3.19.45 \(\int \frac {-6 e^4+8 x+e^x (10 x^3-5 x^4+e^8 (10 x-5 x^2)+e^4 (-20 x^2+10 x^3))}{-2 e^4 x+2 x^2+e^x (5 e^8 x^2-10 e^4 x^3+5 x^4)} \, dx\)

Optimal. Leaf size=29 \[ \log \left (\frac {x^2 \log (5)}{5 e^x-\frac {2}{\left (e^4-x\right ) x}}\right ) \]

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Rubi [F]  time = 2.12, 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 {-6 e^4+8 x+e^x \left (10 x^3-5 x^4+e^8 \left (10 x-5 x^2\right )+e^4 \left (-20 x^2+10 x^3\right )\right )}{-2 e^4 x+2 x^2+e^x \left (5 e^8 x^2-10 e^4 x^3+5 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-6*E^4 + 8*x + E^x*(10*x^3 - 5*x^4 + E^8*(10*x - 5*x^2) + E^4*(-20*x^2 + 10*x^3)))/(-2*E^4*x + 2*x^2 + E^
x*(5*E^8*x^2 - 10*E^4*x^3 + 5*x^4)),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6 e^4-8 x-e^x \left (10 x^3-5 x^4+e^8 \left (10 x-5 x^2\right )+e^4 \left (-20 x^2+10 x^3\right )\right )}{\left (e^4-x\right ) x \left (2-5 e^{4+x} x+5 e^x x^2\right )} \, dx\\ &=\int \left (\frac {2-x}{x}+\frac {2 \left (e^4-\left (2-e^4\right ) x-x^2\right )}{\left (e^4-x\right ) x \left (2-5 e^{4+x} x+5 e^x x^2\right )}\right ) \, dx\\ &=2 \int \frac {e^4-\left (2-e^4\right ) x-x^2}{\left (e^4-x\right ) x \left (2-5 e^{4+x} x+5 e^x x^2\right )} \, dx+\int \frac {2-x}{x} \, dx\\ &=2 \int \left (-\frac {1}{-2+5 e^{4+x} x-5 e^x x^2}+\frac {1}{\left (e^4-x\right ) \left (-2+5 e^{4+x} x-5 e^x x^2\right )}+\frac {1}{x \left (2-5 e^{4+x} x+5 e^x x^2\right )}\right ) \, dx+\int \left (-1+\frac {2}{x}\right ) \, dx\\ &=-x+2 \log (x)-2 \int \frac {1}{-2+5 e^{4+x} x-5 e^x x^2} \, dx+2 \int \frac {1}{\left (e^4-x\right ) \left (-2+5 e^{4+x} x-5 e^x x^2\right )} \, dx+2 \int \frac {1}{x \left (2-5 e^{4+x} x+5 e^x x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.32, size = 36, normalized size = 1.24 \begin {gather*} 2 \log (x)+\log \left (x \left (-e^4+x\right )\right )-\log \left (2-5 e^{4+x} x+5 e^x x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*E^4 + 8*x + E^x*(10*x^3 - 5*x^4 + E^8*(10*x - 5*x^2) + E^4*(-20*x^2 + 10*x^3)))/(-2*E^4*x + 2*x^
2 + E^x*(5*E^8*x^2 - 10*E^4*x^3 + 5*x^4)),x]

[Out]

2*Log[x] + Log[x*(-E^4 + x)] - Log[2 - 5*E^(4 + x)*x + 5*E^x*x^2]

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fricas [A]  time = 0.68, size = 36, normalized size = 1.24 \begin {gather*} 2 \, \log \relax (x) - \log \left (-\frac {5 \, {\left (x^{2} - x e^{4}\right )} e^{x} + 2}{x^{2} - x e^{4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2+10*x)*exp(2)^4+(10*x^3-20*x^2)*exp(2)^2-5*x^4+10*x^3)*exp(x)-6*exp(2)^2+8*x)/((5*x^2*exp(2
)^4-10*x^3*exp(2)^2+5*x^4)*exp(x)-2*x*exp(2)^2+2*x^2),x, algorithm="fricas")

[Out]

2*log(x) - log(-(5*(x^2 - x*e^4)*e^x + 2)/(x^2 - x*e^4))

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giac [A]  time = 0.43, size = 31, normalized size = 1.07 \begin {gather*} -\log \left (-5 \, x^{2} e^{x} + 5 \, x e^{\left (x + 4\right )} - 2\right ) + \log \left (x - e^{4}\right ) + 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2+10*x)*exp(2)^4+(10*x^3-20*x^2)*exp(2)^2-5*x^4+10*x^3)*exp(x)-6*exp(2)^2+8*x)/((5*x^2*exp(2
)^4-10*x^3*exp(2)^2+5*x^4)*exp(x)-2*x*exp(2)^2+2*x^2),x, algorithm="giac")

[Out]

-log(-5*x^2*e^x + 5*x*e^(x + 4) - 2) + log(x - e^4) + 3*log(x)

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maple [A]  time = 0.13, size = 25, normalized size = 0.86

method	result	size
risch	\(2 \ln \relax (x )-\ln \left ({\mathrm e}^{x}-\frac {2}{5 x \left ({\mathrm e}^{4}-x \right )}\right )\)	\(25\)
norman	\(3 \ln \relax (x )-\ln \left (5 x \,{\mathrm e}^{4} {\mathrm e}^{x}-5 \,{\mathrm e}^{x} x^{2}-2\right )+\ln \left ({\mathrm e}^{4}-x \right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-5*x^2+10*x)*exp(2)^4+(10*x^3-20*x^2)*exp(2)^2-5*x^4+10*x^3)*exp(x)-6*exp(2)^2+8*x)/((5*x^2*exp(2)^4-10
*x^3*exp(2)^2+5*x^4)*exp(x)-2*x*exp(2)^2+2*x^2),x,method=_RETURNVERBOSE)

[Out]

2*ln(x)-ln(exp(x)-2/5/x/(exp(4)-x))

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maxima [A]  time = 0.63, size = 36, normalized size = 1.24 \begin {gather*} 2 \, \log \relax (x) - \log \left (\frac {5 \, {\left (x^{2} - x e^{4}\right )} e^{x} + 2}{5 \, {\left (x^{2} - x e^{4}\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2+10*x)*exp(2)^4+(10*x^3-20*x^2)*exp(2)^2-5*x^4+10*x^3)*exp(x)-6*exp(2)^2+8*x)/((5*x^2*exp(2
)^4-10*x^3*exp(2)^2+5*x^4)*exp(x)-2*x*exp(2)^2+2*x^2),x, algorithm="maxima")

[Out]

2*log(x) - log(1/5*(5*(x^2 - x*e^4)*e^x + 2)/(x^2 - x*e^4))

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mupad [B]  time = 1.34, size = 31, normalized size = 1.07 \begin {gather*} \ln \left (x-{\mathrm {e}}^4\right )+3\,\ln \relax (x)-\ln \left (5\,x^2\,{\mathrm {e}}^x-5\,x\,{\mathrm {e}}^4\,{\mathrm {e}}^x+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x - 6*exp(4) + exp(x)*(exp(8)*(10*x - 5*x^2) - exp(4)*(20*x^2 - 10*x^3) + 10*x^3 - 5*x^4))/(exp(x)*(5*x
^2*exp(8) - 10*x^3*exp(4) + 5*x^4) - 2*x*exp(4) + 2*x^2),x)

[Out]

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

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sympy [A]  time = 0.37, size = 22, normalized size = 0.76 \begin {gather*} 2 \log {\relax (x )} - \log {\left (e^{x} + \frac {2}{5 x^{2} - 5 x e^{4}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x**2+10*x)*exp(2)**4+(10*x**3-20*x**2)*exp(2)**2-5*x**4+10*x**3)*exp(x)-6*exp(2)**2+8*x)/((5*x
**2*exp(2)**4-10*x**3*exp(2)**2+5*x**4)*exp(x)-2*x*exp(2)**2+2*x**2),x)

[Out]

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

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