3.102.76 \(\int \frac {-25 x+e^{4 x^2} (-50 x^3-200 x^5)-25 x \log (e^{e^{4 x^2} x^2} x)+(2+45 x) \log ^2(e^{e^{4 x^2} x^2} x)}{5 x^3 \log ^2(e^{e^{4 x^2} x^2} x)} \, dx\)

Optimal. Leaf size=33 \[ \frac {-9-\frac {1}{5 x}+\frac {5}{\log \left (e^{e^{4 x^2} x^2} x\right )}}{x} \]

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Rubi [F]  time = 1.31, 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 {-25 x+e^{4 x^2} \left (-50 x^3-200 x^5\right )-25 x \log \left (e^{e^{4 x^2} x^2} x\right )+(2+45 x) \log ^2\left (e^{e^{4 x^2} x^2} x\right )}{5 x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-25*x + E^(4*x^2)*(-50*x^3 - 200*x^5) - 25*x*Log[E^(E^(4*x^2)*x^2)*x] + (2 + 45*x)*Log[E^(E^(4*x^2)*x^2)*
x]^2)/(5*x^3*Log[E^(E^(4*x^2)*x^2)*x]^2),x]

[Out]

-1/20*(2 + 45*x)^2/x^2 - 10*Defer[Int][E^(4*x^2)/Log[E^(E^(4*x^2)*x^2)*x]^2, x] - 5*Defer[Int][1/(x^2*Log[E^(E
^(4*x^2)*x^2)*x]^2), x] - 40*Defer[Int][(E^(4*x^2)*x^2)/Log[E^(E^(4*x^2)*x^2)*x]^2, x] - 5*Defer[Int][1/(x^2*L
og[E^(E^(4*x^2)*x^2)*x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-25 x+e^{4 x^2} \left (-50 x^3-200 x^5\right )-25 x \log \left (e^{e^{4 x^2} x^2} x\right )+(2+45 x) \log ^2\left (e^{e^{4 x^2} x^2} x\right )}{x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx\\ &=\frac {1}{5} \int \left (-\frac {50 e^{4 x^2} \left (1+4 x^2\right )}{\log ^2\left (e^{e^{4 x^2} x^2} x\right )}+\frac {-25 x-25 x \log \left (e^{e^{4 x^2} x^2} x\right )+2 \log ^2\left (e^{e^{4 x^2} x^2} x\right )+45 x \log ^2\left (e^{e^{4 x^2} x^2} x\right )}{x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )}\right ) \, dx\\ &=\frac {1}{5} \int \frac {-25 x-25 x \log \left (e^{e^{4 x^2} x^2} x\right )+2 \log ^2\left (e^{e^{4 x^2} x^2} x\right )+45 x \log ^2\left (e^{e^{4 x^2} x^2} x\right )}{x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx-10 \int \frac {e^{4 x^2} \left (1+4 x^2\right )}{\log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx\\ &=\frac {1}{5} \int \frac {2+45 x-\frac {25 x}{\log ^2\left (e^{e^{4 x^2} x^2} x\right )}-\frac {25 x}{\log \left (e^{e^{4 x^2} x^2} x\right )}}{x^3} \, dx-10 \int \left (\frac {e^{4 x^2}}{\log ^2\left (e^{e^{4 x^2} x^2} x\right )}+\frac {4 e^{4 x^2} x^2}{\log ^2\left (e^{e^{4 x^2} x^2} x\right )}\right ) \, dx\\ &=\frac {1}{5} \int \left (\frac {2+45 x}{x^3}-\frac {25}{x^2 \log ^2\left (e^{e^{4 x^2} x^2} x\right )}-\frac {25}{x^2 \log \left (e^{e^{4 x^2} x^2} x\right )}\right ) \, dx-10 \int \frac {e^{4 x^2}}{\log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx-40 \int \frac {e^{4 x^2} x^2}{\log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx\\ &=\frac {1}{5} \int \frac {2+45 x}{x^3} \, dx-5 \int \frac {1}{x^2 \log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx-5 \int \frac {1}{x^2 \log \left (e^{e^{4 x^2} x^2} x\right )} \, dx-10 \int \frac {e^{4 x^2}}{\log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx-40 \int \frac {e^{4 x^2} x^2}{\log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx\\ &=-\frac {(2+45 x)^2}{20 x^2}-5 \int \frac {1}{x^2 \log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx-5 \int \frac {1}{x^2 \log \left (e^{e^{4 x^2} x^2} x\right )} \, dx-10 \int \frac {e^{4 x^2}}{\log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx-40 \int \frac {e^{4 x^2} x^2}{\log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 38, normalized size = 1.15 \begin {gather*} \frac {1}{5} \left (-\frac {1}{x^2}-\frac {45}{x}+\frac {25}{x \log \left (e^{e^{4 x^2} x^2} x\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25*x + E^(4*x^2)*(-50*x^3 - 200*x^5) - 25*x*Log[E^(E^(4*x^2)*x^2)*x] + (2 + 45*x)*Log[E^(E^(4*x^2)
*x^2)*x]^2)/(5*x^3*Log[E^(E^(4*x^2)*x^2)*x]^2),x]

[Out]

(-x^(-2) - 45/x + 25/(x*Log[E^(E^(4*x^2)*x^2)*x]))/5

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fricas [A]  time = 1.00, size = 45, normalized size = 1.36 \begin {gather*} -\frac {{\left (45 \, x + 1\right )} \log \left (x e^{\left (x^{2} e^{\left (4 \, x^{2}\right )}\right )}\right ) - 25 \, x}{5 \, x^{2} \log \left (x e^{\left (x^{2} e^{\left (4 \, x^{2}\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((45*x+2)*log(x*exp(x^2*exp(x^2)^4))^2-25*x*log(x*exp(x^2*exp(x^2)^4))+(-200*x^5-50*x^3)*exp(x^2
)^4-25*x)/x^3/log(x*exp(x^2*exp(x^2)^4))^2,x, algorithm="fricas")

[Out]

-1/5*((45*x + 1)*log(x*e^(x^2*e^(4*x^2))) - 25*x)/(x^2*log(x*e^(x^2*e^(4*x^2))))

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giac [A]  time = 0.17, size = 53, normalized size = 1.61 \begin {gather*} -\frac {45 \, x^{3} e^{\left (4 \, x^{2}\right )} + x^{2} e^{\left (4 \, x^{2}\right )} + 45 \, x \log \relax (x) - 25 \, x + \log \relax (x)}{5 \, {\left (x^{4} e^{\left (4 \, x^{2}\right )} + x^{2} \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((45*x+2)*log(x*exp(x^2*exp(x^2)^4))^2-25*x*log(x*exp(x^2*exp(x^2)^4))+(-200*x^5-50*x^3)*exp(x^2
)^4-25*x)/x^3/log(x*exp(x^2*exp(x^2)^4))^2,x, algorithm="giac")

[Out]

-1/5*(45*x^3*e^(4*x^2) + x^2*e^(4*x^2) + 45*x*log(x) - 25*x + log(x))/(x^4*e^(4*x^2) + x^2*log(x))

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maple [C]  time = 0.24, size = 161, normalized size = 4.88

method	result	size
risch	\(-\frac {45 x +1}{5 x^{2}}+\frac {10 i}{x \left (\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{x^{2} {\mathrm e}^{4 x^{2}}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x^{2} {\mathrm e}^{4 x^{2}}}\right )-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x^{2} {\mathrm e}^{4 x^{2}}}\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{x^{2} {\mathrm e}^{4 x^{2}}}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{x^{2} {\mathrm e}^{4 x^{2}}}\right )^{2}+\pi \mathrm {csgn}\left (i x \,{\mathrm e}^{x^{2} {\mathrm e}^{4 x^{2}}}\right )^{3}+2 i \ln \relax (x )+2 i \ln \left ({\mathrm e}^{x^{2} {\mathrm e}^{4 x^{2}}}\right )\right )}\)	\(161\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((45*x+2)*ln(x*exp(x^2*exp(x^2)^4))^2-25*x*ln(x*exp(x^2*exp(x^2)^4))+(-200*x^5-50*x^3)*exp(x^2)^4-25*x
)/x^3/ln(x*exp(x^2*exp(x^2)^4))^2,x,method=_RETURNVERBOSE)

[Out]

-1/5*(45*x+1)/x^2+10*I/x/(Pi*csgn(I*x)*csgn(I*exp(x^2*exp(4*x^2)))*csgn(I*x*exp(x^2*exp(4*x^2)))-Pi*csgn(I*x)*
csgn(I*x*exp(x^2*exp(4*x^2)))^2-Pi*csgn(I*exp(x^2*exp(4*x^2)))*csgn(I*x*exp(x^2*exp(4*x^2)))^2+Pi*csgn(I*x*exp
(x^2*exp(4*x^2)))^3+2*I*ln(x)+2*I*ln(exp(x^2*exp(4*x^2))))

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maxima [A]  time = 0.40, size = 30, normalized size = 0.91 \begin {gather*} \frac {5}{x^{3} e^{\left (4 \, x^{2}\right )} + x \log \relax (x)} - \frac {9}{x} - \frac {1}{5 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((45*x+2)*log(x*exp(x^2*exp(x^2)^4))^2-25*x*log(x*exp(x^2*exp(x^2)^4))+(-200*x^5-50*x^3)*exp(x^2
)^4-25*x)/x^3/log(x*exp(x^2*exp(x^2)^4))^2,x, algorithm="maxima")

[Out]

5/(x^3*e^(4*x^2) + x*log(x)) - 9/x - 1/5/x^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {5\,x+5\,x\,\ln \left (x\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{4\,x^2}}\right )-\frac {{\ln \left (x\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{4\,x^2}}\right )}^2\,\left (45\,x+2\right )}{5}+\frac {{\mathrm {e}}^{4\,x^2}\,\left (200\,x^5+50\,x^3\right )}{5}}{x^3\,{\ln \left (x\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{4\,x^2}}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x + 5*x*log(x*exp(x^2*exp(4*x^2))) - (log(x*exp(x^2*exp(4*x^2)))^2*(45*x + 2))/5 + (exp(4*x^2)*(50*x^3
+ 200*x^5))/5)/(x^3*log(x*exp(x^2*exp(4*x^2)))^2),x)

[Out]

-int((5*x + 5*x*log(x*exp(x^2*exp(4*x^2))) - (log(x*exp(x^2*exp(4*x^2)))^2*(45*x + 2))/5 + (exp(4*x^2)*(50*x^3
+ 200*x^5))/5)/(x^3*log(x*exp(x^2*exp(4*x^2)))^2), x)

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sympy [A]  time = 0.41, size = 29, normalized size = 0.88 \begin {gather*} \frac {5}{x \log {\left (x e^{x^{2} e^{4 x^{2}}} \right )}} + \frac {- 45 x - 1}{5 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((45*x+2)*ln(x*exp(x**2*exp(x**2)**4))**2-25*x*ln(x*exp(x**2*exp(x**2)**4))+(-200*x**5-50*x**3)*
exp(x**2)**4-25*x)/x**3/ln(x*exp(x**2*exp(x**2)**4))**2,x)

[Out]

5/(x*log(x*exp(x**2*exp(4*x**2)))) + (-45*x - 1)/(5*x**2)

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