3.57.42 \(\int \frac {e^{\sqrt {\frac {200-25 \log (e^x x)}{x}}} (16+\sqrt {\frac {200-25 \log (e^x x)}{x}} (9+x-\log (e^x x))-2 \log (e^x x))}{-16 x^2+2 x^2 \log (e^x x)} \, dx\)

Optimal. Leaf size=26 \[ \frac {e^{5 \sqrt {\frac {8-\log \left (e^x x\right )}{x}}}}{x} \]

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Rubi [B]  time = 0.26, antiderivative size = 101, normalized size of antiderivative = 3.88, number of steps used = 1, number of rules used = 1, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {2288} \begin {gather*} \frac {e^{5 \sqrt {\frac {8-\log \left (e^x x\right )}{x}}} \left (8-\log \left (e^x x\right )\right ) \left (x-\log \left (e^x x\right )+9\right )}{x \left (\frac {e^{-x} \left (e^x x+e^x\right )}{x^2}+\frac {8-\log \left (e^x x\right )}{x^2}\right ) \left (8 x^2-x^2 \log \left (e^x x\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^Sqrt[(200 - 25*Log[E^x*x])/x]*(16 + Sqrt[(200 - 25*Log[E^x*x])/x]*(9 + x - Log[E^x*x]) - 2*Log[E^x*x]))
/(-16*x^2 + 2*x^2*Log[E^x*x]),x]

[Out]

(E^(5*Sqrt[(8 - Log[E^x*x])/x])*(8 - Log[E^x*x])*(9 + x - Log[E^x*x]))/(x*((E^x + E^x*x)/(E^x*x^2) + (8 - Log[
E^x*x])/x^2)*(8*x^2 - x^2*Log[E^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 {e^{5 \sqrt {\frac {8-\log \left (e^x x\right )}{x}}} \left (8-\log \left (e^x x\right )\right ) \left (9+x-\log \left (e^x x\right )\right )}{x \left (\frac {e^{-x} \left (e^x+e^x x\right )}{x^2}+\frac {8-\log \left (e^x x\right )}{x^2}\right ) \left (8 x^2-x^2 \log \left (e^x x\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 26, normalized size = 1.00 \begin {gather*} \frac {e^{5 \sqrt {\frac {8-\log \left (e^x x\right )}{x}}}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^Sqrt[(200 - 25*Log[E^x*x])/x]*(16 + Sqrt[(200 - 25*Log[E^x*x])/x]*(9 + x - Log[E^x*x]) - 2*Log[E^
x*x]))/(-16*x^2 + 2*x^2*Log[E^x*x]),x]

[Out]

E^(5*Sqrt[(8 - Log[E^x*x])/x])/x

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fricas [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(((-log(exp(x)*x)+x+9)*((-25*log(exp(x)*x)+200)/x)^(1/2)-2*log(exp(x)*x)+16)*exp(((-25*log(exp(x)*x)+
200)/x)^(1/2))/(2*x^2*log(exp(x)*x)-16*x^2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   do_alg_rde: unimplemented kernel

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (5 \, {\left (x - \log \left (x e^{x}\right ) + 9\right )} \sqrt {-\frac {\log \left (x e^{x}\right ) - 8}{x}} - 2 \, \log \left (x e^{x}\right ) + 16\right )} e^{\left (5 \, \sqrt {-\frac {\log \left (x e^{x}\right ) - 8}{x}}\right )}}{2 \, {\left (x^{2} \log \left (x e^{x}\right ) - 8 \, x^{2}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(exp(x)*x)+x+9)*((-25*log(exp(x)*x)+200)/x)^(1/2)-2*log(exp(x)*x)+16)*exp(((-25*log(exp(x)*x)+
200)/x)^(1/2))/(2*x^2*log(exp(x)*x)-16*x^2),x, algorithm="giac")

[Out]

integrate(1/2*(5*(x - log(x*e^x) + 9)*sqrt(-(log(x*e^x) - 8)/x) - 2*log(x*e^x) + 16)*e^(5*sqrt(-(log(x*e^x) -
8)/x))/(x^2*log(x*e^x) - 8*x^2), x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (-\ln \left ({\mathrm e}^{x} x \right )+x +9\right ) \sqrt {\frac {-25 \ln \left ({\mathrm e}^{x} x \right )+200}{x}}-2 \ln \left ({\mathrm e}^{x} x \right )+16\right ) {\mathrm e}^{\sqrt {\frac {-25 \ln \left ({\mathrm e}^{x} x \right )+200}{x}}}}{2 x^{2} \ln \left ({\mathrm e}^{x} x \right )-16 x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-ln(exp(x)*x)+x+9)*((-25*ln(exp(x)*x)+200)/x)^(1/2)-2*ln(exp(x)*x)+16)*exp(((-25*ln(exp(x)*x)+200)/x)^(1
/2))/(2*x^2*ln(exp(x)*x)-16*x^2),x)

[Out]

int(((-ln(exp(x)*x)+x+9)*((-25*ln(exp(x)*x)+200)/x)^(1/2)-2*ln(exp(x)*x)+16)*exp(((-25*ln(exp(x)*x)+200)/x)^(1
/2))/(2*x^2*ln(exp(x)*x)-16*x^2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {{\left ({\left (x - 1\right )} x - x^{2} + {\left (x - 17\right )} \log \relax (x) + \log \relax (x)^{2} - 8 \, x + 72\right )} e^{\left (\frac {5 \, \sqrt {-x - \log \relax (x) + 8}}{\sqrt {x}}\right )}}{{\left ({\left (x - 17\right )} \log \relax (x) + \log \relax (x)^{2} - 9 \, x + 72\right )} x} + \frac {1}{2} \, \int -\frac {2 \, {\left (x \log \relax (x)^{4} - {\left (\log \relax (x)^{2} - x - 19 \, \log \relax (x) + 89\right )} x^{3} + 2 \, {\left (x^{2} - 17 \, x\right )} \log \relax (x)^{3} + {\left (x \log \relax (x)^{2} - x^{2} - 2 \, {\left (9 \, x - 1\right )} \log \relax (x) + 81 \, x - 17\right )} x^{2} + 81 \, x^{3} + {\left (x^{3} - 52 \, x^{2} + 433 \, x\right )} \log \relax (x)^{2} - {\left (2 \, {\left (x - 17\right )} \log \relax (x)^{3} + \log \relax (x)^{4} + {\left (x^{2} - 52 \, x + 433\right )} \log \relax (x)^{2} + 73 \, x^{2} - {\left (17 \, x^{2} - 452 \, x + 2448\right )} \log \relax (x) - 1313 \, x + 5184\right )} x - 1296 \, x^{2} - 18 \, {\left (x^{3} - 25 \, x^{2} + 136 \, x\right )} \log \relax (x) + 5184 \, x\right )} e^{\left (\frac {5 \, \sqrt {-x - \log \relax (x) + 8}}{\sqrt {x}}\right )}}{{\left (x \log \relax (x)^{4} + 2 \, {\left (x^{2} - 17 \, x\right )} \log \relax (x)^{3} + 81 \, x^{3} + {\left (x^{3} - 52 \, x^{2} + 433 \, x\right )} \log \relax (x)^{2} - 1296 \, x^{2} - 18 \, {\left (x^{3} - 25 \, x^{2} + 136 \, x\right )} \log \relax (x) + 5184 \, x\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(exp(x)*x)+x+9)*((-25*log(exp(x)*x)+200)/x)^(1/2)-2*log(exp(x)*x)+16)*exp(((-25*log(exp(x)*x)+
200)/x)^(1/2))/(2*x^2*log(exp(x)*x)-16*x^2),x, algorithm="maxima")

[Out]

((x - 1)*x - x^2 + (x - 17)*log(x) + log(x)^2 - 8*x + 72)*e^(5*sqrt(-x - log(x) + 8)/sqrt(x))/(((x - 17)*log(x
) + log(x)^2 - 9*x + 72)*x) + 1/2*integrate(-2*(x*log(x)^4 - (log(x)^2 - x - 19*log(x) + 89)*x^3 + 2*(x^2 - 17
*x)*log(x)^3 + (x*log(x)^2 - x^2 - 2*(9*x - 1)*log(x) + 81*x - 17)*x^2 + 81*x^3 + (x^3 - 52*x^2 + 433*x)*log(x
)^2 - (2*(x - 17)*log(x)^3 + log(x)^4 + (x^2 - 52*x + 433)*log(x)^2 + 73*x^2 - (17*x^2 - 452*x + 2448)*log(x)
- 1313*x + 5184)*x - 1296*x^2 - 18*(x^3 - 25*x^2 + 136*x)*log(x) + 5184*x)*e^(5*sqrt(-x - log(x) + 8)/sqrt(x))
/((x*log(x)^4 + 2*(x^2 - 17*x)*log(x)^3 + 81*x^3 + (x^3 - 52*x^2 + 433*x)*log(x)^2 - 1296*x^2 - 18*(x^3 - 25*x
^2 + 136*x)*log(x) + 5184*x)*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{\sqrt {-\frac {25\,\ln \left (x\,{\mathrm {e}}^x\right )-200}{x}}}\,\left (\sqrt {-\frac {25\,\ln \left (x\,{\mathrm {e}}^x\right )-200}{x}}\,\left (x-\ln \left (x\,{\mathrm {e}}^x\right )+9\right )-2\,\ln \left (x\,{\mathrm {e}}^x\right )+16\right )}{2\,x^2\,\ln \left (x\,{\mathrm {e}}^x\right )-16\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((-(25*log(x*exp(x)) - 200)/x)^(1/2))*((-(25*log(x*exp(x)) - 200)/x)^(1/2)*(x - log(x*exp(x)) + 9) - 2
*log(x*exp(x)) + 16))/(2*x^2*log(x*exp(x)) - 16*x^2),x)

[Out]

int((exp((-(25*log(x*exp(x)) - 200)/x)^(1/2))*((-(25*log(x*exp(x)) - 200)/x)^(1/2)*(x - log(x*exp(x)) + 9) - 2
*log(x*exp(x)) + 16))/(2*x^2*log(x*exp(x)) - 16*x^2), x)

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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(((-ln(exp(x)*x)+x+9)*((-25*ln(exp(x)*x)+200)/x)**(1/2)-2*ln(exp(x)*x)+16)*exp(((-25*ln(exp(x)*x)+200
)/x)**(1/2))/(2*x**2*ln(exp(x)*x)-16*x**2),x)

[Out]

Timed out

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