∫ −x2+x3+e−1+x(2x−2x2) log(4)+e−2+2x(−1+x) log2(4)+(9x+3x2+3x3+x4+e−1+x(−6x−11x2−3x3) log(4)+e−2+2x(3x+x2) log2(4)+(−3x−x3+e−1+x(2x+3x2) log(4)−e−2+2xx log2(4)) log(x)) log(3+x−log(x))______________________________________________________________________________________________________________________________________________________ (−3x3−x4+e−1+x(6x2+2x3) log(4)+e−2+2x(−3x−x2) log2(4)+(x3−2e−1+xx2 log(4)+e−2+2xx log2(4)) log(x)) log(3+x−log(x)) dx

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

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 {gather*} \int \frac {-x^2+x^3+e^{-1+x} \left (2 x-2 x^2\right ) \log (4)+e^{-2+2 x} (-1+x) \log ^2(4)+\left (9 x+3 x^2+3 x^3+x^4+e^{-1+x} \left (-6 x-11 x^2-3 x^3\right ) \log (4)+e^{-2+2 x} \left (3 x+x^2\right ) \log ^2(4)+\left (-3 x-x^3+e^{-1+x} \left (2 x+3 x^2\right ) \log (4)-e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))}{\left (-3 x^3-x^4+e^{-1+x} \left (6 x^2+2 x^3\right ) \log (4)+e^{-2+2 x} \left (-3 x-x^2\right ) \log ^2(4)+\left (x^3-2 e^{-1+x} x^2 \log (4)+e^{-2+2 x} x \log ^2(4)\right ) \log (x)\right ) \log (3+x-\log (x))} \, dx \end {gather*}

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

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

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 {gather*} -\frac {2 \, x e^{\left (x - 1\right )} \log \relax (2) - x^{2} + {\left (2 \, e^{\left (x - 1\right )} \log \relax (2) - x\right )} \log \left (\log \left (x - \log \relax (x) + 3\right )\right ) + x + 3}{2 \, e^{\left (x - 1\right )} \log \relax (2) - x} \end {gather*}

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 {gather*} -\frac {x^{2} e - 2 \, x e^{x} \log \relax (2) + x e \log \left (\log \left (x - \log \relax (x) + 3\right )\right ) - 2 \, e^{x} \log \relax (2) \log \left (\log \left (x - \log \relax (x) + 3\right )\right ) - x e - 3 \, e}{x e - 2 \, e^{x} \log \relax (2)} \end {gather*}

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	\(-\frac {2 x \ln \relax (2) {\mathrm e}^{x -1}-x^{2}+x +3}{2 \ln \relax (2) {\mathrm e}^{x -1}-x}-\ln \left (\ln \left (-\ln \relax (x )+3+x \right )\right )\)	\(46\)

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 {gather*} -\frac {x^{2} e - 2 \, x e^{x} \log \relax (2) - x e - 3 \, e}{x e - 2 \, e^{x} \log \relax (2)} - \log \left (\log \left (x - \log \relax (x) + 3\right )\right ) \end {gather*}

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 {gather*} -x-\ln \left (\ln \left (x-\ln \relax (x)+3\right )\right )-\frac {6\,\ln \relax (2)-x\,\ln \left (16\right )+4\,x^2\,\ln \relax (2)-x^2\,\ln \left (64\right )}{4\,\ln \relax (2)\,\left (\ln \relax (2)-x\,\ln \relax (2)\right )\,\left ({\mathrm {e}}^{x-1}-\frac {x}{2\,\ln \relax (2)}\right )} \end {gather*}

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 {gather*} - x + \frac {- x - 3}{- x + 2 e^{x - 1} \log {\relax (2 )}} - \log {\left (\log {\left (x - \log {\relax (x )} + 3 \right )} \right )} \end {gather*}

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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