3.8.60 \(\int \frac {5-5 x+e^{\frac {-x^3+\log (50 x)}{x}} (-x^2+(1-x-2 x^3+2 x^4) \log (1-x)+(-1+x) \log (1-x) \log (50 x))}{-25+25 x+e^{\frac {-x^3+\log (50 x)}{x}} (10 x-10 x^2) \log (1-x)+e^{\frac {2 (-x^3+\log (50 x))}{x}} (-x^2+x^3) \log ^2(1-x)} \, dx\)

Optimal. Leaf size=30 \[ \frac {x}{-5+e^{-x^2+\frac {\log (50 x)}{x}} x \log (1-x)} \]

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Rubi [F]  time = 174.95, 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 {5-5 x+e^{\frac {-x^3+\log (50 x)}{x}} \left (-x^2+\left (1-x-2 x^3+2 x^4\right ) \log (1-x)+(-1+x) \log (1-x) \log (50 x)\right )}{-25+25 x+e^{\frac {-x^3+\log (50 x)}{x}} \left (10 x-10 x^2\right ) \log (1-x)+e^{\frac {2 \left (-x^3+\log (50 x)\right )}{x}} \left (-x^2+x^3\right ) \log ^2(1-x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5 - 5*x + E^((-x^3 + Log[50*x])/x)*(-x^2 + (1 - x - 2*x^3 + 2*x^4)*Log[1 - x] + (-1 + x)*Log[1 - x]*Log[5
0*x]))/(-25 + 25*x + E^((-x^3 + Log[50*x])/x)*(10*x - 10*x^2)*Log[1 - x] + E^((2*(-x^3 + Log[50*x]))/x)*(-x^2
+ x^3)*Log[1 - x]^2),x]

[Out]

-5*Defer[Int][E^(2*x^2)/(5*E^x^2 - 50^x^(-1)*x^(1 + x^(-1))*Log[1 - x])^2, x] - 5*Defer[Int][E^(2*x^2)/(x*(-5*
E^x^2 + 50^x^(-1)*x^(1 + x^(-1))*Log[1 - x])^2), x] + 10*Defer[Int][(E^(2*x^2)*x^2)/(-5*E^x^2 + 50^x^(-1)*x^(1
+ x^(-1))*Log[1 - x])^2, x] - 5*Defer[Int][E^(2*x^2)/(Log[1 - x]*(-5*E^x^2 + 50^x^(-1)*x^(1 + x^(-1))*Log[1 -
x])^2), x] - 5*Defer[Int][E^(2*x^2)/((-1 + x)*Log[1 - x]*(-5*E^x^2 + 50^x^(-1)*x^(1 + x^(-1))*Log[1 - x])^2),
x] - Defer[Int][E^x^2/(x*(-5*E^x^2 + 50^x^(-1)*x^(1 + x^(-1))*Log[1 - x])), x] + 2*Defer[Int][(E^x^2*x^2)/(-5
*E^x^2 + 50^x^(-1)*x^(1 + x^(-1))*Log[1 - x]), x] - Defer[Int][E^x^2/(Log[1 - x]*(-5*E^x^2 + 50^x^(-1)*x^(1 +
x^(-1))*Log[1 - x])), x] - Defer[Int][E^x^2/((-1 + x)*Log[1 - x]*(-5*E^x^2 + 50^x^(-1)*x^(1 + x^(-1))*Log[1 -
x])), x] + 5*Defer[Int][(E^(2*x^2)*Log[50*x])/(x*(-5*E^x^2 + 50^x^(-1)*x^(1 + x^(-1))*Log[1 - x])^2), x] + Def
er[Int][(E^x^2*Log[50*x])/(x*(-5*E^x^2 + 50^x^(-1)*x^(1 + x^(-1))*Log[1 - x])), x]

Rubi steps

Aborted

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Mathematica [A]  time = 0.41, size = 38, normalized size = 1.27 \begin {gather*} -\frac {e^{x^2} x}{5 e^{x^2}-50^{\frac {1}{x}} x^{1+\frac {1}{x}} \log (1-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - 5*x + E^((-x^3 + Log[50*x])/x)*(-x^2 + (1 - x - 2*x^3 + 2*x^4)*Log[1 - x] + (-1 + x)*Log[1 - x]
*Log[50*x]))/(-25 + 25*x + E^((-x^3 + Log[50*x])/x)*(10*x - 10*x^2)*Log[1 - x] + E^((2*(-x^3 + Log[50*x]))/x)*
(-x^2 + x^3)*Log[1 - x]^2),x]

[Out]

-((E^x^2*x)/(5*E^x^2 - 50^x^(-1)*x^(1 + x^(-1))*Log[1 - x]))

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fricas [A]  time = 0.66, size = 30, normalized size = 1.00 \begin {gather*} \frac {x}{x e^{\left (-\frac {x^{3} - \log \left (50 \, x\right )}{x}\right )} \log \left (-x + 1\right ) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x-1)*log(-x+1)*log(50*x)+(2*x^4-2*x^3-x+1)*log(-x+1)-x^2)*exp((log(50*x)-x^3)/x)-5*x+5)/((x^3-x^2
)*log(-x+1)^2*exp((log(50*x)-x^3)/x)^2+(-10*x^2+10*x)*log(-x+1)*exp((log(50*x)-x^3)/x)+25*x-25),x, algorithm="
fricas")

[Out]

x/(x*e^(-(x^3 - log(50*x))/x)*log(-x + 1) - 5)

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giac [A]  time = 2.19, size = 30, normalized size = 1.00 \begin {gather*} \frac {x}{x e^{\left (-\frac {x^{3} - \log \left (50 \, x\right )}{x}\right )} \log \left (-x + 1\right ) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x-1)*log(-x+1)*log(50*x)+(2*x^4-2*x^3-x+1)*log(-x+1)-x^2)*exp((log(50*x)-x^3)/x)-5*x+5)/((x^3-x^2
)*log(-x+1)^2*exp((log(50*x)-x^3)/x)^2+(-10*x^2+10*x)*log(-x+1)*exp((log(50*x)-x^3)/x)+25*x-25),x, algorithm="
giac")

[Out]

x/(x*e^(-(x^3 - log(50*x))/x)*log(-x + 1) - 5)

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maple [A]  time = 0.05, size = 28, normalized size = 0.93

method	result	size
risch	\(\frac {x}{\left (50 x \right )^{\frac {1}{x}} \ln \left (1-x \right ) {\mathrm e}^{-x^{2}} x -5}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x-1)*ln(1-x)*ln(50*x)+(2*x^4-2*x^3-x+1)*ln(1-x)-x^2)*exp((ln(50*x)-x^3)/x)-5*x+5)/((x^3-x^2)*ln(1-x)^2*
exp((ln(50*x)-x^3)/x)^2+(-10*x^2+10*x)*ln(1-x)*exp((ln(50*x)-x^3)/x)+25*x-25),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.89, size = 44, normalized size = 1.47 \begin {gather*} \frac {x e^{\left (x^{2}\right )}}{x e^{\left (\frac {2 \, \log \relax (5)}{x} + \frac {\log \relax (2)}{x} + \frac {\log \relax (x)}{x}\right )} \log \left (-x + 1\right ) - 5 \, e^{\left (x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x-1)*log(-x+1)*log(50*x)+(2*x^4-2*x^3-x+1)*log(-x+1)-x^2)*exp((log(50*x)-x^3)/x)-5*x+5)/((x^3-x^2
)*log(-x+1)^2*exp((log(50*x)-x^3)/x)^2+(-10*x^2+10*x)*log(-x+1)*exp((log(50*x)-x^3)/x)+25*x-25),x, algorithm="
maxima")

[Out]

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

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mupad [B]  time = 1.06, size = 31, normalized size = 1.03 \begin {gather*} \frac {x}{{50}^{1/x}\,x^{\frac {1}{x}+1}\,{\mathrm {e}}^{-x^2}\,\ln \left (1-x\right )-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x + exp((log(50*x) - x^3)/x)*(log(1 - x)*(x + 2*x^3 - 2*x^4 - 1) + x^2 - log(50*x)*log(1 - x)*(x - 1))
- 5)/(25*x - exp((2*(log(50*x) - x^3))/x)*log(1 - x)^2*(x^2 - x^3) + exp((log(50*x) - x^3)/x)*log(1 - x)*(10*
x - 10*x^2) - 25),x)

[Out]

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

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sympy [A]  time = 0.56, size = 20, normalized size = 0.67 \begin {gather*} \frac {x}{x e^{\frac {- x^{3} + \log {\left (50 x \right )}}{x}} \log {\left (1 - x \right )} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x-1)*ln(-x+1)*ln(50*x)+(2*x**4-2*x**3-x+1)*ln(-x+1)-x**2)*exp((ln(50*x)-x**3)/x)-5*x+5)/((x**3-x*
*2)*ln(-x+1)**2*exp((ln(50*x)-x**3)/x)**2+(-10*x**2+10*x)*ln(-x+1)*exp((ln(50*x)-x**3)/x)+25*x-25),x)

[Out]

x/(x*exp((-x**3 + log(50*x))/x)*log(1 - x) - 5)

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