3.46.61 \(\int \frac {-400 x^6+(100 x^3+100 x^4-640 x^5+400 x^6) \log (x)+(100 x^3+640 x^5) \log ^2(x)+(-50-75 x+295 x^2+160 x^3-8 x^5) \log ^3(x)}{8 x^5 \log ^3(x)} \, dx\)

Optimal. Leaf size=32 \[ 4-x+\left (2-\frac {5}{4} \left (-\frac {1+x}{x^2}+4 \left (2+\frac {x}{\log (x)}\right )\right )\right )^2 \]

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Rubi [F]  time = 0.40, 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 {-400 x^6+\left (100 x^3+100 x^4-640 x^5+400 x^6\right ) \log (x)+\left (100 x^3+640 x^5\right ) \log ^2(x)+\left (-50-75 x+295 x^2+160 x^3-8 x^5\right ) \log ^3(x)}{8 x^5 \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-400*x^6 + (100*x^3 + 100*x^4 - 640*x^5 + 400*x^6)*Log[x] + (100*x^3 + 640*x^5)*Log[x]^2 + (-50 - 75*x +
295*x^2 + 160*x^3 - 8*x^5)*Log[x]^3)/(8*x^5*Log[x]^3),x]

[Out]

25/(16*x^4) + 25/(8*x^3) - 295/(16*x^2) - 20/x - x + (25*ExpIntegralEi[-Log[x]])/2 - 100*ExpIntegralEi[2*Log[x
]] + (25*x^2)/Log[x]^2 + (50*x^2)/Log[x] + 80*LogIntegral[x] + (5*Defer[Int][(5 + 5*x - 32*x^2 + 20*x^3)/(x^2*
Log[x]^2), x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} \int \frac {-400 x^6+\left (100 x^3+100 x^4-640 x^5+400 x^6\right ) \log (x)+\left (100 x^3+640 x^5\right ) \log ^2(x)+\left (-50-75 x+295 x^2+160 x^3-8 x^5\right ) \log ^3(x)}{x^5 \log ^3(x)} \, dx\\ &=\frac {1}{8} \int \left (\frac {-50-75 x+295 x^2+160 x^3-8 x^5}{x^5}-\frac {400 x}{\log ^3(x)}+\frac {20 \left (5+5 x-32 x^2+20 x^3\right )}{x^2 \log ^2(x)}+\frac {20 \left (5+32 x^2\right )}{x^2 \log (x)}\right ) \, dx\\ &=\frac {1}{8} \int \frac {-50-75 x+295 x^2+160 x^3-8 x^5}{x^5} \, dx+\frac {5}{2} \int \frac {5+5 x-32 x^2+20 x^3}{x^2 \log ^2(x)} \, dx+\frac {5}{2} \int \frac {5+32 x^2}{x^2 \log (x)} \, dx-50 \int \frac {x}{\log ^3(x)} \, dx\\ &=\frac {25 x^2}{\log ^2(x)}+\frac {1}{8} \int \left (-8-\frac {50}{x^5}-\frac {75}{x^4}+\frac {295}{x^3}+\frac {160}{x^2}\right ) \, dx+\frac {5}{2} \int \left (\frac {32}{\log (x)}+\frac {5}{x^2 \log (x)}\right ) \, dx+\frac {5}{2} \int \frac {5+5 x-32 x^2+20 x^3}{x^2 \log ^2(x)} \, dx-50 \int \frac {x}{\log ^2(x)} \, dx\\ &=\frac {25}{16 x^4}+\frac {25}{8 x^3}-\frac {295}{16 x^2}-\frac {20}{x}-x+\frac {25 x^2}{\log ^2(x)}+\frac {50 x^2}{\log (x)}+\frac {5}{2} \int \frac {5+5 x-32 x^2+20 x^3}{x^2 \log ^2(x)} \, dx+\frac {25}{2} \int \frac {1}{x^2 \log (x)} \, dx+80 \int \frac {1}{\log (x)} \, dx-100 \int \frac {x}{\log (x)} \, dx\\ &=\frac {25}{16 x^4}+\frac {25}{8 x^3}-\frac {295}{16 x^2}-\frac {20}{x}-x+\frac {25 x^2}{\log ^2(x)}+\frac {50 x^2}{\log (x)}+80 \text {li}(x)+\frac {5}{2} \int \frac {5+5 x-32 x^2+20 x^3}{x^2 \log ^2(x)} \, dx+\frac {25}{2} \operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )-100 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=\frac {25}{16 x^4}+\frac {25}{8 x^3}-\frac {295}{16 x^2}-\frac {20}{x}-x+\frac {25 \text {Ei}(-\log (x))}{2}-100 \text {Ei}(2 \log (x))+\frac {25 x^2}{\log ^2(x)}+\frac {50 x^2}{\log (x)}+80 \text {li}(x)+\frac {5}{2} \int \frac {5+5 x-32 x^2+20 x^3}{x^2 \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.04, size = 65, normalized size = 2.03 \begin {gather*} \frac {25}{16 x^4}+\frac {25}{8 x^3}-\frac {295}{16 x^2}-\frac {20}{x}-x+\frac {25 x^2}{\log ^2(x)}-\frac {25}{2 \log (x)}-\frac {25}{2 x \log (x)}+\frac {80 x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-400*x^6 + (100*x^3 + 100*x^4 - 640*x^5 + 400*x^6)*Log[x] + (100*x^3 + 640*x^5)*Log[x]^2 + (-50 - 7
5*x + 295*x^2 + 160*x^3 - 8*x^5)*Log[x]^3)/(8*x^5*Log[x]^3),x]

[Out]

25/(16*x^4) + 25/(8*x^3) - 295/(16*x^2) - 20/x - x + (25*x^2)/Log[x]^2 - 25/(2*Log[x]) - 25/(2*x*Log[x]) + (80
*x)/Log[x]

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fricas [B]  time = 0.69, size = 61, normalized size = 1.91 \begin {gather*} \frac {400 \, x^{6} - {\left (16 \, x^{5} + 320 \, x^{3} + 295 \, x^{2} - 50 \, x - 25\right )} \log \relax (x)^{2} + 40 \, {\left (32 \, x^{5} - 5 \, x^{4} - 5 \, x^{3}\right )} \log \relax (x)}{16 \, x^{4} \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((-8*x^5+160*x^3+295*x^2-75*x-50)*log(x)^3+(640*x^5+100*x^3)*log(x)^2+(400*x^6-640*x^5+100*x^4+1
00*x^3)*log(x)-400*x^6)/x^5/log(x)^3,x, algorithm="fricas")

[Out]

1/16*(400*x^6 - (16*x^5 + 320*x^3 + 295*x^2 - 50*x - 25)*log(x)^2 + 40*(32*x^5 - 5*x^4 - 5*x^3)*log(x))/(x^4*l
og(x)^2)

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giac [B]  time = 0.19, size = 55, normalized size = 1.72 \begin {gather*} -x + \frac {5 \, {\left (10 \, x^{3} + 32 \, x^{2} \log \relax (x) - 5 \, x \log \relax (x) - 5 \, \log \relax (x)\right )}}{2 \, x \log \relax (x)^{2}} - \frac {5 \, {\left (64 \, x^{3} + 59 \, x^{2} - 10 \, x - 5\right )}}{16 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((-8*x^5+160*x^3+295*x^2-75*x-50)*log(x)^3+(640*x^5+100*x^3)*log(x)^2+(400*x^6-640*x^5+100*x^4+1
00*x^3)*log(x)-400*x^6)/x^5/log(x)^3,x, algorithm="giac")

[Out]

-x + 5/2*(10*x^3 + 32*x^2*log(x) - 5*x*log(x) - 5*log(x))/(x*log(x)^2) - 5/16*(64*x^3 + 59*x^2 - 10*x - 5)/x^4

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maple [B]  time = 0.02, size = 56, normalized size = 1.75

method	result	size
default	\(-x -\frac {20}{x}+\frac {80 x}{\ln \relax (x )}+\frac {25 x^{2}}{\ln \relax (x )^{2}}-\frac {295}{16 x^{2}}-\frac {25}{2 \ln \relax (x )}+\frac {25}{8 x^{3}}-\frac {25}{2 x \ln \relax (x )}+\frac {25}{16 x^{4}}\)	\(56\)
risch	\(-\frac {16 x^{5}+320 x^{3}+295 x^{2}-50 x -25}{16 x^{4}}+\frac {25 x^{3}+80 x^{2} \ln \relax (x )-\frac {25 x \ln \relax (x )}{2}-\frac {25 \ln \relax (x )}{2}}{x \ln \relax (x )^{2}}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*((-8*x^5+160*x^3+295*x^2-75*x-50)*ln(x)^3+(640*x^5+100*x^3)*ln(x)^2+(400*x^6-640*x^5+100*x^4+100*x^3)*
ln(x)-400*x^6)/x^5/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

-x-20/x+80*x/ln(x)+25*x^2/ln(x)^2-295/16/x^2-25/2/ln(x)+25/8/x^3-25/2/x/ln(x)+25/16/x^4

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maxima [C]  time = 0.40, size = 72, normalized size = 2.25 \begin {gather*} -x - \frac {20}{x} - \frac {25}{2 \, \log \relax (x)} - \frac {295}{16 \, x^{2}} + \frac {25}{8 \, x^{3}} + \frac {25}{16 \, x^{4}} + \frac {25}{2} \, {\rm Ei}\left (-\log \relax (x)\right ) + 80 \, {\rm Ei}\left (\log \relax (x)\right ) - 80 \, \Gamma \left (-1, -\log \relax (x)\right ) + 100 \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) - \frac {25}{2} \, \Gamma \left (-1, \log \relax (x)\right ) + 200 \, \Gamma \left (-2, -2 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((-8*x^5+160*x^3+295*x^2-75*x-50)*log(x)^3+(640*x^5+100*x^3)*log(x)^2+(400*x^6-640*x^5+100*x^4+1
00*x^3)*log(x)-400*x^6)/x^5/log(x)^3,x, algorithm="maxima")

[Out]

-x - 20/x - 25/2/log(x) - 295/16/x^2 + 25/8/x^3 + 25/16/x^4 + 25/2*Ei(-log(x)) + 80*Ei(log(x)) - 80*gamma(-1,
-log(x)) + 100*gamma(-1, -2*log(x)) - 25/2*gamma(-1, log(x)) + 200*gamma(-2, -2*log(x))

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mupad [B]  time = 3.34, size = 58, normalized size = 1.81 \begin {gather*} -\frac {x^5+20\,x^3+\frac {295\,x^2}{16}-\frac {25\,x}{8}-\frac {25}{16}}{x^4}-\frac {\ln \relax (x)\,\left (-80\,x^5+\frac {25\,x^4}{2}+\frac {25\,x^3}{2}\right )-25\,x^6}{x^4\,{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(x)^2*(100*x^3 + 640*x^5))/8 - (log(x)^3*(75*x - 295*x^2 - 160*x^3 + 8*x^5 + 50))/8 + (log(x)*(100*x^
3 + 100*x^4 - 640*x^5 + 400*x^6))/8 - 50*x^6)/(x^5*log(x)^3),x)

[Out]

- ((295*x^2)/16 - (25*x)/8 + 20*x^3 + x^5 - 25/16)/x^4 - (log(x)*((25*x^3)/2 + (25*x^4)/2 - 80*x^5) - 25*x^6)/
(x^4*log(x)^2)

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sympy [A]  time = 0.18, size = 48, normalized size = 1.50 \begin {gather*} - x + \frac {50 x^{3} + \left (160 x^{2} - 25 x - 25\right ) \log {\relax (x )}}{2 x \log {\relax (x )}^{2}} - \frac {320 x^{3} + 295 x^{2} - 50 x - 25}{16 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((-8*x**5+160*x**3+295*x**2-75*x-50)*ln(x)**3+(640*x**5+100*x**3)*ln(x)**2+(400*x**6-640*x**5+10
0*x**4+100*x**3)*ln(x)-400*x**6)/x**5/ln(x)**3,x)

[Out]

-x + (50*x**3 + (160*x**2 - 25*x - 25)*log(x))/(2*x*log(x)**2) - (320*x**3 + 295*x**2 - 50*x - 25)/(16*x**4)

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