3.22.77 \(\int \frac {e (2-x)+48 x-32 x^2+(16-16 x) \log (4)+(-16+16 x) \log (-3+3 x)+e^{e^x} (-32+16 x+e^x (-16 x+16 x^2+(-16+16 x) \log (4))+e^x (16-16 x) \log (-3+3 x))}{-16+16 x} \, dx\)

Optimal. Leaf size=29 \[ \left (\frac {e}{16}-e^{e^x}+x\right ) (-x-\log (4)+\log (3 (-1+x))) \]

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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 {e (2-x)+48 x-32 x^2+(16-16 x) \log (4)+(-16+16 x) \log (-3+3 x)+e^{e^x} \left (-32+16 x+e^x \left (-16 x+16 x^2+(-16+16 x) \log (4)\right )+e^x (16-16 x) \log (-3+3 x)\right )}{-16+16 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E*(2 - x) + 48*x - 32*x^2 + (16 - 16*x)*Log[4] + (-16 + 16*x)*Log[-3 + 3*x] + E^E^x*(-32 + 16*x + E^x*(-1
6*x + 16*x^2 + (-16 + 16*x)*Log[4]) + E^x*(16 - 16*x)*Log[-3 + 3*x]))/(-16 + 16*x),x]

[Out]

-x - x^2 + ExpIntegralEi[E^x] + (x*(16 - E - 16*Log[4]))/16 + E^E^x*Log[4] - (1 - x)*Log[-3*(1 - x)] + ((16 +
E)*Log[1 - x])/16 - E^E^x*Log[-3 + 3*x] + Defer[Int][E^E^x/(1 - x), x] + Defer[Int][E^E^x/(-1 + x), x] + Defer
[Int][E^(E^x + x)*x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {32 e^{e^x}-16 e^{e^x} x+32 x^2-48 x \left (1-\frac {e}{48}-\frac {2 \log (2)}{3}\right )-2 e \left (1+\frac {8 \log (4)}{e}\right )+16 \log (3 (-1+x))-16 x \log (3 (-1+x))}{16 (1-x)}+e^{e^x+x} (x+\log (4)-\log (-3+3 x))\right ) \, dx\\ &=\frac {1}{16} \int \frac {32 e^{e^x}-16 e^{e^x} x+32 x^2-48 x \left (1-\frac {e}{48}-\frac {2 \log (2)}{3}\right )-2 e \left (1+\frac {8 \log (4)}{e}\right )+16 \log (3 (-1+x))-16 x \log (3 (-1+x))}{1-x} \, dx+\int e^{e^x+x} (x+\log (4)-\log (-3+3 x)) \, dx\\ &=\frac {1}{16} \int \frac {e (-2+x)-16 e^{e^x} (-2+x)-16 \left (-2 x^2-x (-3+\log (4))+\log (4)\right )-16 (-1+x) \log (3 (-1+x))}{1-x} \, dx+\int \left (e^{e^x+x} x+e^{e^x+x} \log (4)-e^{e^x+x} \log (-3+3 x)\right ) \, dx\\ &=\frac {1}{16} \int \left (\frac {16 e^{e^x} (-2+x)}{-1+x}+\frac {32 x^2-48 x \left (1-\frac {e}{48}-\frac {2 \log (2)}{3}\right )-2 e \left (1+\frac {8 \log (4)}{e}\right )+16 \log (3 (-1+x))-16 x \log (3 (-1+x))}{1-x}\right ) \, dx+\log (4) \int e^{e^x+x} \, dx+\int e^{e^x+x} x \, dx-\int e^{e^x+x} \log (-3+3 x) \, dx\\ &=-e^{e^x} \log (-3+3 x)+\frac {1}{16} \int \frac {32 x^2-48 x \left (1-\frac {e}{48}-\frac {2 \log (2)}{3}\right )-2 e \left (1+\frac {8 \log (4)}{e}\right )+16 \log (3 (-1+x))-16 x \log (3 (-1+x))}{1-x} \, dx+\log (4) \operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )+\int \frac {e^{e^x}}{-1+x} \, dx+\int \frac {e^{e^x} (-2+x)}{-1+x} \, dx+\int e^{e^x+x} x \, dx\\ &=e^{e^x} \log (4)-e^{e^x} \log (-3+3 x)+\frac {1}{16} \int \frac {e (-2+x)-16 \left (-2 x^2-x (-3+\log (4))+\log (4)\right )-16 (-1+x) \log (3 (-1+x))}{1-x} \, dx+\int \left (e^{e^x}+\frac {e^{e^x}}{1-x}\right ) \, dx+\int \frac {e^{e^x}}{-1+x} \, dx+\int e^{e^x+x} x \, dx\\ &=e^{e^x} \log (4)-e^{e^x} \log (-3+3 x)+\frac {1}{16} \int \left (\frac {32 x^2-x (48-e-16 \log (4))-2 (e+8 \log (4))}{1-x}+16 \log (-3+3 x)\right ) \, dx+\int e^{e^x} \, dx+\int \frac {e^{e^x}}{1-x} \, dx+\int \frac {e^{e^x}}{-1+x} \, dx+\int e^{e^x+x} x \, dx\\ &=e^{e^x} \log (4)-e^{e^x} \log (-3+3 x)+\frac {1}{16} \int \frac {32 x^2-x (48-e-16 \log (4))-2 (e+8 \log (4))}{1-x} \, dx+\int \frac {e^{e^x}}{1-x} \, dx+\int \frac {e^{e^x}}{-1+x} \, dx+\int e^{e^x+x} x \, dx+\int \log (-3+3 x) \, dx+\operatorname {Subst}\left (\int \frac {e^x}{x} \, dx,x,e^x\right )\\ &=\text {Ei}\left (e^x\right )+e^{e^x} \log (4)-e^{e^x} \log (-3+3 x)+\frac {1}{16} \int \left (\frac {16+e}{-1+x}-32 x+16 \left (1-\frac {e}{16}-\log (4)\right )\right ) \, dx+\frac {1}{3} \operatorname {Subst}(\int \log (x) \, dx,x,-3+3 x)+\int \frac {e^{e^x}}{1-x} \, dx+\int \frac {e^{e^x}}{-1+x} \, dx+\int e^{e^x+x} x \, dx\\ &=-x-x^2+\text {Ei}\left (e^x\right )+\frac {1}{16} x (16-e-16 \log (4))+e^{e^x} \log (4)-(1-x) \log (-3 (1-x))+\frac {1}{16} (16+e) \log (1-x)-e^{e^x} \log (-3+3 x)+\int \frac {e^{e^x}}{1-x} \, dx+\int \frac {e^{e^x}}{-1+x} \, dx+\int e^{e^x+x} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.74, size = 59, normalized size = 2.03 \begin {gather*} -\frac {e x}{16}+e^{e^x} x-x^2+e^{e^x} \log (4)-x \log (4)+\frac {1}{16} e \log (1-x)+\left (-e^{e^x}+x\right ) \log (3 (-1+x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E*(2 - x) + 48*x - 32*x^2 + (16 - 16*x)*Log[4] + (-16 + 16*x)*Log[-3 + 3*x] + E^E^x*(-32 + 16*x + E
^x*(-16*x + 16*x^2 + (-16 + 16*x)*Log[4]) + E^x*(16 - 16*x)*Log[-3 + 3*x]))/(-16 + 16*x),x]

[Out]

-1/16*(E*x) + E^E^x*x - x^2 + E^E^x*Log[4] - x*Log[4] + (E*Log[1 - x])/16 + (-E^E^x + x)*Log[3*(-1 + x)]

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fricas [A]  time = 0.68, size = 48, normalized size = 1.66 \begin {gather*} -x^{2} - \frac {1}{16} \, x e + {\left (x + 2 \, \log \relax (2) - \log \left (3 \, x - 3\right )\right )} e^{\left (e^{x}\right )} - 2 \, x \log \relax (2) + \frac {1}{16} \, {\left (16 \, x + e\right )} \log \left (3 \, x - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x+16)*exp(x)*log(3*x-3)+(2*(16*x-16)*log(2)+16*x^2-16*x)*exp(x)+16*x-32)*exp(exp(x))+(16*x-16
)*log(3*x-3)+2*(-16*x+16)*log(2)+(2-x)*exp(1)-32*x^2+48*x)/(16*x-16),x, algorithm="fricas")

[Out]

-x^2 - 1/16*x*e + (x + 2*log(2) - log(3*x - 3))*e^(e^x) - 2*x*log(2) + 1/16*(16*x + e)*log(3*x - 3)

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giac [B]  time = 0.33, size = 78, normalized size = 2.69 \begin {gather*} -\frac {1}{16} \, {\left (16 \, x^{2} e^{x} + 32 \, x e^{x} \log \relax (2) - 16 \, x e^{x} \log \left (3 \, x - 3\right ) - 16 \, x e^{\left (x + e^{x}\right )} + x e^{\left (x + 1\right )} - 32 \, e^{\left (x + e^{x}\right )} \log \relax (2) + 16 \, e^{\left (x + e^{x}\right )} \log \left (3 \, x - 3\right ) - e^{\left (x + 1\right )} \log \left (x - 1\right )\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x+16)*exp(x)*log(3*x-3)+(2*(16*x-16)*log(2)+16*x^2-16*x)*exp(x)+16*x-32)*exp(exp(x))+(16*x-16
)*log(3*x-3)+2*(-16*x+16)*log(2)+(2-x)*exp(1)-32*x^2+48*x)/(16*x-16),x, algorithm="giac")

[Out]

-1/16*(16*x^2*e^x + 32*x*e^x*log(2) - 16*x*e^x*log(3*x - 3) - 16*x*e^(x + e^x) + x*e^(x + 1) - 32*e^(x + e^x)*
log(2) + 16*e^(x + e^x)*log(3*x - 3) - e^(x + 1)*log(x - 1))*e^(-x)

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maple [A]  time = 0.39, size = 51, normalized size = 1.76

method	result	size
risch	\(\ln \left (3 x -3\right ) x -\frac {x \,{\mathrm e}}{16}-2 x \ln \relax (2)-x^{2}+\frac {{\mathrm e} \ln \left (x -1\right )}{16}+\left (x +2 \ln \relax (2)-\ln \left (3 x -3\right )\right ) {\mathrm e}^{{\mathrm e}^{x}}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-16*x+16)*exp(x)*ln(3*x-3)+(2*(16*x-16)*ln(2)+16*x^2-16*x)*exp(x)+16*x-32)*exp(exp(x))+(16*x-16)*ln(3*x
-3)+2*(-16*x+16)*ln(2)+(2-x)*exp(1)-32*x^2+48*x)/(16*x-16),x,method=_RETURNVERBOSE)

[Out]

ln(3*x-3)*x-1/16*x*exp(1)-2*x*ln(2)-x^2+1/16*exp(1)*ln(x-1)+(x+2*ln(2)-ln(3*x-3))*exp(exp(x))

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maxima [B]  time = 0.55, size = 81, normalized size = 2.79 \begin {gather*} -x^{2} + x {\left (\log \relax (3) - 1\right )} - \frac {1}{16} \, {\left (x + \log \left (x - 1\right )\right )} e + {\left (x - \log \relax (3) + 2 \, \log \relax (2) - \log \left (x - 1\right )\right )} e^{\left (e^{x}\right )} - 2 \, {\left (x + \log \left (x - 1\right )\right )} \log \relax (2) + {\left (x - 1\right )} \log \left (x - 1\right ) + \frac {1}{8} \, e \log \left (x - 1\right ) + 2 \, \log \relax (2) \log \left (x - 1\right ) + x + \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x+16)*exp(x)*log(3*x-3)+(2*(16*x-16)*log(2)+16*x^2-16*x)*exp(x)+16*x-32)*exp(exp(x))+(16*x-16
)*log(3*x-3)+2*(-16*x+16)*log(2)+(2-x)*exp(1)-32*x^2+48*x)/(16*x-16),x, algorithm="maxima")

[Out]

-x^2 + x*(log(3) - 1) - 1/16*(x + log(x - 1))*e + (x - log(3) + 2*log(2) - log(x - 1))*e^(e^x) - 2*(x + log(x
- 1))*log(2) + (x - 1)*log(x - 1) + 1/8*e*log(x - 1) + 2*log(2)*log(x - 1) + x + log(x - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {48\,x-2\,\ln \relax (2)\,\left (16\,x-16\right )+\ln \left (3\,x-3\right )\,\left (16\,x-16\right )-\mathrm {e}\,\left (x-2\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (16\,x+{\mathrm {e}}^x\,\left (2\,\ln \relax (2)\,\left (16\,x-16\right )-16\,x+16\,x^2\right )-{\mathrm {e}}^x\,\ln \left (3\,x-3\right )\,\left (16\,x-16\right )-32\right )-32\,x^2}{16\,x-16} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((48*x - 2*log(2)*(16*x - 16) + log(3*x - 3)*(16*x - 16) - exp(1)*(x - 2) + exp(exp(x))*(16*x + exp(x)*(2*l
og(2)*(16*x - 16) - 16*x + 16*x^2) - exp(x)*log(3*x - 3)*(16*x - 16) - 32) - 32*x^2)/(16*x - 16),x)

[Out]

int((48*x - 2*log(2)*(16*x - 16) + log(3*x - 3)*(16*x - 16) - exp(1)*(x - 2) + exp(exp(x))*(16*x + exp(x)*(2*l
og(2)*(16*x - 16) - 16*x + 16*x^2) - exp(x)*log(3*x - 3)*(16*x - 16) - 32) - 32*x^2)/(16*x - 16), x)

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sympy [B]  time = 33.56, size = 51, normalized size = 1.76 \begin {gather*} - x^{2} + x \log {\left (3 x - 3 \right )} - x \left (\frac {e}{16} + 2 \log {\relax (2 )}\right ) + \left (x - \log {\left (3 x - 3 \right )} + 2 \log {\relax (2 )}\right ) e^{e^{x}} + \frac {e \log {\left (x - 1 \right )}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x+16)*exp(x)*ln(3*x-3)+(2*(16*x-16)*ln(2)+16*x**2-16*x)*exp(x)+16*x-32)*exp(exp(x))+(16*x-16)
*ln(3*x-3)+2*(-16*x+16)*ln(2)+(2-x)*exp(1)-32*x**2+48*x)/(16*x-16),x)

[Out]

-x**2 + x*log(3*x - 3) - x*(E/16 + 2*log(2)) + (x - log(3*x - 3) + 2*log(2))*exp(exp(x)) + E*log(x - 1)/16

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