3.28.2 \(\int \frac {e^{-11+2 \log (x) \log (5 x)} (e^5 x-e^{5+\frac {x}{e^5}} x+(-2 e^5 x+e^{\frac {x}{e^5}} (2 e^5 x+x^2)) \log (x)+(-2 e^5 x+2 e^{5+\frac {x}{e^5}} x) \log ^2(x)+(-2 e^5 x+2 e^{5+\frac {x}{e^5}} x) \log (x) \log (5 x))}{\log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {e^{-6+2 \log (x) \log (5 x)} \left (-1+e^{\frac {x}{e^5}}\right ) x^2}{\log (x)} \]

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Rubi [B]  time = 0.44, antiderivative size = 83, normalized size of antiderivative = 2.86, number of steps used = 1, number of rules used = 1, integrand size = 113, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {2288} \begin {gather*} -\frac {x^{2 \log (5 x)} \left (\left (e^5 x-e^{\frac {x}{e^5}+5} x\right ) \log ^2(x)+\left (e^5 x-e^{\frac {x}{e^5}+5} x\right ) \log (5 x) \log (x)\right )}{e^{11} \log ^2(x) \left (\frac {\log (x)}{x}+\frac {\log (5 x)}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-11 + 2*Log[x]*Log[5*x])*(E^5*x - E^(5 + x/E^5)*x + (-2*E^5*x + E^(x/E^5)*(2*E^5*x + x^2))*Log[x] + (-
2*E^5*x + 2*E^(5 + x/E^5)*x)*Log[x]^2 + (-2*E^5*x + 2*E^(5 + x/E^5)*x)*Log[x]*Log[5*x]))/Log[x]^2,x]

[Out]

-((x^(2*Log[5*x])*((E^5*x - E^(5 + x/E^5)*x)*Log[x]^2 + (E^5*x - E^(5 + x/E^5)*x)*Log[x]*Log[5*x]))/(E^11*Log[
x]^2*(Log[x]/x + Log[5*x]/x)))

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {x^{2 \log (5 x)} \left (\left (e^5 x-e^{5+\frac {x}{e^5}} x\right ) \log ^2(x)+\left (e^5 x-e^{5+\frac {x}{e^5}} x\right ) \log (x) \log (5 x)\right )}{e^{11} \log ^2(x) \left (\frac {\log (x)}{x}+\frac {\log (5 x)}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.47, size = 30, normalized size = 1.03 \begin {gather*} \frac {e^{-6+2 \log ^2(x)} \left (-1+e^{\frac {x}{e^5}}\right ) x^{2+\log (25)}}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-11 + 2*Log[x]*Log[5*x])*(E^5*x - E^(5 + x/E^5)*x + (-2*E^5*x + E^(x/E^5)*(2*E^5*x + x^2))*Log[x
] + (-2*E^5*x + 2*E^(5 + x/E^5)*x)*Log[x]^2 + (-2*E^5*x + 2*E^(5 + x/E^5)*x)*Log[x]*Log[5*x]))/Log[x]^2,x]

[Out]

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

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fricas [A]  time = 0.82, size = 43, normalized size = 1.48 \begin {gather*} -\frac {{\left (x^{2} e^{5} - x^{2} e^{\left ({\left (x + 5 \, e^{5}\right )} e^{\left (-5\right )}\right )}\right )} e^{\left (2 \, \log \relax (5) \log \relax (x) + 2 \, \log \relax (x)^{2} - 11\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(5)*exp(x/exp(5))-2*x*exp(5))*log(x)*log(5*x)+(2*x*exp(5)*exp(x/exp(5))-2*x*exp(5))*log(x)^
2+((2*x*exp(5)+x^2)*exp(x/exp(5))-2*x*exp(5))*log(x)-x*exp(5)*exp(x/exp(5))+x*exp(5))*exp(log(x)*log(5*x)-3)^2
/exp(5)/log(x)^2,x, algorithm="fricas")

[Out]

-(x^2*e^5 - x^2*e^((x + 5*e^5)*e^(-5)))*e^(2*log(5)*log(x) + 2*log(x)^2 - 11)/log(x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (2 \, {\left (x e^{5} - x e^{\left (x e^{\left (-5\right )} + 5\right )}\right )} \log \left (5 \, x\right ) \log \relax (x) + 2 \, {\left (x e^{5} - x e^{\left (x e^{\left (-5\right )} + 5\right )}\right )} \log \relax (x)^{2} - x e^{5} + x e^{\left (x e^{\left (-5\right )} + 5\right )} + {\left (2 \, x e^{5} - {\left (x^{2} + 2 \, x e^{5}\right )} e^{\left (x e^{\left (-5\right )}\right )}\right )} \log \relax (x)\right )} e^{\left (2 \, \log \left (5 \, x\right ) \log \relax (x) - 11\right )}}{\log \relax (x)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(5)*exp(x/exp(5))-2*x*exp(5))*log(x)*log(5*x)+(2*x*exp(5)*exp(x/exp(5))-2*x*exp(5))*log(x)^
2+((2*x*exp(5)+x^2)*exp(x/exp(5))-2*x*exp(5))*log(x)-x*exp(5)*exp(x/exp(5))+x*exp(5))*exp(log(x)*log(5*x)-3)^2
/exp(5)/log(x)^2,x, algorithm="giac")

[Out]

integrate(-(2*(x*e^5 - x*e^(x*e^(-5) + 5))*log(5*x)*log(x) + 2*(x*e^5 - x*e^(x*e^(-5) + 5))*log(x)^2 - x*e^5 +
x*e^(x*e^(-5) + 5) + (2*x*e^5 - (x^2 + 2*x*e^5)*e^(x*e^(-5)))*log(x))*e^(2*log(5*x)*log(x) - 11)/log(x)^2, x)

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maple [A]  time = 0.12, size = 27, normalized size = 0.93

method	result	size
risch	\(\frac {x^{2} \left ({\mathrm e}^{{\mathrm e}^{-5} x}-1\right ) x^{2 \ln \left (5 x \right )} {\mathrm e}^{-6}}{\ln \relax (x )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*exp(5)*exp(x/exp(5))-2*x*exp(5))*ln(x)*ln(5*x)+(2*x*exp(5)*exp(x/exp(5))-2*x*exp(5))*ln(x)^2+((2*x*e
xp(5)+x^2)*exp(x/exp(5))-2*x*exp(5))*ln(x)-x*exp(5)*exp(x/exp(5))+x*exp(5))*exp(ln(x)*ln(5*x)-3)^2/exp(5)/ln(x
)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.62, size = 49, normalized size = 1.69 \begin {gather*} \frac {{\left (x^{2} e^{\left (x e^{\left (-5\right )} + 2 \, \log \relax (5) \log \relax (x) + 2 \, \log \relax (x)^{2}\right )} - x^{2} e^{\left (2 \, \log \relax (5) \log \relax (x) + 2 \, \log \relax (x)^{2}\right )}\right )} e^{\left (-6\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(5)*exp(x/exp(5))-2*x*exp(5))*log(x)*log(5*x)+(2*x*exp(5)*exp(x/exp(5))-2*x*exp(5))*log(x)^
2+((2*x*exp(5)+x^2)*exp(x/exp(5))-2*x*exp(5))*log(x)-x*exp(5)*exp(x/exp(5))+x*exp(5))*exp(log(x)*log(5*x)-3)^2
/exp(5)/log(x)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.93, size = 29, normalized size = 1.00 \begin {gather*} \frac {x^{2\,\ln \relax (5)+2}\,{\mathrm {e}}^{2\,{\ln \relax (x)}^2-6}\,\left ({\mathrm {e}}^{x\,{\mathrm {e}}^{-5}}-1\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*log(5*x)*log(x) - 6)*exp(-5)*(log(x)^2*(2*x*exp(5) - 2*x*exp(5)*exp(x*exp(-5))) - x*exp(5) + log(x
)*(2*x*exp(5) - exp(x*exp(-5))*(2*x*exp(5) + x^2)) + log(5*x)*log(x)*(2*x*exp(5) - 2*x*exp(5)*exp(x*exp(-5)))
+ x*exp(5)*exp(x*exp(-5))))/log(x)^2,x)

[Out]

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

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sympy [A]  time = 0.46, size = 31, normalized size = 1.07 \begin {gather*} \frac {\left (x^{2} e^{\frac {x}{e^{5}}} - x^{2}\right ) e^{2 \left (\log {\relax (x )} + \log {\relax (5 )}\right ) \log {\relax (x )} - 6}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(5)*exp(x/exp(5))-2*x*exp(5))*ln(x)*ln(5*x)+(2*x*exp(5)*exp(x/exp(5))-2*x*exp(5))*ln(x)**2+
((2*x*exp(5)+x**2)*exp(x/exp(5))-2*x*exp(5))*ln(x)-x*exp(5)*exp(x/exp(5))+x*exp(5))*exp(ln(x)*ln(5*x)-3)**2/ex
p(5)/ln(x)**2,x)

[Out]

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

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