3.82.2 \(\int \frac {e^{2 x} (5-10 x+40 \log ^2(5))+e^{2 x+2 x^2} (1-2 x+4 x^2+(8-16 x) \log ^2(5))}{25 x^2-200 x \log ^2(5)+400 \log ^4(5)+e^{4 x^2} (x^2-8 x \log ^2(5)+16 \log ^4(5))+e^{2 x^2} (10 x^2-80 x \log ^2(5)+160 \log ^4(5))} \, dx\)

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

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Rubi [F]  time = 2.63, 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 x} \left (5-10 x+40 \log ^2(5)\right )+e^{2 x+2 x^2} \left (1-2 x+4 x^2+(8-16 x) \log ^2(5)\right )}{25 x^2-200 x \log ^2(5)+400 \log ^4(5)+e^{4 x^2} \left (x^2-8 x \log ^2(5)+16 \log ^4(5)\right )+e^{2 x^2} \left (10 x^2-80 x \log ^2(5)+160 \log ^4(5)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*x)*(5 - 10*x + 40*Log[5]^2) + E^(2*x + 2*x^2)*(1 - 2*x + 4*x^2 + (8 - 16*x)*Log[5]^2))/(25*x^2 - 200
*x*Log[5]^2 + 400*Log[5]^4 + E^(4*x^2)*(x^2 - 8*x*Log[5]^2 + 16*Log[5]^4) + E^(2*x^2)*(10*x^2 - 80*x*Log[5]^2
+ 160*Log[5]^4)),x]

[Out]

-20*Defer[Int][E^(2*x)/(5 + E^(2*x^2))^2, x] + 4*Defer[Int][E^(2*x)/(5 + E^(2*x^2)), x] + Defer[Int][E^(2*x)/(
(5 + E^(2*x^2))*(x - 4*Log[5]^2)^2), x] - 80*Log[5]^2*Defer[Int][E^(2*x)/((5 + E^(2*x^2))^2*(x - 4*Log[5]^2)),
x] - 2*(1 - 8*Log[5]^2)*Defer[Int][E^(2*x)/((5 + E^(2*x^2))*(x - 4*Log[5]^2)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 x} \left (-10 x+5 \left (1+8 \log ^2(5)\right )+e^{2 x^2} \left (1+4 x^2+8 \log ^2(5)-2 x \left (1+8 \log ^2(5)\right )\right )\right )}{\left (5+e^{2 x^2}\right )^2 \left (x-4 \log ^2(5)\right )^2} \, dx\\ &=\int \left (-\frac {20 e^{2 x} x}{\left (5+e^{2 x^2}\right )^2 \left (x-4 \log ^2(5)\right )}+\frac {e^{2 x} \left (1+4 x^2+8 \log ^2(5)-2 x \left (1+8 \log ^2(5)\right )\right )}{\left (5+e^{2 x^2}\right ) \left (x-4 \log ^2(5)\right )^2}\right ) \, dx\\ &=-\left (20 \int \frac {e^{2 x} x}{\left (5+e^{2 x^2}\right )^2 \left (x-4 \log ^2(5)\right )} \, dx\right )+\int \frac {e^{2 x} \left (1+4 x^2+8 \log ^2(5)-2 x \left (1+8 \log ^2(5)\right )\right )}{\left (5+e^{2 x^2}\right ) \left (x-4 \log ^2(5)\right )^2} \, dx\\ &=-\left (20 \int \left (\frac {e^{2 x}}{\left (5+e^{2 x^2}\right )^2}+\frac {4 e^{2 x} \log ^2(5)}{\left (5+e^{2 x^2}\right )^2 \left (x-4 \log ^2(5)\right )}\right ) \, dx\right )+\int \left (\frac {4 e^{2 x}}{5+e^{2 x^2}}+\frac {e^{2 x}}{\left (5+e^{2 x^2}\right ) \left (x-4 \log ^2(5)\right )^2}+\frac {2 e^{2 x} \left (-1+8 \log ^2(5)\right )}{\left (5+e^{2 x^2}\right ) \left (x-4 \log ^2(5)\right )}\right ) \, dx\\ &=4 \int \frac {e^{2 x}}{5+e^{2 x^2}} \, dx-20 \int \frac {e^{2 x}}{\left (5+e^{2 x^2}\right )^2} \, dx-\left (80 \log ^2(5)\right ) \int \frac {e^{2 x}}{\left (5+e^{2 x^2}\right )^2 \left (x-4 \log ^2(5)\right )} \, dx-\left (2 \left (1-8 \log ^2(5)\right )\right ) \int \frac {e^{2 x}}{\left (5+e^{2 x^2}\right ) \left (x-4 \log ^2(5)\right )} \, dx+\int \frac {e^{2 x}}{\left (5+e^{2 x^2}\right ) \left (x-4 \log ^2(5)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.25, size = 28, normalized size = 0.97 \begin {gather*} -\frac {e^{2 x}}{\left (5+e^{2 x^2}\right ) \left (x-4 \log ^2(5)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(5 - 10*x + 40*Log[5]^2) + E^(2*x + 2*x^2)*(1 - 2*x + 4*x^2 + (8 - 16*x)*Log[5]^2))/(25*x^2
- 200*x*Log[5]^2 + 400*Log[5]^4 + E^(4*x^2)*(x^2 - 8*x*Log[5]^2 + 16*Log[5]^4) + E^(2*x^2)*(10*x^2 - 80*x*Log
[5]^2 + 160*Log[5]^4)),x]

[Out]

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

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fricas [A]  time = 0.96, size = 49, normalized size = 1.69 \begin {gather*} \frac {e^{\left (2 \, x^{2} + 2 \, x\right )}}{{\left (4 \, \log \relax (5)^{2} - x\right )} e^{\left (4 \, x^{2}\right )} + 5 \, {\left (4 \, \log \relax (5)^{2} - x\right )} e^{\left (2 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x+8)*log(5)^2+4*x^2-2*x+1)*exp(2*x)*exp(2*x^2)+(40*log(5)^2-10*x+5)*exp(2*x))/((16*log(5)^4-8
*x*log(5)^2+x^2)*exp(2*x^2)^2+(160*log(5)^4-80*x*log(5)^2+10*x^2)*exp(2*x^2)+400*log(5)^4-200*x*log(5)^2+25*x^
2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 38, normalized size = 1.31 \begin {gather*} \frac {e^{\left (2 \, x\right )}}{4 \, e^{\left (2 \, x^{2}\right )} \log \relax (5)^{2} - x e^{\left (2 \, x^{2}\right )} + 20 \, \log \relax (5)^{2} - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x+8)*log(5)^2+4*x^2-2*x+1)*exp(2*x)*exp(2*x^2)+(40*log(5)^2-10*x+5)*exp(2*x))/((16*log(5)^4-8
*x*log(5)^2+x^2)*exp(2*x^2)^2+(160*log(5)^4-80*x*log(5)^2+10*x^2)*exp(2*x^2)+400*log(5)^4-200*x*log(5)^2+25*x^
2),x, algorithm="giac")

[Out]

e^(2*x)/(4*e^(2*x^2)*log(5)^2 - x*e^(2*x^2) + 20*log(5)^2 - 5*x)

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maple [A]  time = 0.21, size = 28, normalized size = 0.97

method	result	size
norman	\(\frac {{\mathrm e}^{2 x}}{\left (4 \ln \relax (5)^{2}-x \right ) \left (5+{\mathrm e}^{2 x^{2}}\right )}\)	\(28\)
risch	\(\frac {{\mathrm e}^{2 x}}{\left (4 \ln \relax (5)^{2}-x \right ) \left (5+{\mathrm e}^{2 x^{2}}\right )}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-16*x+8)*ln(5)^2+4*x^2-2*x+1)*exp(2*x)*exp(2*x^2)+(40*ln(5)^2-10*x+5)*exp(2*x))/((16*ln(5)^4-8*x*ln(5)^
2+x^2)*exp(2*x^2)^2+(160*ln(5)^4-80*x*ln(5)^2+10*x^2)*exp(2*x^2)+400*ln(5)^4-200*x*ln(5)^2+25*x^2),x,method=_R
ETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 34, normalized size = 1.17 \begin {gather*} \frac {e^{\left (2 \, x\right )}}{{\left (4 \, \log \relax (5)^{2} - x\right )} e^{\left (2 \, x^{2}\right )} + 20 \, \log \relax (5)^{2} - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x+8)*log(5)^2+4*x^2-2*x+1)*exp(2*x)*exp(2*x^2)+(40*log(5)^2-10*x+5)*exp(2*x))/((16*log(5)^4-8
*x*log(5)^2+x^2)*exp(2*x^2)^2+(160*log(5)^4-80*x*log(5)^2+10*x^2)*exp(2*x^2)+400*log(5)^4-200*x*log(5)^2+25*x^
2),x, algorithm="maxima")

[Out]

e^(2*x)/((4*log(5)^2 - x)*e^(2*x^2) + 20*log(5)^2 - 5*x)

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mupad [B]  time = 5.96, size = 45, normalized size = 1.55 \begin {gather*} -\frac {x^2\,{\mathrm {e}}^{2\,x}-4\,x\,{\mathrm {e}}^{2\,x}\,{\ln \relax (5)}^2}{x\,{\left (x-4\,{\ln \relax (5)}^2\right )}^2\,\left ({\mathrm {e}}^{2\,x^2}+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*(40*log(5)^2 - 10*x + 5) - exp(2*x)*exp(2*x^2)*(2*x + log(5)^2*(16*x - 8) - 4*x^2 - 1))/(exp(4*x
^2)*(16*log(5)^4 - 8*x*log(5)^2 + x^2) + exp(2*x^2)*(160*log(5)^4 - 80*x*log(5)^2 + 10*x^2) - 200*x*log(5)^2 +
400*log(5)^4 + 25*x^2),x)

[Out]

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

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sympy [A]  time = 0.20, size = 31, normalized size = 1.07 \begin {gather*} - \frac {e^{2 x}}{5 x + \left (x - 4 \log {\relax (5 )}^{2}\right ) e^{2 x^{2}} - 20 \log {\relax (5 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x+8)*ln(5)**2+4*x**2-2*x+1)*exp(2*x)*exp(2*x**2)+(40*ln(5)**2-10*x+5)*exp(2*x))/((16*ln(5)**4
-8*x*ln(5)**2+x**2)*exp(2*x**2)**2+(160*ln(5)**4-80*x*ln(5)**2+10*x**2)*exp(2*x**2)+400*ln(5)**4-200*x*ln(5)**
2+25*x**2),x)

[Out]

-exp(2*x)/(5*x + (x - 4*log(5)**2)*exp(2*x**2) - 20*log(5)**2)

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