3.32.8 \(\int \frac {5 e^2+e^4 x^2+(-5 e^2+e^4 (-x-3 x^2)) \log (x)+(e^4 x-2 e^4 x \log (x)) \log (e^5 x)}{25 x^2+10 e^2 x^4+e^4 x^6+(10 e^2 x^3+2 e^4 x^5) \log (e^5 x)+e^4 x^4 \log ^2(e^5 x)} \, dx\)

Optimal. Leaf size=26 \[ -4+\frac {\log (x)}{x \left (\frac {5}{e^2}+x \left (x+\log \left (e^5 x\right )\right )\right )} \]

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Rubi [F]  time = 1.70, 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 {5 e^2+e^4 x^2+\left (-5 e^2+e^4 \left (-x-3 x^2\right )\right ) \log (x)+\left (e^4 x-2 e^4 x \log (x)\right ) \log \left (e^5 x\right )}{25 x^2+10 e^2 x^4+e^4 x^6+\left (10 e^2 x^3+2 e^4 x^5\right ) \log \left (e^5 x\right )+e^4 x^4 \log ^2\left (e^5 x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5*E^2 + E^4*x^2 + (-5*E^2 + E^4*(-x - 3*x^2))*Log[x] + (E^4*x - 2*E^4*x*Log[x])*Log[E^5*x])/(25*x^2 + 10*
E^2*x^4 + E^4*x^6 + (10*E^2*x^3 + 2*E^4*x^5)*Log[E^5*x] + E^4*x^4*Log[E^5*x]^2),x]

[Out]

x^(-2) + 6*E^4*Defer[Int][(5 + 5*E^2*x + E^2*x^2 + E^2*x*Log[x])^(-2), x] - 25*Defer[Int][1/(x^3*(5 + 5*E^2*x
+ E^2*x^2 + E^2*x*Log[x])^2), x] - 20*E^2*Defer[Int][1/(x^2*(5 + 5*E^2*x + E^2*x^2 + E^2*x*Log[x])^2), x] + 5*
E^4*Defer[Int][1/(x*(5 + 5*E^2*x + E^2*x^2 + E^2*x*Log[x])^2), x] + E^4*Defer[Int][x/(5 + 5*E^2*x + E^2*x^2 +
E^2*x*Log[x])^2, x] + 15*Defer[Int][1/(x^3*(5 + 5*E^2*x + E^2*x^2 + E^2*x*Log[x])), x] + 10*E^2*Defer[Int][1/(
x^2*(5 + 5*E^2*x + E^2*x^2 + E^2*x*Log[x])), x] + E^2*Defer[Int][1/(x*(5 + 5*E^2*x + E^2*x^2 + E^2*x*Log[x])),
x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^2 \left (5+e^2 x (5+x)-\left (5+e^2 x (10+3 x)\right ) \log (x)-2 e^2 x \log ^2(x)\right )}{x^2 \left (5+e^2 x (5+x)+e^2 x \log (x)\right )^2} \, dx\\ &=e^2 \int \frac {5+e^2 x (5+x)-\left (5+e^2 x (10+3 x)\right ) \log (x)-2 e^2 x \log ^2(x)}{x^2 \left (5+e^2 x (5+x)+e^2 x \log (x)\right )^2} \, dx\\ &=e^2 \int \left (-\frac {2}{e^2 x^3}+\frac {-25-20 e^2 x+5 e^4 x^2+6 e^4 x^3+e^4 x^4}{e^2 x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2}+\frac {15+10 e^2 x+e^2 x^2}{e^2 x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )}\right ) \, dx\\ &=\frac {1}{x^2}+\int \frac {-25-20 e^2 x+5 e^4 x^2+6 e^4 x^3+e^4 x^4}{x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2} \, dx+\int \frac {15+10 e^2 x+e^2 x^2}{x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )} \, dx\\ &=\frac {1}{x^2}+\int \left (\frac {6 e^4}{\left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2}-\frac {25}{x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2}-\frac {20 e^2}{x^2 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2}+\frac {5 e^4}{x \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2}+\frac {e^4 x}{\left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2}\right ) \, dx+\int \left (\frac {15}{x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )}+\frac {10 e^2}{x^2 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )}+\frac {e^2}{x \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )}\right ) \, dx\\ &=\frac {1}{x^2}+15 \int \frac {1}{x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )} \, dx-25 \int \frac {1}{x^3 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2} \, dx+e^2 \int \frac {1}{x \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )} \, dx+\left (10 e^2\right ) \int \frac {1}{x^2 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )} \, dx-\left (20 e^2\right ) \int \frac {1}{x^2 \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2} \, dx+e^4 \int \frac {x}{\left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2} \, dx+\left (5 e^4\right ) \int \frac {1}{x \left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2} \, dx+\left (6 e^4\right ) \int \frac {1}{\left (5+5 e^2 x+e^2 x^2+e^2 x \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.61, size = 28, normalized size = 1.08 \begin {gather*} \frac {e^2 \log (x)}{x \left (5+e^2 x (5+x)+e^2 x \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*E^2 + E^4*x^2 + (-5*E^2 + E^4*(-x - 3*x^2))*Log[x] + (E^4*x - 2*E^4*x*Log[x])*Log[E^5*x])/(25*x^2
+ 10*E^2*x^4 + E^4*x^6 + (10*E^2*x^3 + 2*E^4*x^5)*Log[E^5*x] + E^4*x^4*Log[E^5*x]^2),x]

[Out]

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

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fricas [A]  time = 0.53, size = 31, normalized size = 1.19 \begin {gather*} \frac {e^{2} \log \relax (x)}{x^{2} e^{2} \log \relax (x) + {\left (x^{3} + 5 \, x^{2}\right )} e^{2} + 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(2)^2*log(x)+x*exp(2)^2)*log(x*exp(5))+((-3*x^2-x)*exp(2)^2-5*exp(2))*log(x)+x^2*exp(2)^2+
5*exp(2))/(x^4*exp(2)^2*log(x*exp(5))^2+(2*x^5*exp(2)^2+10*x^3*exp(2))*log(x*exp(5))+x^6*exp(2)^2+10*x^4*exp(2
)+25*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.32, size = 32, normalized size = 1.23 \begin {gather*} \frac {e^{2} \log \relax (x)}{x^{3} e^{2} + x^{2} e^{2} \log \relax (x) + 5 \, x^{2} e^{2} + 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(2)^2*log(x)+x*exp(2)^2)*log(x*exp(5))+((-3*x^2-x)*exp(2)^2-5*exp(2))*log(x)+x^2*exp(2)^2+
5*exp(2))/(x^4*exp(2)^2*log(x*exp(5))^2+(2*x^5*exp(2)^2+10*x^3*exp(2))*log(x*exp(5))+x^6*exp(2)^2+10*x^4*exp(2
)+25*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.14, size = 47, normalized size = 1.81

method	result	size
risch	\(\frac {1}{x^{2}}-\frac {10+2 x^{2} {\mathrm e}^{2}+10 \,{\mathrm e}^{2} x}{x^{2} \left (10+2 x^{2} {\mathrm e}^{2}+2 x \,{\mathrm e}^{2} \ln \relax (x )+10 \,{\mathrm e}^{2} x \right )}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*exp(2)^2*ln(x)+x*exp(2)^2)*ln(x*exp(5))+((-3*x^2-x)*exp(2)^2-5*exp(2))*ln(x)+x^2*exp(2)^2+5*exp(2))
/(x^4*exp(2)^2*ln(x*exp(5))^2+(2*x^5*exp(2)^2+10*x^3*exp(2))*ln(x*exp(5))+x^6*exp(2)^2+10*x^4*exp(2)+25*x^2),x
,method=_RETURNVERBOSE)

[Out]

1/x^2-(10+2*x^2*exp(2)+10*exp(2)*x)/x^2/(10+2*x^2*exp(2)+2*x*exp(2)*ln(x)+10*exp(2)*x)

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maxima [A]  time = 0.85, size = 32, normalized size = 1.23 \begin {gather*} \frac {e^{2} \log \relax (x)}{x^{3} e^{2} + x^{2} e^{2} \log \relax (x) + 5 \, x^{2} e^{2} + 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(2)^2*log(x)+x*exp(2)^2)*log(x*exp(5))+((-3*x^2-x)*exp(2)^2-5*exp(2))*log(x)+x^2*exp(2)^2+
5*exp(2))/(x^4*exp(2)^2*log(x*exp(5))^2+(2*x^5*exp(2)^2+10*x^3*exp(2))*log(x*exp(5))+x^6*exp(2)^2+10*x^4*exp(2
)+25*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.60, size = 29, normalized size = 1.12 \begin {gather*} \frac {{\mathrm {e}}^2\,\ln \relax (x)}{x\,\left (5\,x\,{\mathrm {e}}^2+x^2\,{\mathrm {e}}^2+x\,{\mathrm {e}}^2\,\ln \relax (x)+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*exp(2) - log(x)*(5*exp(2) + exp(4)*(x + 3*x^2)) + x^2*exp(4) + log(x*exp(5))*(x*exp(4) - 2*x*exp(4)*log
(x)))/(10*x^4*exp(2) + x^6*exp(4) + log(x*exp(5))*(10*x^3*exp(2) + 2*x^5*exp(4)) + 25*x^2 + x^4*exp(4)*log(x*e
xp(5))^2),x)

[Out]

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

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sympy [B]  time = 0.34, size = 51, normalized size = 1.96 \begin {gather*} \frac {- x^{2} e^{2} - 5 x e^{2} - 5}{x^{4} e^{2} + x^{3} e^{2} \log {\relax (x )} + 5 x^{3} e^{2} + 5 x^{2}} + \frac {1}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(2)**2*ln(x)+x*exp(2)**2)*ln(x*exp(5))+((-3*x**2-x)*exp(2)**2-5*exp(2))*ln(x)+x**2*exp(2)*
*2+5*exp(2))/(x**4*exp(2)**2*ln(x*exp(5))**2+(2*x**5*exp(2)**2+10*x**3*exp(2))*ln(x*exp(5))+x**6*exp(2)**2+10*
x**4*exp(2)+25*x**2),x)

[Out]

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

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