3.85.25 \(\int \frac {4 e^2+e (-16-16 x)+28 x+16 x^2+(4 e^2-8 e x) \log (x)}{18 x+e^2 x+14 x^2+4 x^3+e (-8 x-4 x^2)+(2 e^2 x+e (-8 x-4 x^2)) \log (x)+e^2 x \log ^2(x)} \, dx\)

Optimal. Leaf size=24 \[ \log \left (\left (-2+2 x-(4+2 x-e (1+\log (x)))^2\right )^2\right ) \]

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Rubi [F]  time = 1.64, 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 {4 e^2+e (-16-16 x)+28 x+16 x^2+\left (4 e^2-8 e x\right ) \log (x)}{18 x+e^2 x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(4*E^2 + E*(-16 - 16*x) + 28*x + 16*x^2 + (4*E^2 - 8*E*x)*Log[x])/(18*x + E^2*x + 14*x^2 + 4*x^3 + E*(-8*x
- 4*x^2) + (2*E^2*x + E*(-8*x - 4*x^2))*Log[x] + E^2*x*Log[x]^2),x]

[Out]

-4*(7 - 4*E)*Defer[Int][(-18*(1 + ((-8 + E)*E)/18) - 14*(1 - (2*E)/7)*x - 4*x^2 + 8*(1 - E/4)*E*Log[x] + 4*E*x
*Log[x] - E^2*Log[x]^2)^(-1), x] + 4*(4 - E)*E*Defer[Int][1/(x*(-18*(1 + ((-8 + E)*E)/18) - 14*(1 - (2*E)/7)*x
- 4*x^2 + 8*(1 - E/4)*E*Log[x] + 4*E*x*Log[x] - E^2*Log[x]^2)), x] + 8*E*Defer[Int][Log[x]/(-18*(1 + ((-8 + E
)*E)/18) - 14*(1 - (2*E)/7)*x - 4*x^2 + 8*(1 - E/4)*E*Log[x] + 4*E*x*Log[x] - E^2*Log[x]^2), x] + 16*Defer[Int
][x/(18*(1 + ((-8 + E)*E)/18) + 14*(1 - (2*E)/7)*x + 4*x^2 - 8*(1 - E/4)*E*Log[x] - 4*E*x*Log[x] + E^2*Log[x]^
2), x] + 4*E^2*Defer[Int][Log[x]/(x*(18*(1 + ((-8 + E)*E)/18) + 14*(1 - (2*E)/7)*x + 4*x^2 - 8*(1 - E/4)*E*Log
[x] - 4*E*x*Log[x] + E^2*Log[x]^2)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 e^2+e (-16-16 x)+28 x+16 x^2+\left (4 e^2-8 e x\right ) \log (x)}{\left (18+e^2\right ) x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx\\ &=\int \frac {4 \left (-4 \left (1-\frac {e}{4}\right ) e+7 \left (1-\frac {4 e}{7}\right ) x+4 x^2+e^2 \log (x)-2 e x \log (x)\right )}{\left (18+e^2\right ) x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx\\ &=4 \int \frac {-4 \left (1-\frac {e}{4}\right ) e+7 \left (1-\frac {4 e}{7}\right ) x+4 x^2+e^2 \log (x)-2 e x \log (x)}{\left (18+e^2\right ) x+14 x^2+4 x^3+e \left (-8 x-4 x^2\right )+\left (2 e^2 x+e \left (-8 x-4 x^2\right )\right ) \log (x)+e^2 x \log ^2(x)} \, dx\\ &=4 \int \frac {e^2-4 e (1+x)+x (7+4 x)+e (e-2 x) \log (x)}{x \left (e^2-4 e (2+x)+2 \left (9+7 x+2 x^2\right )+2 e (e-2 (2+x)) \log (x)+e^2 \log ^2(x)\right )} \, dx\\ &=4 \int \left (\frac {4 \left (1-\frac {7}{4 e}\right ) e}{-18 \left (1+\frac {1}{18} (-8+e) e\right )-14 \left (1-\frac {2 e}{7}\right ) x-4 x^2+8 \left (1-\frac {e}{4}\right ) e \log (x)+4 e x \log (x)-e^2 \log ^2(x)}+\frac {4 \left (1-\frac {e}{4}\right ) e}{x \left (-18 \left (1+\frac {1}{18} (-8+e) e\right )-14 \left (1-\frac {2 e}{7}\right ) x-4 x^2+8 \left (1-\frac {e}{4}\right ) e \log (x)+4 e x \log (x)-e^2 \log ^2(x)\right )}+\frac {2 e \log (x)}{-18 \left (1+\frac {1}{18} (-8+e) e\right )-14 \left (1-\frac {2 e}{7}\right ) x-4 x^2+8 \left (1-\frac {e}{4}\right ) e \log (x)+4 e x \log (x)-e^2 \log ^2(x)}+\frac {4 x}{18 \left (1+\frac {1}{18} (-8+e) e\right )+14 \left (1-\frac {2 e}{7}\right ) x+4 x^2-8 \left (1-\frac {e}{4}\right ) e \log (x)-4 e x \log (x)+e^2 \log ^2(x)}+\frac {e^2 \log (x)}{x \left (18 \left (1+\frac {1}{18} (-8+e) e\right )+14 \left (1-\frac {2 e}{7}\right ) x+4 x^2-8 \left (1-\frac {e}{4}\right ) e \log (x)-4 e x \log (x)+e^2 \log ^2(x)\right )}\right ) \, dx\\ &=16 \int \frac {x}{18 \left (1+\frac {1}{18} (-8+e) e\right )+14 \left (1-\frac {2 e}{7}\right ) x+4 x^2-8 \left (1-\frac {e}{4}\right ) e \log (x)-4 e x \log (x)+e^2 \log ^2(x)} \, dx-(4 (7-4 e)) \int \frac {1}{-18 \left (1+\frac {1}{18} (-8+e) e\right )-14 \left (1-\frac {2 e}{7}\right ) x-4 x^2+8 \left (1-\frac {e}{4}\right ) e \log (x)+4 e x \log (x)-e^2 \log ^2(x)} \, dx+(8 e) \int \frac {\log (x)}{-18 \left (1+\frac {1}{18} (-8+e) e\right )-14 \left (1-\frac {2 e}{7}\right ) x-4 x^2+8 \left (1-\frac {e}{4}\right ) e \log (x)+4 e x \log (x)-e^2 \log ^2(x)} \, dx+(4 (4-e) e) \int \frac {1}{x \left (-18 \left (1+\frac {1}{18} (-8+e) e\right )-14 \left (1-\frac {2 e}{7}\right ) x-4 x^2+8 \left (1-\frac {e}{4}\right ) e \log (x)+4 e x \log (x)-e^2 \log ^2(x)\right )} \, dx+\left (4 e^2\right ) \int \frac {\log (x)}{x \left (18 \left (1+\frac {1}{18} (-8+e) e\right )+14 \left (1-\frac {2 e}{7}\right ) x+4 x^2-8 \left (1-\frac {e}{4}\right ) e \log (x)-4 e x \log (x)+e^2 \log ^2(x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 1.47, size = 49, normalized size = 2.04 \begin {gather*} 2 \log \left (18-8 e+e^2+14 x-4 e x+4 x^2-8 e \log (x)+2 e^2 \log (x)-4 e x \log (x)+e^2 \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^2 + E*(-16 - 16*x) + 28*x + 16*x^2 + (4*E^2 - 8*E*x)*Log[x])/(18*x + E^2*x + 14*x^2 + 4*x^3 + E
*(-8*x - 4*x^2) + (2*E^2*x + E*(-8*x - 4*x^2))*Log[x] + E^2*x*Log[x]^2),x]

[Out]

2*Log[18 - 8*E + E^2 + 14*x - 4*E*x + 4*x^2 - 8*E*Log[x] + 2*E^2*Log[x] - 4*E*x*Log[x] + E^2*Log[x]^2]

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fricas [B]  time = 0.65, size = 45, normalized size = 1.88 \begin {gather*} 2 \, \log \left (e^{2} \log \relax (x)^{2} + 4 \, x^{2} - 4 \, {\left (x + 2\right )} e - 2 \, {\left (2 \, {\left (x + 2\right )} e - e^{2}\right )} \log \relax (x) + 14 \, x + e^{2} + 18\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(1)^2-8*x*exp(1))*log(x)+4*exp(1)^2+(-16*x-16)*exp(1)+16*x^2+28*x)/(x*exp(1)^2*log(x)^2+(2*x*
exp(1)^2+(-4*x^2-8*x)*exp(1))*log(x)+x*exp(1)^2+(-4*x^2-8*x)*exp(1)+4*x^3+14*x^2+18*x),x, algorithm="fricas")

[Out]

2*log(e^2*log(x)^2 + 4*x^2 - 4*(x + 2)*e - 2*(2*(x + 2)*e - e^2)*log(x) + 14*x + e^2 + 18)

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giac [B]  time = 0.23, size = 50, normalized size = 2.08 \begin {gather*} 2 \, \log \left (-4 \, x e \log \relax (x) + e^{2} \log \relax (x)^{2} + 4 \, x^{2} - 4 \, x e + 2 \, e^{2} \log \relax (x) - 8 \, e \log \relax (x) + 14 \, x + e^{2} - 8 \, e + 18\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(1)^2-8*x*exp(1))*log(x)+4*exp(1)^2+(-16*x-16)*exp(1)+16*x^2+28*x)/(x*exp(1)^2*log(x)^2+(2*x*
exp(1)^2+(-4*x^2-8*x)*exp(1))*log(x)+x*exp(1)^2+(-4*x^2-8*x)*exp(1)+4*x^3+14*x^2+18*x),x, algorithm="giac")

[Out]

2*log(-4*x*e*log(x) + e^2*log(x)^2 + 4*x^2 - 4*x*e + 2*e^2*log(x) - 8*e*log(x) + 14*x + e^2 - 8*e + 18)

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maple [A]  time = 0.09, size = 46, normalized size = 1.92

method	result	size
risch	\(2 \ln \left (\ln \relax (x )^{2}+2 \,{\mathrm e}^{-1} \left ({\mathrm e}-2 x -4\right ) \ln \relax (x )+\left ({\mathrm e}^{2}-4 x \,{\mathrm e}+4 x^{2}-8 \,{\mathrm e}+14 x +18\right ) {\mathrm e}^{-2}\right )\)	\(46\)
norman	\(2 \ln \left ({\mathrm e}^{2} \ln \relax (x )^{2}+2 \,{\mathrm e}^{2} \ln \relax (x )-4 x \,{\mathrm e} \ln \relax (x )-8 \,{\mathrm e} \ln \relax (x )+{\mathrm e}^{2}-4 x \,{\mathrm e}+4 x^{2}-8 \,{\mathrm e}+14 x +18\right )\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*exp(1)^2-8*x*exp(1))*ln(x)+4*exp(1)^2+(-16*x-16)*exp(1)+16*x^2+28*x)/(x*exp(1)^2*ln(x)^2+(2*x*exp(1)^2
+(-4*x^2-8*x)*exp(1))*ln(x)+x*exp(1)^2+(-4*x^2-8*x)*exp(1)+4*x^3+14*x^2+18*x),x,method=_RETURNVERBOSE)

[Out]

2*ln(ln(x)^2+2*exp(-1)*(exp(1)-2*x-4)*ln(x)+(exp(2)-4*x*exp(1)+4*x^2-8*exp(1)+14*x+18)*exp(-2))

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maxima [B]  time = 0.84, size = 53, normalized size = 2.21 \begin {gather*} 2 \, \log \left ({\left (e^{2} \log \relax (x)^{2} + 4 \, x^{2} - 2 \, x {\left (2 \, e - 7\right )} - 2 \, {\left (2 \, x e - e^{2} + 4 \, e\right )} \log \relax (x) + e^{2} - 8 \, e + 18\right )} e^{\left (-2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(1)^2-8*x*exp(1))*log(x)+4*exp(1)^2+(-16*x-16)*exp(1)+16*x^2+28*x)/(x*exp(1)^2*log(x)^2+(2*x*
exp(1)^2+(-4*x^2-8*x)*exp(1))*log(x)+x*exp(1)^2+(-4*x^2-8*x)*exp(1)+4*x^3+14*x^2+18*x),x, algorithm="maxima")

[Out]

2*log((e^2*log(x)^2 + 4*x^2 - 2*x*(2*e - 7) - 2*(2*x*e - e^2 + 4*e)*log(x) + e^2 - 8*e + 18)*e^(-2))

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mupad [B]  time = 5.67, size = 50, normalized size = 2.08 \begin {gather*} 2\,\ln \left (14\,x-8\,\mathrm {e}+{\mathrm {e}}^2+{\mathrm {e}}^2\,{\ln \relax (x)}^2-4\,x\,\mathrm {e}-8\,\mathrm {e}\,\ln \relax (x)+2\,{\mathrm {e}}^2\,\ln \relax (x)+4\,x^2-4\,x\,\mathrm {e}\,\ln \relax (x)+18\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((28*x + 4*exp(2) + log(x)*(4*exp(2) - 8*x*exp(1)) + 16*x^2 - exp(1)*(16*x + 16))/(18*x - log(x)*(exp(1)*(8
*x + 4*x^2) - 2*x*exp(2)) - exp(1)*(8*x + 4*x^2) + x*exp(2) + 14*x^2 + 4*x^3 + x*exp(2)*log(x)^2),x)

[Out]

2*log(14*x - 8*exp(1) + exp(2) + exp(2)*log(x)^2 - 4*x*exp(1) - 8*exp(1)*log(x) + 2*exp(2)*log(x) + 4*x^2 - 4*
x*exp(1)*log(x) + 18)

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sympy [B]  time = 0.31, size = 53, normalized size = 2.21 \begin {gather*} 2 \log {\left (\frac {\left (- 4 x - 8 + 2 e\right ) \log {\relax (x )}}{e} + \frac {4 x^{2} - 4 e x + 14 x - 8 e + e^{2} + 18}{e^{2}} + \log {\relax (x )}^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(1)**2-8*x*exp(1))*ln(x)+4*exp(1)**2+(-16*x-16)*exp(1)+16*x**2+28*x)/(x*exp(1)**2*ln(x)**2+(2
*x*exp(1)**2+(-4*x**2-8*x)*exp(1))*ln(x)+x*exp(1)**2+(-4*x**2-8*x)*exp(1)+4*x**3+14*x**2+18*x),x)

[Out]

2*log((-4*x - 8 + 2*E)*exp(-1)*log(x) + (4*x**2 - 4*E*x + 14*x - 8*E + exp(2) + 18)*exp(-2) + log(x)**2)

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