3.44.29 \(\int \frac {-625 e^x x+e^x (625-625 x) \log (3 e^x)+e^x (-100+100 x) \log ^2(3 e^x)}{10000 x^2-3200 x^2 \log (3 e^x)+256 x^2 \log ^2(3 e^x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {e^x}{\frac {64 x}{25}-\frac {16 x}{\log \left (3 e^x\right )}} \]

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Rubi [F]  time = 1.27, 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 {-625 e^x x+e^x (625-625 x) \log \left (3 e^x\right )+e^x (-100+100 x) \log ^2\left (3 e^x\right )}{10000 x^2-3200 x^2 \log \left (3 e^x\right )+256 x^2 \log ^2\left (3 e^x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-625*E^x*x + E^x*(625 - 625*x)*Log[3*E^x] + E^x*(-100 + 100*x)*Log[3*E^x]^2)/(10000*x^2 - 3200*x^2*Log[3*
E^x] + 256*x^2*Log[3*E^x]^2),x]

[Out]

(25*E^x)/(64*x) - (625*Defer[Int][E^x/(x*(-25 + 4*Log[3*E^x])^2), x])/16 - (625*Defer[Int][E^x/(x^2*(-25 + 4*L
og[3*E^x])), x])/64 + (625*Defer[Int][E^x/(x*(-25 + 4*Log[3*E^x])), x])/64

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 e^x \left (-25 x-25 (-1+x) \log \left (3 e^x\right )+4 (-1+x) \log ^2\left (3 e^x\right )\right )}{16 x^2 \left (25-4 \log \left (3 e^x\right )\right )^2} \, dx\\ &=\frac {25}{16} \int \frac {e^x \left (-25 x-25 (-1+x) \log \left (3 e^x\right )+4 (-1+x) \log ^2\left (3 e^x\right )\right )}{x^2 \left (25-4 \log \left (3 e^x\right )\right )^2} \, dx\\ &=\frac {25}{16} \int \left (\frac {e^x (-1+x)}{4 x^2}-\frac {25 e^x}{x \left (-25+4 \log \left (3 e^x\right )\right )^2}+\frac {25 e^x (-1+x)}{4 x^2 \left (-25+4 \log \left (3 e^x\right )\right )}\right ) \, dx\\ &=\frac {25}{64} \int \frac {e^x (-1+x)}{x^2} \, dx+\frac {625}{64} \int \frac {e^x (-1+x)}{x^2 \left (-25+4 \log \left (3 e^x\right )\right )} \, dx-\frac {625}{16} \int \frac {e^x}{x \left (-25+4 \log \left (3 e^x\right )\right )^2} \, dx\\ &=\frac {25 e^x}{64 x}+\frac {625}{64} \int \left (-\frac {e^x}{x^2 \left (-25+4 \log \left (3 e^x\right )\right )}+\frac {e^x}{x \left (-25+4 \log \left (3 e^x\right )\right )}\right ) \, dx-\frac {625}{16} \int \frac {e^x}{x \left (-25+4 \log \left (3 e^x\right )\right )^2} \, dx\\ &=\frac {25 e^x}{64 x}-\frac {625}{64} \int \frac {e^x}{x^2 \left (-25+4 \log \left (3 e^x\right )\right )} \, dx+\frac {625}{64} \int \frac {e^x}{x \left (-25+4 \log \left (3 e^x\right )\right )} \, dx-\frac {625}{16} \int \frac {e^x}{x \left (-25+4 \log \left (3 e^x\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.42, size = 28, normalized size = 1.22 \begin {gather*} \frac {25 e^x \log \left (3 e^x\right )}{16 x \left (-25+4 \log \left (3 e^x\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-625*E^x*x + E^x*(625 - 625*x)*Log[3*E^x] + E^x*(-100 + 100*x)*Log[3*E^x]^2)/(10000*x^2 - 3200*x^2*
Log[3*E^x] + 256*x^2*Log[3*E^x]^2),x]

[Out]

(25*E^x*Log[3*E^x])/(16*x*(-25 + 4*Log[3*E^x]))

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fricas [A]  time = 0.50, size = 24, normalized size = 1.04 \begin {gather*} \frac {25 \, {\left (x + \log \relax (3)\right )} e^{x}}{16 \, {\left (4 \, x^{2} + 4 \, x \log \relax (3) - 25 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x-100)*exp(x)*log(3*exp(x))^2+(-625*x+625)*exp(x)*log(3*exp(x))-625*exp(x)*x)/(256*x^2*log(3*e
xp(x))^2-3200*x^2*log(3*exp(x))+10000*x^2),x, algorithm="fricas")

[Out]

25/16*(x + log(3))*e^x/(4*x^2 + 4*x*log(3) - 25*x)

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giac [A]  time = 0.15, size = 28, normalized size = 1.22 \begin {gather*} \frac {25 \, {\left (x e^{x} + e^{x} \log \relax (3)\right )}}{16 \, {\left (4 \, x^{2} + 4 \, x \log \relax (3) - 25 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x-100)*exp(x)*log(3*exp(x))^2+(-625*x+625)*exp(x)*log(3*exp(x))-625*exp(x)*x)/(256*x^2*log(3*e
xp(x))^2-3200*x^2*log(3*exp(x))+10000*x^2),x, algorithm="giac")

[Out]

25/16*(x*e^x + e^x*log(3))/(4*x^2 + 4*x*log(3) - 25*x)

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maple [A]  time = 0.10, size = 24, normalized size = 1.04

method	result	size
norman	\(\frac {25 \,{\mathrm e}^{x} \ln \left (3 \,{\mathrm e}^{x}\right )}{16 x \left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-25\right )}\)	\(24\)
risch	\(\frac {25 \,{\mathrm e}^{x}}{64 x}+\frac {625 i {\mathrm e}^{x}}{64 x \left (4 i \ln \relax (3)+4 i \ln \left ({\mathrm e}^{x}\right )-25 i\right )}\)	\(33\)
default	\(\frac {25 \left (-4 \ln \left (3 \,{\mathrm e}^{x}\right )+4 x +25\right ) \left (-\frac {{\mathrm e}^{x}}{16 \left (\ln \left (3 \,{\mathrm e}^{x}\right )-\frac {25}{4}\right )}-\frac {{\mathrm e}^{-\ln \left (3 \,{\mathrm e}^{x}\right )+x +\frac {25}{4}} \expIntegralEi \left (1, -\ln \left (3 \,{\mathrm e}^{x}\right )+\frac {25}{4}\right )}{16}\right )}{16}+\frac {625 \,{\mathrm e}^{-\ln \left (3 \,{\mathrm e}^{x}\right )+x +\frac {25}{4}} \expIntegralEi \left (1, -\ln \left (3 \,{\mathrm e}^{x}\right )+\frac {25}{4}\right )}{256}+\frac {725 \,{\mathrm e}^{x}}{256 \left (\ln \left (3 \,{\mathrm e}^{x}\right )-\frac {25}{4}\right )}+\frac {625 \left (\ln \left (3 \,{\mathrm e}^{x}\right )-x \right ) \left (\frac {-\frac {{\mathrm e}^{x}}{\ln \left (3 \,{\mathrm e}^{x}\right )-\frac {25}{4}}-{\mathrm e}^{-\ln \left (3 \,{\mathrm e}^{x}\right )+x +\frac {25}{4}} \expIntegralEi \left (1, -\ln \left (3 \,{\mathrm e}^{x}\right )+\frac {25}{4}\right )}{\left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-4 x -25\right )^{2}}+\frac {-\frac {{\mathrm e}^{x}}{x}-\expIntegralEi \left (1, -x \right )}{\left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-4 x -25\right )^{2}}-\frac {8 \,{\mathrm e}^{-\ln \left (3 \,{\mathrm e}^{x}\right )+x +\frac {25}{4}} \expIntegralEi \left (1, -\ln \left (3 \,{\mathrm e}^{x}\right )+\frac {25}{4}\right )}{\left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-4 x -25\right )^{3}}+\frac {8 \expIntegralEi \left (1, -x \right )}{\left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-4 x -25\right )^{3}}\right )}{16}-\frac {25 \left (\ln \left (3 \,{\mathrm e}^{x}\right )-x \right )^{2} \left (\frac {-\frac {{\mathrm e}^{x}}{\ln \left (3 \,{\mathrm e}^{x}\right )-\frac {25}{4}}-{\mathrm e}^{-\ln \left (3 \,{\mathrm e}^{x}\right )+x +\frac {25}{4}} \expIntegralEi \left (1, -\ln \left (3 \,{\mathrm e}^{x}\right )+\frac {25}{4}\right )}{\left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-4 x -25\right )^{2}}+\frac {-\frac {{\mathrm e}^{x}}{x}-\expIntegralEi \left (1, -x \right )}{\left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-4 x -25\right )^{2}}-\frac {8 \,{\mathrm e}^{-\ln \left (3 \,{\mathrm e}^{x}\right )+x +\frac {25}{4}} \expIntegralEi \left (1, -\ln \left (3 \,{\mathrm e}^{x}\right )+\frac {25}{4}\right )}{\left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-4 x -25\right )^{3}}+\frac {8 \expIntegralEi \left (1, -x \right )}{\left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-4 x -25\right )^{3}}\right )}{4}-\frac {825 \left (\ln \left (3 \,{\mathrm e}^{x}\right )-x \right ) \left (-\frac {4 \left (-\frac {{\mathrm e}^{x}}{16 \left (\ln \left (3 \,{\mathrm e}^{x}\right )-\frac {25}{4}\right )}-\frac {{\mathrm e}^{-\ln \left (3 \,{\mathrm e}^{x}\right )+x +\frac {25}{4}} \expIntegralEi \left (1, -\ln \left (3 \,{\mathrm e}^{x}\right )+\frac {25}{4}\right )}{16}\right )}{4 \ln \left (3 \,{\mathrm e}^{x}\right )-4 x -25}+\frac {{\mathrm e}^{-\ln \left (3 \,{\mathrm e}^{x}\right )+x +\frac {25}{4}} \expIntegralEi \left (1, -\ln \left (3 \,{\mathrm e}^{x}\right )+\frac {25}{4}\right )}{\left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-4 x -25\right )^{2}}-\frac {\expIntegralEi \left (1, -x \right )}{\left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-4 x -25\right )^{2}}\right )}{16}+\frac {25 \left (\ln \left (3 \,{\mathrm e}^{x}\right )-x \right )^{2} \left (-\frac {4 \left (-\frac {{\mathrm e}^{x}}{16 \left (\ln \left (3 \,{\mathrm e}^{x}\right )-\frac {25}{4}\right )}-\frac {{\mathrm e}^{-\ln \left (3 \,{\mathrm e}^{x}\right )+x +\frac {25}{4}} \expIntegralEi \left (1, -\ln \left (3 \,{\mathrm e}^{x}\right )+\frac {25}{4}\right )}{16}\right )}{4 \ln \left (3 \,{\mathrm e}^{x}\right )-4 x -25}+\frac {{\mathrm e}^{-\ln \left (3 \,{\mathrm e}^{x}\right )+x +\frac {25}{4}} \expIntegralEi \left (1, -\ln \left (3 \,{\mathrm e}^{x}\right )+\frac {25}{4}\right )}{\left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-4 x -25\right )^{2}}-\frac {\expIntegralEi \left (1, -x \right )}{\left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-4 x -25\right )^{2}}\right )}{4}+\frac {25 \left (\ln \left (3 \,{\mathrm e}^{x}\right )-x \right ) \left (-\frac {{\mathrm e}^{x}}{16 \left (\ln \left (3 \,{\mathrm e}^{x}\right )-\frac {25}{4}\right )}-\frac {{\mathrm e}^{-\ln \left (3 \,{\mathrm e}^{x}\right )+x +\frac {25}{4}} \expIntegralEi \left (1, -\ln \left (3 \,{\mathrm e}^{x}\right )+\frac {25}{4}\right )}{16}\right )}{2}\)	\(702\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((100*x-100)*exp(x)*ln(3*exp(x))^2+(-625*x+625)*exp(x)*ln(3*exp(x))-625*exp(x)*x)/(256*x^2*ln(3*exp(x))^2-
3200*x^2*ln(3*exp(x))+10000*x^2),x,method=_RETURNVERBOSE)

[Out]

25/16*exp(x)*ln(3*exp(x))/x/(4*ln(3*exp(x))-25)

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maxima [A]  time = 0.49, size = 24, normalized size = 1.04 \begin {gather*} \frac {25 \, {\left (x + \log \relax (3)\right )} e^{x}}{16 \, {\left (4 \, x^{2} + x {\left (4 \, \log \relax (3) - 25\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x-100)*exp(x)*log(3*exp(x))^2+(-625*x+625)*exp(x)*log(3*exp(x))-625*exp(x)*x)/(256*x^2*log(3*e
xp(x))^2-3200*x^2*log(3*exp(x))+10000*x^2),x, algorithm="maxima")

[Out]

25/16*(x + log(3))*e^x/(4*x^2 + x*(4*log(3) - 25))

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mupad [B]  time = 3.23, size = 23, normalized size = 1.00 \begin {gather*} \frac {{\mathrm {e}}^x\,\left (\frac {25\,x}{64}+\frac {25\,\ln \relax (3)}{64}\right )}{x^2+\left (\ln \relax (3)-\frac {25}{4}\right )\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(625*x*exp(x) + exp(x)*log(3*exp(x))*(625*x - 625) - exp(x)*log(3*exp(x))^2*(100*x - 100))/(256*x^2*log(3
*exp(x))^2 + 10000*x^2 - 3200*x^2*log(3*exp(x))),x)

[Out]

(exp(x)*((25*x)/64 + (25*log(3))/64))/(x*(log(3) - 25/4) + x^2)

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sympy [A]  time = 0.22, size = 26, normalized size = 1.13 \begin {gather*} \frac {\left (25 x + 25 \log {\relax (3 )}\right ) e^{x}}{64 x^{2} - 400 x + 64 x \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((100*x-100)*exp(x)*ln(3*exp(x))**2+(-625*x+625)*exp(x)*ln(3*exp(x))-625*exp(x)*x)/(256*x**2*ln(3*ex
p(x))**2-3200*x**2*ln(3*exp(x))+10000*x**2),x)

[Out]

(25*x + 25*log(3))*exp(x)/(64*x**2 - 400*x + 64*x*log(3))

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