3.53.57 \(\int \frac {e^{\log (5-e^{x+x^2}) \log (-\frac {x}{4})} ((-5+e^{x+x^2}) \log (5-e^{x+x^2})+e^{x+x^2} (x+2 x^2) \log (-\frac {x}{4}))}{-5 x+e^{x+x^2} x} \, dx\)

Optimal. Leaf size=21 \[ e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \]

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Rubi [F]  time = 4.81, 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^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \left (\left (-5+e^{x+x^2}\right ) \log \left (5-e^{x+x^2}\right )+e^{x+x^2} \left (x+2 x^2\right ) \log \left (-\frac {x}{4}\right )\right )}{-5 x+e^{x+x^2} x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(Log[5 - E^(x + x^2)]*Log[-1/4*x])*((-5 + E^(x + x^2))*Log[5 - E^(x + x^2)] + E^(x + x^2)*(x + 2*x^2)*L
og[-1/4*x]))/(-5*x + E^(x + x^2)*x),x]

[Out]

Defer[Int][(E^(Log[5 - E^(x + x^2)]*Log[-1/4*x])*Log[5 - E^(x + x^2)])/x, x] + 4*Defer[Subst][Defer[Int][E^(Lo
g[5 - E^(4*x*(1 + 4*x))]*Log[-x])*Log[-x], x], x, x/4] - 20*Defer[Subst][Defer[Int][(5 - E^(4*x*(1 + 4*x)))^(-
1 + Log[-x])*Log[-x], x], x, x/4] + 32*Defer[Subst][Defer[Int][E^(Log[5 - E^(4*x*(1 + 4*x))]*Log[-x])*x*Log[-x
], x], x, x/4] - 160*Defer[Subst][Defer[Int][(5 - E^(4*x*(1 + 4*x)))^(-1 + Log[-x])*x*Log[-x], x], x, x/4]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {5 e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} (1+2 x) \log \left (-\frac {x}{4}\right )}{-5+e^{x+x^2}}+\frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \left (\log \left (5-e^{x+x^2}\right )+x \log \left (-\frac {x}{4}\right )+2 x^2 \log \left (-\frac {x}{4}\right )\right )}{x}\right ) \, dx\\ &=5 \int \frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} (1+2 x) \log \left (-\frac {x}{4}\right )}{-5+e^{x+x^2}} \, dx+\int \frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \left (\log \left (5-e^{x+x^2}\right )+x \log \left (-\frac {x}{4}\right )+2 x^2 \log \left (-\frac {x}{4}\right )\right )}{x} \, dx\\ &=5 \int \left (5-e^{x+x^2}\right )^{-1+\log \left (-\frac {x}{4}\right )} (-1-2 x) \log \left (-\frac {x}{4}\right ) \, dx+\int \left (\frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \log \left (5-e^{x+x^2}\right )}{x}+e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} (1+2 x) \log \left (-\frac {x}{4}\right )\right ) \, dx\\ &=20 \operatorname {Subst}\left (\int \left (5-e^{4 x (1+4 x)}\right )^{-1+\log (-x)} (-1-8 x) \log (-x) \, dx,x,\frac {x}{4}\right )+\int \frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \log \left (5-e^{x+x^2}\right )}{x} \, dx+\int e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} (1+2 x) \log \left (-\frac {x}{4}\right ) \, dx\\ &=4 \operatorname {Subst}\left (\int e^{\log \left (5-e^{4 x (1+4 x)}\right ) \log (-x)} (1+8 x) \log (-x) \, dx,x,\frac {x}{4}\right )+20 \operatorname {Subst}\left (\int \left (-\left (5-e^{4 x (1+4 x)}\right )^{-1+\log (-x)} \log (-x)-8 \left (5-e^{4 x (1+4 x)}\right )^{-1+\log (-x)} x \log (-x)\right ) \, dx,x,\frac {x}{4}\right )+\int \frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \log \left (5-e^{x+x^2}\right )}{x} \, dx\\ &=4 \operatorname {Subst}\left (\int \left (e^{\log \left (5-e^{4 x (1+4 x)}\right ) \log (-x)} \log (-x)+8 e^{\log \left (5-e^{4 x (1+4 x)}\right ) \log (-x)} x \log (-x)\right ) \, dx,x,\frac {x}{4}\right )-20 \operatorname {Subst}\left (\int \left (5-e^{4 x (1+4 x)}\right )^{-1+\log (-x)} \log (-x) \, dx,x,\frac {x}{4}\right )-160 \operatorname {Subst}\left (\int \left (5-e^{4 x (1+4 x)}\right )^{-1+\log (-x)} x \log (-x) \, dx,x,\frac {x}{4}\right )+\int \frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \log \left (5-e^{x+x^2}\right )}{x} \, dx\\ &=4 \operatorname {Subst}\left (\int e^{\log \left (5-e^{4 x (1+4 x)}\right ) \log (-x)} \log (-x) \, dx,x,\frac {x}{4}\right )-20 \operatorname {Subst}\left (\int \left (5-e^{4 x (1+4 x)}\right )^{-1+\log (-x)} \log (-x) \, dx,x,\frac {x}{4}\right )+32 \operatorname {Subst}\left (\int e^{\log \left (5-e^{4 x (1+4 x)}\right ) \log (-x)} x \log (-x) \, dx,x,\frac {x}{4}\right )-160 \operatorname {Subst}\left (\int \left (5-e^{4 x (1+4 x)}\right )^{-1+\log (-x)} x \log (-x) \, dx,x,\frac {x}{4}\right )+\int \frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \log \left (5-e^{x+x^2}\right )}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.78, size = 18, normalized size = 0.86 \begin {gather*} \left (5-e^{x+x^2}\right )^{\log \left (-\frac {x}{4}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5 - E^(x + x^2))^Log[-1/4*x]

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fricas [A]  time = 0.46, size = 17, normalized size = 0.81 \begin {gather*} e^{\left (\log \left (-\frac {1}{4} \, x\right ) \log \left (-e^{\left (x^{2} + x\right )} + 5\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(log(-1/4*x)*log(-e^(x^2 + x) + 5))

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giac [A]  time = 0.25, size = 17, normalized size = 0.81 \begin {gather*} e^{\left (\log \left (-\frac {1}{4} \, x\right ) \log \left (-e^{\left (x^{2} + x\right )} + 5\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(log(-1/4*x)*log(-e^(x^2 + x) + 5))

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maple [A]  time = 0.04, size = 16, normalized size = 0.76

method	result	size
risch	\(\left (-\frac {x}{4}\right )^{\ln \left (-{\mathrm e}^{\left (x +1\right ) x}+5\right )}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x^2+x)-5)*ln(-exp(x^2+x)+5)+(2*x^2+x)*exp(x^2+x)*ln(-1/4*x))*exp(ln(-1/4*x)*ln(-exp(x^2+x)+5))/(x*ex
p(x^2+x)-5*x),x,method=_RETURNVERBOSE)

[Out]

(-1/4*x)^ln(-exp((x+1)*x)+5)

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maxima [A]  time = 0.64, size = 33, normalized size = 1.57 \begin {gather*} e^{\left (-2 \, \log \relax (2) \log \left (-e^{\left (x^{2} + x\right )} + 5\right ) + \log \left (-x\right ) \log \left (-e^{\left (x^{2} + x\right )} + 5\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.52, size = 15, normalized size = 0.71 \begin {gather*} {\left (5-{\mathrm {e}}^{x^2+x}\right )}^{\ln \left (-\frac {x}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(5 - exp(x + x^2))*log(-x/4))*(log(5 - exp(x + x^2))*(exp(x + x^2) - 5) + log(-x/4)*exp(x + x^2)*
(x + 2*x^2)))/(5*x - x*exp(x + x^2)),x)

[Out]

(5 - exp(x + x^2))^log(-x/4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x**2+x)-5)*ln(-exp(x**2+x)+5)+(2*x**2+x)*exp(x**2+x)*ln(-1/4*x))*exp(ln(-1/4*x)*ln(-exp(x**2+x
)+5))/(x*exp(x**2+x)-5*x),x)

[Out]

Timed out

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