3.19.23 \(\int \frac {-875 x+25 e^x x^3+125 x \log (x)}{-54000-5400 e^x x-180 e^{2 x} x^2-2 e^{3 x} x^3+(27000+1800 e^x x+30 e^{2 x} x^2) \log (x)+(-4500-150 e^x x) \log ^2(x)+250 \log ^3(x)} \, dx\)

Optimal. Leaf size=21 \[ \frac {x^2}{4 \left (-6-\frac {e^x x}{5}+\log (x)\right )^2} \]

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Rubi [F]  time = 1.65, 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 {-875 x+25 e^x x^3+125 x \log (x)}{-54000-5400 e^x x-180 e^{2 x} x^2-2 e^{3 x} x^3+\left (27000+1800 e^x x+30 e^{2 x} x^2\right ) \log (x)+\left (-4500-150 e^x x\right ) \log ^2(x)+250 \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-875*x + 25*E^x*x^3 + 125*x*Log[x])/(-54000 - 5400*E^x*x - 180*E^(2*x)*x^2 - 2*E^(3*x)*x^3 + (27000 + 180
0*E^x*x + 30*E^(2*x)*x^2)*Log[x] + (-4500 - 150*E^x*x)*Log[x]^2 + 250*Log[x]^3),x]

[Out]

(875*Defer[Int][x/(30 + E^x*x - 5*Log[x])^3, x])/2 + 375*Defer[Int][x^2/(30 + E^x*x - 5*Log[x])^3, x] - (25*De
fer[Int][x^2/(30 + E^x*x - 5*Log[x])^2, x])/2 - (125*Defer[Int][(x*Log[x])/(30 + E^x*x - 5*Log[x])^3, x])/2 -
(125*Defer[Int][(x^2*Log[x])/(30 + E^x*x - 5*Log[x])^3, x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 x \left (35-e^x x^2-5 \log (x)\right )}{2 \left (30+e^x x-5 \log (x)\right )^3} \, dx\\ &=\frac {25}{2} \int \frac {x \left (35-e^x x^2-5 \log (x)\right )}{\left (30+e^x x-5 \log (x)\right )^3} \, dx\\ &=\frac {25}{2} \int \left (-\frac {x^2}{\left (30+e^x x-5 \log (x)\right )^2}-\frac {5 x (-7-6 x+\log (x)+x \log (x))}{\left (30+e^x x-5 \log (x)\right )^3}\right ) \, dx\\ &=-\left (\frac {25}{2} \int \frac {x^2}{\left (30+e^x x-5 \log (x)\right )^2} \, dx\right )-\frac {125}{2} \int \frac {x (-7-6 x+\log (x)+x \log (x))}{\left (30+e^x x-5 \log (x)\right )^3} \, dx\\ &=-\left (\frac {25}{2} \int \frac {x^2}{\left (30+e^x x-5 \log (x)\right )^2} \, dx\right )-\frac {125}{2} \int \left (-\frac {7 x}{\left (30+e^x x-5 \log (x)\right )^3}-\frac {6 x^2}{\left (30+e^x x-5 \log (x)\right )^3}+\frac {x \log (x)}{\left (30+e^x x-5 \log (x)\right )^3}+\frac {x^2 \log (x)}{\left (30+e^x x-5 \log (x)\right )^3}\right ) \, dx\\ &=-\left (\frac {25}{2} \int \frac {x^2}{\left (30+e^x x-5 \log (x)\right )^2} \, dx\right )-\frac {125}{2} \int \frac {x \log (x)}{\left (30+e^x x-5 \log (x)\right )^3} \, dx-\frac {125}{2} \int \frac {x^2 \log (x)}{\left (30+e^x x-5 \log (x)\right )^3} \, dx+375 \int \frac {x^2}{\left (30+e^x x-5 \log (x)\right )^3} \, dx+\frac {875}{2} \int \frac {x}{\left (30+e^x x-5 \log (x)\right )^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 21, normalized size = 1.00 \begin {gather*} \frac {25 x^2}{4 \left (-30-e^x x+5 \log (x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-875*x + 25*E^x*x^3 + 125*x*Log[x])/(-54000 - 5400*E^x*x - 180*E^(2*x)*x^2 - 2*E^(3*x)*x^3 + (27000
+ 1800*E^x*x + 30*E^(2*x)*x^2)*Log[x] + (-4500 - 150*E^x*x)*Log[x]^2 + 250*Log[x]^3),x]

[Out]

(25*x^2)/(4*(-30 - E^x*x + 5*Log[x])^2)

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fricas [B]  time = 0.87, size = 38, normalized size = 1.81 \begin {gather*} \frac {25 \, x^{2}}{4 \, {\left (x^{2} e^{\left (2 \, x\right )} + 60 \, x e^{x} - 10 \, {\left (x e^{x} + 30\right )} \log \relax (x) + 25 \, \log \relax (x)^{2} + 900\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((125*x*log(x)+25*exp(x)*x^3-875*x)/(250*log(x)^3+(-150*exp(x)*x-4500)*log(x)^2+(30*exp(x)^2*x^2+1800
*exp(x)*x+27000)*log(x)-2*x^3*exp(x)^3-180*exp(x)^2*x^2-5400*exp(x)*x-54000),x, algorithm="fricas")

[Out]

25/4*x^2/(x^2*e^(2*x) + 60*x*e^x - 10*(x*e^x + 30)*log(x) + 25*log(x)^2 + 900)

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giac [B]  time = 0.43, size = 39, normalized size = 1.86 \begin {gather*} \frac {25 \, x^{2}}{4 \, {\left (x^{2} e^{\left (2 \, x\right )} - 10 \, x e^{x} \log \relax (x) + 60 \, x e^{x} + 25 \, \log \relax (x)^{2} - 300 \, \log \relax (x) + 900\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((125*x*log(x)+25*exp(x)*x^3-875*x)/(250*log(x)^3+(-150*exp(x)*x-4500)*log(x)^2+(30*exp(x)^2*x^2+1800
*exp(x)*x+27000)*log(x)-2*x^3*exp(x)^3-180*exp(x)^2*x^2-5400*exp(x)*x-54000),x, algorithm="giac")

[Out]

25/4*x^2/(x^2*e^(2*x) - 10*x*e^x*log(x) + 60*x*e^x + 25*log(x)^2 - 300*log(x) + 900)

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maple [A]  time = 0.03, size = 18, normalized size = 0.86

method	result	size
risch	\(\frac {25 x^{2}}{4 \left ({\mathrm e}^{x} x -5 \ln \relax (x )+30\right )^{2}}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((125*x*ln(x)+25*exp(x)*x^3-875*x)/(250*ln(x)^3+(-150*exp(x)*x-4500)*ln(x)^2+(30*exp(x)^2*x^2+1800*exp(x)*x
+27000)*ln(x)-2*x^3*exp(x)^3-180*exp(x)^2*x^2-5400*exp(x)*x-54000),x,method=_RETURNVERBOSE)

[Out]

25/4*x^2/(exp(x)*x-5*ln(x)+30)^2

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maxima [B]  time = 0.58, size = 39, normalized size = 1.86 \begin {gather*} \frac {25 \, x^{2}}{4 \, {\left (x^{2} e^{\left (2 \, x\right )} - 10 \, {\left (x \log \relax (x) - 6 \, x\right )} e^{x} + 25 \, \log \relax (x)^{2} - 300 \, \log \relax (x) + 900\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((125*x*log(x)+25*exp(x)*x^3-875*x)/(250*log(x)^3+(-150*exp(x)*x-4500)*log(x)^2+(30*exp(x)^2*x^2+1800
*exp(x)*x+27000)*log(x)-2*x^3*exp(x)^3-180*exp(x)^2*x^2-5400*exp(x)*x-54000),x, algorithm="maxima")

[Out]

25/4*x^2/(x^2*e^(2*x) - 10*(x*log(x) - 6*x)*e^x + 25*log(x)^2 - 300*log(x) + 900)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {25\,x^3\,{\mathrm {e}}^x-875\,x+125\,x\,\ln \relax (x)}{180\,x^2\,{\mathrm {e}}^{2\,x}-250\,{\ln \relax (x)}^3+2\,x^3\,{\mathrm {e}}^{3\,x}+5400\,x\,{\mathrm {e}}^x+{\ln \relax (x)}^2\,\left (150\,x\,{\mathrm {e}}^x+4500\right )-\ln \relax (x)\,\left (30\,x^2\,{\mathrm {e}}^{2\,x}+1800\,x\,{\mathrm {e}}^x+27000\right )+54000} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(25*x^3*exp(x) - 875*x + 125*x*log(x))/(180*x^2*exp(2*x) - 250*log(x)^3 + 2*x^3*exp(3*x) + 5400*x*exp(x)
+ log(x)^2*(150*x*exp(x) + 4500) - log(x)*(30*x^2*exp(2*x) + 1800*x*exp(x) + 27000) + 54000),x)

[Out]

int(-(25*x^3*exp(x) - 875*x + 125*x*log(x))/(180*x^2*exp(2*x) - 250*log(x)^3 + 2*x^3*exp(3*x) + 5400*x*exp(x)
+ log(x)^2*(150*x*exp(x) + 4500) - log(x)*(30*x^2*exp(2*x) + 1800*x*exp(x) + 27000) + 54000), x)

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sympy [B]  time = 0.37, size = 41, normalized size = 1.95 \begin {gather*} \frac {25 x^{2}}{4 x^{2} e^{2 x} + \left (- 40 x \log {\relax (x )} + 240 x\right ) e^{x} + 100 \log {\relax (x )}^{2} - 1200 \log {\relax (x )} + 3600} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((125*x*ln(x)+25*exp(x)*x**3-875*x)/(250*ln(x)**3+(-150*exp(x)*x-4500)*ln(x)**2+(30*exp(x)**2*x**2+18
00*exp(x)*x+27000)*ln(x)-2*x**3*exp(x)**3-180*exp(x)**2*x**2-5400*exp(x)*x-54000),x)

[Out]

25*x**2/(4*x**2*exp(2*x) + (-40*x*log(x) + 240*x)*exp(x) + 100*log(x)**2 - 1200*log(x) + 3600)

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