3.29.21 \(\int \frac {e^{-\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} (e^x (-1+x+(-2 x-2 x^2) \log (5)+(-x^2-x^3) \log ^2(5))+e^{\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} (-6+2 x+(-12 x+4 x^2) \log (5)+(-6 x^2+2 x^3) \log ^2(5)))}{1+2 x \log (5)+x^2 \log ^2(5)} \, dx\)

Optimal. Leaf size=25 \[ (-3+x)^2-e^{-50+x-\frac {2 x}{1+x \log (5)}} x \]

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Rubi [F]  time = 4.58, 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 {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (e^x \left (-1+x+\left (-2 x-2 x^2\right ) \log (5)+\left (-x^2-x^3\right ) \log ^2(5)\right )+e^{\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (-6+2 x+\left (-12 x+4 x^2\right ) \log (5)+\left (-6 x^2+2 x^3\right ) \log ^2(5)\right )\right )}{1+2 x \log (5)+x^2 \log ^2(5)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-1 + x + (-2*x - 2*x^2)*Log[5] + (-x^2 - x^3)*Log[5]^2) + E^((2*(25 + x + 25*x*Log[5]))/(1 + x*Log[5
]))*(-6 + 2*x + (-12*x + 4*x^2)*Log[5] + (-6*x^2 + 2*x^3)*Log[5]^2))/(E^((2*(25 + x + 25*x*Log[5]))/(1 + x*Log
[5]))*(1 + 2*x*Log[5] + x^2*Log[5]^2)),x]

[Out]

x^2 + (4*x)/Log[5] + 2/(Log[5]^2*(1 + x*Log[5])) + 6/(Log[5]*(1 + x*Log[5])) - (2*(1 + Log[125]))/(Log[5]^2*(1
+ x*Log[5])) - (2*x*(2 + Log[125]))/Log[5] + (2*Log[1 + x*Log[5]])/Log[5]^2 + (6*(1 + Log[25])*Log[1 + x*Log[
5]])/Log[5]^2 - (4*(2 + Log[125])*Log[1 + x*Log[5]])/Log[5]^2 - Defer[Int][5^(((-50 + x)*x)/(1 + x*Log[5]))*E^
((-50 - x)/(1 + x*Log[5])), x] - Defer[Int][5^(((-50 + x)*x)/(1 + x*Log[5]))*E^((-50 - x)/(1 + x*Log[5]))*x, x
] - (2*Defer[Int][(5^(((-50 + x)*x)/(1 + x*Log[5]))*E^((-50 - x)/(1 + x*Log[5])))/(1 + x*Log[5])^2, x])/Log[5]
+ (2*Defer[Int][(5^(((-50 + x)*x)/(1 + x*Log[5]))*E^((-50 - x)/(1 + x*Log[5])))/(1 + x*Log[5]), x])/Log[5]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (e^x \left (-1+x+\left (-2 x-2 x^2\right ) \log (5)+\left (-x^2-x^3\right ) \log ^2(5)\right )+e^{\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (-6+2 x+\left (-12 x+4 x^2\right ) \log (5)+\left (-6 x^2+2 x^3\right ) \log ^2(5)\right )\right )}{(1+x \log (5))^2} \, dx\\ &=\int \frac {-6+2 x+4 (-3+x) x \log (5)+2 (-3+x) x^2 \log ^2(5)-e^{x-\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}} \left (1+x^3 \log ^2(5)+x^2 \log (5) (2+\log (5))+x (-1+\log (25))\right )}{(1+x \log (5))^2} \, dx\\ &=\int \left (-\frac {6}{(1+x \log (5))^2}+\frac {2 x}{(1+x \log (5))^2}+\frac {4 (-3+x) x \log (5)}{(1+x \log (5))^2}+\frac {2 (-3+x) x^2 \log ^2(5)}{(1+x \log (5))^2}+\frac {5^{\frac {(-50+x) x}{1+x \log (5)}} e^{\frac {-50-x}{1+x \log (5)}} \left (-1-x^3 \log ^2(5)-x^2 \log (5) (2+\log (5))+x (1-\log (25))\right )}{(1+x \log (5))^2}\right ) \, dx\\ &=\frac {6}{\log (5) (1+x \log (5))}+2 \int \frac {x}{(1+x \log (5))^2} \, dx+(4 \log (5)) \int \frac {(-3+x) x}{(1+x \log (5))^2} \, dx+\left (2 \log ^2(5)\right ) \int \frac {(-3+x) x^2}{(1+x \log (5))^2} \, dx+\int \frac {5^{\frac {(-50+x) x}{1+x \log (5)}} e^{\frac {-50-x}{1+x \log (5)}} \left (-1-x^3 \log ^2(5)-x^2 \log (5) (2+\log (5))+x (1-\log (25))\right )}{(1+x \log (5))^2} \, dx\\ &=\frac {6}{\log (5) (1+x \log (5))}+2 \int \left (-\frac {1}{\log (5) (1+x \log (5))^2}+\frac {1}{\log (5) (1+x \log (5))}\right ) \, dx+(4 \log (5)) \int \left (\frac {1}{\log ^2(5)}+\frac {-2-\log (125)}{\log ^2(5) (1+x \log (5))}+\frac {1+\log (125)}{\log ^2(5) (1+x \log (5))^2}\right ) \, dx+\left (2 \log ^2(5)\right ) \int \left (\frac {x}{\log ^2(5)}+\frac {-1-3 \log (5)}{\log ^3(5) (1+x \log (5))^2}+\frac {3 (1+\log (25))}{\log ^3(5) (1+x \log (5))}+\frac {-2-\log (125)}{\log ^3(5)}\right ) \, dx+\int \left (-5^{\frac {(-50+x) x}{1+x \log (5)}} e^{\frac {-50-x}{1+x \log (5)}}-5^{\frac {(-50+x) x}{1+x \log (5)}} e^{\frac {-50-x}{1+x \log (5)}} x-\frac {2\ 5^{\frac {(-50+x) x}{1+x \log (5)}} e^{\frac {-50-x}{1+x \log (5)}}}{\log (5) (1+x \log (5))^2}+\frac {2\ 5^{\frac {(-50+x) x}{1+x \log (5)}} e^{\frac {-50-x}{1+x \log (5)}}}{\log (5) (1+x \log (5))}\right ) \, dx\\ &=x^2+\frac {4 x}{\log (5)}+\frac {2}{\log ^2(5) (1+x \log (5))}+\frac {6}{\log (5) (1+x \log (5))}-\frac {2 (1+\log (125))}{\log ^2(5) (1+x \log (5))}-\frac {2 x (2+\log (125))}{\log (5)}+\frac {2 \log (1+x \log (5))}{\log ^2(5)}+\frac {6 (1+\log (25)) \log (1+x \log (5))}{\log ^2(5)}-\frac {4 (2+\log (125)) \log (1+x \log (5))}{\log ^2(5)}-\frac {2 \int \frac {5^{\frac {(-50+x) x}{1+x \log (5)}} e^{\frac {-50-x}{1+x \log (5)}}}{(1+x \log (5))^2} \, dx}{\log (5)}+\frac {2 \int \frac {5^{\frac {(-50+x) x}{1+x \log (5)}} e^{\frac {-50-x}{1+x \log (5)}}}{1+x \log (5)} \, dx}{\log (5)}-\int 5^{\frac {(-50+x) x}{1+x \log (5)}} e^{\frac {-50-x}{1+x \log (5)}} \, dx-\int 5^{\frac {(-50+x) x}{1+x \log (5)}} e^{\frac {-50-x}{1+x \log (5)}} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.46, size = 29, normalized size = 1.16 \begin {gather*} x \left (-6-e^{x-\frac {2 (25+x+25 x \log (5))}{1+x \log (5)}}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-1 + x + (-2*x - 2*x^2)*Log[5] + (-x^2 - x^3)*Log[5]^2) + E^((2*(25 + x + 25*x*Log[5]))/(1 + x
*Log[5]))*(-6 + 2*x + (-12*x + 4*x^2)*Log[5] + (-6*x^2 + 2*x^3)*Log[5]^2))/(E^((2*(25 + x + 25*x*Log[5]))/(1 +
x*Log[5]))*(1 + 2*x*Log[5] + x^2*Log[5]^2)),x]

[Out]

x*(-6 - E^(x - (2*(25 + x + 25*x*Log[5]))/(1 + x*Log[5])) + x)

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fricas [B]  time = 0.78, size = 54, normalized size = 2.16 \begin {gather*} -{\left (x e^{x} - {\left (x^{2} - 6 \, x\right )} e^{\left (\frac {2 \, {\left (25 \, x \log \relax (5) + x + 25\right )}}{x \log \relax (5) + 1}\right )}\right )} e^{\left (-\frac {2 \, {\left (25 \, x \log \relax (5) + x + 25\right )}}{x \log \relax (5) + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3-x^2)*log(5)^2+(-2*x^2-2*x)*log(5)+x-1)*exp(x)+((2*x^3-6*x^2)*log(5)^2+(4*x^2-12*x)*log(5)+2*
x-6)*exp((25*x*log(5)+x+25)/(x*log(5)+1))^2)/(x^2*log(5)^2+2*x*log(5)+1)/exp((25*x*log(5)+x+25)/(x*log(5)+1))^
2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.20, size = 38, normalized size = 1.52 \begin {gather*} {\left (x^{2} e^{50} - 6 \, x e^{50} - x e^{\left (\frac {x^{2} \log \relax (5) - x}{x \log \relax (5) + 1}\right )}\right )} e^{\left (-50\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3-x^2)*log(5)^2+(-2*x^2-2*x)*log(5)+x-1)*exp(x)+((2*x^3-6*x^2)*log(5)^2+(4*x^2-12*x)*log(5)+2*
x-6)*exp((25*x*log(5)+x+25)/(x*log(5)+1))^2)/(x^2*log(5)^2+2*x*log(5)+1)/exp((25*x*log(5)+x+25)/(x*log(5)+1))^
2,x, algorithm="giac")

[Out]

(x^2*e^50 - 6*x*e^50 - x*e^((x^2*log(5) - x)/(x*log(5) + 1)))*e^(-50)

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maple [A]  time = 0.49, size = 37, normalized size = 1.48

method	result	size
risch	\(x^{2}-6 x -x \,{\mathrm e}^{\frac {x^{2} \ln \relax (5)-50 x \ln \relax (5)-x -50}{x \ln \relax (5)+1}}\)	\(37\)
norman	\(\frac {\left (x^{3} \ln \relax (5) {\mathrm e}^{\frac {50 x \ln \relax (5)+2 x +50}{x \ln \relax (5)+1}}-6 \,{\mathrm e}^{\frac {50 x \ln \relax (5)+2 x +50}{x \ln \relax (5)+1}} x +\left (-6 \ln \relax (5)+1\right ) x^{2} {\mathrm e}^{\frac {50 x \ln \relax (5)+2 x +50}{x \ln \relax (5)+1}}-{\mathrm e}^{x} x -x^{2} \ln \relax (5) {\mathrm e}^{x}\right ) {\mathrm e}^{-\frac {2 \left (25 x \ln \relax (5)+x +25\right )}{x \ln \relax (5)+1}}}{x \ln \relax (5)+1}\)	\(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^3-x^2)*ln(5)^2+(-2*x^2-2*x)*ln(5)+x-1)*exp(x)+((2*x^3-6*x^2)*ln(5)^2+(4*x^2-12*x)*ln(5)+2*x-6)*exp((
25*x*ln(5)+x+25)/(x*ln(5)+1))^2)/(x^2*ln(5)^2+2*x*ln(5)+1)/exp((25*x*ln(5)+x+25)/(x*ln(5)+1))^2,x,method=_RETU
RNVERBOSE)

[Out]

x^2-6*x-x*exp((x^2*ln(5)-50*x*ln(5)-x-50)/(x*ln(5)+1))

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maxima [B]  time = 0.95, size = 225, normalized size = 9.00 \begin {gather*} {\left (\frac {2}{x \log \relax (5)^{5} + \log \relax (5)^{4}} + \frac {x^{2} \log \relax (5) - 4 \, x}{\log \relax (5)^{3}} + \frac {6 \, \log \left (x \log \relax (5) + 1\right )}{\log \relax (5)^{4}}\right )} \log \relax (5)^{2} + 6 \, {\left (\frac {1}{x \log \relax (5)^{4} + \log \relax (5)^{3}} - \frac {x}{\log \relax (5)^{2}} + \frac {2 \, \log \left (x \log \relax (5) + 1\right )}{\log \relax (5)^{3}}\right )} \log \relax (5)^{2} - x e^{\left (x + \frac {2}{x \log \relax (5)^{2} + \log \relax (5)} - \frac {2}{\log \relax (5)} - 50\right )} - 4 \, {\left (\frac {1}{x \log \relax (5)^{4} + \log \relax (5)^{3}} - \frac {x}{\log \relax (5)^{2}} + \frac {2 \, \log \left (x \log \relax (5) + 1\right )}{\log \relax (5)^{3}}\right )} \log \relax (5) - 12 \, {\left (\frac {1}{x \log \relax (5)^{3} + \log \relax (5)^{2}} + \frac {\log \left (x \log \relax (5) + 1\right )}{\log \relax (5)^{2}}\right )} \log \relax (5) + \frac {2}{x \log \relax (5)^{3} + \log \relax (5)^{2}} + \frac {6}{x \log \relax (5)^{2} + \log \relax (5)} + \frac {2 \, \log \left (x \log \relax (5) + 1\right )}{\log \relax (5)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3-x^2)*log(5)^2+(-2*x^2-2*x)*log(5)+x-1)*exp(x)+((2*x^3-6*x^2)*log(5)^2+(4*x^2-12*x)*log(5)+2*
x-6)*exp((25*x*log(5)+x+25)/(x*log(5)+1))^2)/(x^2*log(5)^2+2*x*log(5)+1)/exp((25*x*log(5)+x+25)/(x*log(5)+1))^
2,x, algorithm="maxima")

[Out]

(2/(x*log(5)^5 + log(5)^4) + (x^2*log(5) - 4*x)/log(5)^3 + 6*log(x*log(5) + 1)/log(5)^4)*log(5)^2 + 6*(1/(x*lo
g(5)^4 + log(5)^3) - x/log(5)^2 + 2*log(x*log(5) + 1)/log(5)^3)*log(5)^2 - x*e^(x + 2/(x*log(5)^2 + log(5)) -
2/log(5) - 50) - 4*(1/(x*log(5)^4 + log(5)^3) - x/log(5)^2 + 2*log(x*log(5) + 1)/log(5)^3)*log(5) - 12*(1/(x*l
og(5)^3 + log(5)^2) + log(x*log(5) + 1)/log(5)^2)*log(5) + 2/(x*log(5)^3 + log(5)^2) + 6/(x*log(5)^2 + log(5))
+ 2*log(x*log(5) + 1)/log(5)^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{-\frac {2\,\left (x+25\,x\,\ln \relax (5)+25\right )}{x\,\ln \relax (5)+1}}\,\left ({\mathrm {e}}^{\frac {2\,\left (x+25\,x\,\ln \relax (5)+25\right )}{x\,\ln \relax (5)+1}}\,\left (\ln \relax (5)\,\left (12\,x-4\,x^2\right )-2\,x+{\ln \relax (5)}^2\,\left (6\,x^2-2\,x^3\right )+6\right )+{\mathrm {e}}^x\,\left (\ln \relax (5)\,\left (2\,x^2+2\,x\right )-x+{\ln \relax (5)}^2\,\left (x^3+x^2\right )+1\right )\right )}{{\ln \relax (5)}^2\,x^2+2\,\ln \relax (5)\,x+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(2*(x + 25*x*log(5) + 25))/(x*log(5) + 1))*(exp((2*(x + 25*x*log(5) + 25))/(x*log(5) + 1))*(log(5)*
(12*x - 4*x^2) - 2*x + log(5)^2*(6*x^2 - 2*x^3) + 6) + exp(x)*(log(5)*(2*x + 2*x^2) - x + log(5)^2*(x^2 + x^3)
+ 1)))/(x^2*log(5)^2 + 2*x*log(5) + 1),x)

[Out]

int(-(exp(-(2*(x + 25*x*log(5) + 25))/(x*log(5) + 1))*(exp((2*(x + 25*x*log(5) + 25))/(x*log(5) + 1))*(log(5)*
(12*x - 4*x^2) - 2*x + log(5)^2*(6*x^2 - 2*x^3) + 6) + exp(x)*(log(5)*(2*x + 2*x^2) - x + log(5)^2*(x^2 + x^3)
+ 1)))/(x^2*log(5)^2 + 2*x*log(5) + 1), x)

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sympy [A]  time = 13.58, size = 31, normalized size = 1.24 \begin {gather*} x^{2} - x e^{x} e^{- \frac {2 \left (x + 25 x \log {\relax (5 )} + 25\right )}{x \log {\relax (5 )} + 1}} - 6 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**3-x**2)*ln(5)**2+(-2*x**2-2*x)*ln(5)+x-1)*exp(x)+((2*x**3-6*x**2)*ln(5)**2+(4*x**2-12*x)*ln(5
)+2*x-6)*exp((25*x*ln(5)+x+25)/(x*ln(5)+1))**2)/(x**2*ln(5)**2+2*x*ln(5)+1)/exp((25*x*ln(5)+x+25)/(x*ln(5)+1))
**2,x)

[Out]

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

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