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

Optimal. Leaf size=20 \[ \frac {5 (1+x)}{x \left (-2+x-e^x \log (x)\right )} \]

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Rubi [F]  time = 4.28, 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 {10-10 x-5 x^2+e^x (5+5 x)+e^x \left (5+5 x+5 x^2\right ) \log (x)}{4 x^2-4 x^3+x^4+e^x \left (4 x^2-2 x^3\right ) \log (x)+e^{2 x} x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.60, size = 20, normalized size = 1.00 \begin {gather*} \frac {5 (1+x)}{x \left (-2+x-e^x \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(1 + x))/(x*(-2 + x - E^x*Log[x]))

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fricas [A]  time = 0.62, size = 22, normalized size = 1.10 \begin {gather*} -\frac {5 \, {\left (x + 1\right )}}{x e^{x} \log \relax (x) - x^{2} + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*(x + 1)/(x*e^x*log(x) - x^2 + 2*x)

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giac [A]  time = 0.30, size = 22, normalized size = 1.10 \begin {gather*} -\frac {5 \, {\left (x + 1\right )}}{x e^{x} \log \relax (x) - x^{2} + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*(x + 1)/(x*e^x*log(x) - x^2 + 2*x)

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maple [A]  time = 0.03, size = 20, normalized size = 1.00

method	result	size
risch	\(\frac {5 x +5}{\left (x -2-{\mathrm e}^{x} \ln \relax (x )\right ) x}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/(x-2-exp(x)*ln(x))*(x+1)/x

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maxima [A]  time = 0.54, size = 22, normalized size = 1.10 \begin {gather*} -\frac {5 \, {\left (x + 1\right )}}{x e^{x} \log \relax (x) - x^{2} + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*(x + 1)/(x*e^x*log(x) - x^2 + 2*x)

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mupad [B]  time = 1.16, size = 68, normalized size = 3.40 \begin {gather*} -\frac {5\,{\mathrm {e}}^{2\,x}+5\,x\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (-5\,x^3+10\,x^2+15\,x\right )}{\left ({\mathrm {e}}^x\,\ln \relax (x)-x+2\right )\,\left (x\,{\mathrm {e}}^{2\,x}-3\,x^2\,{\mathrm {e}}^x+x^3\,{\mathrm {e}}^x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.39, size = 20, normalized size = 1.00 \begin {gather*} \frac {- 5 x - 5}{- x^{2} + x e^{x} \log {\relax (x )} + 2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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