3.36.52 \(\int \frac {e^2 (-1+x)-4 x^2+2 e^x x^2+3 x^3+x^4-x^2 \log (25)}{e^2 x-5 x^2+e^x x^2+x^3+x^4-x^2 \log (25)} \, dx\)

Optimal. Leaf size=24 \[ 5+x+\log \left (-5+e^x+\frac {e^2}{x}+x+x^2-\log (25)\right ) \]

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Rubi [F]  time = 1.48, 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 {e^2 (-1+x)-4 x^2+2 e^x x^2+3 x^3+x^4-x^2 \log (25)}{e^2 x-5 x^2+e^x x^2+x^3+x^4-x^2 \log (25)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.48, size = 30, normalized size = 1.25 \begin {gather*} x-\log (x)+\log \left (e^2-5 x+e^x x+x^2+x^3-x \log (25)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(-1 + x) - 4*x^2 + 2*E^x*x^2 + 3*x^3 + x^4 - x^2*Log[25])/(E^2*x - 5*x^2 + E^x*x^2 + x^3 + x^4
- x^2*Log[25]),x]

[Out]

x - Log[x] + Log[E^2 - 5*x + E^x*x + x^2 + x^3 - x*Log[25]]

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fricas [A]  time = 0.62, size = 28, normalized size = 1.17 \begin {gather*} x + \log \left (\frac {x^{3} + x^{2} + x e^{x} - 2 \, x \log \relax (5) - 5 \, x + e^{2}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)*x^2-2*x^2*log(5)+(x-1)*exp(2)+x^4+3*x^3-4*x^2)/(exp(x)*x^2-2*x^2*log(5)+exp(2)*x+x^4+x^3-5
*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 28, normalized size = 1.17 \begin {gather*} x + \log \left (x^{3} + x^{2} + x e^{x} - 2 \, x \log \relax (5) - 5 \, x + e^{2}\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)*x^2-2*x^2*log(5)+(x-1)*exp(2)+x^4+3*x^3-4*x^2)/(exp(x)*x^2-2*x^2*log(5)+exp(2)*x+x^4+x^3-5
*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.14, size = 35, normalized size = 1.46

method	result	size
risch	\(x +\ln \left ({\mathrm e}^{x}-\frac {-x^{3}+2 x \ln \relax (5)-x^{2}-{\mathrm e}^{2}+5 x}{x}\right )\)	\(35\)
norman	\(x -\ln \relax (x )+\ln \left (-x^{3}+2 x \ln \relax (5)-{\mathrm e}^{x} x -x^{2}-{\mathrm e}^{2}+5 x \right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(x)*x^2-2*x^2*ln(5)+(x-1)*exp(2)+x^4+3*x^3-4*x^2)/(exp(x)*x^2-2*x^2*ln(5)+exp(2)*x+x^4+x^3-5*x^2),x,
method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.62, size = 29, normalized size = 1.21 \begin {gather*} x + \log \left (\frac {x^{3} + x^{2} - x {\left (2 \, \log \relax (5) + 5\right )} + x e^{x} + e^{2}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)*x^2-2*x^2*log(5)+(x-1)*exp(2)+x^4+3*x^3-4*x^2)/(exp(x)*x^2-2*x^2*log(5)+exp(2)*x+x^4+x^3-5
*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.27, size = 28, normalized size = 1.17 \begin {gather*} x+\ln \left (\frac {{\mathrm {e}}^2-5\,x-2\,x\,\ln \relax (5)+x\,{\mathrm {e}}^x+x^2+x^3}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2*exp(x) + exp(2)*(x - 1) - 2*x^2*log(5) - 4*x^2 + 3*x^3 + x^4)/(x^2*exp(x) + x*exp(2) - 2*x^2*log(5)
- 5*x^2 + x^3 + x^4),x)

[Out]

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

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sympy [A]  time = 0.23, size = 27, normalized size = 1.12 \begin {gather*} x + \log {\left (e^{x} + \frac {x^{3} + x^{2} - 5 x - 2 x \log {\relax (5 )} + e^{2}}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*exp(x)*x**2-2*x**2*ln(5)+(x-1)*exp(2)+x**4+3*x**3-4*x**2)/(exp(x)*x**2-2*x**2*ln(5)+exp(2)*x+x**4
+x**3-5*x**2),x)

[Out]

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

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