3.98.41 \(\int \frac {2+e^x (3-x)+(-120-120 x+30 x^2) \log (5)}{4+e^{2 x}+(-240 x-120 x^2) \log (5)+(3600 x^2+3600 x^3+900 x^4) \log ^2(5)+e^x (4+(-120 x-60 x^2) \log (5))} \, dx\)

Optimal. Leaf size=26 \[ \frac {2-x}{-2-e^x+30 \left (2 x+x^2\right ) \log (5)} \]

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Rubi [F]  time = 1.25, 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 {2+e^x (3-x)+\left (-120-120 x+30 x^2\right ) \log (5)}{4+e^{2 x}+\left (-240 x-120 x^2\right ) \log (5)+\left (3600 x^2+3600 x^3+900 x^4\right ) \log ^2(5)+e^x \left (4+\left (-120 x-60 x^2\right ) \log (5)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2 + E^x*(3 - x) + (-120 - 120*x + 30*x^2)*Log[5])/(4 + E^(2*x) + (-240*x - 120*x^2)*Log[5] + (3600*x^2 +
3600*x^3 + 900*x^4)*Log[5]^2 + E^x*(4 + (-120*x - 60*x^2)*Log[5])),x]

[Out]

3*Defer[Int][(2 + E^x - 60*x*Log[5] - 30*x^2*Log[5])^(-1), x] - 4*(1 + 30*Log[5])*Defer[Int][(-2 - E^x + 60*x*
Log[5] + 30*x^2*Log[5])^(-2), x] + 2*(1 + 30*Log[5])*Defer[Int][x/(-2 - E^x + 60*x*Log[5] + 30*x^2*Log[5])^2,
x] + 60*Log[5]*Defer[Int][x^2/(-2 - E^x + 60*x*Log[5] + 30*x^2*Log[5])^2, x] - 30*Log[5]*Defer[Int][x^3/(-2 -
E^x + 60*x*Log[5] + 30*x^2*Log[5])^2, x] + Defer[Int][x/(-2 - E^x + 60*x*Log[5] + 30*x^2*Log[5]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^x (-3+x)+2 \left (1-60 \log (5)-60 x \log (5)+15 x^2 \log (5)\right )}{\left (2+e^x-60 x \log (5)-30 x^2 \log (5)\right )^2} \, dx\\ &=\int \left (-\frac {2 (-2+x) \left (-1-30 \log (5)+15 x^2 \log (5)\right )}{\left (-2-e^x+60 x \log (5)+30 x^2 \log (5)\right )^2}+\frac {-3+x}{-2-e^x+60 x \log (5)+30 x^2 \log (5)}\right ) \, dx\\ &=-\left (2 \int \frac {(-2+x) \left (-1-30 \log (5)+15 x^2 \log (5)\right )}{\left (-2-e^x+60 x \log (5)+30 x^2 \log (5)\right )^2} \, dx\right )+\int \frac {-3+x}{-2-e^x+60 x \log (5)+30 x^2 \log (5)} \, dx\\ &=-\left (2 \int \left (-\frac {30 x^2 \log (5)}{\left (-2-e^x+60 x \log (5)+30 x^2 \log (5)\right )^2}+\frac {15 x^3 \log (5)}{\left (-2-e^x+60 x \log (5)+30 x^2 \log (5)\right )^2}+\frac {2 (1+30 \log (5))}{\left (-2-e^x+60 x \log (5)+30 x^2 \log (5)\right )^2}-\frac {x (1+30 \log (5))}{\left (-2-e^x+60 x \log (5)+30 x^2 \log (5)\right )^2}\right ) \, dx\right )+\int \left (\frac {3}{2+e^x-60 x \log (5)-30 x^2 \log (5)}+\frac {x}{-2-e^x+60 x \log (5)+30 x^2 \log (5)}\right ) \, dx\\ &=3 \int \frac {1}{2+e^x-60 x \log (5)-30 x^2 \log (5)} \, dx-(30 \log (5)) \int \frac {x^3}{\left (-2-e^x+60 x \log (5)+30 x^2 \log (5)\right )^2} \, dx+(60 \log (5)) \int \frac {x^2}{\left (-2-e^x+60 x \log (5)+30 x^2 \log (5)\right )^2} \, dx+(2 (1+30 \log (5))) \int \frac {x}{\left (-2-e^x+60 x \log (5)+30 x^2 \log (5)\right )^2} \, dx-(4 (1+30 \log (5))) \int \frac {1}{\left (-2-e^x+60 x \log (5)+30 x^2 \log (5)\right )^2} \, dx+\int \frac {x}{-2-e^x+60 x \log (5)+30 x^2 \log (5)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 23, normalized size = 0.88 \begin {gather*} \frac {-2+x}{2+e^x-60 x \log (5)-30 x^2 \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + E^x*(3 - x) + (-120 - 120*x + 30*x^2)*Log[5])/(4 + E^(2*x) + (-240*x - 120*x^2)*Log[5] + (3600*
x^2 + 3600*x^3 + 900*x^4)*Log[5]^2 + E^x*(4 + (-120*x - 60*x^2)*Log[5])),x]

[Out]

(-2 + x)/(2 + E^x - 60*x*Log[5] - 30*x^2*Log[5])

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fricas [A]  time = 0.93, size = 24, normalized size = 0.92 \begin {gather*} -\frac {x - 2}{30 \, {\left (x^{2} + 2 \, x\right )} \log \relax (5) - e^{x} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3-x)*exp(x)+(30*x^2-120*x-120)*log(5)+2)/(exp(x)^2+((-60*x^2-120*x)*log(5)+4)*exp(x)+(900*x^4+3600
*x^3+3600*x^2)*log(5)^2+(-120*x^2-240*x)*log(5)+4),x, algorithm="fricas")

[Out]

-(x - 2)/(30*(x^2 + 2*x)*log(5) - e^x - 2)

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giac [A]  time = 0.22, size = 25, normalized size = 0.96 \begin {gather*} -\frac {x - 2}{30 \, x^{2} \log \relax (5) + 60 \, x \log \relax (5) - e^{x} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3-x)*exp(x)+(30*x^2-120*x-120)*log(5)+2)/(exp(x)^2+((-60*x^2-120*x)*log(5)+4)*exp(x)+(900*x^4+3600
*x^3+3600*x^2)*log(5)^2+(-120*x^2-240*x)*log(5)+4),x, algorithm="giac")

[Out]

-(x - 2)/(30*x^2*log(5) + 60*x*log(5) - e^x - 2)

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maple [A]  time = 0.22, size = 26, normalized size = 1.00

method	result	size
risch	\(-\frac {x -2}{30 x^{2} \ln \relax (5)+60 x \ln \relax (5)-{\mathrm e}^{x}-2}\)	\(26\)
norman	\(\frac {2-x}{30 x^{2} \ln \relax (5)+60 x \ln \relax (5)-{\mathrm e}^{x}-2}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3-x)*exp(x)+(30*x^2-120*x-120)*ln(5)+2)/(exp(x)^2+((-60*x^2-120*x)*ln(5)+4)*exp(x)+(900*x^4+3600*x^3+360
0*x^2)*ln(5)^2+(-120*x^2-240*x)*ln(5)+4),x,method=_RETURNVERBOSE)

[Out]

-(x-2)/(30*x^2*ln(5)+60*x*ln(5)-exp(x)-2)

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maxima [A]  time = 0.47, size = 25, normalized size = 0.96 \begin {gather*} -\frac {x - 2}{30 \, x^{2} \log \relax (5) + 60 \, x \log \relax (5) - e^{x} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3-x)*exp(x)+(30*x^2-120*x-120)*log(5)+2)/(exp(x)^2+((-60*x^2-120*x)*log(5)+4)*exp(x)+(900*x^4+3600
*x^3+3600*x^2)*log(5)^2+(-120*x^2-240*x)*log(5)+4),x, algorithm="maxima")

[Out]

-(x - 2)/(30*x^2*log(5) + 60*x*log(5) - e^x - 2)

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mupad [B]  time = 0.63, size = 22, normalized size = 0.85 \begin {gather*} \frac {x-2}{{\mathrm {e}}^x-60\,x\,\ln \relax (5)-30\,x^2\,\ln \relax (5)+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(x - 3) + log(5)*(120*x - 30*x^2 + 120) - 2)/(exp(2*x) + log(5)^2*(3600*x^2 + 3600*x^3 + 900*x^4)
- log(5)*(240*x + 120*x^2) - exp(x)*(log(5)*(120*x + 60*x^2) - 4) + 4),x)

[Out]

(x - 2)/(exp(x) - 60*x*log(5) - 30*x^2*log(5) + 2)

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sympy [A]  time = 0.16, size = 22, normalized size = 0.85 \begin {gather*} \frac {x - 2}{- 30 x^{2} \log {\relax (5 )} - 60 x \log {\relax (5 )} + e^{x} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3-x)*exp(x)+(30*x**2-120*x-120)*ln(5)+2)/(exp(x)**2+((-60*x**2-120*x)*ln(5)+4)*exp(x)+(900*x**4+36
00*x**3+3600*x**2)*ln(5)**2+(-120*x**2-240*x)*ln(5)+4),x)

[Out]

(x - 2)/(-30*x**2*log(5) - 60*x*log(5) + exp(x) + 2)

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