out]

(exp((5*x*exp(10))/(5*x - 10))*exp(-6/(5*x - 10)))/(5^((10*x)/(5*x - 10))*x^((5*x)/(5*x - 10)))

________________________________________________________________________________________

sympy [A]  time = 0.58, size = 22, normalized size = 0.85 \begin {gather*} e^{\frac {- 5 x \log {\left (25 x \right )} + 5 x e^{10} - 6}{5 x - 10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*ln(25*x)-10*exp(5)**2-5*x+16)*exp((-5*x*ln(25*x)+5*x*exp(5)**2-6)/(5*x-10))/(5*x**2-20*x+20),x)

[Out]

exp((-5*x*log(25*x) + 5*x*exp(10) - 6)/(5*x - 10))

________________________________________________________________________________________