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

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

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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 = {} $$\begin{array}{c}\int \frac{4e^2+e(-16-16x)+28x+16x^2+(4e^2-8ex)\log (x)}{18x+e^{2}x+14x^2+4x^3+e(-8x-4x^2)+(2e^{2}x+e(-8x-4x^2))\log (x)+e^{2}x{\log }^2(x)}dx\end{array}$$

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{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{4e^2+e(-16-16x)+28x+16x^2+(4e^2-8ex)\log (x)}{(18+e^2)x+14x^2+4x^3+e(-8x-4x^2)+(2e^{2}x+e(-8x-4x^2))\log (x)+e^{2}x{\log }^2(x)}dx\\ & =\int \frac{4(-4(1-\frac{e}{4})e+7(1-\frac{4e}{7})x+4x^2+e^2\log (x)-2ex\log (x))}{(18+e^2)x+14x^2+4x^3+e(-8x-4x^2)+(2e^{2}x+e(-8x-4x^2))\log (x)+e^{2}x{\log }^2(x)}dx\\ & =4\int \frac{-4(1-\frac{e}{4})e+7(1-\frac{4e}{7})x+4x^2+e^2\log (x)-2ex\log (x)}{(18+e^2)x+14x^2+4x^3+e(-8x-4x^2)+(2e^{2}x+e(-8x-4x^2))\log (x)+e^{2}x{\log }^2(x)}dx\\ & =4\int \frac{e^2-4e(1+x)+x(7+4x)+e(e-2x)\log (x)}{x(e^2-4e(2+x)+2(9+7x+2x^2)+2e(e-2(2+x))\log (x)+e^2{\log }^2(x))}dx\\ & =4\int (\frac{4(1-\frac{7}{4e})e}{-18(1+\frac{1}{18}(-8+e)e)-14(1-\frac{2e}{7})x-4x^2+8(1-\frac{e}{4})e\log (x)+4ex\log (x)-e^2{\log }^2(x)}+\frac{4(1-\frac{e}{4})e}{x(-18(1+\frac{1}{18}(-8+e)e)-14(1-\frac{2e}{7})x-4x^2+8(1-\frac{e}{4})e\log (x)+4ex\log (x)-e^2{\log }^2(x))}+\frac{2e\log (x)}{-18(1+\frac{1}{18}(-8+e)e)-14(1-\frac{2e}{7})x-4x^2+8(1-\frac{e}{4})e\log (x)+4ex\log (x)-e^2{\log }^2(x)}+\frac{4x}{18(1+\frac{1}{18}(-8+e)e)+14(1-\frac{2e}{7})x+4x^2-8(1-\frac{e}{4})e\log (x)-4ex\log (x)+e^2{\log }^2(x)}+\frac{e^2\log (x)}{x(18(1+\frac{1}{18}(-8+e)e)+14(1-\frac{2e}{7})x+4x^2-8(1-\frac{e}{4})e\log (x)-4ex\log (x)+e^2{\log }^2(x))})dx\\ & =16\int \frac{x}{18(1+\frac{1}{18}(-8+e)e)+14(1-\frac{2e}{7})x+4x^2-8(1-\frac{e}{4})e\log (x)-4ex\log (x)+e^2{\log }^2(x)}dx-(4(7-4e))\int \frac{1}{-18(1+\frac{1}{18}(-8+e)e)-14(1-\frac{2e}{7})x-4x^2+8(1-\frac{e}{4})e\log (x)+4ex\log (x)-e^2{\log }^2(x)}dx+(8e)\int \frac{\log (x)}{-18(1+\frac{1}{18}(-8+e)e)-14(1-\frac{2e}{7})x-4x^2+8(1-\frac{e}{4})e\log (x)+4ex\log (x)-e^2{\log }^2(x)}dx+(4(4-e)e)\int \frac{1}{x(-18(1+\frac{1}{18}(-8+e)e)-14(1-\frac{2e}{7})x-4x^2+8(1-\frac{e}{4})e\log (x)+4ex\log (x)-e^2{\log }^2(x))}dx+\left(4e^2\right)\int \frac{\log (x)}{x(18(1+\frac{1}{18}(-8+e)e)+14(1-\frac{2e}{7})x+4x^2-8(1-\frac{e}{4})e\log (x)-4ex\log (x)+e^2{\log }^2(x))}dx\end{array}\end{array}$$

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Mathematica [B]  time = 1.47, size = 49, normalized size = 2.04 $$\begin{array}{c}2\log (18-8e+e^2+14x-4ex+4x^2-8e\log (x)+2e^2\log (x)-4ex\log (x)+e^2{\log }^2(x))\end{array}$$

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 $$\begin{array}{c}2\log (e^2\log (x{)}^2+4x^2-4(x+2)e-2(2(x+2)e-e^2)\log (x)+14x+e^2+18)\end{array}$$

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 $$\begin{array}{c}2\log (-4xe\log (x)+e^2\log (x{)}^2+4x^2-4xe+2e^2\log (x)-8e\log (x)+14x+e^2-8e+18)\end{array}$$

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	$2\ln (\ln (x{)}^2+2{\mathrm{e}}^{-1}(\mathrm{e}-2x-4)\ln (x)+({\mathrm{e}}^2-4x\mathrm{e}+4x^2-8\mathrm{e}+14x+18){\mathrm{e}}^{-2})$	$46$
norman	$2\ln ({\mathrm{e}}^2\ln (x{)}^2+2{\mathrm{e}}^2\ln (x)-4x\mathrm{e}\ln (x)-8\mathrm{e}\ln (x)+{\mathrm{e}}^2-4x\mathrm{e}+4x^2-8\mathrm{e}+14x+18)$	$57$

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 $$\begin{array}{c}2\log \left((e^2\log (x{)}^2+4x^2-2x(2e-7)-2(2xe-e^2+4e)\log (x)+e^2-8e+18)e^{(-2)}\right)\end{array}$$

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 $$\begin{array}{c}2\ln (14x-8\mathrm{e}+{\mathrm{e}}^2+{\mathrm{e}}^2{\ln (x)}^2-4x\mathrm{e}-8\mathrm{e}\ln (x)+2{\mathrm{e}}^2\ln (x)+4x^2-4x\mathrm{e}\ln (x)+18)\end{array}$$

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 $$\begin{array}{c}2\log (\frac{(-4x-8+2e)\log (x)}{e}+\frac{4x^2-4ex+14x-8e+e^2+18}{e^2}+\log {(x)}^2)\end{array}$$

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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