3.8.46 \(\int \frac {8 x+40 e^{2 x^2} x+(8 x+e^{2 x^2} (40 x-80 x^3)) \log (x) \log (\log ^2(x))}{(3+30 e^{2 x^2}+75 e^{4 x^2}) \log (x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {4 x^2 \log \left (\log ^2(x)\right )}{3 \left (1+5 e^{2 x^2}\right )} \]

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Rubi [F]  time = 2.02, 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 {8 x+40 e^{2 x^2} x+\left (8 x+e^{2 x^2} \left (40 x-80 x^3\right )\right ) \log (x) \log \left (\log ^2(x)\right )}{\left (3+30 e^{2 x^2}+75 e^{4 x^2}\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(8*x + 40*E^(2*x^2)*x + (8*x + E^(2*x^2)*(40*x - 80*x^3))*Log[x]*Log[Log[x]^2])/((3 + 30*E^(2*x^2) + 75*E^
(4*x^2))*Log[x]),x]

[Out]

(8*Defer[Int][x/((1 + 5*E^(2*x^2))*Log[x]), x])/3 + (8*Defer[Int][(x*Log[Log[x]^2])/(1 + 5*E^(2*x^2)), x])/3 +
(16*Defer[Int][(x^3*Log[Log[x]^2])/(1 + 5*E^(2*x^2))^2, x])/3 - (16*Defer[Int][(x^3*Log[Log[x]^2])/(1 + 5*E^(
2*x^2)), x])/3

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 25, normalized size = 1.00 \begin {gather*} \frac {4 x^2 \log \left (\log ^2(x)\right )}{3 \left (1+5 e^{2 x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*x + 40*E^(2*x^2)*x + (8*x + E^(2*x^2)*(40*x - 80*x^3))*Log[x]*Log[Log[x]^2])/((3 + 30*E^(2*x^2) +
75*E^(4*x^2))*Log[x]),x]

[Out]

(4*x^2*Log[Log[x]^2])/(3*(1 + 5*E^(2*x^2)))

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fricas [A]  time = 0.74, size = 22, normalized size = 0.88 \begin {gather*} \frac {4 \, x^{2} \log \left (\log \relax (x)^{2}\right )}{3 \, {\left (5 \, e^{\left (2 \, x^{2}\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-80*x^3+40*x)*exp(2*x^2)+8*x)*log(x)*log(log(x)^2)+40*x*exp(2*x^2)+8*x)/(75*exp(2*x^2)^2+30*exp(2
*x^2)+3)/log(x),x, algorithm="fricas")

[Out]

4/3*x^2*log(log(x)^2)/(5*e^(2*x^2) + 1)

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giac [A]  time = 0.58, size = 22, normalized size = 0.88 \begin {gather*} \frac {4 \, x^{2} \log \left (\log \relax (x)^{2}\right )}{3 \, {\left (5 \, e^{\left (2 \, x^{2}\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-80*x^3+40*x)*exp(2*x^2)+8*x)*log(x)*log(log(x)^2)+40*x*exp(2*x^2)+8*x)/(75*exp(2*x^2)^2+30*exp(2
*x^2)+3)/log(x),x, algorithm="giac")

[Out]

4/3*x^2*log(log(x)^2)/(5*e^(2*x^2) + 1)

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maple [C]  time = 0.20, size = 84, normalized size = 3.36

method	result	size
risch	\(\frac {8 x^{2} \ln \left (\ln \relax (x )\right )}{3 \left (1+5 \,{\mathrm e}^{2 x^{2}}\right )}-\frac {2 i x^{2} \pi \,\mathrm {csgn}\left (i \ln \relax (x )^{2}\right ) \left (\mathrm {csgn}\left (i \ln \relax (x )\right )^{2}-2 \,\mathrm {csgn}\left (i \ln \relax (x )^{2}\right ) \mathrm {csgn}\left (i \ln \relax (x )\right )+\mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2}\right )}{3 \left (1+5 \,{\mathrm e}^{2 x^{2}}\right )}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-80*x^3+40*x)*exp(2*x^2)+8*x)*ln(x)*ln(ln(x)^2)+40*x*exp(2*x^2)+8*x)/(75*exp(2*x^2)^2+30*exp(2*x^2)+3)/
ln(x),x,method=_RETURNVERBOSE)

[Out]

8/3*x^2/(1+5*exp(2*x^2))*ln(ln(x))-2/3*I*x^2*Pi*csgn(I*ln(x)^2)*(csgn(I*ln(x))^2-2*csgn(I*ln(x)^2)*csgn(I*ln(x
))+csgn(I*ln(x)^2)^2)/(1+5*exp(2*x^2))

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maxima [A]  time = 0.51, size = 20, normalized size = 0.80 \begin {gather*} \frac {8 \, x^{2} \log \left (\log \relax (x)\right )}{3 \, {\left (5 \, e^{\left (2 \, x^{2}\right )} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-80*x^3+40*x)*exp(2*x^2)+8*x)*log(x)*log(log(x)^2)+40*x*exp(2*x^2)+8*x)/(75*exp(2*x^2)^2+30*exp(2
*x^2)+3)/log(x),x, algorithm="maxima")

[Out]

8/3*x^2*log(log(x))/(5*e^(2*x^2) + 1)

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mupad [B]  time = 0.77, size = 22, normalized size = 0.88 \begin {gather*} \frac {4\,x^2\,\ln \left ({\ln \relax (x)}^2\right )}{15\,\left ({\mathrm {e}}^{2\,x^2}+\frac {1}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + 40*x*exp(2*x^2) + log(log(x)^2)*log(x)*(8*x + exp(2*x^2)*(40*x - 80*x^3)))/(log(x)*(30*exp(2*x^2) +
75*exp(4*x^2) + 3)),x)

[Out]

(4*x^2*log(log(x)^2))/(15*(exp(2*x^2) + 1/5))

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sympy [A]  time = 0.29, size = 20, normalized size = 0.80 \begin {gather*} \frac {4 x^{2} \log {\left (\log {\relax (x )}^{2} \right )}}{15 e^{2 x^{2}} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-80*x**3+40*x)*exp(2*x**2)+8*x)*ln(x)*ln(ln(x)**2)+40*x*exp(2*x**2)+8*x)/(75*exp(2*x**2)**2+30*ex
p(2*x**2)+3)/ln(x),x)

[Out]

4*x**2*log(log(x)**2)/(15*exp(2*x**2) + 3)

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