3.15.64 \(\int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} (-50 x-40 x^2)+e^{e^{\frac {2}{e^{2 x}-x}}} (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} (10 x-20 e^{2 x} x))+(5 e^{4 x}-10 e^{2 x} x+5 x^2) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx\)

Optimal. Leaf size=30 \[ 5 \left (3+x \left (4+e^{e^{\frac {2}{e^{2 x}-x}}}+2 x+\log (x)\right )\right ) \]

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Rubi [F]  time = 5.59, 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 {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{e^{4 x}-2 e^{2 x} x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(25*x^2 + 20*x^3 + E^(4*x)*(25 + 20*x) + E^(2*x)*(-50*x - 40*x^2) + E^E^(2/(E^(2*x) - x))*(5*E^(4*x) - 10*
E^(2*x)*x + 5*x^2 + E^(2/(E^(2*x) - x))*(10*x - 20*E^(2*x)*x)) + (5*E^(4*x) - 10*E^(2*x)*x + 5*x^2)*Log[x])/(E
^(4*x) - 2*E^(2*x)*x + x^2),x]

[Out]

20*x + 10*x^2 + 5*x*Log[x] + 5*Defer[Int][E^E^(2/(E^(2*x) - x)), x] + 10*Defer[Int][(E^(E^(2/(E^(2*x) - x)) +
2/(E^(2*x) - x))*x)/(E^(2*x) - x)^2, x] - 20*Defer[Int][(E^(E^(2/(E^(2*x) - x)) + 2/(E^(2*x) - x))*x)/(E^(2*x)
- x), x] - 20*Defer[Int][(E^(E^(2/(E^(2*x) - x)) + 2/(E^(2*x) - x))*x^2)/(E^(2*x) - x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 x^2+20 x^3+e^{4 x} (25+20 x)+e^{2 x} \left (-50 x-40 x^2\right )+e^{e^{\frac {2}{e^{2 x}-x}}} \left (5 e^{4 x}-10 e^{2 x} x+5 x^2+e^{\frac {2}{e^{2 x}-x}} \left (10 x-20 e^{2 x} x\right )\right )+\left (5 e^{4 x}-10 e^{2 x} x+5 x^2\right ) \log (x)}{\left (e^{2 x}-x\right )^2} \, dx\\ &=\int \left (-\frac {20 e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{e^{2 x}-x}-\frac {10 e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x (-1+2 x)}{\left (e^{2 x}-x\right )^2}+5 \left (5+e^{e^{\frac {2}{e^{2 x}-x}}}+4 x+\log (x)\right )\right ) \, dx\\ &=5 \int \left (5+e^{e^{\frac {2}{e^{2 x}-x}}}+4 x+\log (x)\right ) \, dx-10 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x (-1+2 x)}{\left (e^{2 x}-x\right )^2} \, dx-20 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{e^{2 x}-x} \, dx\\ &=25 x+10 x^2+5 \int e^{e^{\frac {2}{e^{2 x}-x}}} \, dx+5 \int \log (x) \, dx-10 \int \left (-\frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{\left (e^{2 x}-x\right )^2}+\frac {2 e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x^2}{\left (e^{2 x}-x\right )^2}\right ) \, dx-20 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{e^{2 x}-x} \, dx\\ &=20 x+10 x^2+5 x \log (x)+5 \int e^{e^{\frac {2}{e^{2 x}-x}}} \, dx+10 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{\left (e^{2 x}-x\right )^2} \, dx-20 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x}{e^{2 x}-x} \, dx-20 \int \frac {e^{e^{\frac {2}{e^{2 x}-x}}+\frac {2}{e^{2 x}-x}} x^2}{\left (e^{2 x}-x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.64, size = 27, normalized size = 0.90 \begin {gather*} 5 x \left (4+e^{e^{\frac {2}{e^{2 x}-x}}}+2 x+\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*x^2 + 20*x^3 + E^(4*x)*(25 + 20*x) + E^(2*x)*(-50*x - 40*x^2) + E^E^(2/(E^(2*x) - x))*(5*E^(4*x)
- 10*E^(2*x)*x + 5*x^2 + E^(2/(E^(2*x) - x))*(10*x - 20*E^(2*x)*x)) + (5*E^(4*x) - 10*E^(2*x)*x + 5*x^2)*Log[
x])/(E^(4*x) - 2*E^(2*x)*x + x^2),x]

[Out]

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

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fricas [A]  time = 0.77, size = 31, normalized size = 1.03 \begin {gather*} 10 \, x^{2} + 5 \, x e^{\left (e^{\left (-\frac {2}{x - e^{\left (2 \, x\right )}}\right )}\right )} + 5 \, x \log \relax (x) + 20 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-20*x*exp(x)^2+10*x)*exp(2/(exp(x)^2-x))+5*exp(x)^4-10*x*exp(x)^2+5*x^2)*exp(exp(2/(exp(x)^2-x)))
+(5*exp(x)^4-10*x*exp(x)^2+5*x^2)*log(x)+(20*x+25)*exp(x)^4+(-40*x^2-50*x)*exp(x)^2+20*x^3+25*x^2)/(exp(x)^4-2
*x*exp(x)^2+x^2),x, algorithm="fricas")

[Out]

10*x^2 + 5*x*e^(e^(-2/(x - e^(2*x)))) + 5*x*log(x) + 20*x

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-20*x*exp(x)^2+10*x)*exp(2/(exp(x)^2-x))+5*exp(x)^4-10*x*exp(x)^2+5*x^2)*exp(exp(2/(exp(x)^2-x)))
+(5*exp(x)^4-10*x*exp(x)^2+5*x^2)*log(x)+(20*x+25)*exp(x)^4+(-40*x^2-50*x)*exp(x)^2+20*x^3+25*x^2)/(exp(x)^4-2
*x*exp(x)^2+x^2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.04, size = 32, normalized size = 1.07

method	result	size
risch	\(10 x^{2}+5 x \ln \relax (x )+5 x \,{\mathrm e}^{{\mathrm e}^{-\frac {2}{x -{\mathrm e}^{2 x}}}}+20 x\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-20*x*exp(x)^2+10*x)*exp(2/(exp(x)^2-x))+5*exp(x)^4-10*x*exp(x)^2+5*x^2)*exp(exp(2/(exp(x)^2-x)))+(5*ex
p(x)^4-10*x*exp(x)^2+5*x^2)*ln(x)+(20*x+25)*exp(x)^4+(-40*x^2-50*x)*exp(x)^2+20*x^3+25*x^2)/(exp(x)^4-2*x*exp(
x)^2+x^2),x,method=_RETURNVERBOSE)

[Out]

10*x^2+5*x*ln(x)+5*x*exp(exp(-2/(x-exp(2*x))))+20*x

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maxima [A]  time = 0.57, size = 31, normalized size = 1.03 \begin {gather*} 10 \, x^{2} + 5 \, x e^{\left (e^{\left (-\frac {2}{x - e^{\left (2 \, x\right )}}\right )}\right )} + 5 \, x \log \relax (x) + 20 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-20*x*exp(x)^2+10*x)*exp(2/(exp(x)^2-x))+5*exp(x)^4-10*x*exp(x)^2+5*x^2)*exp(exp(2/(exp(x)^2-x)))
+(5*exp(x)^4-10*x*exp(x)^2+5*x^2)*log(x)+(20*x+25)*exp(x)^4+(-40*x^2-50*x)*exp(x)^2+20*x^3+25*x^2)/(exp(x)^4-2
*x*exp(x)^2+x^2),x, algorithm="maxima")

[Out]

10*x^2 + 5*x*e^(e^(-2/(x - e^(2*x)))) + 5*x*log(x) + 20*x

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mupad [B]  time = 1.24, size = 31, normalized size = 1.03 \begin {gather*} 20\,x+5\,x\,\ln \relax (x)+5\,x\,{\mathrm {e}}^{{\mathrm {e}}^{-\frac {2}{x-{\mathrm {e}}^{2\,x}}}}+10\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(5*exp(4*x) - 10*x*exp(2*x) + 5*x^2) - exp(2*x)*(50*x + 40*x^2) + exp(4*x)*(20*x + 25) + 25*x^2 +
20*x^3 + exp(exp(-2/(x - exp(2*x))))*(5*exp(4*x) - 10*x*exp(2*x) + exp(-2/(x - exp(2*x)))*(10*x - 20*x*exp(2*x
)) + 5*x^2))/(exp(4*x) - 2*x*exp(2*x) + x^2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-20*x*exp(x)**2+10*x)*exp(2/(exp(x)**2-x))+5*exp(x)**4-10*x*exp(x)**2+5*x**2)*exp(exp(2/(exp(x)**
2-x)))+(5*exp(x)**4-10*x*exp(x)**2+5*x**2)*ln(x)+(20*x+25)*exp(x)**4+(-40*x**2-50*x)*exp(x)**2+20*x**3+25*x**2
)/(exp(x)**4-2*x*exp(x)**2+x**2),x)

[Out]

Timed out

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