3.102.71 \(\int \frac {9 e^{10} x^3 \log (4)+(-27-36 x-18 x^2) \log (5)+(18 e^{10} x^2 \log (4)-9 x \log (5)) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+(18 x^2+12 x^3+2 x^4) \log (x)+(9 x+6 x^2+x^3) \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ e-\frac {9 \left (e^{10} \log (4)-\frac {\log (5)}{x+\log (x)}\right )}{3+x} \]

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Rubi [F]  time = 0.92, 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 {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(9*E^10*x^3*Log[4] + (-27 - 36*x - 18*x^2)*Log[5] + (18*E^10*x^2*Log[4] - 9*x*Log[5])*Log[x] + 9*E^10*x*Lo
g[4]*Log[x]^2)/(9*x^3 + 6*x^4 + x^5 + (18*x^2 + 12*x^3 + 2*x^4)*Log[x] + (9*x + 6*x^2 + x^3)*Log[x]^2),x]

[Out]

(-9*E^10*Log[4])/(3 + x) - 3*Log[5]*Defer[Int][1/(x*(x + Log[x])^2), x] - 6*Log[5]*Defer[Int][1/((3 + x)*(x +
Log[x])^2), x] - 9*Log[5]*Defer[Int][1/((3 + x)^2*(x + Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9 \left (e^{10} x^3 \log (4)-\left (3+4 x+2 x^2\right ) \log (5)+x \left (-\log (5)+e^{10} x \log (16)\right ) \log (x)+e^{10} x \log (4) \log ^2(x)\right )}{x (3+x)^2 (x+\log (x))^2} \, dx\\ &=9 \int \frac {e^{10} x^3 \log (4)-\left (3+4 x+2 x^2\right ) \log (5)+x \left (-\log (5)+e^{10} x \log (16)\right ) \log (x)+e^{10} x \log (4) \log ^2(x)}{x (3+x)^2 (x+\log (x))^2} \, dx\\ &=9 \int \left (\frac {e^{10} \log (4)}{(3+x)^2}-\frac {(1+x) \log (5)}{x (3+x) (x+\log (x))^2}-\frac {\log (5)}{(3+x)^2 (x+\log (x))}\right ) \, dx\\ &=-\frac {9 e^{10} \log (4)}{3+x}-(9 \log (5)) \int \frac {1+x}{x (3+x) (x+\log (x))^2} \, dx-(9 \log (5)) \int \frac {1}{(3+x)^2 (x+\log (x))} \, dx\\ &=-\frac {9 e^{10} \log (4)}{3+x}-(9 \log (5)) \int \frac {1}{(3+x)^2 (x+\log (x))} \, dx-(9 \log (5)) \int \left (\frac {1}{3 x (x+\log (x))^2}+\frac {2}{3 (3+x) (x+\log (x))^2}\right ) \, dx\\ &=-\frac {9 e^{10} \log (4)}{3+x}-(3 \log (5)) \int \frac {1}{x (x+\log (x))^2} \, dx-(6 \log (5)) \int \frac {1}{(3+x) (x+\log (x))^2} \, dx-(9 \log (5)) \int \frac {1}{(3+x)^2 (x+\log (x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.60, size = 33, normalized size = 1.27 \begin {gather*} \frac {9 \left (-e^{10} x \log (4)+\log (5)-e^{10} \log (4) \log (x)\right )}{(3+x) (x+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(9*E^10*x^3*Log[4] + (-27 - 36*x - 18*x^2)*Log[5] + (18*E^10*x^2*Log[4] - 9*x*Log[5])*Log[x] + 9*E^1
0*x*Log[4]*Log[x]^2)/(9*x^3 + 6*x^4 + x^5 + (18*x^2 + 12*x^3 + 2*x^4)*Log[x] + (9*x + 6*x^2 + x^3)*Log[x]^2),x
]

[Out]

(9*(-(E^10*x*Log[4]) + Log[5] - E^10*Log[4]*Log[x]))/((3 + x)*(x + Log[x]))

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fricas [A]  time = 1.18, size = 37, normalized size = 1.42 \begin {gather*} -\frac {9 \, {\left (2 \, x e^{10} \log \relax (2) + 2 \, e^{10} \log \relax (2) \log \relax (x) - \log \relax (5)\right )}}{x^{2} + {\left (x + 3\right )} \log \relax (x) + 3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x*exp(5)^2*log(2)*log(x)^2+(-9*x*log(5)+36*x^2*exp(5)^2*log(2))*log(x)+(-18*x^2-36*x-27)*log(5)+
18*x^3*exp(5)^2*log(2))/((x^3+6*x^2+9*x)*log(x)^2+(2*x^4+12*x^3+18*x^2)*log(x)+x^5+6*x^4+9*x^3),x, algorithm="
fricas")

[Out]

-9*(2*x*e^10*log(2) + 2*e^10*log(2)*log(x) - log(5))/(x^2 + (x + 3)*log(x) + 3*x)

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giac [A]  time = 0.22, size = 39, normalized size = 1.50 \begin {gather*} -\frac {9 \, {\left (2 \, x e^{10} \log \relax (2) + 2 \, e^{10} \log \relax (2) \log \relax (x) - \log \relax (5)\right )}}{x^{2} + x \log \relax (x) + 3 \, x + 3 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x*exp(5)^2*log(2)*log(x)^2+(-9*x*log(5)+36*x^2*exp(5)^2*log(2))*log(x)+(-18*x^2-36*x-27)*log(5)+
18*x^3*exp(5)^2*log(2))/((x^3+6*x^2+9*x)*log(x)^2+(2*x^4+12*x^3+18*x^2)*log(x)+x^5+6*x^4+9*x^3),x, algorithm="
giac")

[Out]

-9*(2*x*e^10*log(2) + 2*e^10*log(2)*log(x) - log(5))/(x^2 + x*log(x) + 3*x + 3*log(x))

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maple [A]  time = 0.31, size = 28, normalized size = 1.08

method	result	size
risch	\(-\frac {18 \,{\mathrm e}^{10} \ln \relax (2)}{3+x}+\frac {9 \ln \relax (5)}{\left (3+x \right ) \left (x +\ln \relax (x )\right )}\)	\(28\)
default	\(-\frac {18 \,{\mathrm e}^{10} \ln \relax (2)}{3+x}+\frac {9 \ln \relax (5)}{x \ln \relax (x )+x^{2}+3 \ln \relax (x )+3 x}\)	\(36\)
norman	\(\frac {-18 x \,{\mathrm e}^{10} \ln \relax (2)-18 \,{\mathrm e}^{10} \ln \relax (2) \ln \relax (x )+9 \ln \relax (5)}{x \ln \relax (x )+x^{2}+3 \ln \relax (x )+3 x}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((18*x*exp(5)^2*ln(2)*ln(x)^2+(-9*x*ln(5)+36*x^2*exp(5)^2*ln(2))*ln(x)+(-18*x^2-36*x-27)*ln(5)+18*x^3*exp(5
)^2*ln(2))/((x^3+6*x^2+9*x)*ln(x)^2+(2*x^4+12*x^3+18*x^2)*ln(x)+x^5+6*x^4+9*x^3),x,method=_RETURNVERBOSE)

[Out]

-18*exp(10)*ln(2)/(3+x)+9*ln(5)/(3+x)/(x+ln(x))

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maxima [A]  time = 0.47, size = 37, normalized size = 1.42 \begin {gather*} -\frac {9 \, {\left (2 \, x e^{10} \log \relax (2) + 2 \, e^{10} \log \relax (2) \log \relax (x) - \log \relax (5)\right )}}{x^{2} + {\left (x + 3\right )} \log \relax (x) + 3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x*exp(5)^2*log(2)*log(x)^2+(-9*x*log(5)+36*x^2*exp(5)^2*log(2))*log(x)+(-18*x^2-36*x-27)*log(5)+
18*x^3*exp(5)^2*log(2))/((x^3+6*x^2+9*x)*log(x)^2+(2*x^4+12*x^3+18*x^2)*log(x)+x^5+6*x^4+9*x^3),x, algorithm="
maxima")

[Out]

-9*(2*x*e^10*log(2) + 2*e^10*log(2)*log(x) - log(5))/(x^2 + (x + 3)*log(x) + 3*x)

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mupad [B]  time = 7.13, size = 27, normalized size = 1.04 \begin {gather*} \frac {9\,\ln \relax (5)}{\left (x+\ln \relax (x)\right )\,\left (x+3\right )}-\frac {18\,{\mathrm {e}}^{10}\,\ln \relax (2)}{x+3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5)*(36*x + 18*x^2 + 27) + log(x)*(9*x*log(5) - 36*x^2*exp(10)*log(2)) - 18*x^3*exp(10)*log(2) - 18*x
*exp(10)*log(2)*log(x)^2)/(log(x)*(18*x^2 + 12*x^3 + 2*x^4) + 9*x^3 + 6*x^4 + x^5 + log(x)^2*(9*x + 6*x^2 + x^
3)),x)

[Out]

(9*log(5))/((x + log(x))*(x + 3)) - (18*exp(10)*log(2))/(x + 3)

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sympy [A]  time = 0.17, size = 29, normalized size = 1.12 \begin {gather*} \frac {9 \log {\relax (5 )}}{x^{2} + 3 x + \left (x + 3\right ) \log {\relax (x )}} - \frac {18 e^{10} \log {\relax (2 )}}{x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((18*x*exp(5)**2*ln(2)*ln(x)**2+(-9*x*ln(5)+36*x**2*exp(5)**2*ln(2))*ln(x)+(-18*x**2-36*x-27)*ln(5)+1
8*x**3*exp(5)**2*ln(2))/((x**3+6*x**2+9*x)*ln(x)**2+(2*x**4+12*x**3+18*x**2)*ln(x)+x**5+6*x**4+9*x**3),x)

[Out]

9*log(5)/(x**2 + 3*x + (x + 3)*log(x)) - 18*exp(10)*log(2)/(x + 3)

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