3.79.2 \(\int \frac {(1-32 x+256 x^2) \log ^2(x)+e^{\frac {-16 x^3+(-64 x-16 x^2) \log (x)}{(-1+16 x) \log (x)}} (-16 x^2+256 x^3+(48 x^2-512 x^3) \log (x)+(64+32 x-256 x^2) \log ^2(x))}{(1-32 x+256 x^2) \log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ e^{\frac {4+x+\frac {x^2}{\log (x)}}{-1+\frac {1}{16 x}}}+x \]

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Rubi [F]  time = 5.43, 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 {\left (1-32 x+256 x^2\right ) \log ^2(x)+\exp \left (\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}\right ) \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{\left (1-32 x+256 x^2\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((1 - 32*x + 256*x^2)*Log[x]^2 + E^((-16*x^3 + (-64*x - 16*x^2)*Log[x])/((-1 + 16*x)*Log[x]))*(-16*x^2 + 2
56*x^3 + (48*x^2 - 512*x^3)*Log[x] + (64 + 32*x - 256*x^2)*Log[x]^2))/((1 - 32*x + 256*x^2)*Log[x]^2),x]

[Out]

x - Defer[Int][E^((-16*x*(x^2 + 4*Log[x] + x*Log[x]))/((-1 + 16*x)*Log[x])), x] + 65*Defer[Int][1/(E^((16*x*(x
^2 + 4*Log[x] + x*Log[x]))/((-1 + 16*x)*Log[x]))*(-1 + 16*x)^2), x] + Defer[Int][1/(E^((16*x*(x^2 + 4*Log[x] +
x*Log[x]))/((-1 + 16*x)*Log[x]))*Log[x]^2), x]/16 + Defer[Int][x/(E^((16*x*(x^2 + 4*Log[x] + x*Log[x]))/((-1
+ 16*x)*Log[x]))*Log[x]^2), x] + Defer[Int][1/(E^((16*x*(x^2 + 4*Log[x] + x*Log[x]))/((-1 + 16*x)*Log[x]))*(-1
+ 16*x)*Log[x]^2), x]/16 - Defer[Int][1/(E^((16*x*(x^2 + 4*Log[x] + x*Log[x]))/((-1 + 16*x)*Log[x]))*Log[x]),
x]/16 - 2*Defer[Int][x/(E^((16*x*(x^2 + 4*Log[x] + x*Log[x]))/((-1 + 16*x)*Log[x]))*Log[x]), x] + Defer[Int][
1/(E^((16*x*(x^2 + 4*Log[x] + x*Log[x]))/((-1 + 16*x)*Log[x]))*(-1 + 16*x)^2*Log[x]), x]/16

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (1-32 x+256 x^2\right ) \log ^2(x)+\exp \left (\frac {-16 x^3+\left (-64 x-16 x^2\right ) \log (x)}{(-1+16 x) \log (x)}\right ) \left (-16 x^2+256 x^3+\left (48 x^2-512 x^3\right ) \log (x)+\left (64+32 x-256 x^2\right ) \log ^2(x)\right )}{(-1+16 x)^2 \log ^2(x)} \, dx\\ &=\int \left (1+\frac {16 \exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) \left (-x^2+16 x^3+3 x^2 \log (x)-32 x^3 \log (x)+4 \log ^2(x)+2 x \log ^2(x)-16 x^2 \log ^2(x)\right )}{(1-16 x)^2 \log ^2(x)}\right ) \, dx\\ &=x+16 \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) \left (-x^2+16 x^3+3 x^2 \log (x)-32 x^3 \log (x)+4 \log ^2(x)+2 x \log ^2(x)-16 x^2 \log ^2(x)\right )}{(1-16 x)^2 \log ^2(x)} \, dx\\ &=x+16 \int \left (-\frac {2 \exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) \left (-2-x+8 x^2\right )}{(-1+16 x)^2}+\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x^2}{(-1+16 x) \log ^2(x)}-\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x^2 (-3+32 x)}{(-1+16 x)^2 \log (x)}\right ) \, dx\\ &=x+16 \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x^2}{(-1+16 x) \log ^2(x)} \, dx-16 \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x^2 (-3+32 x)}{(-1+16 x)^2 \log (x)} \, dx-32 \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) \left (-2-x+8 x^2\right )}{(-1+16 x)^2} \, dx\\ &=x+16 \int \left (\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{256 \log ^2(x)}+\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x}{16 \log ^2(x)}+\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{256 (-1+16 x) \log ^2(x)}\right ) \, dx-16 \int \left (\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{256 \log (x)}+\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x}{8 \log (x)}-\frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{256 (-1+16 x)^2 \log (x)}\right ) \, dx-32 \int \left (\frac {1}{32} \exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )-\frac {65 \exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{32 (-1+16 x)^2}\right ) \, dx\\ &=x+\frac {1}{16} \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{\log ^2(x)} \, dx+\frac {1}{16} \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{(-1+16 x) \log ^2(x)} \, dx-\frac {1}{16} \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{\log (x)} \, dx+\frac {1}{16} \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{(-1+16 x)^2 \log (x)} \, dx-2 \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x}{\log (x)} \, dx+65 \int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right )}{(-1+16 x)^2} \, dx-\int \exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) \, dx+\int \frac {\exp \left (-\frac {16 x \left (x^2+4 \log (x)+x \log (x)\right )}{(-1+16 x) \log (x)}\right ) x}{\log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.73, size = 38, normalized size = 1.41 \begin {gather*} e^{-\frac {65}{16}-x-\frac {65}{16 (-1+16 x)}-\frac {16 x^3}{(-1+16 x) \log (x)}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 - 32*x + 256*x^2)*Log[x]^2 + E^((-16*x^3 + (-64*x - 16*x^2)*Log[x])/((-1 + 16*x)*Log[x]))*(-16*x
^2 + 256*x^3 + (48*x^2 - 512*x^3)*Log[x] + (64 + 32*x - 256*x^2)*Log[x]^2))/((1 - 32*x + 256*x^2)*Log[x]^2),x]

[Out]

E^(-65/16 - x - 65/(16*(-1 + 16*x)) - (16*x^3)/((-1 + 16*x)*Log[x])) + x

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fricas [A]  time = 0.64, size = 30, normalized size = 1.11 \begin {gather*} x + e^{\left (-\frac {16 \, {\left (x^{3} + {\left (x^{2} + 4 \, x\right )} \log \relax (x)\right )}}{{\left (16 \, x - 1\right )} \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-256*x^2+32*x+64)*log(x)^2+(-512*x^3+48*x^2)*log(x)+256*x^3-16*x^2)*exp(((-16*x^2-64*x)*log(x)-16
*x^3)/(16*x-1)/log(x))+(256*x^2-32*x+1)*log(x)^2)/(256*x^2-32*x+1)/log(x)^2,x, algorithm="fricas")

[Out]

x + e^(-16*(x^3 + (x^2 + 4*x)*log(x))/((16*x - 1)*log(x)))

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giac [A]  time = 1.00, size = 32, normalized size = 1.19 \begin {gather*} x + e^{\left (-\frac {16 \, {\left (x^{3} + x^{2} \log \relax (x) + 4 \, x \log \relax (x)\right )}}{16 \, x \log \relax (x) - \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-256*x^2+32*x+64)*log(x)^2+(-512*x^3+48*x^2)*log(x)+256*x^3-16*x^2)*exp(((-16*x^2-64*x)*log(x)-16
*x^3)/(16*x-1)/log(x))+(256*x^2-32*x+1)*log(x)^2)/(256*x^2-32*x+1)/log(x)^2,x, algorithm="giac")

[Out]

x + e^(-16*(x^3 + x^2*log(x) + 4*x*log(x))/(16*x*log(x) - log(x)))

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maple [A]  time = 0.26, size = 30, normalized size = 1.11

method	result	size
risch	\(x +{\mathrm e}^{-\frac {16 x \left (x \ln \relax (x )+x^{2}+4 \ln \relax (x )\right )}{\left (16 x -1\right ) \ln \relax (x )}}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-256*x^2+32*x+64)*ln(x)^2+(-512*x^3+48*x^2)*ln(x)+256*x^3-16*x^2)*exp(((-16*x^2-64*x)*ln(x)-16*x^3)/(16
*x-1)/ln(x))+(256*x^2-32*x+1)*ln(x)^2)/(256*x^2-32*x+1)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

x+exp(-16*x*(x*ln(x)+x^2+4*ln(x))/(16*x-1)/ln(x))

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maxima [B]  time = 0.54, size = 89, normalized size = 3.30 \begin {gather*} {\left (x e^{\left (x + \frac {x^{2}}{\log \relax (x)} + \frac {x}{16 \, \log \relax (x)} + \frac {65}{16 \, {\left (16 \, x - 1\right )}} + \frac {1}{256 \, \log \relax (x)} + \frac {65}{16}\right )} + e^{\left (-\frac {1}{256 \, {\left (16 \, x - 1\right )} \log \relax (x)}\right )}\right )} e^{\left (-x - \frac {x^{2}}{\log \relax (x)} - \frac {x}{16 \, \log \relax (x)} - \frac {65}{16 \, {\left (16 \, x - 1\right )}} - \frac {1}{256 \, \log \relax (x)} - \frac {65}{16}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-256*x^2+32*x+64)*log(x)^2+(-512*x^3+48*x^2)*log(x)+256*x^3-16*x^2)*exp(((-16*x^2-64*x)*log(x)-16
*x^3)/(16*x-1)/log(x))+(256*x^2-32*x+1)*log(x)^2)/(256*x^2-32*x+1)/log(x)^2,x, algorithm="maxima")

[Out]

(x*e^(x + x^2/log(x) + 1/16*x/log(x) + 65/16/(16*x - 1) + 1/256/log(x) + 65/16) + e^(-1/256/((16*x - 1)*log(x)
)))*e^(-x - x^2/log(x) - 1/16*x/log(x) - 65/16/(16*x - 1) - 1/256/log(x) - 65/16)

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mupad [B]  time = 5.17, size = 53, normalized size = 1.96 \begin {gather*} x+{\mathrm {e}}^{\frac {64\,x\,\ln \relax (x)}{\ln \relax (x)-16\,x\,\ln \relax (x)}}\,{\mathrm {e}}^{\frac {16\,x^2\,\ln \relax (x)}{\ln \relax (x)-16\,x\,\ln \relax (x)}}\,{\mathrm {e}}^{\frac {16\,x^3}{\ln \relax (x)-16\,x\,\ln \relax (x)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)^2*(256*x^2 - 32*x + 1) + exp(-(log(x)*(64*x + 16*x^2) + 16*x^3)/(log(x)*(16*x - 1)))*(log(x)^2*(32
*x - 256*x^2 + 64) + log(x)*(48*x^2 - 512*x^3) - 16*x^2 + 256*x^3))/(log(x)^2*(256*x^2 - 32*x + 1)),x)

[Out]

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

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sympy [A]  time = 0.62, size = 29, normalized size = 1.07 \begin {gather*} x + e^{\frac {- 16 x^{3} + \left (- 16 x^{2} - 64 x\right ) \log {\relax (x )}}{\left (16 x - 1\right ) \log {\relax (x )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-256*x**2+32*x+64)*ln(x)**2+(-512*x**3+48*x**2)*ln(x)+256*x**3-16*x**2)*exp(((-16*x**2-64*x)*ln(x
)-16*x**3)/(16*x-1)/ln(x))+(256*x**2-32*x+1)*ln(x)**2)/(256*x**2-32*x+1)/ln(x)**2,x)

[Out]

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

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