3.51.89 \(\int \frac {e^{\frac {1}{-4+e \log (x)}} (800-320 x+e (-50+10 x)+e (-400+160 x) \log (x)+e^2 (50-20 x) \log ^2(x))}{16-8 e \log (x)+e^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=23 \[ 5+10 e^{-\frac {1}{4-e \log (x)}} (5-x) x \]

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Rubi [A]  time = 0.10, antiderivative size = 44, normalized size of antiderivative = 1.91, number of steps used = 1, number of rules used = 1, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {2288} \begin {gather*} \frac {10 (5-x) x e^{\frac {1}{e \log (x)-4}} (4-e \log (x))^2}{e^2 \log ^2(x)-8 e \log (x)+16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-4 + E*Log[x])^(-1)*(800 - 320*x + E*(-50 + 10*x) + E*(-400 + 160*x)*Log[x] + E^2*(50 - 20*x)*Log[x]^2
))/(16 - 8*E*Log[x] + E^2*Log[x]^2),x]

[Out]

(10*E^(-4 + E*Log[x])^(-1)*(5 - x)*x*(4 - E*Log[x])^2)/(16 - 8*E*Log[x] + 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 {10 e^{\frac {1}{-4+e \log (x)}} (5-x) x (4-e \log (x))^2}{16-8 e \log (x)+e^2 \log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 16, normalized size = 0.70 \begin {gather*} -10 e^{\frac {1}{-4+e \log (x)}} (-5+x) x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-4 + E*Log[x])^(-1)*(800 - 320*x + E*(-50 + 10*x) + E*(-400 + 160*x)*Log[x] + E^2*(50 - 20*x)*Lo
g[x]^2))/(16 - 8*E*Log[x] + E^2*Log[x]^2),x]

[Out]

-10*E^(-4 + E*Log[x])^(-1)*(-5 + x)*x

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fricas [A]  time = 0.70, size = 19, normalized size = 0.83 \begin {gather*} -10 \, {\left (x^{2} - 5 \, x\right )} e^{\left (\frac {1}{e \log \relax (x) - 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x+50)*exp(1)^2*log(x)^2+(160*x-400)*exp(1)*log(x)+(10*x-50)*exp(1)-320*x+800)/(exp(1)^2*log(x)
^2-8*exp(1)*log(x)+16)/exp(-1/(exp(1)*log(x)-4)),x, algorithm="fricas")

[Out]

-10*(x^2 - 5*x)*e^(1/(e*log(x) - 4))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {10 \, {\left ({\left (2 \, x - 5\right )} e^{2} \log \relax (x)^{2} - 8 \, {\left (2 \, x - 5\right )} e \log \relax (x) - {\left (x - 5\right )} e + 32 \, x - 80\right )} e^{\left (\frac {1}{e \log \relax (x) - 4}\right )}}{e^{2} \log \relax (x)^{2} - 8 \, e \log \relax (x) + 16}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x+50)*exp(1)^2*log(x)^2+(160*x-400)*exp(1)*log(x)+(10*x-50)*exp(1)-320*x+800)/(exp(1)^2*log(x)
^2-8*exp(1)*log(x)+16)/exp(-1/(exp(1)*log(x)-4)),x, algorithm="giac")

[Out]

integrate(-10*((2*x - 5)*e^2*log(x)^2 - 8*(2*x - 5)*e*log(x) - (x - 5)*e + 32*x - 80)*e^(1/(e*log(x) - 4))/(e^
2*log(x)^2 - 8*e*log(x) + 16), x)

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maple [A]  time = 0.13, size = 21, normalized size = 0.91

method	result	size
risch	\(\left (-10 x^{2}+50 x \right ) {\mathrm e}^{\frac {1}{{\mathrm e} \ln \relax (x )-4}}\)	\(21\)
norman	\(\frac {\left (-200 x +40 x^{2}+50 x \,{\mathrm e} \ln \relax (x )-10 x^{2} {\mathrm e} \ln \relax (x )\right ) {\mathrm e}^{\frac {1}{{\mathrm e} \ln \relax (x )-4}}}{{\mathrm e} \ln \relax (x )-4}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-20*x+50)*exp(1)^2*ln(x)^2+(160*x-400)*exp(1)*ln(x)+(10*x-50)*exp(1)-320*x+800)/(exp(1)^2*ln(x)^2-8*exp(
1)*ln(x)+16)/exp(-1/(exp(1)*ln(x)-4)),x,method=_RETURNVERBOSE)

[Out]

(-10*x^2+50*x)*exp(1/(exp(1)*ln(x)-4))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 320 \, x^{2} e^{\left (\frac {1}{e \log \relax (x) - 4} - 1\right )} - 10 \, x^{2} e^{\left (\frac {1}{e \log \relax (x) - 4}\right )} - 800 \, x e^{\left (\frac {1}{e \log \relax (x) - 4} - 1\right )} + 50 \, x e^{\left (\frac {1}{e \log \relax (x) - 4}\right )} - 10 \, \int \frac {{\left ({\left (2 \, x e^{2} - 5 \, e^{2}\right )} \log \relax (x)^{2} - 8 \, {\left (2 \, x e - 5 \, e\right )} \log \relax (x)\right )} e^{\left (\frac {1}{e \log \relax (x) - 4}\right )}}{e^{2} \log \relax (x)^{2} - 8 \, e \log \relax (x) + 16}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x+50)*exp(1)^2*log(x)^2+(160*x-400)*exp(1)*log(x)+(10*x-50)*exp(1)-320*x+800)/(exp(1)^2*log(x)
^2-8*exp(1)*log(x)+16)/exp(-1/(exp(1)*log(x)-4)),x, algorithm="maxima")

[Out]

320*x^2*e^(1/(e*log(x) - 4) - 1) - 10*x^2*e^(1/(e*log(x) - 4)) - 800*x*e^(1/(e*log(x) - 4) - 1) + 50*x*e^(1/(e
*log(x) - 4)) - 10*integrate(((2*x*e^2 - 5*e^2)*log(x)^2 - 8*(2*x*e - 5*e)*log(x))*e^(1/(e*log(x) - 4))/(e^2*l
og(x)^2 - 8*e*log(x) + 16), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {1}{\mathrm {e}\,\ln \relax (x)-4}}\,\left (-{\mathrm {e}}^2\,\left (20\,x-50\right )\,{\ln \relax (x)}^2+\mathrm {e}\,\left (160\,x-400\right )\,\ln \relax (x)-320\,x+\mathrm {e}\,\left (10\,x-50\right )+800\right )}{{\mathrm {e}}^2\,{\ln \relax (x)}^2-8\,\mathrm {e}\,\ln \relax (x)+16} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1/(exp(1)*log(x) - 4))*(exp(1)*(10*x - 50) - 320*x - exp(2)*log(x)^2*(20*x - 50) + exp(1)*log(x)*(160
*x - 400) + 800))/(exp(2)*log(x)^2 - 8*exp(1)*log(x) + 16),x)

[Out]

int((exp(1/(exp(1)*log(x) - 4))*(exp(1)*(10*x - 50) - 320*x - exp(2)*log(x)^2*(20*x - 50) + exp(1)*log(x)*(160
*x - 400) + 800))/(exp(2)*log(x)^2 - 8*exp(1)*log(x) + 16), x)

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sympy [A]  time = 4.41, size = 19, normalized size = 0.83 \begin {gather*} \left (- 10 x^{2} + 50 x\right ) e^{\frac {1}{e \log {\relax (x )} - 4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x+50)*exp(1)**2*ln(x)**2+(160*x-400)*exp(1)*ln(x)+(10*x-50)*exp(1)-320*x+800)/(exp(1)**2*ln(x)
**2-8*exp(1)*ln(x)+16)/exp(-1/(exp(1)*ln(x)-4)),x)

[Out]

(-10*x**2 + 50*x)*exp(1/(E*log(x) - 4))

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