3.69.62 $\int \frac{-x^2+x^3+e^{-1+x}(2x-2x^2)\log (4)+e^{-2+2x}(-1+x){\log }^2(4)+(9x+3x^2+3x^3+x^4+e^{-1+x}(-6x-11x^2-3x^3)\log (4)+e^{-2+2x}(3x+x^2){\log }^2(4)+(-3x-x^3+e^{-1+x}(2x+3x^2)\log (4)-e^{-2+2x}x{\log }^2(4))\log (x))\log (3+x-\log (x))}{(-3x^3-x^4+e^{-1+x}(6x^2+2x^3)\log (4)+e^{-2+2x}(-3x-x^2){\log }^2(4)+(x^3-2e^{-1+x}{x}^2\log (4)+e^{-2+2x}x{\log }^2(4))\log (x))\log (3+x-\log (x))}dx$

Optimal. Leaf size=33 $$5-x+\frac{3+x}{x-e^{-1+x}\log (4)}-\log (\log (3+x-\log (x)))$$

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Rubi [F]  time = 1.74, 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 = {} $$\begin{array}{c}\int \frac{-x^2+x^3+e^{-1+x}(2x-2x^2)\log (4)+e^{-2+2x}(-1+x){\log }^2(4)+(9x+3x^2+3x^3+x^4+e^{-1+x}(-6x-11x^2-3x^3)\log (4)+e^{-2+2x}(3x+x^2){\log }^2(4)+(-3x-x^3+e^{-1+x}(2x+3x^2)\log (4)-e^{-2+2x}x{\log }^2(4))\log (x))\log (3+x-\log (x))}{(-3x^3-x^4+e^{-1+x}(6x^2+2x^3)\log (4)+e^{-2+2x}(-3x-x^2){\log }^2(4)+(x^3-2e^{-1+x}{x}^2\log (4)+e^{-2+2x}x{\log }^2(4))\log (x))\log (3+x-\log (x))}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-x^2 + x^3 + E^(-1 + x)*(2*x - 2*x^2)*Log[4] + E^(-2 + 2*x)*(-1 + x)*Log[4]^2 + (9*x + 3*x^2 + 3*x^3 + x^
4 + E^(-1 + x)*(-6*x - 11*x^2 - 3*x^3)*Log[4] + E^(-2 + 2*x)*(3*x + x^2)*Log[4]^2 + (-3*x - x^3 + E^(-1 + x)*(
2*x + 3*x^2)*Log[4] - E^(-2 + 2*x)*x*Log[4]^2)*Log[x])*Log[3 + x - Log[x]])/((-3*x^3 - x^4 + E^(-1 + x)*(6*x^2
+ 2*x^3)*Log[4] + E^(-2 + 2*x)*(-3*x - x^2)*Log[4]^2 + (x^3 - 2*E^(-1 + x)*x^2*Log[4] + E^(-2 + 2*x)*x*Log[4]
^2)*Log[x])*Log[3 + x - Log[x]]),x]

[Out]

-x - Log[Log[3 + x - Log[x]]] - 3*E^2*Defer[Int][(E*x - E^x*Log[4])^(-2), x] + 2*E^2*Defer[Int][x/(E*x - E^x*L
og[4])^2, x] + E^2*Defer[Int][x^2/(E*x - E^x*Log[4])^2, x] - 2*E*Defer[Int][(E*x - E^x*Log[4])^(-1), x] - E*De
fer[Int][x/(E*x - E^x*Log[4]), x]

Rubi steps

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

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Mathematica [A]  time = 0.45, size = 37, normalized size = 1.12 $$\begin{array}{c}-x+\frac{-3e-ex}{-ex+e^x\log (4)}-\log (\log (3+x-\log (x)))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-x^2 + x^3 + E^(-1 + x)*(2*x - 2*x^2)*Log[4] + E^(-2 + 2*x)*(-1 + x)*Log[4]^2 + (9*x + 3*x^2 + 3*x^
3 + x^4 + E^(-1 + x)*(-6*x - 11*x^2 - 3*x^3)*Log[4] + E^(-2 + 2*x)*(3*x + x^2)*Log[4]^2 + (-3*x - x^3 + E^(-1
+ x)*(2*x + 3*x^2)*Log[4] - E^(-2 + 2*x)*x*Log[4]^2)*Log[x])*Log[3 + x - Log[x]])/((-3*x^3 - x^4 + E^(-1 + x)*
(6*x^2 + 2*x^3)*Log[4] + E^(-2 + 2*x)*(-3*x - x^2)*Log[4]^2 + (x^3 - 2*E^(-1 + x)*x^2*Log[4] + E^(-2 + 2*x)*x*
Log[4]^2)*Log[x])*Log[3 + x - Log[x]]),x]

[Out]

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

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fricas [A]  time = 0.59, size = 55, normalized size = 1.67 $$\begin{array}{c}-\frac{2x{e}^{(x-1)}\log (2)-x^2+(2e^{(x-1)}\log (2)-x)\log (\log (x-\log (x)+3))+x+3}{2e^{(x-1)}\log (2)-x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x*log(2)^2*exp(x-1)^2+2*(3*x^2+2*x)*log(2)*exp(x-1)-x^3-3*x)*log(x)+4*(x^2+3*x)*log(2)^2*exp(x
-1)^2+2*(-3*x^3-11*x^2-6*x)*log(2)*exp(x-1)+x^4+3*x^3+3*x^2+9*x)*log(-log(x)+3+x)+4*(x-1)*log(2)^2*exp(x-1)^2+
2*(-2*x^2+2*x)*log(2)*exp(x-1)+x^3-x^2)/((4*x*log(2)^2*exp(x-1)^2-4*x^2*log(2)*exp(x-1)+x^3)*log(x)+4*(-x^2-3*
x)*log(2)^2*exp(x-1)^2+2*(2*x^3+6*x^2)*log(2)*exp(x-1)-x^4-3*x^3)/log(-log(x)+3+x),x, algorithm="fricas")

[Out]

-(2*x*e^(x - 1)*log(2) - x^2 + (2*e^(x - 1)*log(2) - x)*log(log(x - log(x) + 3)) + x + 3)/(2*e^(x - 1)*log(2)
- x)

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giac [A]  time = 0.45, size = 66, normalized size = 2.00 $$\begin{array}{c}-\frac{x^{2}e-2x{e}^x\log (2)+xe\log (\log (x-\log (x)+3))-2e^x\log (2)\log (\log (x-\log (x)+3))-xe-3e}{xe-2e^x\log (2)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x*log(2)^2*exp(x-1)^2+2*(3*x^2+2*x)*log(2)*exp(x-1)-x^3-3*x)*log(x)+4*(x^2+3*x)*log(2)^2*exp(x
-1)^2+2*(-3*x^3-11*x^2-6*x)*log(2)*exp(x-1)+x^4+3*x^3+3*x^2+9*x)*log(-log(x)+3+x)+4*(x-1)*log(2)^2*exp(x-1)^2+
2*(-2*x^2+2*x)*log(2)*exp(x-1)+x^3-x^2)/((4*x*log(2)^2*exp(x-1)^2-4*x^2*log(2)*exp(x-1)+x^3)*log(x)+4*(-x^2-3*
x)*log(2)^2*exp(x-1)^2+2*(2*x^3+6*x^2)*log(2)*exp(x-1)-x^4-3*x^3)/log(-log(x)+3+x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 46, normalized size = 1.39

method	result	size
risch	$-\frac{2x\ln (2){\mathrm{e}}^{x-1}-x^2+x+3}{2\ln (2){\mathrm{e}}^{x-1}-x}-\ln (\ln (-\ln (x)+3+x))$	$46$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x*ln(2)^2*exp(x-1)^2+2*(3*x^2+2*x)*ln(2)*exp(x-1)-x^3-3*x)*ln(x)+4*(x^2+3*x)*ln(2)^2*exp(x-1)^2+2*(-
3*x^3-11*x^2-6*x)*ln(2)*exp(x-1)+x^4+3*x^3+3*x^2+9*x)*ln(-ln(x)+3+x)+4*(x-1)*ln(2)^2*exp(x-1)^2+2*(-2*x^2+2*x)
*ln(2)*exp(x-1)+x^3-x^2)/((4*x*ln(2)^2*exp(x-1)^2-4*x^2*ln(2)*exp(x-1)+x^3)*ln(x)+4*(-x^2-3*x)*ln(2)^2*exp(x-1
)^2+2*(2*x^3+6*x^2)*ln(2)*exp(x-1)-x^4-3*x^3)/ln(-ln(x)+3+x),x,method=_RETURNVERBOSE)

[Out]

-(2*x*ln(2)*exp(x-1)-x^2+x+3)/(2*ln(2)*exp(x-1)-x)-ln(ln(-ln(x)+3+x))

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maxima [A]  time = 0.55, size = 50, normalized size = 1.52 $$\begin{array}{c}-\frac{x^{2}e-2x{e}^x\log (2)-xe-3e}{xe-2e^x\log (2)}-\log (\log (x-\log (x)+3))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x*log(2)^2*exp(x-1)^2+2*(3*x^2+2*x)*log(2)*exp(x-1)-x^3-3*x)*log(x)+4*(x^2+3*x)*log(2)^2*exp(x
-1)^2+2*(-3*x^3-11*x^2-6*x)*log(2)*exp(x-1)+x^4+3*x^3+3*x^2+9*x)*log(-log(x)+3+x)+4*(x-1)*log(2)^2*exp(x-1)^2+
2*(-2*x^2+2*x)*log(2)*exp(x-1)+x^3-x^2)/((4*x*log(2)^2*exp(x-1)^2-4*x^2*log(2)*exp(x-1)+x^3)*log(x)+4*(-x^2-3*
x)*log(2)^2*exp(x-1)^2+2*(2*x^3+6*x^2)*log(2)*exp(x-1)-x^4-3*x^3)/log(-log(x)+3+x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.66, size = 69, normalized size = 2.09 $$\begin{array}{c}-x-\ln (\ln (x-\ln (x)+3))-\frac{6\ln (2)-x\ln \left(16\right)+4x^2\ln (2)-x^2\ln \left(64\right)}{4\ln (2)(\ln (2)-x\ln (2))({\mathrm{e}}^{x-1}-\frac{x}{2\ln (2)})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^3 - x^2 + log(x - log(x) + 3)*(9*x - log(x)*(3*x + x^3 + 4*x*exp(2*x - 2)*log(2)^2 - 2*exp(x - 1)*log(
2)*(2*x + 3*x^2)) + 3*x^2 + 3*x^3 + x^4 - 2*exp(x - 1)*log(2)*(6*x + 11*x^2 + 3*x^3) + 4*exp(2*x - 2)*log(2)^2
*(3*x + x^2)) + 2*exp(x - 1)*log(2)*(2*x - 2*x^2) + 4*exp(2*x - 2)*log(2)^2*(x - 1))/(log(x - log(x) + 3)*(3*x
^3 - log(x)*(x^3 + 4*x*exp(2*x - 2)*log(2)^2 - 4*x^2*exp(x - 1)*log(2)) + x^4 + 4*exp(2*x - 2)*log(2)^2*(3*x +
x^2) - 2*exp(x - 1)*log(2)*(6*x^2 + 2*x^3))),x)

[Out]

- x - log(log(x - log(x) + 3)) - (6*log(2) - x*log(16) + 4*x^2*log(2) - x^2*log(64))/(4*log(2)*(log(2) - x*log
(2))*(exp(x - 1) - x/(2*log(2))))

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sympy [A]  time = 0.74, size = 27, normalized size = 0.82 $$\begin{array}{c}-x+\frac{-x-3}{-x+2e^{x-1}\log (2)}-\log (\log (x-\log (x)+3))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x*ln(2)**2*exp(x-1)**2+2*(3*x**2+2*x)*ln(2)*exp(x-1)-x**3-3*x)*ln(x)+4*(x**2+3*x)*ln(2)**2*exp
(x-1)**2+2*(-3*x**3-11*x**2-6*x)*ln(2)*exp(x-1)+x**4+3*x**3+3*x**2+9*x)*ln(-ln(x)+3+x)+4*(x-1)*ln(2)**2*exp(x-
1)**2+2*(-2*x**2+2*x)*ln(2)*exp(x-1)+x**3-x**2)/((4*x*ln(2)**2*exp(x-1)**2-4*x**2*ln(2)*exp(x-1)+x**3)*ln(x)+4
*(-x**2-3*x)*ln(2)**2*exp(x-1)**2+2*(2*x**3+6*x**2)*ln(2)*exp(x-1)-x**4-3*x**3)/ln(-ln(x)+3+x),x)

[Out]

-x + (-x - 3)/(-x + 2*exp(x - 1)*log(2)) - log(log(x - log(x) + 3))

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