3.27.58 \(\int \frac {e^x (-3+x+x^2)-4 e^{2 x} \log (3)}{18 x^2+12 x^3+2 x^4+e^x (144 x+96 x^2+16 x^3) \log (3)+e^{2 x} (288+192 x+32 x^2) \log ^2(3)} \, dx\)

Optimal. Leaf size=24 \[ \frac {1}{8 (3+x) \left (\frac {e^{-x} x}{4}+\log (3)\right )} \]

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Rubi [F]  time = 1.56, 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^x \left (-3+x+x^2\right )-4 e^{2 x} \log (3)}{18 x^2+12 x^3+2 x^4+e^x \left (144 x+96 x^2+16 x^3\right ) \log (3)+e^{2 x} \left (288+192 x+32 x^2\right ) \log ^2(3)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-3 + x + x^2) - 4*E^(2*x)*Log[3])/(18*x^2 + 12*x^3 + 2*x^4 + E^x*(144*x + 96*x^2 + 16*x^3)*Log[3] +
E^(2*x)*(288 + 192*x + 32*x^2)*Log[3]^2),x]

[Out]

-1/2*1/(Log[81]*(x + E^x*Log[81])) - Defer[Int][(x + E^x*Log[81])^(-2), x]/(2*Log[81]) - 2*Defer[Int][E^x/((3
+ x)*(x + E^x*Log[81])^2), x] - (2*Log[3]*Defer[Int][E^x/((3 + x)^2*(x + E^x*Log[81])), x])/Log[81]

Rubi steps

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

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Mathematica [A]  time = 0.51, size = 46, normalized size = 1.92 \begin {gather*} \frac {e^x \left (x^2 \log (81)+x \log (6561)-\log (531441)\right )}{2 (-1+x) (3+x)^2 \log (81) \left (x+e^x \log (81)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-3 + x + x^2) - 4*E^(2*x)*Log[3])/(18*x^2 + 12*x^3 + 2*x^4 + E^x*(144*x + 96*x^2 + 16*x^3)*Log
[3] + E^(2*x)*(288 + 192*x + 32*x^2)*Log[3]^2),x]

[Out]

(E^x*(x^2*Log[81] + x*Log[6561] - Log[531441]))/(2*(-1 + x)*(3 + x)^2*Log[81]*(x + E^x*Log[81]))

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fricas [A]  time = 1.22, size = 22, normalized size = 0.92 \begin {gather*} \frac {e^{x}}{2 \, {\left (4 \, {\left (x + 3\right )} e^{x} \log \relax (3) + x^{2} + 3 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(3)*exp(x)^2+(x^2+x-3)*exp(x))/((32*x^2+192*x+288)*log(3)^2*exp(x)^2+(16*x^3+96*x^2+144*x)*lo
g(3)*exp(x)+2*x^4+12*x^3+18*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.24, size = 26, normalized size = 1.08 \begin {gather*} \frac {e^{x}}{2 \, {\left (4 \, x e^{x} \log \relax (3) + x^{2} + 12 \, e^{x} \log \relax (3) + 3 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(3)*exp(x)^2+(x^2+x-3)*exp(x))/((32*x^2+192*x+288)*log(3)^2*exp(x)^2+(16*x^3+96*x^2+144*x)*lo
g(3)*exp(x)+2*x^4+12*x^3+18*x^2),x, algorithm="giac")

[Out]

1/2*e^x/(4*x*e^x*log(3) + x^2 + 12*e^x*log(3) + 3*x)

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maple [A]  time = 0.18, size = 20, normalized size = 0.83

method	result	size
norman	\(\frac {{\mathrm e}^{x}}{2 \left (3+x \right ) \left (4 \ln \relax (3) {\mathrm e}^{x}+x \right )}\)	\(20\)
risch	\(\frac {1}{8 \ln \relax (3) \left (3+x \right )}-\frac {x}{8 \ln \relax (3) \left (3+x \right ) \left (4 \ln \relax (3) {\mathrm e}^{x}+x \right )}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*ln(3)*exp(x)^2+(x^2+x-3)*exp(x))/((32*x^2+192*x+288)*ln(3)^2*exp(x)^2+(16*x^3+96*x^2+144*x)*ln(3)*exp(
x)+2*x^4+12*x^3+18*x^2),x,method=_RETURNVERBOSE)

[Out]

1/2*exp(x)/(3+x)/(4*ln(3)*exp(x)+x)

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maxima [A]  time = 0.81, size = 26, normalized size = 1.08 \begin {gather*} \frac {e^{x}}{2 \, {\left (x^{2} + 4 \, {\left (x \log \relax (3) + 3 \, \log \relax (3)\right )} e^{x} + 3 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(3)*exp(x)^2+(x^2+x-3)*exp(x))/((32*x^2+192*x+288)*log(3)^2*exp(x)^2+(16*x^3+96*x^2+144*x)*lo
g(3)*exp(x)+2*x^4+12*x^3+18*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.69, size = 19, normalized size = 0.79 \begin {gather*} \frac {{\mathrm {e}}^x}{2\,\left (x+4\,{\mathrm {e}}^x\,\ln \relax (3)\right )\,\left (x+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(2*x)*log(3) - exp(x)*(x + x^2 - 3))/(18*x^2 + 12*x^3 + 2*x^4 + exp(x)*log(3)*(144*x + 96*x^2 + 16*
x^3) + exp(2*x)*log(3)^2*(192*x + 32*x^2 + 288)),x)

[Out]

exp(x)/(2*(x + 4*exp(x)*log(3))*(x + 3))

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sympy [B]  time = 0.21, size = 48, normalized size = 2.00 \begin {gather*} - \frac {x}{8 x^{2} \log {\relax (3 )} + 24 x \log {\relax (3 )} + \left (32 x \log {\relax (3 )}^{2} + 96 \log {\relax (3 )}^{2}\right ) e^{x}} + \frac {1}{8 x \log {\relax (3 )} + 24 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*ln(3)*exp(x)**2+(x**2+x-3)*exp(x))/((32*x**2+192*x+288)*ln(3)**2*exp(x)**2+(16*x**3+96*x**2+144*
x)*ln(3)*exp(x)+2*x**4+12*x**3+18*x**2),x)

[Out]

-x/(8*x**2*log(3) + 24*x*log(3) + (32*x*log(3)**2 + 96*log(3)**2)*exp(x)) + 1/(8*x*log(3) + 24*log(3))

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