3.82.6 \(\int \frac {e^2 (-6 x+24 x^2)+e^4 (9 x^2-48 x^3) \log (x)+(6-24 x+e^2 (-12 x+72 x^2) \log (x)) \log (\log (x))+(3-24 x) \log (x) \log ^2(\log (x))}{64 \log (x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {3}{16} \left (\frac {1}{4}-x\right ) x \left (e^2 x-\log (\log (x))\right )^2 \]

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Rubi [F]  time = 0.65, 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 \left (-6 x+24 x^2\right )+e^4 \left (9 x^2-48 x^3\right ) \log (x)+\left (6-24 x+e^2 \left (-12 x+72 x^2\right ) \log (x)\right ) \log (\log (x))+(3-24 x) \log (x) \log ^2(\log (x))}{64 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^2*(-6*x + 24*x^2) + E^4*(9*x^2 - 48*x^3)*Log[x] + (6 - 24*x + E^2*(-12*x + 72*x^2)*Log[x])*Log[Log[x]]
+ (3 - 24*x)*Log[x]*Log[Log[x]]^2)/(64*Log[x]),x]

[Out]

(3*E^4*x^3)/64 - (3*E^4*x^4)/16 - (3*E^2*x^2*Log[Log[x]])/32 + (3*E^2*x^3*Log[Log[x]])/8 + (3*Defer[Int][Log[L
og[x]]/Log[x], x])/32 - (3*Defer[Int][(x*Log[Log[x]])/Log[x], x])/8 + (3*Defer[Int][Log[Log[x]]^2, x])/64 - (3
*Defer[Int][x*Log[Log[x]]^2, x])/8

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{64} \int \frac {e^2 \left (-6 x+24 x^2\right )+e^4 \left (9 x^2-48 x^3\right ) \log (x)+\left (6-24 x+e^2 \left (-12 x+72 x^2\right ) \log (x)\right ) \log (\log (x))+(3-24 x) \log (x) \log ^2(\log (x))}{\log (x)} \, dx\\ &=\frac {1}{64} \int \frac {3 \left (e^2 x-\log (\log (x))\right ) \left (-2+8 x-\log (x) \left (e^2 x (-3+16 x)+(1-8 x) \log (\log (x))\right )\right )}{\log (x)} \, dx\\ &=\frac {3}{64} \int \frac {\left (e^2 x-\log (\log (x))\right ) \left (-2+8 x-\log (x) \left (e^2 x (-3+16 x)+(1-8 x) \log (\log (x))\right )\right )}{\log (x)} \, dx\\ &=\frac {3}{64} \int \left (-\frac {e^2 x \left (2-8 x-3 e^2 x \log (x)+16 e^2 x^2 \log (x)\right )}{\log (x)}+\frac {2 \left (1-4 x-2 e^2 x \log (x)+12 e^2 x^2 \log (x)\right ) \log (\log (x))}{\log (x)}-(-1+8 x) \log ^2(\log (x))\right ) \, dx\\ &=-\left (\frac {3}{64} \int (-1+8 x) \log ^2(\log (x)) \, dx\right )+\frac {3}{32} \int \frac {\left (1-4 x-2 e^2 x \log (x)+12 e^2 x^2 \log (x)\right ) \log (\log (x))}{\log (x)} \, dx-\frac {1}{64} \left (3 e^2\right ) \int \frac {x \left (2-8 x-3 e^2 x \log (x)+16 e^2 x^2 \log (x)\right )}{\log (x)} \, dx\\ &=-\left (\frac {3}{64} \int \left (-\log ^2(\log (x))+8 x \log ^2(\log (x))\right ) \, dx\right )+\frac {3}{32} \int \left (-2 e^2 x \log (\log (x))+12 e^2 x^2 \log (\log (x))+\frac {\log (\log (x))}{\log (x)}-\frac {4 x \log (\log (x))}{\log (x)}\right ) \, dx-\frac {1}{64} \left (3 e^2\right ) \int \frac {x \left (2-8 x+e^2 x (-3+16 x) \log (x)\right )}{\log (x)} \, dx\\ &=\frac {3}{64} \int \log ^2(\log (x)) \, dx+\frac {3}{32} \int \frac {\log (\log (x))}{\log (x)} \, dx-\frac {3}{8} \int \frac {x \log (\log (x))}{\log (x)} \, dx-\frac {3}{8} \int x \log ^2(\log (x)) \, dx-\frac {1}{64} \left (3 e^2\right ) \int \left (e^2 x^2 (-3+16 x)-\frac {2 x (-1+4 x)}{\log (x)}\right ) \, dx-\frac {1}{16} \left (3 e^2\right ) \int x \log (\log (x)) \, dx+\frac {1}{8} \left (9 e^2\right ) \int x^2 \log (\log (x)) \, dx\\ &=-\frac {3}{32} e^2 x^2 \log (\log (x))+\frac {3}{8} e^2 x^3 \log (\log (x))+\frac {3}{64} \int \log ^2(\log (x)) \, dx+\frac {3}{32} \int \frac {\log (\log (x))}{\log (x)} \, dx-\frac {3}{8} \int \frac {x \log (\log (x))}{\log (x)} \, dx-\frac {3}{8} \int x \log ^2(\log (x)) \, dx+\frac {1}{32} \left (3 e^2\right ) \int \frac {x}{\log (x)} \, dx+\frac {1}{32} \left (3 e^2\right ) \int \frac {x (-1+4 x)}{\log (x)} \, dx-\frac {1}{8} \left (3 e^2\right ) \int \frac {x^2}{\log (x)} \, dx-\frac {1}{64} \left (3 e^4\right ) \int x^2 (-3+16 x) \, dx\\ &=-\frac {3}{32} e^2 x^2 \log (\log (x))+\frac {3}{8} e^2 x^3 \log (\log (x))+\frac {3}{64} \int \log ^2(\log (x)) \, dx+\frac {3}{32} \int \frac {\log (\log (x))}{\log (x)} \, dx-\frac {3}{8} \int \frac {x \log (\log (x))}{\log (x)} \, dx-\frac {3}{8} \int x \log ^2(\log (x)) \, dx+\frac {1}{32} \left (3 e^2\right ) \int \left (-\frac {x}{\log (x)}+\frac {4 x^2}{\log (x)}\right ) \, dx+\frac {1}{32} \left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )-\frac {1}{8} \left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )-\frac {1}{64} \left (3 e^4\right ) \int \left (-3 x^2+16 x^3\right ) \, dx\\ &=\frac {3 e^4 x^3}{64}-\frac {3 e^4 x^4}{16}+\frac {3}{32} e^2 \text {Ei}(2 \log (x))-\frac {3}{8} e^2 \text {Ei}(3 \log (x))-\frac {3}{32} e^2 x^2 \log (\log (x))+\frac {3}{8} e^2 x^3 \log (\log (x))+\frac {3}{64} \int \log ^2(\log (x)) \, dx+\frac {3}{32} \int \frac {\log (\log (x))}{\log (x)} \, dx-\frac {3}{8} \int \frac {x \log (\log (x))}{\log (x)} \, dx-\frac {3}{8} \int x \log ^2(\log (x)) \, dx-\frac {1}{32} \left (3 e^2\right ) \int \frac {x}{\log (x)} \, dx+\frac {1}{8} \left (3 e^2\right ) \int \frac {x^2}{\log (x)} \, dx\\ &=\frac {3 e^4 x^3}{64}-\frac {3 e^4 x^4}{16}+\frac {3}{32} e^2 \text {Ei}(2 \log (x))-\frac {3}{8} e^2 \text {Ei}(3 \log (x))-\frac {3}{32} e^2 x^2 \log (\log (x))+\frac {3}{8} e^2 x^3 \log (\log (x))+\frac {3}{64} \int \log ^2(\log (x)) \, dx+\frac {3}{32} \int \frac {\log (\log (x))}{\log (x)} \, dx-\frac {3}{8} \int \frac {x \log (\log (x))}{\log (x)} \, dx-\frac {3}{8} \int x \log ^2(\log (x)) \, dx-\frac {1}{32} \left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )+\frac {1}{8} \left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {3 e^4 x^3}{64}-\frac {3 e^4 x^4}{16}-\frac {3}{32} e^2 x^2 \log (\log (x))+\frac {3}{8} e^2 x^3 \log (\log (x))+\frac {3}{64} \int \log ^2(\log (x)) \, dx+\frac {3}{32} \int \frac {\log (\log (x))}{\log (x)} \, dx-\frac {3}{8} \int \frac {x \log (\log (x))}{\log (x)} \, dx-\frac {3}{8} \int x \log ^2(\log (x)) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 22, normalized size = 0.88 \begin {gather*} -\frac {3}{64} x (-1+4 x) \left (-e^2 x+\log (\log (x))\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(-6*x + 24*x^2) + E^4*(9*x^2 - 48*x^3)*Log[x] + (6 - 24*x + E^2*(-12*x + 72*x^2)*Log[x])*Log[Lo
g[x]] + (3 - 24*x)*Log[x]*Log[Log[x]]^2)/(64*Log[x]),x]

[Out]

(-3*x*(-1 + 4*x)*(-(E^2*x) + Log[Log[x]])^2)/64

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fricas [B]  time = 0.68, size = 50, normalized size = 2.00 \begin {gather*} \frac {3}{32} \, {\left (4 \, x^{3} - x^{2}\right )} e^{2} \log \left (\log \relax (x)\right ) - \frac {3}{64} \, {\left (4 \, x^{2} - x\right )} \log \left (\log \relax (x)\right )^{2} - \frac {3}{64} \, {\left (4 \, x^{4} - x^{3}\right )} e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*((-24*x+3)*log(x)*log(log(x))^2+((72*x^2-12*x)*exp(2)*log(x)-24*x+6)*log(log(x))+(-48*x^3+9*x^2
)*exp(2)^2*log(x)+(24*x^2-6*x)*exp(2))/log(x),x, algorithm="fricas")

[Out]

3/32*(4*x^3 - x^2)*e^2*log(log(x)) - 3/64*(4*x^2 - x)*log(log(x))^2 - 3/64*(4*x^4 - x^3)*e^4

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giac [B]  time = 0.16, size = 53, normalized size = 2.12 \begin {gather*} -\frac {3}{16} \, x^{4} e^{4} + \frac {3}{8} \, x^{3} e^{2} \log \left (\log \relax (x)\right ) + \frac {3}{64} \, x^{3} e^{4} - \frac {3}{32} \, x^{2} e^{2} \log \left (\log \relax (x)\right ) - \frac {3}{16} \, x^{2} \log \left (\log \relax (x)\right )^{2} + \frac {3}{64} \, x \log \left (\log \relax (x)\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*((-24*x+3)*log(x)*log(log(x))^2+((72*x^2-12*x)*exp(2)*log(x)-24*x+6)*log(log(x))+(-48*x^3+9*x^2
)*exp(2)^2*log(x)+(24*x^2-6*x)*exp(2))/log(x),x, algorithm="giac")

[Out]

-3/16*x^4*e^4 + 3/8*x^3*e^2*log(log(x)) + 3/64*x^3*e^4 - 3/32*x^2*e^2*log(log(x)) - 3/16*x^2*log(log(x))^2 + 3
/64*x*log(log(x))^2

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maple [A]  time = 0.04, size = 45, normalized size = 1.80

method	result	size
risch	\(\frac {\left (-12 x^{2}+3 x \right ) \ln \left (\ln \relax (x )\right )^{2}}{64}+\frac {3 \,{\mathrm e}^{2} x^{2} \left (4 x -1\right ) \ln \left (\ln \relax (x )\right )}{32}-\frac {3 \,{\mathrm e}^{4} x^{3} \left (4 x -1\right )}{64}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/64*((-24*x+3)*ln(x)*ln(ln(x))^2+((72*x^2-12*x)*exp(2)*ln(x)-24*x+6)*ln(ln(x))+(-48*x^3+9*x^2)*exp(2)^2*l
n(x)+(24*x^2-6*x)*exp(2))/ln(x),x,method=_RETURNVERBOSE)

[Out]

1/64*(-12*x^2+3*x)*ln(ln(x))^2+3/32*exp(2)*x^2*(4*x-1)*ln(ln(x))-3/64*exp(4)*x^3*(4*x-1)

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maxima [C]  time = 0.44, size = 87, normalized size = 3.48 \begin {gather*} -\frac {3}{16} \, x^{4} e^{4} + \frac {3}{64} \, x^{3} e^{4} - \frac {3}{64} \, {\left (4 \, x^{2} - x\right )} \log \left (\log \relax (x)\right )^{2} + \frac {3}{8} \, {\left (x^{3} \log \left (\log \relax (x)\right ) - {\rm Ei}\left (3 \, \log \relax (x)\right )\right )} e^{2} - \frac {3}{32} \, {\left (x^{2} \log \left (\log \relax (x)\right ) - {\rm Ei}\left (2 \, \log \relax (x)\right )\right )} e^{2} + \frac {3}{8} \, {\rm Ei}\left (3 \, \log \relax (x)\right ) e^{2} - \frac {3}{32} \, {\rm Ei}\left (2 \, \log \relax (x)\right ) e^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*((-24*x+3)*log(x)*log(log(x))^2+((72*x^2-12*x)*exp(2)*log(x)-24*x+6)*log(log(x))+(-48*x^3+9*x^2
)*exp(2)^2*log(x)+(24*x^2-6*x)*exp(2))/log(x),x, algorithm="maxima")

[Out]

-3/16*x^4*e^4 + 3/64*x^3*e^4 - 3/64*(4*x^2 - x)*log(log(x))^2 + 3/8*(x^3*log(log(x)) - Ei(3*log(x)))*e^2 - 3/3
2*(x^2*log(log(x)) - Ei(2*log(x)))*e^2 + 3/8*Ei(3*log(x))*e^2 - 3/32*Ei(2*log(x))*e^2

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mupad [B]  time = 5.63, size = 19, normalized size = 0.76 \begin {gather*} -\frac {3\,x\,\left (4\,x-1\right )\,{\left (\ln \left (\ln \relax (x)\right )-x\,{\mathrm {e}}^2\right )}^2}{64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(2)*(6*x - 24*x^2))/64 + (log(log(x))*(24*x + exp(2)*log(x)*(12*x - 72*x^2) - 6))/64 + (log(log(x))^
2*log(x)*(24*x - 3))/64 - (exp(4)*log(x)*(9*x^2 - 48*x^3))/64)/log(x),x)

[Out]

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

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sympy [A]  time = 0.57, size = 63, normalized size = 2.52 \begin {gather*} - \frac {3 x^{4} e^{4}}{16} + \frac {3 x^{3} e^{4}}{64} + \left (- \frac {3 x^{2}}{16} + \frac {3 x}{64}\right ) \log {\left (\log {\relax (x )} \right )}^{2} + \left (\frac {3 x^{3} e^{2}}{8} - \frac {3 x^{2} e^{2}}{32}\right ) \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*((-24*x+3)*ln(x)*ln(ln(x))**2+((72*x**2-12*x)*exp(2)*ln(x)-24*x+6)*ln(ln(x))+(-48*x**3+9*x**2)*
exp(2)**2*ln(x)+(24*x**2-6*x)*exp(2))/ln(x),x)

[Out]

-3*x**4*exp(4)/16 + 3*x**3*exp(4)/64 + (-3*x**2/16 + 3*x/64)*log(log(x))**2 + (3*x**3*exp(2)/8 - 3*x**2*exp(2)
/32)*log(log(x))

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