3.15.51 \(\int \frac {-80 e^{\frac {2}{5} (10-7 x)}+e^{\frac {2}{5} (10-7 x)} (-40-56 x) \log (x^2)}{5 x^3 \log ^3(x^2) \log (\log (5))} \, dx\)

Optimal. Leaf size=25 \[ \frac {4 e^{4-\frac {14 x}{5}}}{x^2 \log ^2\left (x^2\right ) \log (\log (5))} \]

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Rubi [F]  time = 1.08, 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 {-80 e^{\frac {2}{5} (10-7 x)}+e^{\frac {2}{5} (10-7 x)} (-40-56 x) \log \left (x^2\right )}{5 x^3 \log ^3\left (x^2\right ) \log (\log (5))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-80*E^((2*(10 - 7*x))/5) + E^((2*(10 - 7*x))/5)*(-40 - 56*x)*Log[x^2])/(5*x^3*Log[x^2]^3*Log[Log[5]]),x]

[Out]

(-16*Defer[Int][E^(4 - (14*x)/5)/(x^3*Log[x^2]^3), x])/Log[Log[5]] - (8*Defer[Int][E^(4 - (14*x)/5)/(x^3*Log[x
^2]^2), x])/Log[Log[5]] - (56*Defer[Int][E^(4 - (14*x)/5)/(x^2*Log[x^2]^2), x])/(5*Log[Log[5]])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-80 e^{\frac {2}{5} (10-7 x)}+e^{\frac {2}{5} (10-7 x)} (-40-56 x) \log \left (x^2\right )}{x^3 \log ^3\left (x^2\right )} \, dx}{5 \log (\log (5))}\\ &=\frac {\int \frac {8 e^{4-\frac {14 x}{5}} \left (-10-(5+7 x) \log \left (x^2\right )\right )}{x^3 \log ^3\left (x^2\right )} \, dx}{5 \log (\log (5))}\\ &=\frac {8 \int \frac {e^{4-\frac {14 x}{5}} \left (-10-(5+7 x) \log \left (x^2\right )\right )}{x^3 \log ^3\left (x^2\right )} \, dx}{5 \log (\log (5))}\\ &=\frac {8 \int \left (-\frac {10 e^{4-\frac {14 x}{5}}}{x^3 \log ^3\left (x^2\right )}+\frac {e^{4-\frac {14 x}{5}} (-5-7 x)}{x^3 \log ^2\left (x^2\right )}\right ) \, dx}{5 \log (\log (5))}\\ &=\frac {8 \int \frac {e^{4-\frac {14 x}{5}} (-5-7 x)}{x^3 \log ^2\left (x^2\right )} \, dx}{5 \log (\log (5))}-\frac {16 \int \frac {e^{4-\frac {14 x}{5}}}{x^3 \log ^3\left (x^2\right )} \, dx}{\log (\log (5))}\\ &=\frac {8 \int \left (-\frac {5 e^{4-\frac {14 x}{5}}}{x^3 \log ^2\left (x^2\right )}-\frac {7 e^{4-\frac {14 x}{5}}}{x^2 \log ^2\left (x^2\right )}\right ) \, dx}{5 \log (\log (5))}-\frac {16 \int \frac {e^{4-\frac {14 x}{5}}}{x^3 \log ^3\left (x^2\right )} \, dx}{\log (\log (5))}\\ &=-\frac {8 \int \frac {e^{4-\frac {14 x}{5}}}{x^3 \log ^2\left (x^2\right )} \, dx}{\log (\log (5))}-\frac {56 \int \frac {e^{4-\frac {14 x}{5}}}{x^2 \log ^2\left (x^2\right )} \, dx}{5 \log (\log (5))}-\frac {16 \int \frac {e^{4-\frac {14 x}{5}}}{x^3 \log ^3\left (x^2\right )} \, dx}{\log (\log (5))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 25, normalized size = 1.00 \begin {gather*} \frac {4 e^{4-\frac {14 x}{5}}}{x^2 \log ^2\left (x^2\right ) \log (\log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-80*E^((2*(10 - 7*x))/5) + E^((2*(10 - 7*x))/5)*(-40 - 56*x)*Log[x^2])/(5*x^3*Log[x^2]^3*Log[Log[5]
]),x]

[Out]

(4*E^(4 - (14*x)/5))/(x^2*Log[x^2]^2*Log[Log[5]])

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fricas [A]  time = 0.61, size = 22, normalized size = 0.88 \begin {gather*} \frac {4 \, e^{\left (-\frac {14}{5} \, x + 4\right )}}{x^{2} \log \left (x^{2}\right )^{2} \log \left (\log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-56*x-40)*exp(-7/5*x+2)^2*log(x^2)-80*exp(-7/5*x+2)^2)/x^3/log(x^2)^3/log(log(5)),x, algorithm
="fricas")

[Out]

4*e^(-14/5*x + 4)/(x^2*log(x^2)^2*log(log(5)))

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giac [A]  time = 0.42, size = 22, normalized size = 0.88 \begin {gather*} \frac {4 \, e^{\left (-\frac {14}{5} \, x + 4\right )}}{x^{2} \log \left (x^{2}\right )^{2} \log \left (\log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-56*x-40)*exp(-7/5*x+2)^2*log(x^2)-80*exp(-7/5*x+2)^2)/x^3/log(x^2)^3/log(log(5)),x, algorithm
="giac")

[Out]

4*e^(-14/5*x + 4)/(x^2*log(x^2)^2*log(log(5)))

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maple [C]  time = 0.08, size = 69, normalized size = 2.76

method	result	size
risch	\(-\frac {16 \,{\mathrm e}^{-\frac {14 x}{5}+4}}{\ln \left (\ln \relax (5)\right ) x^{2} \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )\right )^{2}}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-56*x-40)*exp(-7/5*x+2)^2*ln(x^2)-80*exp(-7/5*x+2)^2)/x^3/ln(x^2)^3/ln(ln(5)),x,method=_RETURNVERBOS
E)

[Out]

-16/ln(ln(5))*exp(-14/5*x+4)/x^2/(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+4*I
*ln(x))^2

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maxima [A]  time = 0.59, size = 19, normalized size = 0.76 \begin {gather*} \frac {e^{\left (-\frac {14}{5} \, x + 4\right )}}{x^{2} \log \relax (x)^{2} \log \left (\log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-56*x-40)*exp(-7/5*x+2)^2*log(x^2)-80*exp(-7/5*x+2)^2)/x^3/log(x^2)^3/log(log(5)),x, algorithm
="maxima")

[Out]

e^(-14/5*x + 4)/(x^2*log(x)^2*log(log(5)))

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mupad [B]  time = 1.08, size = 22, normalized size = 0.88 \begin {gather*} \frac {4\,{\mathrm {e}}^{-\frac {14\,x}{5}}\,{\mathrm {e}}^4}{x^2\,{\ln \left (x^2\right )}^2\,\ln \left (\ln \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*exp(4 - (14*x)/5) + (log(x^2)*exp(4 - (14*x)/5)*(56*x + 40))/5)/(x^3*log(x^2)^3*log(log(5))),x)

[Out]

(4*exp(-(14*x)/5)*exp(4))/(x^2*log(x^2)^2*log(log(5)))

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sympy [A]  time = 0.27, size = 24, normalized size = 0.96 \begin {gather*} \frac {4 e^{4 - \frac {14 x}{5}}}{x^{2} \log {\left (x^{2} \right )}^{2} \log {\left (\log {\relax (5 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-56*x-40)*exp(-7/5*x+2)**2*ln(x**2)-80*exp(-7/5*x+2)**2)/x**3/ln(x**2)**3/ln(ln(5)),x)

[Out]

4*exp(4 - 14*x/5)/(x**2*log(x**2)**2*log(log(5)))

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