3.71.51 \(\int \frac {120 x-40 x^2+(-960+440 x) \log (x)-640 \log ^2(x)}{x^3 \log ^2(5) \log ^2(x)-8 x^2 \log ^2(5) \log ^3(x)+16 x \log ^2(5) \log ^4(x)} \, dx\)

Optimal. Leaf size=24 \[ \frac {10 \left (4+\frac {3}{-\frac {x}{4}+\log (x)}\right )}{\log ^2(5) \log (x)} \]

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Rubi [F]  time = 0.76, 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 {120 x-40 x^2+(-960+440 x) \log (x)-640 \log ^2(x)}{x^3 \log ^2(5) \log ^2(x)-8 x^2 \log ^2(5) \log ^3(x)+16 x \log ^2(5) \log ^4(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(120*x - 40*x^2 + (-960 + 440*x)*Log[x] - 640*Log[x]^2)/(x^3*Log[5]^2*Log[x]^2 - 8*x^2*Log[5]^2*Log[x]^3 +
16*x*Log[5]^2*Log[x]^4),x]

[Out]

40/(Log[5]^2*Log[x]) - 120/(x*Log[5]^2*Log[x]) - (1920*Defer[Int][1/(x^2*(x - 4*Log[x])^2), x])/Log[5]^2 + (48
0*Defer[Int][1/(x*(x - 4*Log[x])^2), x])/Log[5]^2 + (480*Defer[Int][1/(x^2*(x - 4*Log[x])), x])/Log[5]^2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {40 \left (-((-3+x) x)-(24-11 x) \log (x)-16 \log ^2(x)\right )}{x \log ^2(5) (x-4 \log (x))^2 \log ^2(x)} \, dx\\ &=\frac {40 \int \frac {-((-3+x) x)-(24-11 x) \log (x)-16 \log ^2(x)}{x (x-4 \log (x))^2 \log ^2(x)} \, dx}{\log ^2(5)}\\ &=\frac {40 \int \left (\frac {12 (-4+x)}{x^2 (x-4 \log (x))^2}+\frac {12}{x^2 (x-4 \log (x))}+\frac {3-x}{x^2 \log ^2(x)}+\frac {3}{x^2 \log (x)}\right ) \, dx}{\log ^2(5)}\\ &=\frac {40 \int \frac {3-x}{x^2 \log ^2(x)} \, dx}{\log ^2(5)}+\frac {120 \int \frac {1}{x^2 \log (x)} \, dx}{\log ^2(5)}+\frac {480 \int \frac {-4+x}{x^2 (x-4 \log (x))^2} \, dx}{\log ^2(5)}+\frac {480 \int \frac {1}{x^2 (x-4 \log (x))} \, dx}{\log ^2(5)}\\ &=\frac {40 \int \left (\frac {3}{x^2 \log ^2(x)}-\frac {1}{x \log ^2(x)}\right ) \, dx}{\log ^2(5)}+\frac {120 \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )}{\log ^2(5)}+\frac {480 \int \left (-\frac {4}{x^2 (x-4 \log (x))^2}+\frac {1}{x (x-4 \log (x))^2}\right ) \, dx}{\log ^2(5)}+\frac {480 \int \frac {1}{x^2 (x-4 \log (x))} \, dx}{\log ^2(5)}\\ &=\frac {120 \text {Ei}(-\log (x))}{\log ^2(5)}-\frac {40 \int \frac {1}{x \log ^2(x)} \, dx}{\log ^2(5)}+\frac {120 \int \frac {1}{x^2 \log ^2(x)} \, dx}{\log ^2(5)}+\frac {480 \int \frac {1}{x (x-4 \log (x))^2} \, dx}{\log ^2(5)}+\frac {480 \int \frac {1}{x^2 (x-4 \log (x))} \, dx}{\log ^2(5)}-\frac {1920 \int \frac {1}{x^2 (x-4 \log (x))^2} \, dx}{\log ^2(5)}\\ &=\frac {120 \text {Ei}(-\log (x))}{\log ^2(5)}-\frac {120}{x \log ^2(5) \log (x)}-\frac {40 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )}{\log ^2(5)}-\frac {120 \int \frac {1}{x^2 \log (x)} \, dx}{\log ^2(5)}+\frac {480 \int \frac {1}{x (x-4 \log (x))^2} \, dx}{\log ^2(5)}+\frac {480 \int \frac {1}{x^2 (x-4 \log (x))} \, dx}{\log ^2(5)}-\frac {1920 \int \frac {1}{x^2 (x-4 \log (x))^2} \, dx}{\log ^2(5)}\\ &=\frac {120 \text {Ei}(-\log (x))}{\log ^2(5)}+\frac {40}{\log ^2(5) \log (x)}-\frac {120}{x \log ^2(5) \log (x)}-\frac {120 \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )}{\log ^2(5)}+\frac {480 \int \frac {1}{x (x-4 \log (x))^2} \, dx}{\log ^2(5)}+\frac {480 \int \frac {1}{x^2 (x-4 \log (x))} \, dx}{\log ^2(5)}-\frac {1920 \int \frac {1}{x^2 (x-4 \log (x))^2} \, dx}{\log ^2(5)}\\ &=\frac {40}{\log ^2(5) \log (x)}-\frac {120}{x \log ^2(5) \log (x)}+\frac {480 \int \frac {1}{x (x-4 \log (x))^2} \, dx}{\log ^2(5)}+\frac {480 \int \frac {1}{x^2 (x-4 \log (x))} \, dx}{\log ^2(5)}-\frac {1920 \int \frac {1}{x^2 (x-4 \log (x))^2} \, dx}{\log ^2(5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 28, normalized size = 1.17 \begin {gather*} -\frac {40 (3-x+4 \log (x))}{\log ^2(5) \left (x \log (x)-4 \log ^2(x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(120*x - 40*x^2 + (-960 + 440*x)*Log[x] - 640*Log[x]^2)/(x^3*Log[5]^2*Log[x]^2 - 8*x^2*Log[5]^2*Log[
x]^3 + 16*x*Log[5]^2*Log[x]^4),x]

[Out]

(-40*(3 - x + 4*Log[x]))/(Log[5]^2*(x*Log[x] - 4*Log[x]^2))

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fricas [A]  time = 0.88, size = 30, normalized size = 1.25 \begin {gather*} \frac {40 \, {\left (x - 4 \, \log \relax (x) - 3\right )}}{x \log \relax (5)^{2} \log \relax (x) - 4 \, \log \relax (5)^{2} \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-640*log(x)^2+(440*x-960)*log(x)-40*x^2+120*x)/(16*x*log(5)^2*log(x)^4-8*x^2*log(5)^2*log(x)^3+x^3*
log(5)^2*log(x)^2),x, algorithm="fricas")

[Out]

40*(x - 4*log(x) - 3)/(x*log(5)^2*log(x) - 4*log(5)^2*log(x)^2)

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giac [A]  time = 0.16, size = 39, normalized size = 1.62 \begin {gather*} -\frac {480}{x^{2} \log \relax (5)^{2} - 4 \, x \log \relax (5)^{2} \log \relax (x)} + \frac {40 \, {\left (x - 3\right )}}{x \log \relax (5)^{2} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-640*log(x)^2+(440*x-960)*log(x)-40*x^2+120*x)/(16*x*log(5)^2*log(x)^4-8*x^2*log(5)^2*log(x)^3+x^3*
log(5)^2*log(x)^2),x, algorithm="giac")

[Out]

-480/(x^2*log(5)^2 - 4*x*log(5)^2*log(x)) + 40*(x - 3)/(x*log(5)^2*log(x))

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maple [A]  time = 0.09, size = 26, normalized size = 1.08

method	result	size
risch	\(\frac {40 x -160 \ln \relax (x )-120}{\ln \relax (5)^{2} \ln \relax (x ) \left (x -4 \ln \relax (x )\right )}\)	\(26\)
norman	\(\frac {-\frac {120}{\ln \relax (5)}+\frac {40 x}{\ln \relax (5)}-\frac {160 \ln \relax (x )}{\ln \relax (5)}}{\ln \relax (5) \ln \relax (x ) \left (x -4 \ln \relax (x )\right )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-640*ln(x)^2+(440*x-960)*ln(x)-40*x^2+120*x)/(16*x*ln(5)^2*ln(x)^4-8*x^2*ln(5)^2*ln(x)^3+x^3*ln(5)^2*ln(x
)^2),x,method=_RETURNVERBOSE)

[Out]

40*(x-4*ln(x)-3)/ln(5)^2/ln(x)/(x-4*ln(x))

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maxima [A]  time = 0.51, size = 30, normalized size = 1.25 \begin {gather*} \frac {40 \, {\left (x - 4 \, \log \relax (x) - 3\right )}}{x \log \relax (5)^{2} \log \relax (x) - 4 \, \log \relax (5)^{2} \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-640*log(x)^2+(440*x-960)*log(x)-40*x^2+120*x)/(16*x*log(5)^2*log(x)^4-8*x^2*log(5)^2*log(x)^3+x^3*
log(5)^2*log(x)^2),x, algorithm="maxima")

[Out]

40*(x - 4*log(x) - 3)/(x*log(5)^2*log(x) - 4*log(5)^2*log(x)^2)

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mupad [B]  time = 4.35, size = 27, normalized size = 1.12 \begin {gather*} -\frac {40\,\left (4\,\ln \relax (x)-x+3\right )}{{\ln \relax (5)}^2\,\ln \relax (x)\,\left (x-4\,\ln \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((120*x - 640*log(x)^2 + log(x)*(440*x - 960) - 40*x^2)/(x^3*log(5)^2*log(x)^2 - 8*x^2*log(5)^2*log(x)^3 +
16*x*log(5)^2*log(x)^4),x)

[Out]

-(40*(4*log(x) - x + 3))/(log(5)^2*log(x)*(x - 4*log(x)))

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sympy [A]  time = 0.16, size = 31, normalized size = 1.29 \begin {gather*} \frac {- 40 x + 160 \log {\relax (x )} + 120}{- x \log {\relax (5 )}^{2} \log {\relax (x )} + 4 \log {\relax (5 )}^{2} \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-640*ln(x)**2+(440*x-960)*ln(x)-40*x**2+120*x)/(16*x*ln(5)**2*ln(x)**4-8*x**2*ln(5)**2*ln(x)**3+x**
3*ln(5)**2*ln(x)**2),x)

[Out]

(-40*x + 160*log(x) + 120)/(-x*log(5)**2*log(x) + 4*log(5)**2*log(x)**2)

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