3.68.20 \(\int \frac {(15 e^5-15 x) \log ^2(x)+((30 e^5-30 x) \log (x)+(-15 e^5+45 x) \log ^2(x)) \log (2 x)}{e^{15} x^2-3 e^{10} x^3+3 e^5 x^4-x^5} \, dx\)

Optimal. Leaf size=24 \[ -5+\frac {15 \log ^2(x) \log (2 x)}{\left (e^5-x\right )^2 x} \]

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Rubi [F]  time = 25.59, 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 {\left (15 e^5-15 x\right ) \log ^2(x)+\left (\left (30 e^5-30 x\right ) \log (x)+\left (-15 e^5+45 x\right ) \log ^2(x)\right ) \log (2 x)}{e^{15} x^2-3 e^{10} x^3+3 e^5 x^4-x^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((15*E^5 - 15*x)*Log[x]^2 + ((30*E^5 - 30*x)*Log[x] + (-15*E^5 + 45*x)*Log[x]^2)*Log[2*x])/(E^15*x^2 - 3*E
^10*x^3 + 3*E^5*x^4 - x^5),x]

[Out]

30/(E^10*x) + (225*Log[E^5 - x])/E^15 + (15*(5 + Log[2])*Log[E^5 - x])/E^15 + (30*Log[x])/(E^10*x) - (15*Log[x
]^2)/(2*E^15) + (15*x*Log[x]^2)/(E^15*(E^5 - x)) - (30*(1 + Log[x]))/(E^10*x) + (15*Log[x]*Log[2*x])/(E^5*(E^5
 - x)^2) - (15*x^2*Log[x]*Log[2*x])/(E^15*(E^5 - x)^2) + (30*Log[x]^2*Log[2*x])/E^15 + (15*Log[x]^2*Log[2*x])/
(E^10*x) - (15*Log[2*x]^2)/(2*E^15) + (75*Log[-E^5 + x])/E^15 + (15*(5 + Log[2])*Log[-E^5 + x])/E^15 - (30*Log
[x]^2*Log[1 - x/E^5])/E^15 - (60*Log[x]*PolyLog[2, x/E^5])/E^15 - (90*PolyLog[2, 1 - x/E^5])/E^15 + (60*PolyLo
g[3, x/E^5])/E^15 + (60*Defer[Int][(Log[x]*Log[2*x])/(E^5 - x), x])/E^15 + (30*Defer[Int][(Log[x]^2*Log[2*x])/
(E^5 - x)^3, x])/E^5 + (15*Defer[Int][(Log[x]^2*Log[2*x])/(E^5 - x)^2, x])/E^10

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {15 \log (x) \left (-2 \left (-e^5+x\right ) \log (2 x)-\log (x) \left (-e^5+x+\left (e^5-3 x\right ) \log (2 x)\right )\right )}{\left (e^5-x\right )^3 x^2} \, dx\\ &=15 \int \frac {\log (x) \left (-2 \left (-e^5+x\right ) \log (2 x)-\log (x) \left (-e^5+x+\left (e^5-3 x\right ) \log (2 x)\right )\right )}{\left (e^5-x\right )^3 x^2} \, dx\\ &=15 \int \left (\frac {\log ^2(x)}{\left (e^5-x\right )^2 x^2}-\frac {\log (x) \left (-2 e^5+2 x+e^5 \log (x)-3 x \log (x)\right ) \log (2 x)}{\left (e^5-x\right )^3 x^2}\right ) \, dx\\ &=15 \int \frac {\log ^2(x)}{\left (e^5-x\right )^2 x^2} \, dx-15 \int \frac {\log (x) \left (-2 e^5+2 x+e^5 \log (x)-3 x \log (x)\right ) \log (2 x)}{\left (e^5-x\right )^3 x^2} \, dx\\ &=15 \int \left (\frac {\log ^2(x)}{e^{10} \left (e^5-x\right )^2}+\frac {2 \log ^2(x)}{e^{15} \left (e^5-x\right )}+\frac {\log ^2(x)}{e^{10} x^2}+\frac {2 \log ^2(x)}{e^{15} x}\right ) \, dx-15 \int \left (\frac {\log (x) \left (-2 e^5+2 x+e^5 \log (x)-3 x \log (x)\right ) \log (2 x)}{e^{10} \left (e^5-x\right )^3}+\frac {2 \log (x) \left (-2 e^5+2 x+e^5 \log (x)-3 x \log (x)\right ) \log (2 x)}{e^{15} \left (e^5-x\right )^2}+\frac {3 \log (x) \left (-2 e^5+2 x+e^5 \log (x)-3 x \log (x)\right ) \log (2 x)}{e^{20} \left (e^5-x\right )}+\frac {\log (x) \left (-2 e^5+2 x+e^5 \log (x)-3 x \log (x)\right ) \log (2 x)}{e^{15} x^2}+\frac {3 \log (x) \left (-2 e^5+2 x+e^5 \log (x)-3 x \log (x)\right ) \log (2 x)}{e^{20} x}\right ) \, dx\\ &=-\frac {45 \int \frac {\log (x) \left (-2 e^5+2 x+e^5 \log (x)-3 x \log (x)\right ) \log (2 x)}{e^5-x} \, dx}{e^{20}}-\frac {45 \int \frac {\log (x) \left (-2 e^5+2 x+e^5 \log (x)-3 x \log (x)\right ) \log (2 x)}{x} \, dx}{e^{20}}-\frac {15 \int \frac {\log (x) \left (-2 e^5+2 x+e^5 \log (x)-3 x \log (x)\right ) \log (2 x)}{x^2} \, dx}{e^{15}}+\frac {30 \int \frac {\log ^2(x)}{e^5-x} \, dx}{e^{15}}+\frac {30 \int \frac {\log ^2(x)}{x} \, dx}{e^{15}}-\frac {30 \int \frac {\log (x) \left (-2 e^5+2 x+e^5 \log (x)-3 x \log (x)\right ) \log (2 x)}{\left (e^5-x\right )^2} \, dx}{e^{15}}+\frac {15 \int \frac {\log ^2(x)}{\left (e^5-x\right )^2} \, dx}{e^{10}}+\frac {15 \int \frac {\log ^2(x)}{x^2} \, dx}{e^{10}}-\frac {15 \int \frac {\log (x) \left (-2 e^5+2 x+e^5 \log (x)-3 x \log (x)\right ) \log (2 x)}{\left (e^5-x\right )^3} \, dx}{e^{10}}\\ &=-\frac {15 \log ^2(x)}{e^{10} x}+\frac {15 x \log ^2(x)}{e^{15} \left (e^5-x\right )}-\frac {30 \log ^2(x) \log \left (1-\frac {x}{e^5}\right )}{e^{15}}-\frac {45 \int \left (2 \log (x) \log (2 x)-\frac {2 e^5 \log (x) \log (2 x)}{x}-3 \log ^2(x) \log (2 x)+\frac {e^5 \log ^2(x) \log (2 x)}{x}\right ) \, dx}{e^{20}}-\frac {45 \int \left (-\frac {2 e^5 \log (x) \log (2 x)}{e^5-x}+\frac {2 x \log (x) \log (2 x)}{e^5-x}+\frac {e^5 \log ^2(x) \log (2 x)}{e^5-x}-\frac {3 x \log ^2(x) \log (2 x)}{e^5-x}\right ) \, dx}{e^{20}}-\frac {15 \int \left (-\frac {2 e^5 \log (x) \log (2 x)}{x^2}+\frac {2 \log (x) \log (2 x)}{x}+\frac {e^5 \log ^2(x) \log (2 x)}{x^2}-\frac {3 \log ^2(x) \log (2 x)}{x}\right ) \, dx}{e^{15}}-\frac {30 \int \frac {\log (x)}{e^5-x} \, dx}{e^{15}}-\frac {30 \int \left (-\frac {2 e^5 \log (x) \log (2 x)}{\left (e^5-x\right )^2}+\frac {2 x \log (x) \log (2 x)}{\left (e^5-x\right )^2}+\frac {e^5 \log ^2(x) \log (2 x)}{\left (e^5-x\right )^2}-\frac {3 x \log ^2(x) \log (2 x)}{\left (e^5-x\right )^2}\right ) \, dx}{e^{15}}+\frac {30 \operatorname {Subst}\left (\int x^2 \, dx,x,\log (x)\right )}{e^{15}}+\frac {60 \int \frac {\log (x) \log \left (1-\frac {x}{e^5}\right )}{x} \, dx}{e^{15}}-\frac {15 \int \left (-\frac {2 e^5 \log (x) \log (2 x)}{\left (e^5-x\right )^3}+\frac {2 x \log (x) \log (2 x)}{\left (e^5-x\right )^3}+\frac {e^5 \log ^2(x) \log (2 x)}{\left (e^5-x\right )^3}-\frac {3 x \log ^2(x) \log (2 x)}{\left (e^5-x\right )^3}\right ) \, dx}{e^{10}}+\frac {30 \int \frac {\log (x)}{x^2} \, dx}{e^{10}}\\ &=-\frac {30}{e^{10} x}+\frac {150 \log \left (e^5-x\right )}{e^{15}}-\frac {30 \log (x)}{e^{10} x}-\frac {15 \log ^2(x)}{e^{10} x}+\frac {15 x \log ^2(x)}{e^{15} \left (e^5-x\right )}+\frac {10 \log ^3(x)}{e^{15}}-\frac {30 \log ^2(x) \log \left (1-\frac {x}{e^5}\right )}{e^{15}}-\frac {60 \log (x) \text {Li}_2\left (\frac {x}{e^5}\right )}{e^{15}}-\frac {90 \int \log (x) \log (2 x) \, dx}{e^{20}}-\frac {90 \int \frac {x \log (x) \log (2 x)}{e^5-x} \, dx}{e^{20}}+\frac {135 \int \log ^2(x) \log (2 x) \, dx}{e^{20}}+\frac {135 \int \frac {x \log ^2(x) \log (2 x)}{e^5-x} \, dx}{e^{20}}-\frac {30 \int \frac {\log (x) \log (2 x)}{x} \, dx}{e^{15}}-\frac {30 \int \frac {\log \left (\frac {x}{e^5}\right )}{e^5-x} \, dx}{e^{15}}-\frac {45 \int \frac {\log ^2(x) \log (2 x)}{e^5-x} \, dx}{e^{15}}-\frac {60 \int \frac {x \log (x) \log (2 x)}{\left (e^5-x\right )^2} \, dx}{e^{15}}+\frac {60 \int \frac {\text {Li}_2\left (\frac {x}{e^5}\right )}{x} \, dx}{e^{15}}+\frac {90 \int \frac {\log (x) \log (2 x)}{e^5-x} \, dx}{e^{15}}+\frac {90 \int \frac {\log (x) \log (2 x)}{x} \, dx}{e^{15}}+\frac {90 \int \frac {x \log ^2(x) \log (2 x)}{\left (e^5-x\right )^2} \, dx}{e^{15}}-\frac {15 \int \frac {\log ^2(x) \log (2 x)}{x^2} \, dx}{e^{10}}+\frac {30 \int \frac {\log (x) \log (2 x)}{x^2} \, dx}{e^{10}}-\frac {30 \int \frac {x \log (x) \log (2 x)}{\left (e^5-x\right )^3} \, dx}{e^{10}}-\frac {30 \int \frac {\log ^2(x) \log (2 x)}{\left (e^5-x\right )^2} \, dx}{e^{10}}+\frac {45 \int \frac {x \log ^2(x) \log (2 x)}{\left (e^5-x\right )^3} \, dx}{e^{10}}+\frac {60 \int \frac {\log (x) \log (2 x)}{\left (e^5-x\right )^2} \, dx}{e^{10}}-\frac {15 \int \frac {\log ^2(x) \log (2 x)}{\left (e^5-x\right )^3} \, dx}{e^5}+\frac {30 \int \frac {\log (x) \log (2 x)}{\left (e^5-x\right )^3} \, dx}{e^5}\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 22, normalized size = 0.92 \begin {gather*} \frac {15 \log ^2(x) \log (2 x)}{\left (e^5-x\right )^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((15*E^5 - 15*x)*Log[x]^2 + ((30*E^5 - 30*x)*Log[x] + (-15*E^5 + 45*x)*Log[x]^2)*Log[2*x])/(E^15*x^2
 - 3*E^10*x^3 + 3*E^5*x^4 - x^5),x]

[Out]

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

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fricas [A]  time = 0.53, size = 31, normalized size = 1.29 \begin {gather*} \frac {15 \, {\left (\log \relax (2) \log \relax (x)^{2} + \log \relax (x)^{3}\right )}}{x^{3} - 2 \, x^{2} e^{5} + x e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-15*exp(5)+45*x)*log(x)^2+(30*exp(5)-30*x)*log(x))*log(2*x)+(15*exp(5)-15*x)*log(x)^2)/(x^2*exp(5
)^3-3*x^3*exp(5)^2+3*x^4*exp(5)-x^5),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 31, normalized size = 1.29 \begin {gather*} \frac {15 \, {\left (\log \relax (2) \log \relax (x)^{2} + \log \relax (x)^{3}\right )}}{x^{3} - 2 \, x^{2} e^{5} + x e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-15*exp(5)+45*x)*log(x)^2+(30*exp(5)-30*x)*log(x))*log(2*x)+(15*exp(5)-15*x)*log(x)^2)/(x^2*exp(5
)^3-3*x^3*exp(5)^2+3*x^4*exp(5)-x^5),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.22, size = 48, normalized size = 2.00




method result size



risch \(\frac {15 \ln \relax (x )^{3}}{x \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}\right )}+\frac {15 \ln \relax (2) \ln \relax (x )^{2}}{x \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}\right )}\) \(48\)
default error in convert/parfrac: numeric exception: division by zero\ N/A



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-15*exp(5)+45*x)*ln(x)^2+(30*exp(5)-30*x)*ln(x))*ln(2*x)+(15*exp(5)-15*x)*ln(x)^2)/(x^2*exp(5)^3-3*x^3*
exp(5)^2+3*x^4*exp(5)-x^5),x,method=_RETURNVERBOSE)

[Out]

15/x/(exp(10)-2*x*exp(5)+x^2)*ln(x)^3+15*ln(2)/x/(exp(10)-2*x*exp(5)+x^2)*ln(x)^2

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maxima [A]  time = 0.53, size = 31, normalized size = 1.29 \begin {gather*} \frac {15 \, {\left (\log \relax (2) \log \relax (x)^{2} + \log \relax (x)^{3}\right )}}{x^{3} - 2 \, x^{2} e^{5} + x e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-15*exp(5)+45*x)*log(x)^2+(30*exp(5)-30*x)*log(x))*log(2*x)+(15*exp(5)-15*x)*log(x)^2)/(x^2*exp(5
)^3-3*x^3*exp(5)^2+3*x^4*exp(5)-x^5),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.43, size = 22, normalized size = 0.92 \begin {gather*} \frac {15\,{\ln \relax (x)}^2\,\left (\ln \relax (2)+\ln \relax (x)\right )}{x\,{\left (x-{\mathrm {e}}^5\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)^2*(15*x - 15*exp(5)) + log(2*x)*(log(x)*(30*x - 30*exp(5)) - log(x)^2*(45*x - 15*exp(5))))/(3*x^4
*exp(5) - 3*x^3*exp(10) + x^2*exp(15) - x^5),x)

[Out]

(15*log(x)^2*(log(2) + log(x)))/(x*(x - exp(5))^2)

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sympy [B]  time = 0.41, size = 49, normalized size = 2.04 \begin {gather*} \frac {15 \log {\relax (x )}^{3}}{x^{3} - 2 x^{2} e^{5} + x e^{10}} + \frac {15 \log {\relax (2 )} \log {\relax (x )}^{2}}{x^{3} - 2 x^{2} e^{5} + x e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-15*exp(5)+45*x)*ln(x)**2+(30*exp(5)-30*x)*ln(x))*ln(2*x)+(15*exp(5)-15*x)*ln(x)**2)/(x**2*exp(5)
**3-3*x**3*exp(5)**2+3*x**4*exp(5)-x**5),x)

[Out]

15*log(x)**3/(x**3 - 2*x**2*exp(5) + x*exp(10)) + 15*log(2)*log(x)**2/(x**3 - 2*x**2*exp(5) + x*exp(10))

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