3.95.55 \(\int \frac {-48 x+16 e^5 x+(6-2 x) \log ^2(5)+48 x \log (x)}{e^{10} x-2 e^5 x^2+x^3+(6 e^5 x-6 x^2) \log (x)+9 x \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {2 \left (8 x-\log ^2(5)\right )}{e^5-x+3 \log (x)} \]

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Rubi [F]  time = 0.71, 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 {-48 x+16 e^5 x+(6-2 x) \log ^2(5)+48 x \log (x)}{e^{10} x-2 e^5 x^2+x^3+\left (6 e^5 x-6 x^2\right ) \log (x)+9 x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-48*x + 16*E^5*x + (6 - 2*x)*Log[5]^2 + 48*x*Log[x])/(E^10*x - 2*E^5*x^2 + x^3 + (6*E^5*x - 6*x^2)*Log[x]
+ 9*x*Log[x]^2),x]

[Out]

6*Log[5]^2*Defer[Int][1/(x*(-E^5 + x - 3*Log[x])^2), x] - 2*(24 + Log[5]^2)*Defer[Int][(E^5 - x + 3*Log[x])^(-
2), x] + 16*Defer[Int][x/(E^5 - x + 3*Log[x])^2, x] + 16*Defer[Int][(E^5 - x + 3*Log[x])^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-48+16 e^5\right ) x+(6-2 x) \log ^2(5)+48 x \log (x)}{e^{10} x-2 e^5 x^2+x^3+\left (6 e^5 x-6 x^2\right ) \log (x)+9 x \log ^2(x)} \, dx\\ &=\int \frac {2 \left (3 \log ^2(5)-24 x \left (1+\frac {1}{24} \left (-8 e^5+\log ^2(5)\right )\right )+24 x \log (x)\right )}{x \left (e^5-x+3 \log (x)\right )^2} \, dx\\ &=2 \int \frac {3 \log ^2(5)-24 x \left (1+\frac {1}{24} \left (-8 e^5+\log ^2(5)\right )\right )+24 x \log (x)}{x \left (e^5-x+3 \log (x)\right )^2} \, dx\\ &=2 \int \left (\frac {(-3+x) \left (8 x-\log ^2(5)\right )}{x \left (-e^5+x-3 \log (x)\right )^2}+\frac {8}{e^5-x+3 \log (x)}\right ) \, dx\\ &=2 \int \frac {(-3+x) \left (8 x-\log ^2(5)\right )}{x \left (-e^5+x-3 \log (x)\right )^2} \, dx+16 \int \frac {1}{e^5-x+3 \log (x)} \, dx\\ &=2 \int \left (\frac {3 \log ^2(5)}{x \left (-e^5+x-3 \log (x)\right )^2}+\frac {8 x}{\left (e^5-x+3 \log (x)\right )^2}-\frac {24 \left (1+\frac {\log ^2(5)}{24}\right )}{\left (e^5-x+3 \log (x)\right )^2}\right ) \, dx+16 \int \frac {1}{e^5-x+3 \log (x)} \, dx\\ &=16 \int \frac {x}{\left (e^5-x+3 \log (x)\right )^2} \, dx+16 \int \frac {1}{e^5-x+3 \log (x)} \, dx+\left (6 \log ^2(5)\right ) \int \frac {1}{x \left (-e^5+x-3 \log (x)\right )^2} \, dx-\left (2 \left (24+\log ^2(5)\right )\right ) \int \frac {1}{\left (e^5-x+3 \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.33, size = 25, normalized size = 1.00 \begin {gather*} \frac {2 \left (8 x-\log ^2(5)\right )}{e^5-x+3 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-48*x + 16*E^5*x + (6 - 2*x)*Log[5]^2 + 48*x*Log[x])/(E^10*x - 2*E^5*x^2 + x^3 + (6*E^5*x - 6*x^2)*
Log[x] + 9*x*Log[x]^2),x]

[Out]

(2*(8*x - Log[5]^2))/(E^5 - x + 3*Log[x])

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fricas [A]  time = 0.57, size = 22, normalized size = 0.88 \begin {gather*} \frac {2 \, {\left (\log \relax (5)^{2} - 8 \, x\right )}}{x - e^{5} - 3 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((48*x*log(x)+(6-2*x)*log(5)^2+16*x*exp(5)-48*x)/(9*x*log(x)^2+(6*x*exp(5)-6*x^2)*log(x)+x*exp(5)^2-2
*x^2*exp(5)+x^3),x, algorithm="fricas")

[Out]

2*(log(5)^2 - 8*x)/(x - e^5 - 3*log(x))

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giac [A]  time = 0.27, size = 22, normalized size = 0.88 \begin {gather*} \frac {2 \, {\left (\log \relax (5)^{2} - 8 \, x\right )}}{x - e^{5} - 3 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((48*x*log(x)+(6-2*x)*log(5)^2+16*x*exp(5)-48*x)/(9*x*log(x)^2+(6*x*exp(5)-6*x^2)*log(x)+x*exp(5)^2-2
*x^2*exp(5)+x^3),x, algorithm="giac")

[Out]

2*(log(5)^2 - 8*x)/(x - e^5 - 3*log(x))

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maple [A]  time = 0.35, size = 23, normalized size = 0.92

method	result	size
risch	\(-\frac {2 \left (\ln \relax (5)^{2}-8 x \right )}{3 \ln \relax (x )+{\mathrm e}^{5}-x}\)	\(23\)
norman	\(\frac {48 \ln \relax (x )-2 \ln \relax (5)^{2}+16 \,{\mathrm e}^{5}}{3 \ln \relax (x )+{\mathrm e}^{5}-x}\)	\(29\)
default	\(\frac {48 \ln \relax (x )+16 \,{\mathrm e}^{5}}{3 \ln \relax (x )+{\mathrm e}^{5}-x}-\frac {2 \ln \relax (5)^{2}}{3 \ln \relax (x )+{\mathrm e}^{5}-x}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((48*x*ln(x)+(6-2*x)*ln(5)^2+16*x*exp(5)-48*x)/(9*x*ln(x)^2+(6*x*exp(5)-6*x^2)*ln(x)+x*exp(5)^2-2*x^2*exp(5
)+x^3),x,method=_RETURNVERBOSE)

[Out]

-2*(ln(5)^2-8*x)/(3*ln(x)+exp(5)-x)

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maxima [A]  time = 0.48, size = 22, normalized size = 0.88 \begin {gather*} \frac {2 \, {\left (\log \relax (5)^{2} - 8 \, x\right )}}{x - e^{5} - 3 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((48*x*log(x)+(6-2*x)*log(5)^2+16*x*exp(5)-48*x)/(9*x*log(x)^2+(6*x*exp(5)-6*x^2)*log(x)+x*exp(5)^2-2
*x^2*exp(5)+x^3),x, algorithm="maxima")

[Out]

2*(log(5)^2 - 8*x)/(x - e^5 - 3*log(x))

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mupad [B]  time = 7.50, size = 47, normalized size = 1.88 \begin {gather*} \frac {2\,{\mathrm {e}}^{-5}\,\left (8\,x\,{\mathrm {e}}^5-x\,{\ln \relax (5)}^2\right )+6\,{\mathrm {e}}^{-5}\,{\ln \relax (5)}^2\,\left (\frac {x}{3}-\frac {{\mathrm {e}}^5}{3}\right )}{{\mathrm {e}}^5-x+3\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(48*x - 16*x*exp(5) + log(5)^2*(2*x - 6) - 48*x*log(x))/(log(x)*(6*x*exp(5) - 6*x^2) + 9*x*log(x)^2 + x*e
xp(10) - 2*x^2*exp(5) + x^3),x)

[Out]

(2*exp(-5)*(8*x*exp(5) - x*log(5)^2) + 6*exp(-5)*log(5)^2*(x/3 - exp(5)/3))/(exp(5) - x + 3*log(x))

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sympy [A]  time = 0.15, size = 19, normalized size = 0.76 \begin {gather*} \frac {16 x - 2 \log {\relax (5 )}^{2}}{- x + 3 \log {\relax (x )} + e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((48*x*ln(x)+(6-2*x)*ln(5)**2+16*x*exp(5)-48*x)/(9*x*ln(x)**2+(6*x*exp(5)-6*x**2)*ln(x)+x*exp(5)**2-2
*x**2*exp(5)+x**3),x)

[Out]

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

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