3.20.67 \(\int \frac {-25 e^6-10 e^3 \log (-5 x)-\log ^2(-5 x)+(10 e^3+2 \log (-5 x)) \log (x)+2 e^6 x^2 \log ^2(x)}{e^6 x \log ^2(x)} \, dx\)

Optimal. Leaf size=21 \[ x^2+\frac {\left (5+\frac {\log (-5 x)}{e^3}\right )^2}{\log (x)} \]

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Rubi [F]  time = 0.33, 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 e^6-10 e^3 \log (-5 x)-\log ^2(-5 x)+\left (10 e^3+2 \log (-5 x)\right ) \log (x)+2 e^6 x^2 \log ^2(x)}{e^6 x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-25*E^6 - 10*E^3*Log[-5*x] - Log[-5*x]^2 + (10*E^3 + 2*Log[-5*x])*Log[x] + 2*E^6*x^2*Log[x]^2)/(E^6*x*Log
[x]^2),x]

[Out]

x^2 + (2*Log[x])/E^6 + (2*(5*E^3 + Log[-5*x])*Log[Log[x]])/E^6 - (2*Log[x]*Log[Log[x]])/E^6 - Defer[Int][(5*E^
3 + Log[-5*x])^2/(x*Log[x]^2), x]/E^6

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-25 e^6-10 e^3 \log (-5 x)-\log ^2(-5 x)+\left (10 e^3+2 \log (-5 x)\right ) \log (x)+2 e^6 x^2 \log ^2(x)}{x \log ^2(x)} \, dx}{e^6}\\ &=\frac {\int \left (2 e^6 x-\frac {\left (5 e^3+\log (-5 x)\right )^2}{x \log ^2(x)}+\frac {2 \left (5 e^3+\log (-5 x)\right )}{x \log (x)}\right ) \, dx}{e^6}\\ &=x^2-\frac {\int \frac {\left (5 e^3+\log (-5 x)\right )^2}{x \log ^2(x)} \, dx}{e^6}+\frac {2 \int \frac {5 e^3+\log (-5 x)}{x \log (x)} \, dx}{e^6}\\ &=x^2+\frac {2 \left (5 e^3+\log (-5 x)\right ) \log (\log (x))}{e^6}-\frac {\int \frac {\left (5 e^3+\log (-5 x)\right )^2}{x \log ^2(x)} \, dx}{e^6}-\frac {2 \int \frac {\log (\log (x))}{x} \, dx}{e^6}\\ &=x^2+\frac {2 \log (x)}{e^6}+\frac {2 \left (5 e^3+\log (-5 x)\right ) \log (\log (x))}{e^6}-\frac {2 \log (x) \log (\log (x))}{e^6}-\frac {\int \frac {\left (5 e^3+\log (-5 x)\right )^2}{x \log ^2(x)} \, dx}{e^6}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 35, normalized size = 1.67 \begin {gather*} \frac {e^6 x^2+\frac {\left (5 e^3+\log (-5 x)-\log (x)\right )^2}{\log (x)}+\log (x)}{e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25*E^6 - 10*E^3*Log[-5*x] - Log[-5*x]^2 + (10*E^3 + 2*Log[-5*x])*Log[x] + 2*E^6*x^2*Log[x]^2)/(E^6
*x*Log[x]^2),x]

[Out]

(E^6*x^2 + (5*E^3 + Log[-5*x] - Log[x])^2/Log[x] + Log[x])/E^6

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fricas [B]  time = 0.73, size = 143, normalized size = 6.81 \begin {gather*} \frac {\pi ^{2} x^{2} e^{6} - 10 \, \pi ^{2} e^{3} + {\left (x^{2} e^{6} - 10 \, e^{3}\right )} \log \relax (5)^{2} - \log \relax (5)^{3} + {\left (x^{2} e^{6} - 2 \, \log \relax (5)\right )} \log \left (-5 \, x\right )^{2} + \log \left (-5 \, x\right )^{3} - {\left (\pi ^{2} + 25 \, e^{6}\right )} \log \relax (5) - {\left (2 \, {\left (x^{2} e^{6} - 5 \, e^{3}\right )} \log \relax (5) - 2 \, \log \relax (5)^{2} - 25 \, e^{6}\right )} \log \left (-5 \, x\right )}{\pi ^{2} e^{6} + e^{6} \log \relax (5)^{2} - 2 \, e^{6} \log \relax (5) \log \left (-5 \, x\right ) + e^{6} \log \left (-5 \, x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2*exp(3)^2*log(x)^2+(2*log(-5*x)+10*exp(3))*log(x)-log(-5*x)^2-10*exp(3)*log(-5*x)-25*exp(3)^2)
/x/exp(3)^2/log(x)^2,x, algorithm="fricas")

[Out]

(pi^2*x^2*e^6 - 10*pi^2*e^3 + (x^2*e^6 - 10*e^3)*log(5)^2 - log(5)^3 + (x^2*e^6 - 2*log(5))*log(-5*x)^2 + log(
-5*x)^3 - (pi^2 + 25*e^6)*log(5) - (2*(x^2*e^6 - 5*e^3)*log(5) - 2*log(5)^2 - 25*e^6)*log(-5*x))/(pi^2*e^6 + e
^6*log(5)^2 - 2*e^6*log(5)*log(-5*x) + e^6*log(-5*x)^2)

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giac [B]  time = 0.24, size = 139, normalized size = 6.62 \begin {gather*} \frac {{\left (\pi ^{2} x^{2} e^{6} \mathrm {sgn}\relax (x) - \pi ^{2} x^{2} e^{6} - 2 \, x^{2} e^{6} \log \left ({\left | x \right |}\right )^{2} + 20 \, \pi ^{2} e^{3} \mathrm {sgn}\relax (x) + 4 \, \pi ^{2} \log \relax (5) \mathrm {sgn}\relax (x) + \pi ^{2} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\relax (x) - 20 \, \pi ^{2} e^{3} - 4 \, \pi ^{2} \log \relax (5) + 7 \, \pi ^{2} \log \left ({\left | x \right |}\right ) - 20 \, e^{3} \log \relax (5) \log \left ({\left | x \right |}\right ) - 2 \, \log \relax (5)^{2} \log \left ({\left | x \right |}\right ) - 2 \, \log \left ({\left | x \right |}\right )^{3} - 50 \, e^{6} \log \left ({\left | x \right |}\right )\right )} e^{\left (-6\right )}}{\pi ^{2} \mathrm {sgn}\relax (x) - \pi ^{2} - 2 \, \log \left ({\left | x \right |}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2*exp(3)^2*log(x)^2+(2*log(-5*x)+10*exp(3))*log(x)-log(-5*x)^2-10*exp(3)*log(-5*x)-25*exp(3)^2)
/x/exp(3)^2/log(x)^2,x, algorithm="giac")

[Out]

(pi^2*x^2*e^6*sgn(x) - pi^2*x^2*e^6 - 2*x^2*e^6*log(abs(x))^2 + 20*pi^2*e^3*sgn(x) + 4*pi^2*log(5)*sgn(x) + pi
^2*log(abs(x))*sgn(x) - 20*pi^2*e^3 - 4*pi^2*log(5) + 7*pi^2*log(abs(x)) - 20*e^3*log(5)*log(abs(x)) - 2*log(5
)^2*log(abs(x)) - 2*log(abs(x))^3 - 50*e^6*log(abs(x)))*e^(-6)/(pi^2*sgn(x) - pi^2 - 2*log(abs(x))^2)

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maple [C]  time = 0.10, size = 164, normalized size = 7.81

method	result	size
risch	\(x^{2}+{\mathrm e}^{-6} \ln \relax (x )+\frac {{\mathrm e}^{-6} \left (-4 \pi ^{2}+100 \,{\mathrm e}^{6}+4 \ln \relax (5)^{2}+40 \,{\mathrm e}^{3} \ln \relax (5)-8 \pi ^{2} \mathrm {csgn}\left (i x \right )^{3}+8 i \ln \relax (5) \pi -8 i \ln \relax (5) \pi \mathrm {csgn}\left (i x \right )^{2}+40 i {\mathrm e}^{3} \pi \mathrm {csgn}\left (i x \right )^{3}-40 i {\mathrm e}^{3} \pi \mathrm {csgn}\left (i x \right )^{2}+40 i {\mathrm e}^{3} \pi +8 i \ln \relax (5) \pi \mathrm {csgn}\left (i x \right )^{3}-4 \pi ^{2} \mathrm {csgn}\left (i x \right )^{6}+8 \pi ^{2} \mathrm {csgn}\left (i x \right )^{2}-4 \pi ^{2} \mathrm {csgn}\left (i x \right )^{4}+8 \pi ^{2} \mathrm {csgn}\left (i x \right )^{5}\right )}{4 \ln \relax (x )}\)	\(164\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2*exp(3)^2*ln(x)^2+(2*ln(-5*x)+10*exp(3))*ln(x)-ln(-5*x)^2-10*exp(3)*ln(-5*x)-25*exp(3)^2)/x/exp(3)^2
/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

x^2+exp(-6)*ln(x)+1/4*exp(-6)*(-4*Pi^2+100*exp(6)+4*ln(5)^2+40*exp(3)*ln(5)-8*Pi^2*csgn(I*x)^3+8*I*ln(5)*Pi-8*
I*ln(5)*Pi*csgn(I*x)^2+40*I*exp(3)*Pi*csgn(I*x)^3-40*I*exp(3)*Pi*csgn(I*x)^2+40*I*exp(3)*Pi+8*I*ln(5)*Pi*csgn(
I*x)^3-4*Pi^2*csgn(I*x)^6+8*Pi^2*csgn(I*x)^2-4*Pi^2*csgn(I*x)^4+8*Pi^2*csgn(I*x)^5)/ln(x)

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maxima [B]  time = 0.43, size = 41, normalized size = 1.95 \begin {gather*} {\left (x^{2} e^{6} + \frac {10 \, e^{3} \log \left (-5 \, x\right )}{\log \relax (x)} + \frac {\log \left (-5 \, x\right )^{2}}{\log \relax (x)} + \frac {25 \, e^{6}}{\log \relax (x)}\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2*exp(3)^2*log(x)^2+(2*log(-5*x)+10*exp(3))*log(x)-log(-5*x)^2-10*exp(3)*log(-5*x)-25*exp(3)^2)
/x/exp(3)^2/log(x)^2,x, algorithm="maxima")

[Out]

(x^2*e^6 + 10*e^3*log(-5*x)/log(x) + log(-5*x)^2/log(x) + 25*e^6/log(x))*e^(-6)

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mupad [B]  time = 1.23, size = 49, normalized size = 2.33 \begin {gather*} 2\,{\mathrm {e}}^{-6}\,\ln \relax (x)-{\mathrm {e}}^{-6}\,\left (2\,\ln \relax (x)-x^2\,{\mathrm {e}}^6\right )+\frac {{\mathrm {e}}^{-6}\,\left ({\ln \left (-5\,x\right )}^2+10\,{\mathrm {e}}^3\,\ln \left (-5\,x\right )+25\,{\mathrm {e}}^6\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-6)*(25*exp(6) + 10*log(-5*x)*exp(3) - log(x)*(2*log(-5*x) + 10*exp(3)) + log(-5*x)^2 - 2*x^2*exp(6)
*log(x)^2))/(x*log(x)^2),x)

[Out]

2*exp(-6)*log(x) - exp(-6)*(2*log(x) - x^2*exp(6)) + (exp(-6)*(25*exp(6) + 10*log(-5*x)*exp(3) + log(-5*x)^2))
/log(x)

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sympy [C]  time = 0.28, size = 58, normalized size = 2.76 \begin {gather*} \frac {x^{2} e^{6} + \log {\relax (x )}}{e^{6}} + \frac {- \pi ^{2} + \log {\relax (5 )}^{2} + 10 e^{3} \log {\relax (5 )} + 25 e^{6} + 2 i \pi \log {\relax (5 )} + 10 i \pi e^{3}}{e^{6} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2*exp(3)**2*ln(x)**2+(2*ln(-5*x)+10*exp(3))*ln(x)-ln(-5*x)**2-10*exp(3)*ln(-5*x)-25*exp(3)**2)
/x/exp(3)**2/ln(x)**2,x)

[Out]

(x**2*exp(6) + log(x))*exp(-6) + (-pi**2 + log(5)**2 + 10*exp(3)*log(5) + 25*exp(6) + 2*I*pi*log(5) + 10*I*pi*
exp(3))*exp(-6)/log(x)

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