3.95.84 \(\int \frac {e^{-2-\frac {-4 x+e^2 x \log (x^2)}{e^2}} (-e^2 \log (\frac {25}{2})+(-20+e^2 (10-2 x)+4 x) \log (\frac {25}{2}) \log (5-x)+e^2 (5-x) \log (\frac {25}{2}) \log (5-x) \log (x^2))}{(-5+x) \log ^2(5-x)} \, dx\)

Optimal. Leaf size=28 \[ \frac {e^{-x \left (-\frac {4}{e^2}+\log \left (x^2\right )\right )} \log \left (\frac {25}{2}\right )}{\log (5-x)} \]

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Rubi [B]  time = 0.49, antiderivative size = 104, normalized size of antiderivative = 3.71, number of steps used = 1, number of rules used = 1, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {2288} \begin {gather*} \frac {e^{\frac {4 x}{e^2}} \left (x^2\right )^{-x} \left (2 \left (-e^2 (5-x)-2 x+10\right ) \log \left (\frac {25}{2}\right ) \log (5-x)-e^2 (5-x) \log \left (\frac {25}{2}\right ) \log (5-x) \log \left (x^2\right )\right )}{(5-x) \log ^2(5-x) \left (2 \left (2-e^2\right )-e^2 \log \left (x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-2 - (-4*x + E^2*x*Log[x^2])/E^2)*(-(E^2*Log[25/2]) + (-20 + E^2*(10 - 2*x) + 4*x)*Log[25/2]*Log[5 - x
] + E^2*(5 - x)*Log[25/2]*Log[5 - x]*Log[x^2]))/((-5 + x)*Log[5 - x]^2),x]

[Out]

(E^((4*x)/E^2)*(2*(10 - E^2*(5 - x) - 2*x)*Log[25/2]*Log[5 - x] - E^2*(5 - x)*Log[25/2]*Log[5 - x]*Log[x^2]))/
((5 - x)*(x^2)^x*Log[5 - x]^2*(2*(2 - E^2) - E^2*Log[x^2]))

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 {e^{\frac {4 x}{e^2}} \left (x^2\right )^{-x} \left (2 \left (10-e^2 (5-x)-2 x\right ) \log \left (\frac {25}{2}\right ) \log (5-x)-e^2 (5-x) \log \left (\frac {25}{2}\right ) \log (5-x) \log \left (x^2\right )\right )}{(5-x) \log ^2(5-x) \left (2 \left (2-e^2\right )-e^2 \log \left (x^2\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 28, normalized size = 1.00 \begin {gather*} \frac {e^{\frac {4 x}{e^2}} \left (x^2\right )^{-x} \log \left (\frac {25}{2}\right )}{\log (5-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-2 - (-4*x + E^2*x*Log[x^2])/E^2)*(-(E^2*Log[25/2]) + (-20 + E^2*(10 - 2*x) + 4*x)*Log[25/2]*Log
[5 - x] + E^2*(5 - x)*Log[25/2]*Log[5 - x]*Log[x^2]))/((-5 + x)*Log[5 - x]^2),x]

[Out]

(E^((4*x)/E^2)*Log[25/2])/((x^2)^x*Log[5 - x])

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fricas [A]  time = 0.58, size = 34, normalized size = 1.21 \begin {gather*} \frac {e^{\left (-{\left (x e^{2} \log \left (x^{2}\right ) - 4 \, x + 2 \, e^{2}\right )} e^{\left (-2\right )} + 2\right )} \log \left (\frac {25}{2}\right )}{\log \left (-x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5-x)*exp(2)*log(25/2)*log(5-x)*log(x^2)+((-2*x+10)*exp(2)+4*x-20)*log(25/2)*log(5-x)-exp(2)*log(25
/2))/(x-5)/exp(2)/log(5-x)^2/exp((x*exp(2)*log(x^2)-4*x)/exp(2)),x, algorithm="fricas")

[Out]

e^(-(x*e^2*log(x^2) - 4*x + 2*e^2)*e^(-2) + 2)*log(25/2)/log(-x + 5)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5-x)*exp(2)*log(25/2)*log(5-x)*log(x^2)+((-2*x+10)*exp(2)+4*x-20)*log(25/2)*log(5-x)-exp(2)*log(25
/2))/(x-5)/exp(2)/log(5-x)^2/exp((x*exp(2)*log(x^2)-4*x)/exp(2)),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{-4,[0,0,5,4,1,0]%%%}+%%%{4,[0,0,5,4,0,1]%%%}+%%%{1,[0,0,
4,5,1,0]%%%

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maple [C]  time = 0.18, size = 87, normalized size = 3.11

method	result	size
risch	\(-\frac {\left (-2 \ln \relax (5)+\ln \relax (2)\right ) {\mathrm e}^{-\frac {x \left (-i {\mathrm e}^{2} \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i {\mathrm e}^{2} \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-i {\mathrm e}^{2} \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+4 \,{\mathrm e}^{2} \ln \relax (x )-8\right ) {\mathrm e}^{-2}}{2}}}{\ln \left (5-x \right )}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5-x)*exp(2)*ln(25/2)*ln(5-x)*ln(x^2)+((-2*x+10)*exp(2)+4*x-20)*ln(25/2)*ln(5-x)-exp(2)*ln(25/2))/(x-5)/e
xp(2)/ln(5-x)^2/exp((x*exp(2)*ln(x^2)-4*x)/exp(2)),x,method=_RETURNVERBOSE)

[Out]

-(-2*ln(5)+ln(2))/ln(5-x)*exp(-1/2*x*(-I*exp(2)*Pi*csgn(I*x^2)^3+2*I*exp(2)*Pi*csgn(I*x^2)^2*csgn(I*x)-I*exp(2
)*Pi*csgn(I*x^2)*csgn(I*x)^2+4*exp(2)*ln(x)-8)*exp(-2))

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maxima [A]  time = 0.60, size = 30, normalized size = 1.07 \begin {gather*} \frac {{\left (2 \, \log \relax (5) - \log \relax (2)\right )} e^{\left (4 \, x e^{\left (-2\right )} - 2 \, x \log \relax (x)\right )}}{\log \left (-x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5-x)*exp(2)*log(25/2)*log(5-x)*log(x^2)+((-2*x+10)*exp(2)+4*x-20)*log(25/2)*log(5-x)-exp(2)*log(25
/2))/(x-5)/exp(2)/log(5-x)^2/exp((x*exp(2)*log(x^2)-4*x)/exp(2)),x, algorithm="maxima")

[Out]

(2*log(5) - log(2))*e^(4*x*e^(-2) - 2*x*log(x))/log(-x + 5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{-2}\,{\mathrm {e}}^{{\mathrm {e}}^{-2}\,\left (4\,x-x\,\ln \left (x^2\right )\,{\mathrm {e}}^2\right )}\,\left ({\mathrm {e}}^2\,\ln \left (\frac {25}{2}\right )+\ln \left (\frac {25}{2}\right )\,\ln \left (5-x\right )\,\left ({\mathrm {e}}^2\,\left (2\,x-10\right )-4\,x+20\right )+\ln \left (x^2\right )\,{\mathrm {e}}^2\,\ln \left (\frac {25}{2}\right )\,\ln \left (5-x\right )\,\left (x-5\right )\right )}{{\ln \left (5-x\right )}^2\,\left (x-5\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2)*exp(exp(-2)*(4*x - x*log(x^2)*exp(2)))*(exp(2)*log(25/2) + log(25/2)*log(5 - x)*(exp(2)*(2*x - 1
0) - 4*x + 20) + log(x^2)*exp(2)*log(25/2)*log(5 - x)*(x - 5)))/(log(5 - x)^2*(x - 5)),x)

[Out]

int(-(exp(-2)*exp(exp(-2)*(4*x - x*log(x^2)*exp(2)))*(exp(2)*log(25/2) + log(25/2)*log(5 - x)*(exp(2)*(2*x - 1
0) - 4*x + 20) + log(x^2)*exp(2)*log(25/2)*log(5 - x)*(x - 5)))/(log(5 - x)^2*(x - 5)), x)

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sympy [A]  time = 0.53, size = 31, normalized size = 1.11 \begin {gather*} \frac {\left (- \log {\relax (2 )} + 2 \log {\relax (5 )}\right ) e^{- \frac {x e^{2} \log {\left (x^{2} \right )} - 4 x}{e^{2}}}}{\log {\left (5 - x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5-x)*exp(2)*ln(25/2)*ln(5-x)*ln(x**2)+((-2*x+10)*exp(2)+4*x-20)*ln(25/2)*ln(5-x)-exp(2)*ln(25/2))/
(x-5)/exp(2)/ln(5-x)**2/exp((x*exp(2)*ln(x**2)-4*x)/exp(2)),x)

[Out]

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

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