3.40.47 \(\int \frac {21+e^{-8+50 x^2} (3-300 x^2)-3 \log (5)-3 \log (x^2)}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+(-10-2 e^{-8+50 x^2}-2 x+2 \log (5)) \log (x^2)+\log ^2(x^2)} \, dx\)

Optimal. Leaf size=27 \[ \frac {3 x}{5+e^{-8+50 x^2}+x-\log (5)-\log \left (x^2\right )} \]

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Rubi [F]  time = 2.52, 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 {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(21 + E^(-8 + 50*x^2)*(3 - 300*x^2) - 3*Log[5] - 3*Log[x^2])/(25 + E^(-16 + 100*x^2) + 10*x + x^2 + E^(-8
+ 50*x^2)*(10 + 2*x - 2*Log[5]) + (-10 - 2*x)*Log[5] + Log[5]^2 + (-10 - 2*E^(-8 + 50*x^2) - 2*x + 2*Log[5])*L
og[x^2] + Log[x^2]^2),x]

[Out]

6*E^16*Defer[Int][(E^(50*x^2) + E^8*x + 5*E^8*(1 - Log[5]/5) - E^8*Log[x^2])^(-2), x] - 3*E^16*Defer[Int][x/(E
^(50*x^2) + E^8*x + 5*E^8*(1 - Log[5]/5) - E^8*Log[x^2])^2, x] + 300*E^16*(5 - Log[5])*Defer[Int][x^2/(E^(50*x
^2) + E^8*x + 5*E^8*(1 - Log[5]/5) - E^8*Log[x^2])^2, x] + 300*E^16*Defer[Int][x^3/(E^(50*x^2) + E^8*x + 5*E^8
*(1 - Log[5]/5) - E^8*Log[x^2])^2, x] - 300*E^16*Defer[Int][(x^2*Log[x^2])/(E^(50*x^2) + E^8*x + 5*E^8*(1 - Lo
g[5]/5) - E^8*Log[x^2])^2, x] + 3*E^8*Defer[Int][(E^(50*x^2) + E^8*x + 5*E^8*(1 - Log[5]/5) - E^8*Log[x^2])^(-
1), x] + 300*E^8*Defer[Int][x^2/(-E^(50*x^2) - E^8*x - 5*E^8*(1 - Log[5]/5) + E^8*Log[x^2]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 e^8 \left (-e^{50 x^2} \left (-1+100 x^2\right )-e^8 (-7+\log (5))-e^8 \log \left (x^2\right )\right )}{\left (e^{50 x^2}+e^8 (5+x-\log (5))-e^8 \log \left (x^2\right )\right )^2} \, dx\\ &=\left (3 e^8\right ) \int \frac {-e^{50 x^2} \left (-1+100 x^2\right )-e^8 (-7+\log (5))-e^8 \log \left (x^2\right )}{\left (e^{50 x^2}+e^8 (5+x-\log (5))-e^8 \log \left (x^2\right )\right )^2} \, dx\\ &=\left (3 e^8\right ) \int \left (\frac {1-100 x^2}{e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )}+\frac {e^8 \left (2-x+100 x^3+500 x^2 \left (1-\frac {\log (5)}{5}\right )-100 x^2 \log \left (x^2\right )\right )}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=\left (3 e^8\right ) \int \frac {1-100 x^2}{e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )} \, dx+\left (3 e^{16}\right ) \int \frac {2-x+100 x^3+500 x^2 \left (1-\frac {\log (5)}{5}\right )-100 x^2 \log \left (x^2\right )}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx\\ &=\left (3 e^8\right ) \int \left (\frac {1}{e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )}+\frac {100 x^2}{-e^{50 x^2}-e^8 x-5 e^8 \left (1-\frac {\log (5)}{5}\right )+e^8 \log \left (x^2\right )}\right ) \, dx+\left (3 e^{16}\right ) \int \left (\frac {2}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}-\frac {x}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}+\frac {100 x^3}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}+\frac {100 x^2 (5-\log (5))}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}-\frac {100 x^2 \log \left (x^2\right )}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=\left (3 e^8\right ) \int \frac {1}{e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )} \, dx+\left (300 e^8\right ) \int \frac {x^2}{-e^{50 x^2}-e^8 x-5 e^8 \left (1-\frac {\log (5)}{5}\right )+e^8 \log \left (x^2\right )} \, dx-\left (3 e^{16}\right ) \int \frac {x}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx+\left (6 e^{16}\right ) \int \frac {1}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx+\left (300 e^{16}\right ) \int \frac {x^3}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx-\left (300 e^{16}\right ) \int \frac {x^2 \log \left (x^2\right )}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx+\left (300 e^{16} (5-\log (5))\right ) \int \frac {x^2}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.69, size = 36, normalized size = 1.33 \begin {gather*} \frac {3 e^8 x}{e^{50 x^2}+e^8 (5+x-\log (5))-e^8 \log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(21 + E^(-8 + 50*x^2)*(3 - 300*x^2) - 3*Log[5] - 3*Log[x^2])/(25 + E^(-16 + 100*x^2) + 10*x + x^2 +
E^(-8 + 50*x^2)*(10 + 2*x - 2*Log[5]) + (-10 - 2*x)*Log[5] + Log[5]^2 + (-10 - 2*E^(-8 + 50*x^2) - 2*x + 2*Log
[5])*Log[x^2] + Log[x^2]^2),x]

[Out]

(3*E^8*x)/(E^(50*x^2) + E^8*(5 + x - Log[5]) - E^8*Log[x^2])

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fricas [A]  time = 0.81, size = 26, normalized size = 0.96 \begin {gather*} \frac {3 \, x}{x + e^{\left (50 \, x^{2} - 8\right )} - \log \relax (5) - \log \left (x^{2}\right ) + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(x^2)+(-300*x^2+3)*exp(25*x^2-4)^2-3*log(5)+21)/(log(x^2)^2+(-2*exp(25*x^2-4)^2+2*log(5)-2*x-
10)*log(x^2)+exp(25*x^2-4)^4+(-2*log(5)+2*x+10)*exp(25*x^2-4)^2+log(5)^2+(-2*x-10)*log(5)+x^2+10*x+25),x, algo
rithm="fricas")

[Out]

3*x/(x + e^(50*x^2 - 8) - log(5) - log(x^2) + 5)

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giac [A]  time = 0.31, size = 36, normalized size = 1.33 \begin {gather*} \frac {3 \, x e^{8}}{x e^{8} - e^{8} \log \relax (5) - e^{8} \log \left (x^{2}\right ) + 5 \, e^{8} + e^{\left (50 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(x^2)+(-300*x^2+3)*exp(25*x^2-4)^2-3*log(5)+21)/(log(x^2)^2+(-2*exp(25*x^2-4)^2+2*log(5)-2*x-
10)*log(x^2)+exp(25*x^2-4)^4+(-2*log(5)+2*x+10)*exp(25*x^2-4)^2+log(5)^2+(-2*x-10)*log(5)+x^2+10*x+25),x, algo
rithm="giac")

[Out]

3*x*e^8/(x*e^8 - e^8*log(5) - e^8*log(x^2) + 5*e^8 + e^(50*x^2))

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maple [C]  time = 0.11, size = 83, normalized size = 3.07

method	result	size
risch	\(\frac {6 x}{i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 \,{\mathrm e}^{2 \left (5 x -2\right ) \left (5 x +2\right )}-2 \ln \relax (5)+2 x -4 \ln \relax (x )+10}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*ln(x^2)+(-300*x^2+3)*exp(25*x^2-4)^2-3*ln(5)+21)/(ln(x^2)^2+(-2*exp(25*x^2-4)^2+2*ln(5)-2*x-10)*ln(x^2
)+exp(25*x^2-4)^4+(-2*ln(5)+2*x+10)*exp(25*x^2-4)^2+ln(5)^2+(-2*x-10)*ln(5)+x^2+10*x+25),x,method=_RETURNVERBO
SE)

[Out]

6*x/(I*Pi*csgn(I*x^2)*csgn(I*x)^2-2*I*Pi*csgn(I*x^2)^2*csgn(I*x)+I*Pi*csgn(I*x^2)^3+2*exp(2*(5*x-2)*(5*x+2))-2
*ln(5)+2*x-4*ln(x)+10)

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maxima [A]  time = 0.49, size = 32, normalized size = 1.19 \begin {gather*} \frac {3 \, x e^{8}}{x e^{8} - {\left (\log \relax (5) - 5\right )} e^{8} - 2 \, e^{8} \log \relax (x) + e^{\left (50 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(x^2)+(-300*x^2+3)*exp(25*x^2-4)^2-3*log(5)+21)/(log(x^2)^2+(-2*exp(25*x^2-4)^2+2*log(5)-2*x-
10)*log(x^2)+exp(25*x^2-4)^4+(-2*log(5)+2*x+10)*exp(25*x^2-4)^2+log(5)^2+(-2*x-10)*log(5)+x^2+10*x+25),x, algo
rithm="maxima")

[Out]

3*x*e^8/(x*e^8 - (log(5) - 5)*e^8 - 2*e^8*log(x) + e^(50*x^2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {3\,\ln \left (x^2\right )+3\,\ln \relax (5)+{\mathrm {e}}^{50\,x^2-8}\,\left (300\,x^2-3\right )-21}{10\,x+{\mathrm {e}}^{100\,x^2-16}-\ln \left (x^2\right )\,\left (2\,x-2\,\ln \relax (5)+2\,{\mathrm {e}}^{50\,x^2-8}+10\right )-\ln \relax (5)\,\left (2\,x+10\right )+{\mathrm {e}}^{50\,x^2-8}\,\left (2\,x-2\,\ln \relax (5)+10\right )+{\ln \left (x^2\right )}^2+{\ln \relax (5)}^2+x^2+25} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*log(x^2) + 3*log(5) + exp(50*x^2 - 8)*(300*x^2 - 3) - 21)/(10*x + exp(100*x^2 - 16) - log(x^2)*(2*x -
2*log(5) + 2*exp(50*x^2 - 8) + 10) - log(5)*(2*x + 10) + exp(50*x^2 - 8)*(2*x - 2*log(5) + 10) + log(x^2)^2 +
log(5)^2 + x^2 + 25),x)

[Out]

int(-(3*log(x^2) + 3*log(5) + exp(50*x^2 - 8)*(300*x^2 - 3) - 21)/(10*x + exp(100*x^2 - 16) - log(x^2)*(2*x -
2*log(5) + 2*exp(50*x^2 - 8) + 10) - log(5)*(2*x + 10) + exp(50*x^2 - 8)*(2*x - 2*log(5) + 10) + log(x^2)^2 +
log(5)^2 + x^2 + 25), x)

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sympy [A]  time = 0.33, size = 22, normalized size = 0.81 \begin {gather*} \frac {3 x}{x + e^{50 x^{2} - 8} - \log {\left (x^{2} \right )} - \log {\relax (5 )} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*ln(x**2)+(-300*x**2+3)*exp(25*x**2-4)**2-3*ln(5)+21)/(ln(x**2)**2+(-2*exp(25*x**2-4)**2+2*ln(5)-
2*x-10)*ln(x**2)+exp(25*x**2-4)**4+(-2*ln(5)+2*x+10)*exp(25*x**2-4)**2+ln(5)**2+(-2*x-10)*ln(5)+x**2+10*x+25),
x)

[Out]

3*x/(x + exp(50*x**2 - 8) - log(x**2) - log(5) + 5)

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