∫ e 1 ___16 (80+x2)(−32x+8x2) log(x)+(e 1_16(80+x2)(−32+40x−8x2) log(−1+x)+e 1_16(80+x2)(−8x+12x2−5x3+x4) log(−1+x) log(x)) log(log(−1+x)) _______________________________________________________________________________________________________________________________________________________________________________________

3.100.34 \(\int \frac {e^{\frac {1}{16} (80+x^2)} (-32 x+8 x^2) \log (x)+(e^{\frac {1}{16} (80+x^2)} (-32+40 x-8 x^2) \log (-1+x)+e^{\frac {1}{16} (80+x^2)} (-8 x+12 x^2-5 x^3+x^4) \log (-1+x) \log (x)) \log (\log (-1+x))}{(-5 x+5 x^2) \log (-1+x) \log ^2(x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {8 e^{5+\frac {x^2}{16}} (-4+x) \log (\log (-1+x))}{5 \log (x)} \]

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Rubi [F] time = 14.73, 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^{\frac {1}{16} \left (80+x^2\right )} \left (-32 x+8 x^2\right ) \log (x)+\left (e^{\frac {1}{16} \left (80+x^2\right )} \left (-32+40 x-8 x^2\right ) \log (-1+x)+e^{\frac {1}{16} \left (80+x^2\right )} \left (-8 x+12 x^2-5 x^3+x^4\right ) \log (-1+x) \log (x)\right ) \log (\log (-1+x))}{\left (-5 x+5 x^2\right ) \log (-1+x) \log ^2(x)} \, dx \end {gather*}

Veriﬁcation is not applicable to the result.

[In]

Int[(E^((80 + x^2)/16)*(-32*x + 8*x^2)*Log[x] + (E^((80 + x^2)/16)*(-32 + 40*x - 8*x^2)*Log[-1 + x] + E^((80 +

x^2)/16)*(-8*x + 12*x^2 - 5*x^3 + x^4)*Log[-1 + x]*Log[x])*Log[Log[-1 + x]])/((-5*x + 5*x^2)*Log[-1 + x]*Log[

x]^2),x]

[Out]

(8*Defer[Int][E^(5 + x^2/16)/(Log[-1 + x]*Log[x]), x])/5 - (24*Defer[Int][E^(5 + x^2/16)/((-1 + x)*Log[-1 + x]

*Log[x]), x])/5 - (8*Defer[Int][(E^(5 + x^2/16)*Log[Log[-1 + x]])/Log[x]^2, x])/5 + (32*Defer[Int][(E^(5 + x^2

/16)*Log[Log[-1 + x]])/(x*Log[x]^2), x])/5 + (8*Defer[Int][(E^(5 + x^2/16)*Log[Log[-1 + x]])/Log[x], x])/5 - (

4*Defer[Int][(E^(5 + x^2/16)*x*Log[Log[-1 + x]])/Log[x], x])/5 + Defer[Int][(E^(5 + x^2/16)*x^2*Log[Log[-1 + x

]])/Log[x], x]/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {1}{16} \left (80+x^2\right )} \left (-32 x+8 x^2\right ) \log (x)+\left (e^{\frac {1}{16} \left (80+x^2\right )} \left (-32+40 x-8 x^2\right ) \log (-1+x)+e^{\frac {1}{16} \left (80+x^2\right )} \left (-8 x+12 x^2-5 x^3+x^4\right ) \log (-1+x) \log (x)\right ) \log (\log (-1+x))}{x (-5+5 x) \log (-1+x) \log ^2(x)} \, dx\\ &=\int \left (\frac {e^{5+\frac {x^2}{16}} \left (-32 x \log (x)+8 x^2 \log (x)-32 \log (-1+x) \log (\log (-1+x))+40 x \log (-1+x) \log (\log (-1+x))-8 x^2 \log (-1+x) \log (\log (-1+x))-8 x \log (-1+x) \log (x) \log (\log (-1+x))+12 x^2 \log (-1+x) \log (x) \log (\log (-1+x))-5 x^3 \log (-1+x) \log (x) \log (\log (-1+x))+x^4 \log (-1+x) \log (x) \log (\log (-1+x))\right )}{5 (-1+x) \log (-1+x) \log ^2(x)}-\frac {e^{5+\frac {x^2}{16}} \left (-32 x \log (x)+8 x^2 \log (x)-32 \log (-1+x) \log (\log (-1+x))+40 x \log (-1+x) \log (\log (-1+x))-8 x^2 \log (-1+x) \log (\log (-1+x))-8 x \log (-1+x) \log (x) \log (\log (-1+x))+12 x^2 \log (-1+x) \log (x) \log (\log (-1+x))-5 x^3 \log (-1+x) \log (x) \log (\log (-1+x))+x^4 \log (-1+x) \log (x) \log (\log (-1+x))\right )}{5 x \log (-1+x) \log ^2(x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} \left (-32 x \log (x)+8 x^2 \log (x)-32 \log (-1+x) \log (\log (-1+x))+40 x \log (-1+x) \log (\log (-1+x))-8 x^2 \log (-1+x) \log (\log (-1+x))-8 x \log (-1+x) \log (x) \log (\log (-1+x))+12 x^2 \log (-1+x) \log (x) \log (\log (-1+x))-5 x^3 \log (-1+x) \log (x) \log (\log (-1+x))+x^4 \log (-1+x) \log (x) \log (\log (-1+x))\right )}{(-1+x) \log (-1+x) \log ^2(x)} \, dx-\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} \left (-32 x \log (x)+8 x^2 \log (x)-32 \log (-1+x) \log (\log (-1+x))+40 x \log (-1+x) \log (\log (-1+x))-8 x^2 \log (-1+x) \log (\log (-1+x))-8 x \log (-1+x) \log (x) \log (\log (-1+x))+12 x^2 \log (-1+x) \log (x) \log (\log (-1+x))-5 x^3 \log (-1+x) \log (x) \log (\log (-1+x))+x^4 \log (-1+x) \log (x) \log (\log (-1+x))\right )}{x \log (-1+x) \log ^2(x)} \, dx\\ &=\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} \left (8 \left (4-5 x+x^2\right ) \log (-1+x) \log (\log (-1+x))-x \log (x) \left (8 (-4+x)+\left (-8+12 x-5 x^2+x^3\right ) \log (-1+x) \log (\log (-1+x))\right )\right )}{(1-x) \log (-1+x) \log ^2(x)} \, dx-\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} \left (-8 \left (4-5 x+x^2\right ) \log (-1+x) \log (\log (-1+x))+x \log (x) \left (8 (-4+x)+\left (-8+12 x-5 x^2+x^3\right ) \log (-1+x) \log (\log (-1+x))\right )\right )}{x \log (-1+x) \log ^2(x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {8 e^{5+\frac {x^2}{16}} (-4+x) x}{(-1+x) \log (-1+x) \log (x)}+\frac {e^{5+\frac {x^2}{16}} \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{\log ^2(x)}\right ) \, dx-\frac {1}{5} \int \left (\frac {8 e^{5+\frac {x^2}{16}} (-4+x)}{\log (-1+x) \log (x)}+\frac {e^{5+\frac {x^2}{16}} (-1+x) \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{x \log ^2(x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{\log ^2(x)} \, dx-\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} (-1+x) \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{x \log ^2(x)} \, dx-\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} (-4+x)}{\log (-1+x) \log (x)} \, dx+\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} (-4+x) x}{(-1+x) \log (-1+x) \log (x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {32 e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log ^2(x)}-\frac {8 e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log ^2(x)}+\frac {8 e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log (x)}-\frac {4 e^{5+\frac {x^2}{16}} x^2 \log (\log (-1+x))}{\log (x)}+\frac {e^{5+\frac {x^2}{16}} x^3 \log (\log (-1+x))}{\log (x)}\right ) \, dx-\frac {1}{5} \int \left (\frac {e^{5+\frac {x^2}{16}} \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{\log ^2(x)}-\frac {e^{5+\frac {x^2}{16}} \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{x \log ^2(x)}\right ) \, dx-\frac {8}{5} \int \left (-\frac {4 e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)}+\frac {e^{5+\frac {x^2}{16}} x}{\log (-1+x) \log (x)}\right ) \, dx+\frac {8}{5} \int \left (-\frac {3 e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)}-\frac {3 e^{5+\frac {x^2}{16}}}{(-1+x) \log (-1+x) \log (x)}+\frac {e^{5+\frac {x^2}{16}} x}{\log (-1+x) \log (x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} x^3 \log (\log (-1+x))}{\log (x)} \, dx-\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{\log ^2(x)} \, dx+\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} \left (32-8 x+8 x \log (x)-4 x^2 \log (x)+x^3 \log (x)\right ) \log (\log (-1+x))}{x \log ^2(x)} \, dx-\frac {4}{5} \int \frac {e^{5+\frac {x^2}{16}} x^2 \log (\log (-1+x))}{\log (x)} \, dx-\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log ^2(x)} \, dx+\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log (x)} \, dx-\frac {24}{5} \int \frac {e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)} \, dx-\frac {24}{5} \int \frac {e^{5+\frac {x^2}{16}}}{(-1+x) \log (-1+x) \log (x)} \, dx+\frac {32}{5} \int \frac {e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)} \, dx+\frac {32}{5} \int \frac {e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log ^2(x)} \, dx\\ &=\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} x^3 \log (\log (-1+x))}{\log (x)} \, dx+\frac {1}{5} \int \left (-\frac {8 e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log ^2(x)}+\frac {32 e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{x \log ^2(x)}+\frac {8 e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log (x)}-\frac {4 e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log (x)}+\frac {e^{5+\frac {x^2}{16}} x^2 \log (\log (-1+x))}{\log (x)}\right ) \, dx-\frac {1}{5} \int \left (\frac {32 e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log ^2(x)}-\frac {8 e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log ^2(x)}+\frac {8 e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log (x)}-\frac {4 e^{5+\frac {x^2}{16}} x^2 \log (\log (-1+x))}{\log (x)}+\frac {e^{5+\frac {x^2}{16}} x^3 \log (\log (-1+x))}{\log (x)}\right ) \, dx-\frac {4}{5} \int \frac {e^{5+\frac {x^2}{16}} x^2 \log (\log (-1+x))}{\log (x)} \, dx-\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log ^2(x)} \, dx+\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log (x)} \, dx-\frac {24}{5} \int \frac {e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)} \, dx-\frac {24}{5} \int \frac {e^{5+\frac {x^2}{16}}}{(-1+x) \log (-1+x) \log (x)} \, dx+\frac {32}{5} \int \frac {e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)} \, dx+\frac {32}{5} \int \frac {e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log ^2(x)} \, dx\\ &=\frac {1}{5} \int \frac {e^{5+\frac {x^2}{16}} x^2 \log (\log (-1+x))}{\log (x)} \, dx-\frac {4}{5} \int \frac {e^{5+\frac {x^2}{16}} x \log (\log (-1+x))}{\log (x)} \, dx-\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log ^2(x)} \, dx+\frac {8}{5} \int \frac {e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{\log (x)} \, dx-\frac {24}{5} \int \frac {e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)} \, dx-\frac {24}{5} \int \frac {e^{5+\frac {x^2}{16}}}{(-1+x) \log (-1+x) \log (x)} \, dx+\frac {32}{5} \int \frac {e^{5+\frac {x^2}{16}}}{\log (-1+x) \log (x)} \, dx+\frac {32}{5} \int \frac {e^{5+\frac {x^2}{16}} \log (\log (-1+x))}{x \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.09, size = 27, normalized size = 1.00 \begin {gather*} \frac {8 e^{5+\frac {x^2}{16}} (-4+x) \log (\log (-1+x))}{5 \log (x)} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(E^((80 + x^2)/16)*(-32*x + 8*x^2)*Log[x] + (E^((80 + x^2)/16)*(-32 + 40*x - 8*x^2)*Log[-1 + x] + E^

((80 + x^2)/16)*(-8*x + 12*x^2 - 5*x^3 + x^4)*Log[-1 + x]*Log[x])*Log[Log[-1 + x]])/((-5*x + 5*x^2)*Log[-1 + x

]*Log[x]^2),x]

[Out]

(8*E^(5 + x^2/16)*(-4 + x)*Log[Log[-1 + x]])/(5*Log[x])

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fricas [A] time = 0.98, size = 22, normalized size = 0.81 \begin {gather*} \frac {8 \, {\left (x - 4\right )} e^{\left (\frac {1}{16} \, x^{2} + 5\right )} \log \left (\log \left (x - 1\right )\right )}{5 \, \log \relax (x)} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^4-5*x^3+12*x^2-8*x)*exp(1/16*x^2+5)*log(x-1)*log(x)+(-8*x^2+40*x-32)*exp(1/16*x^2+5)*log(x-1))*

log(log(x-1))+(8*x^2-32*x)*exp(1/16*x^2+5)*log(x))/(5*x^2-5*x)/log(x-1)/log(x)^2,x, algorithm="fricas")

[Out]

8/5*(x - 4)*e^(1/16*x^2 + 5)*log(log(x - 1))/log(x)

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giac [A] time = 0.33, size = 37, normalized size = 1.37 \begin {gather*} \frac {8 \, {\left (x e^{\left (\frac {1}{16} \, x^{2} + 5\right )} \log \left (\log \left (x - 1\right )\right ) - 4 \, e^{\left (\frac {1}{16} \, x^{2} + 5\right )} \log \left (\log \left (x - 1\right )\right )\right )}}{5 \, \log \relax (x)} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^4-5*x^3+12*x^2-8*x)*exp(1/16*x^2+5)*log(x-1)*log(x)+(-8*x^2+40*x-32)*exp(1/16*x^2+5)*log(x-1))*

log(log(x-1))+(8*x^2-32*x)*exp(1/16*x^2+5)*log(x))/(5*x^2-5*x)/log(x-1)/log(x)^2,x, algorithm="giac")

[Out]

8/5*(x*e^(1/16*x^2 + 5)*log(log(x - 1)) - 4*e^(1/16*x^2 + 5)*log(log(x - 1)))/log(x)

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maple [A] time = 0.11, size = 23, normalized size = 0.85


method	result	size

risch	\(\frac {8 \ln \left (\ln \left (x -1\right )\right ) {\mathrm e}^{\frac {x^{2}}{16}+5} \left (x -4\right )}{5 \ln \relax (x )}\)	\(23\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((((x^4-5*x^3+12*x^2-8*x)*exp(1/16*x^2+5)*ln(x-1)*ln(x)+(-8*x^2+40*x-32)*exp(1/16*x^2+5)*ln(x-1))*ln(ln(x-1

))+(8*x^2-32*x)*exp(1/16*x^2+5)*ln(x))/(5*x^2-5*x)/ln(x-1)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

8/5*ln(ln(x-1))*exp(1/16*x^2+5)/ln(x)*(x-4)

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maxima [A] time = 0.52, size = 26, normalized size = 0.96 \begin {gather*} \frac {8 \, {\left (x e^{5} - 4 \, e^{5}\right )} e^{\left (\frac {1}{16} \, x^{2}\right )} \log \left (\log \left (x - 1\right )\right )}{5 \, \log \relax (x)} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^4-5*x^3+12*x^2-8*x)*exp(1/16*x^2+5)*log(x-1)*log(x)+(-8*x^2+40*x-32)*exp(1/16*x^2+5)*log(x-1))*

log(log(x-1))+(8*x^2-32*x)*exp(1/16*x^2+5)*log(x))/(5*x^2-5*x)/log(x-1)/log(x)^2,x, algorithm="maxima")

[Out]

8/5*(x*e^5 - 4*e^5)*e^(1/16*x^2)*log(log(x - 1))/log(x)

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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\ln \left (\ln \left (x-1\right )\right )\,\left (\ln \left (x-1\right )\,{\mathrm {e}}^{\frac {x^2}{16}+5}\,\left (8\,x^2-40\,x+32\right )+\ln \left (x-1\right )\,{\mathrm {e}}^{\frac {x^2}{16}+5}\,\ln \relax (x)\,\left (-x^4+5\,x^3-12\,x^2+8\,x\right )\right )+{\mathrm {e}}^{\frac {x^2}{16}+5}\,\ln \relax (x)\,\left (32\,x-8\,x^2\right )}{\ln \left (x-1\right )\,{\ln \relax (x)}^2\,\left (5\,x-5\,x^2\right )} \,d x \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(x - 1))*(log(x - 1)*exp(x^2/16 + 5)*(8*x^2 - 40*x + 32) + log(x - 1)*exp(x^2/16 + 5)*log(x)*(8*x

- 12*x^2 + 5*x^3 - x^4)) + exp(x^2/16 + 5)*log(x)*(32*x - 8*x^2))/(log(x - 1)*log(x)^2*(5*x - 5*x^2)),x)

[Out]

int((log(log(x - 1))*(log(x - 1)*exp(x^2/16 + 5)*(8*x^2 - 40*x + 32) + log(x - 1)*exp(x^2/16 + 5)*log(x)*(8*x

- 12*x^2 + 5*x^3 - x^4)) + exp(x^2/16 + 5)*log(x)*(32*x - 8*x^2))/(log(x - 1)*log(x)^2*(5*x - 5*x^2)), x)

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sympy [A] time = 0.86, size = 31, normalized size = 1.15 \begin {gather*} \frac {\left (8 x \log {\left (\log {\left (x - 1 \right )} \right )} - 32 \log {\left (\log {\left (x - 1 \right )} \right )}\right ) e^{\frac {x^{2}}{16} + 5}}{5 \log {\relax (x )}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**4-5*x**3+12*x**2-8*x)*exp(1/16*x**2+5)*ln(x-1)*ln(x)+(-8*x**2+40*x-32)*exp(1/16*x**2+5)*ln(x-1

))*ln(ln(x-1))+(8*x**2-32*x)*exp(1/16*x**2+5)*ln(x))/(5*x**2-5*x)/ln(x-1)/ln(x)**2,x)

[Out]

(8*x*log(log(x - 1)) - 32*log(log(x - 1)))*exp(x**2/16 + 5)/(5*log(x))

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