3.96.23 \(\int \frac {e^{-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}} (3 x^2+2 x^3) \log (5)}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} (-10 x^3-2 x^3 \log (3))} \, dx\)

Optimal. Leaf size=27 \[ e^{\frac {2}{5}+\frac {\log (5)}{5-e^{-2+2 x} x^3+\log (3)}} \]

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Rubi [F]  time = 20.05, 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 (-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) \left (3 x^2+2 x^3\right ) \log (5)}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} \left (-10 x^3-2 x^3 \log (3)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-2 + 2*x + (-10 + 2*E^(-2 + 2*x)*x^3 - 2*Log[3] - 5*Log[5])/(-25 + 5*E^(-2 + 2*x)*x^3 - 5*Log[3]))*(3*
x^2 + 2*x^3)*Log[5])/(25 + E^(-4 + 4*x)*x^6 + 10*Log[3] + Log[3]^2 + E^(-2 + 2*x)*(-10*x^3 - 2*x^3*Log[3])),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (5) \int \frac {\exp \left (-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) \left (3 x^2+2 x^3\right )}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} \left (-10 x^3-2 x^3 \log (3)\right )} \, dx\\ &=\log (5) \int \frac {\exp \left (-2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) x^2 (3+2 x)}{25+e^{-4+4 x} x^6+10 \log (3)+\log ^2(3)+e^{-2+2 x} \left (-10 x^3-2 x^3 \log (3)\right )} \, dx\\ &=\log (5) \int \frac {\exp \left (2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) x^2 (3+2 x)}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2} \, dx\\ &=\log (5) \int \left (\frac {3 \exp \left (2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) x^2}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2}+\frac {2 \exp \left (2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) x^3}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2}\right ) \, dx\\ &=(2 \log (5)) \int \frac {\exp \left (2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) x^3}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2} \, dx+(3 \log (5)) \int \frac {\exp \left (2+2 x+\frac {-10+2 e^{-2+2 x} x^3-2 \log (3)-5 \log (5)}{-25+5 e^{-2+2 x} x^3-5 \log (3)}\right ) x^2}{\left (e^{2 x} x^3-5 e^2 \left (1+\frac {\log (3)}{5}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.48, size = 33, normalized size = 1.22 \begin {gather*} 5^{\frac {e^2}{-e^{2 x} x^3+e^2 (5+\log (3))}} e^{2/5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-2 + 2*x + (-10 + 2*E^(-2 + 2*x)*x^3 - 2*Log[3] - 5*Log[5])/(-25 + 5*E^(-2 + 2*x)*x^3 - 5*Log[3]
))*(3*x^2 + 2*x^3)*Log[5])/(25 + E^(-4 + 4*x)*x^6 + 10*Log[3] + Log[3]^2 + E^(-2 + 2*x)*(-10*x^3 - 2*x^3*Log[3
])),x]

[Out]

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

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fricas [B]  time = 0.56, size = 63, normalized size = 2.33 \begin {gather*} e^{\left (-2 \, x + \frac {2 \, {\left (5 \, x^{4} - 4 \, x^{3}\right )} e^{\left (2 \, x - 2\right )} - 2 \, {\left (5 \, x - 4\right )} \log \relax (3) - 50 \, x - 5 \, \log \relax (5) + 40}{5 \, {\left (x^{3} e^{\left (2 \, x - 2\right )} - \log \relax (3) - 5\right )}} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+3*x^2)*log(5)*exp(x-1)^2*exp((2*x^3*exp(x-1)^2-5*log(5)-2*log(3)-10)/(5*x^3*exp(x-1)^2-5*log(
3)-25))/(x^6*exp(x-1)^4+(-2*x^3*log(3)-10*x^3)*exp(x-1)^2+log(3)^2+10*log(3)+25),x, algorithm="fricas")

[Out]

e^(-2*x + 1/5*(2*(5*x^4 - 4*x^3)*e^(2*x - 2) - 2*(5*x - 4)*log(3) - 50*x - 5*log(5) + 40)/(x^3*e^(2*x - 2) - l
og(3) - 5) + 2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (2 \, x^{3} + 3 \, x^{2}\right )} e^{\left (2 \, x + \frac {2 \, x^{3} e^{\left (2 \, x - 2\right )} - 5 \, \log \relax (5) - 2 \, \log \relax (3) - 10}{5 \, {\left (x^{3} e^{\left (2 \, x - 2\right )} - \log \relax (3) - 5\right )}} - 2\right )} \log \relax (5)}{x^{6} e^{\left (4 \, x - 4\right )} - 2 \, {\left (x^{3} \log \relax (3) + 5 \, x^{3}\right )} e^{\left (2 \, x - 2\right )} + \log \relax (3)^{2} + 10 \, \log \relax (3) + 25}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+3*x^2)*log(5)*exp(x-1)^2*exp((2*x^3*exp(x-1)^2-5*log(5)-2*log(3)-10)/(5*x^3*exp(x-1)^2-5*log(
3)-25))/(x^6*exp(x-1)^4+(-2*x^3*log(3)-10*x^3)*exp(x-1)^2+log(3)^2+10*log(3)+25),x, algorithm="giac")

[Out]

integrate((2*x^3 + 3*x^2)*e^(2*x + 1/5*(2*x^3*e^(2*x - 2) - 5*log(5) - 2*log(3) - 10)/(x^3*e^(2*x - 2) - log(3
) - 5) - 2)*log(5)/(x^6*e^(4*x - 4) - 2*(x^3*log(3) + 5*x^3)*e^(2*x - 2) + log(3)^2 + 10*log(3) + 25), x)

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maple [A]  time = 0.19, size = 42, normalized size = 1.56

method	result	size
risch	\({\mathrm e}^{\frac {-2 x^{3} {\mathrm e}^{2 x -2}+2 \ln \relax (3)+5 \ln \relax (5)+10}{-5 x^{3} {\mathrm e}^{2 x -2}+5 \ln \relax (3)+25}}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^3+3*x^2)*ln(5)*exp(x-1)^2*exp((2*x^3*exp(x-1)^2-5*ln(5)-2*ln(3)-10)/(5*x^3*exp(x-1)^2-5*ln(3)-25))/(x
^6*exp(x-1)^4+(-2*x^3*ln(3)-10*x^3)*exp(x-1)^2+ln(3)^2+10*ln(3)+25),x,method=_RETURNVERBOSE)

[Out]

exp(1/5*(-2*x^3*exp(2*x-2)+2*ln(3)+5*ln(5)+10)/(-x^3*exp(2*x-2)+ln(3)+5))

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maxima [A]  time = 0.74, size = 30, normalized size = 1.11 \begin {gather*} e^{\left (-\frac {e^{2} \log \relax (5)}{x^{3} e^{\left (2 \, x\right )} - e^{2} \log \relax (3) - 5 \, e^{2}} + \frac {2}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3+3*x^2)*log(5)*exp(x-1)^2*exp((2*x^3*exp(x-1)^2-5*log(5)-2*log(3)-10)/(5*x^3*exp(x-1)^2-5*log(
3)-25))/(x^6*exp(x-1)^4+(-2*x^3*log(3)-10*x^3)*exp(x-1)^2+log(3)^2+10*log(3)+25),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{2\,x-2}\,{\mathrm {e}}^{\frac {2\,\ln \relax (3)+5\,\ln \relax (5)-2\,x^3\,{\mathrm {e}}^{2\,x-2}+10}{5\,\ln \relax (3)-5\,x^3\,{\mathrm {e}}^{2\,x-2}+25}}\,\ln \relax (5)\,\left (2\,x^3+3\,x^2\right )}{10\,\ln \relax (3)-{\mathrm {e}}^{2\,x-2}\,\left (2\,x^3\,\ln \relax (3)+10\,x^3\right )+{\ln \relax (3)}^2+x^6\,{\mathrm {e}}^{4\,x-4}+25} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x - 2)*exp((2*log(3) + 5*log(5) - 2*x^3*exp(2*x - 2) + 10)/(5*log(3) - 5*x^3*exp(2*x - 2) + 25))*lo
g(5)*(3*x^2 + 2*x^3))/(10*log(3) - exp(2*x - 2)*(2*x^3*log(3) + 10*x^3) + log(3)^2 + x^6*exp(4*x - 4) + 25),x)

[Out]

int((exp(2*x - 2)*exp((2*log(3) + 5*log(5) - 2*x^3*exp(2*x - 2) + 10)/(5*log(3) - 5*x^3*exp(2*x - 2) + 25))*lo
g(5)*(3*x^2 + 2*x^3))/(10*log(3) - exp(2*x - 2)*(2*x^3*log(3) + 10*x^3) + log(3)^2 + x^6*exp(4*x - 4) + 25), x
)

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sympy [A]  time = 0.47, size = 42, normalized size = 1.56 \begin {gather*} e^{\frac {2 x^{3} e^{2 x - 2} - 10 - 5 \log {\relax (5 )} - 2 \log {\relax (3 )}}{5 x^{3} e^{2 x - 2} - 25 - 5 \log {\relax (3 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**3+3*x**2)*ln(5)*exp(x-1)**2*exp((2*x**3*exp(x-1)**2-5*ln(5)-2*ln(3)-10)/(5*x**3*exp(x-1)**2-5*
ln(3)-25))/(x**6*exp(x-1)**4+(-2*x**3*ln(3)-10*x**3)*exp(x-1)**2+ln(3)**2+10*ln(3)+25),x)

[Out]

exp((2*x**3*exp(2*x - 2) - 10 - 5*log(5) - 2*log(3))/(5*x**3*exp(2*x - 2) - 25 - 5*log(3)))

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