3.45.35 \(\int \frac {-75 e^{4 x} x-150 e^{4 x} x \log (x)+(6 e^{4 x} x^2+e^{2 x} (-18 x-18 x^2)) \log ^2(x)}{1250 e^{4 x}+(300 e^{2 x}-200 e^{4 x} x) \log (x)+(18-24 e^{2 x} x+8 e^{4 x} x^2) \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ \frac {3 x^2}{2 \left (-3 e^{-2 x}+2 x-\frac {25}{\log (x)}\right )} \]

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Rubi [F]  time = 23.99, 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 {-75 e^{4 x} x-150 e^{4 x} x \log (x)+\left (6 e^{4 x} x^2+e^{2 x} \left (-18 x-18 x^2\right )\right ) \log ^2(x)}{1250 e^{4 x}+\left (300 e^{2 x}-200 e^{4 x} x\right ) \log (x)+\left (18-24 e^{2 x} x+8 e^{4 x} x^2\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-75*E^(4*x)*x - 150*E^(4*x)*x*Log[x] + (6*E^(4*x)*x^2 + E^(2*x)*(-18*x - 18*x^2))*Log[x]^2)/(1250*E^(4*x)
+ (300*E^(2*x) - 200*E^(4*x)*x)*Log[x] + (18 - 24*E^(2*x)*x + 8*E^(4*x)*x^2)*Log[x]^2),x]

[Out]

(-225*Defer[Int][(E^(2*x)*x*Log[x])/((-25 + 2*x*Log[x])*(-25*E^(2*x) - 3*Log[x] + 2*E^(2*x)*x*Log[x])^2), x])/
2 + 225*Defer[Int][(E^(2*x)*x^2*Log[x]^2)/((-25 + 2*x*Log[x])*(-25*E^(2*x) - 3*Log[x] + 2*E^(2*x)*x*Log[x])^2)
, x] - 9*Defer[Int][(E^(2*x)*x^2*Log[x]^3)/((-25 + 2*x*Log[x])*(-25*E^(2*x) - 3*Log[x] + 2*E^(2*x)*x*Log[x])^2
), x] - 18*Defer[Int][(E^(2*x)*x^3*Log[x]^3)/((-25 + 2*x*Log[x])*(-25*E^(2*x) - 3*Log[x] + 2*E^(2*x)*x*Log[x])
^2), x] - (75*Defer[Int][(E^(2*x)*x)/((-25 + 2*x*Log[x])*(-25*E^(2*x) - 3*Log[x] + 2*E^(2*x)*x*Log[x])), x])/2
- 75*Defer[Int][(E^(2*x)*x*Log[x])/((-25 + 2*x*Log[x])*(-25*E^(2*x) - 3*Log[x] + 2*E^(2*x)*x*Log[x])), x] + 3
*Defer[Int][(E^(2*x)*x^2*Log[x]^2)/((-25 + 2*x*Log[x])*(-25*E^(2*x) - 3*Log[x] + 2*E^(2*x)*x*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 e^{2 x} x \left (-25 e^{2 x}-50 e^{2 x} \log (x)+2 \left (-3+\left (-3+e^{2 x}\right ) x\right ) \log ^2(x)\right )}{2 \left (25 e^{2 x}-\left (-3+2 e^{2 x} x\right ) \log (x)\right )^2} \, dx\\ &=\frac {3}{2} \int \frac {e^{2 x} x \left (-25 e^{2 x}-50 e^{2 x} \log (x)+2 \left (-3+\left (-3+e^{2 x}\right ) x\right ) \log ^2(x)\right )}{\left (25 e^{2 x}-\left (-3+2 e^{2 x} x\right ) \log (x)\right )^2} \, dx\\ &=\frac {3}{2} \int \left (\frac {e^{2 x} x \left (-25-50 \log (x)+2 x \log ^2(x)\right )}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )}-\frac {3 e^{2 x} x \log (x) \left (25-50 x \log (x)+2 x \log ^2(x)+4 x^2 \log ^2(x)\right )}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )^2}\right ) \, dx\\ &=\frac {3}{2} \int \frac {e^{2 x} x \left (-25-50 \log (x)+2 x \log ^2(x)\right )}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )} \, dx-\frac {9}{2} \int \frac {e^{2 x} x \log (x) \left (25-50 x \log (x)+2 x \log ^2(x)+4 x^2 \log ^2(x)\right )}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )^2} \, dx\\ &=\frac {3}{2} \int \left (-\frac {25 e^{2 x} x}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )}-\frac {50 e^{2 x} x \log (x)}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )}+\frac {2 e^{2 x} x^2 \log ^2(x)}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )}\right ) \, dx-\frac {9}{2} \int \left (\frac {25 e^{2 x} x \log (x)}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )^2}-\frac {50 e^{2 x} x^2 \log ^2(x)}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )^2}+\frac {2 e^{2 x} x^2 \log ^3(x)}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )^2}+\frac {4 e^{2 x} x^3 \log ^3(x)}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )^2}\right ) \, dx\\ &=3 \int \frac {e^{2 x} x^2 \log ^2(x)}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )} \, dx-9 \int \frac {e^{2 x} x^2 \log ^3(x)}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )^2} \, dx-18 \int \frac {e^{2 x} x^3 \log ^3(x)}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )^2} \, dx-\frac {75}{2} \int \frac {e^{2 x} x}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )} \, dx-75 \int \frac {e^{2 x} x \log (x)}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )} \, dx-\frac {225}{2} \int \frac {e^{2 x} x \log (x)}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )^2} \, dx+225 \int \frac {e^{2 x} x^2 \log ^2(x)}{(-25+2 x \log (x)) \left (-25 e^{2 x}-3 \log (x)+2 e^{2 x} x \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.67, size = 37, normalized size = 1.42 \begin {gather*} \frac {3 e^{2 x} x^2 \log (x)}{2 \left (-25 e^{2 x}+\left (-3+2 e^{2 x} x\right ) \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-75*E^(4*x)*x - 150*E^(4*x)*x*Log[x] + (6*E^(4*x)*x^2 + E^(2*x)*(-18*x - 18*x^2))*Log[x]^2)/(1250*E
^(4*x) + (300*E^(2*x) - 200*E^(4*x)*x)*Log[x] + (18 - 24*E^(2*x)*x + 8*E^(4*x)*x^2)*Log[x]^2),x]

[Out]

(3*E^(2*x)*x^2*Log[x])/(2*(-25*E^(2*x) + (-3 + 2*E^(2*x)*x)*Log[x]))

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fricas [A]  time = 0.51, size = 32, normalized size = 1.23 \begin {gather*} \frac {3 \, x^{2} e^{\left (2 \, x\right )} \log \relax (x)}{2 \, {\left ({\left (2 \, x e^{\left (2 \, x\right )} - 3\right )} \log \relax (x) - 25 \, e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2*exp(x)^4+(-18*x^2-18*x)*exp(x)^2)*log(x)^2-150*x*exp(x)^4*log(x)-75*x*exp(x)^4)/((8*x^2*exp(
x)^4-24*x*exp(x)^2+18)*log(x)^2+(-200*x*exp(x)^4+300*exp(x)^2)*log(x)+1250*exp(x)^4),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.29, size = 33, normalized size = 1.27 \begin {gather*} \frac {3 \, x^{2} e^{\left (2 \, x\right )} \log \relax (x)}{2 \, {\left (2 \, x e^{\left (2 \, x\right )} \log \relax (x) - 25 \, e^{\left (2 \, x\right )} - 3 \, \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2*exp(x)^4+(-18*x^2-18*x)*exp(x)^2)*log(x)^2-150*x*exp(x)^4*log(x)-75*x*exp(x)^4)/((8*x^2*exp(
x)^4-24*x*exp(x)^2+18)*log(x)^2+(-200*x*exp(x)^4+300*exp(x)^2)*log(x)+1250*exp(x)^4),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 64, normalized size = 2.46

method	result	size
risch	\(\frac {3 x^{2} {\mathrm e}^{2 x}}{2 \left (2 x \,{\mathrm e}^{2 x}-3\right )}+\frac {75 x^{2} {\mathrm e}^{4 x}}{2 \left (2 x \,{\mathrm e}^{2 x}-3\right ) \left (2 \ln \relax (x ) {\mathrm e}^{2 x} x -25 \,{\mathrm e}^{2 x}-3 \ln \relax (x )\right )}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x^2*exp(x)^4+(-18*x^2-18*x)*exp(x)^2)*ln(x)^2-150*x*exp(x)^4*ln(x)-75*x*exp(x)^4)/((8*x^2*exp(x)^4-24*
x*exp(x)^2+18)*ln(x)^2+(-200*x*exp(x)^4+300*exp(x)^2)*ln(x)+1250*exp(x)^4),x,method=_RETURNVERBOSE)

[Out]

3/2*x^2*exp(2*x)/(2*x*exp(2*x)-3)+75/2*x^2*exp(4*x)/(2*x*exp(2*x)-3)/(2*ln(x)*exp(2*x)*x-25*exp(2*x)-3*ln(x))

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maxima [A]  time = 0.41, size = 30, normalized size = 1.15 \begin {gather*} \frac {3 \, x^{2} e^{\left (2 \, x\right )} \log \relax (x)}{2 \, {\left ({\left (2 \, x \log \relax (x) - 25\right )} e^{\left (2 \, x\right )} - 3 \, \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x^2*exp(x)^4+(-18*x^2-18*x)*exp(x)^2)*log(x)^2-150*x*exp(x)^4*log(x)-75*x*exp(x)^4)/((8*x^2*exp(
x)^4-24*x*exp(x)^2+18)*log(x)^2+(-200*x*exp(x)^4+300*exp(x)^2)*log(x)+1250*exp(x)^4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.48, size = 33, normalized size = 1.27 \begin {gather*} -\frac {3\,x^2\,{\mathrm {e}}^{2\,x}\,\ln \relax (x)}{2\,\left (25\,{\mathrm {e}}^{2\,x}+3\,\ln \relax (x)-2\,x\,{\mathrm {e}}^{2\,x}\,\ln \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(75*x*exp(4*x) + log(x)^2*(exp(2*x)*(18*x + 18*x^2) - 6*x^2*exp(4*x)) + 150*x*exp(4*x)*log(x))/(1250*exp(
4*x) + log(x)*(300*exp(2*x) - 200*x*exp(4*x)) + log(x)^2*(8*x^2*exp(4*x) - 24*x*exp(2*x) + 18)),x)

[Out]

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

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sympy [B]  time = 0.50, size = 63, normalized size = 2.42 \begin {gather*} \frac {9 x^{2} \log {\relax (x )}^{2}}{- 12 x \log {\relax (x )}^{2} + \left (8 x^{2} \log {\relax (x )}^{2} - 200 x \log {\relax (x )} + 1250\right ) e^{2 x} + 150 \log {\relax (x )}} + \frac {3 x}{4} + \frac {75 x}{8 x \log {\relax (x )} - 100} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x**2*exp(x)**4+(-18*x**2-18*x)*exp(x)**2)*ln(x)**2-150*x*exp(x)**4*ln(x)-75*x*exp(x)**4)/((8*x**
2*exp(x)**4-24*x*exp(x)**2+18)*ln(x)**2+(-200*x*exp(x)**4+300*exp(x)**2)*ln(x)+1250*exp(x)**4),x)

[Out]

9*x**2*log(x)**2/(-12*x*log(x)**2 + (8*x**2*log(x)**2 - 200*x*log(x) + 1250)*exp(2*x) + 150*log(x)) + 3*x/4 +
75*x/(8*x*log(x) - 100)

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