3.97.79 \(\int \frac {20-18 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} (e^2 (-9-50 x^2) \log (9)+4 e^2 \log ^2(9))}{25-20 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} (-10 e^2 \log (9)+4 e^2 \log ^2(9))} \, dx\)

Optimal. Leaf size=26 \[ 4+x-\frac {x}{5+\left (-2-e^{2+25 x^2}\right ) \log (9)} \]

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Rubi [F]  time = 1.23, 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 {20-18 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (e^2 \left (-9-50 x^2\right ) \log (9)+4 e^2 \log ^2(9)\right )}{25-20 \log (9)+4 \log ^2(9)+e^{4+50 x^2} \log ^2(9)+e^{25 x^2} \left (-10 e^2 \log (9)+4 e^2 \log ^2(9)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(20 - 18*Log[9] + 4*Log[9]^2 + E^(4 + 50*x^2)*Log[9]^2 + E^(25*x^2)*(E^2*(-9 - 50*x^2)*Log[9] + 4*E^2*Log[
9]^2))/(25 - 20*Log[9] + 4*Log[9]^2 + E^(4 + 50*x^2)*Log[9]^2 + E^(25*x^2)*(-10*E^2*Log[9] + 4*E^2*Log[9]^2)),
x]

[Out]

x - 50*(5 - Log[81])*Defer[Int][x^2/(5*(1 - (4*Log[3])/5) - E^(2 + 25*x^2)*Log[9])^2, x] + 50*Defer[Int][x^2/(
5*(1 - (4*Log[3])/5) - E^(2 + 25*x^2)*Log[9]), x] + Defer[Int][(-5*(1 - (4*Log[3])/5) + E^(2 + 25*x^2)*Log[9])
^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{4+50 x^2} \log ^2(9)+e^{2+25 x^2} \log (9) \left (-9-50 x^2+4 \log (9)\right )+20 \left (1+\frac {1}{10} \log (9) (-9+\log (81))\right )}{\left (5 \left (1-\frac {4 \log (3)}{5}\right )-e^{2+25 x^2} \log (9)\right )^2} \, dx\\ &=\int \left (1+\frac {-1+50 x^2}{5 \left (1-\frac {4 \log (3)}{5}\right )-e^{2+25 x^2} \log (9)}+\frac {2 \left (8 \log ^2(3)-8 \log (3) \log (9)-25 x^2 (5-\log (81))+\log (9) \log (81)\right )}{\left (5 \left (1-\frac {4 \log (3)}{5}\right )-e^{2+25 x^2} \log (9)\right )^2}\right ) \, dx\\ &=x+2 \int \frac {8 \log ^2(3)-8 \log (3) \log (9)-25 x^2 (5-\log (81))+\log (9) \log (81)}{\left (5 \left (1-\frac {4 \log (3)}{5}\right )-e^{2+25 x^2} \log (9)\right )^2} \, dx+\int \frac {-1+50 x^2}{5 \left (1-\frac {4 \log (3)}{5}\right )-e^{2+25 x^2} \log (9)} \, dx\\ &=x+2 \int -\frac {25 x^2 (5-\log (81))}{\left (5 \left (1-\frac {4 \log (3)}{5}\right )-e^{2+25 x^2} \log (9)\right )^2} \, dx+\int \left (\frac {50 x^2}{5 \left (1-\frac {4 \log (3)}{5}\right )-e^{2+25 x^2} \log (9)}+\frac {1}{-5 \left (1-\frac {4 \log (3)}{5}\right )+e^{2+25 x^2} \log (9)}\right ) \, dx\\ &=x+50 \int \frac {x^2}{5 \left (1-\frac {4 \log (3)}{5}\right )-e^{2+25 x^2} \log (9)} \, dx-(50 (5-\log (81))) \int \frac {x^2}{\left (5 \left (1-\frac {4 \log (3)}{5}\right )-e^{2+25 x^2} \log (9)\right )^2} \, dx+\int \frac {1}{-5 \left (1-\frac {4 \log (3)}{5}\right )+e^{2+25 x^2} \log (9)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.57, size = 36, normalized size = 1.38 \begin {gather*} \frac {x \left (-4+e^{2+25 x^2} \log (9)+\log (81)\right )}{-5+e^{2+25 x^2} \log (9)+\log (81)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 - 18*Log[9] + 4*Log[9]^2 + E^(4 + 50*x^2)*Log[9]^2 + E^(25*x^2)*(E^2*(-9 - 50*x^2)*Log[9] + 4*E^
2*Log[9]^2))/(25 - 20*Log[9] + 4*Log[9]^2 + E^(4 + 50*x^2)*Log[9]^2 + E^(25*x^2)*(-10*E^2*Log[9] + 4*E^2*Log[9
]^2)),x]

[Out]

(x*(-4 + E^(2 + 25*x^2)*Log[9] + Log[81]))/(-5 + E^(2 + 25*x^2)*Log[9] + Log[81])

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fricas [A]  time = 0.55, size = 43, normalized size = 1.65 \begin {gather*} \frac {2 \, {\left (x e^{\left (25 \, x^{2} + 2\right )} \log \relax (3) + 2 \, x \log \relax (3) - 2 \, x\right )}}{2 \, e^{\left (25 \, x^{2} + 2\right )} \log \relax (3) + 4 \, \log \relax (3) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(2)^2*log(3)^2*exp(25*x^2)^2+(16*exp(2)*log(3)^2+2*(-50*x^2-9)*exp(2)*log(3))*exp(25*x^2)+16*l
og(3)^2-36*log(3)+20)/(4*exp(2)^2*log(3)^2*exp(25*x^2)^2+(16*exp(2)*log(3)^2-20*exp(2)*log(3))*exp(25*x^2)+16*
log(3)^2-40*log(3)+25),x, algorithm="fricas")

[Out]

2*(x*e^(25*x^2 + 2)*log(3) + 2*x*log(3) - 2*x)/(2*e^(25*x^2 + 2)*log(3) + 4*log(3) - 5)

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giac [A]  time = 0.15, size = 43, normalized size = 1.65 \begin {gather*} \frac {2 \, {\left (x e^{\left (25 \, x^{2} + 2\right )} \log \relax (3) + 2 \, x \log \relax (3) - 2 \, x\right )}}{2 \, e^{\left (25 \, x^{2} + 2\right )} \log \relax (3) + 4 \, \log \relax (3) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(2)^2*log(3)^2*exp(25*x^2)^2+(16*exp(2)*log(3)^2+2*(-50*x^2-9)*exp(2)*log(3))*exp(25*x^2)+16*l
og(3)^2-36*log(3)+20)/(4*exp(2)^2*log(3)^2*exp(25*x^2)^2+(16*exp(2)*log(3)^2-20*exp(2)*log(3))*exp(25*x^2)+16*
log(3)^2-40*log(3)+25),x, algorithm="giac")

[Out]

2*(x*e^(25*x^2 + 2)*log(3) + 2*x*log(3) - 2*x)/(2*e^(25*x^2 + 2)*log(3) + 4*log(3) - 5)

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maple [A]  time = 0.12, size = 25, normalized size = 0.96

method	result	size
risch	\(x +\frac {x}{2 \ln \relax (3) {\mathrm e}^{25 x^{2}+2}+4 \ln \relax (3)-5}\)	\(25\)
norman	\(\frac {\left (4 \ln \relax (3)-4\right ) x +2 x \,{\mathrm e}^{2} \ln \relax (3) {\mathrm e}^{25 x^{2}}}{2 \,{\mathrm e}^{2} \ln \relax (3) {\mathrm e}^{25 x^{2}}+4 \ln \relax (3)-5}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(2)^2*ln(3)^2*exp(25*x^2)^2+(16*exp(2)*ln(3)^2+2*(-50*x^2-9)*exp(2)*ln(3))*exp(25*x^2)+16*ln(3)^2-36
*ln(3)+20)/(4*exp(2)^2*ln(3)^2*exp(25*x^2)^2+(16*exp(2)*ln(3)^2-20*exp(2)*ln(3))*exp(25*x^2)+16*ln(3)^2-40*ln(
3)+25),x,method=_RETURNVERBOSE)

[Out]

x+x/(2*ln(3)*exp(25*x^2+2)+4*ln(3)-5)

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maxima [A]  time = 0.49, size = 42, normalized size = 1.62 \begin {gather*} \frac {2 \, {\left (x e^{\left (25 \, x^{2} + 2\right )} \log \relax (3) + 2 \, x {\left (\log \relax (3) - 1\right )}\right )}}{2 \, e^{\left (25 \, x^{2} + 2\right )} \log \relax (3) + 4 \, \log \relax (3) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(2)^2*log(3)^2*exp(25*x^2)^2+(16*exp(2)*log(3)^2+2*(-50*x^2-9)*exp(2)*log(3))*exp(25*x^2)+16*l
og(3)^2-36*log(3)+20)/(4*exp(2)^2*log(3)^2*exp(25*x^2)^2+(16*exp(2)*log(3)^2-20*exp(2)*log(3))*exp(25*x^2)+16*
log(3)^2-40*log(3)+25),x, algorithm="maxima")

[Out]

2*(x*e^(25*x^2 + 2)*log(3) + 2*x*(log(3) - 1))/(2*e^(25*x^2 + 2)*log(3) + 4*log(3) - 5)

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mupad [B]  time = 0.67, size = 24, normalized size = 0.92 \begin {gather*} x+\frac {x}{4\,\ln \relax (3)+2\,{\mathrm {e}}^{25\,x^2+2}\,\ln \relax (3)-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(25*x^2)*(16*exp(2)*log(3)^2 - 2*exp(2)*log(3)*(50*x^2 + 9)) - 36*log(3) + 16*log(3)^2 + 4*exp(4)*exp(
50*x^2)*log(3)^2 + 20)/(16*log(3)^2 - exp(25*x^2)*(20*exp(2)*log(3) - 16*exp(2)*log(3)^2) - 40*log(3) + 4*exp(
4)*exp(50*x^2)*log(3)^2 + 25),x)

[Out]

x + x/(4*log(3) + 2*exp(25*x^2 + 2)*log(3) - 5)

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sympy [A]  time = 0.27, size = 24, normalized size = 0.92 \begin {gather*} x + \frac {x}{2 e^{2} e^{25 x^{2}} \log {\relax (3 )} - 5 + 4 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(2)**2*ln(3)**2*exp(25*x**2)**2+(16*exp(2)*ln(3)**2+2*(-50*x**2-9)*exp(2)*ln(3))*exp(25*x**2)+
16*ln(3)**2-36*ln(3)+20)/(4*exp(2)**2*ln(3)**2*exp(25*x**2)**2+(16*exp(2)*ln(3)**2-20*exp(2)*ln(3))*exp(25*x**
2)+16*ln(3)**2-40*ln(3)+25),x)

[Out]

x + x/(2*exp(2)*exp(25*x**2)*log(3) - 5 + 4*log(3))

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