3.94.45 \(\int \frac {e^{-2 x} (e^{2 x} (-2 x+x^2) \log ^2(2-x)+e^{\frac {9+3 x}{\log (2-x)}} (-18-6 x+(-12+6 x) \log (2-x)+(8-4 x) \log ^2(2-x))+e^{\frac {9+3 x}{2 \log (2-x)}} (e^x (9 x+3 x^2)+e^x (6 x-3 x^2) \log (2-x)+e^x (4-6 x+2 x^2) \log ^2(2-x)))}{(-4+2 x) \log ^2(2-x)} \, dx\)

Optimal. Leaf size=35 \[ \left (-e^{-x+\frac {9 \left (2+\frac {2 x}{3}\right )}{4 \log (2-x)}}+\frac {x}{2}\right )^2 \]

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Rubi [F]  time = 7.35, 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 (e^{2 x} \left (-2 x+x^2\right ) \log ^2(2-x)+e^{\frac {9+3 x}{\log (2-x)}} \left (-18-6 x+(-12+6 x) \log (2-x)+(8-4 x) \log ^2(2-x)\right )+e^{\frac {9+3 x}{2 \log (2-x)}} \left (e^x \left (9 x+3 x^2\right )+e^x \left (6 x-3 x^2\right ) \log (2-x)+e^x \left (4-6 x+2 x^2\right ) \log ^2(2-x)\right )\right )}{(-4+2 x) \log ^2(2-x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*x)*(-2*x + x^2)*Log[2 - x]^2 + E^((9 + 3*x)/Log[2 - x])*(-18 - 6*x + (-12 + 6*x)*Log[2 - x] + (8 - 4
*x)*Log[2 - x]^2) + E^((9 + 3*x)/(2*Log[2 - x]))*(E^x*(9*x + 3*x^2) + E^x*(6*x - 3*x^2)*Log[2 - x] + E^x*(4 -
6*x + 2*x^2)*Log[2 - x]^2))/(E^(2*x)*(-4 + 2*x)*Log[2 - x]^2),x]

[Out]

x^2/4 - 2*Defer[Int][E^(-2*x + (3*(3 + x))/Log[2 - x]), x] - Defer[Int][E^((9 + 3*x - 2*x*Log[2 - x])/(2*Log[2
- x])), x] + Defer[Int][E^((9 + 3*x - 2*x*Log[2 - x])/(2*Log[2 - x]))*x, x] - 3*Defer[Int][E^(-2*x + (3*(3 +
x))/Log[2 - x])/Log[2 - x]^2, x] + (21*Defer[Int][E^((9 + 3*x - 2*x*Log[2 - x])/(2*Log[2 - x]))/Log[2 - x]^2,
x])/2 - (3*Defer[Int][(E^((9 + 3*x - 2*x*Log[2 - x])/(2*Log[2 - x]))*(2 - x))/Log[2 - x]^2, x])/2 - 15*Defer[I
nt][E^(-2*x + (3*(3 + x))/Log[2 - x])/((-2 + x)*Log[2 - x]^2), x] + 15*Defer[Int][E^((9 + 3*x - 2*x*Log[2 - x]
)/(2*Log[2 - x]))/((-2 + x)*Log[2 - x]^2), x] + 3*Defer[Int][E^(-2*x + (3*(3 + x))/Log[2 - x])/Log[2 - x], x]
- 3*Defer[Int][E^((9 + 3*x - 2*x*Log[2 - x])/(2*Log[2 - x]))/Log[2 - x], x] + (3*Defer[Int][(E^((9 + 3*x - 2*x
*Log[2 - x])/(2*Log[2 - x]))*(2 - x))/Log[2 - x], x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {x}{2}+\frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \left (9+3 x+6 \log (2-x)-3 x \log (2-x)-4 \log ^2(2-x)+2 x \log ^2(2-x)\right )}{(2-x) \log ^2(2-x)}+\frac {\exp \left (-2 x+\frac {9+3 x+2 x \log (2-x)}{2 \log (2-x)}\right ) \left (-9 x-3 x^2-6 x \log (2-x)+3 x^2 \log (2-x)-4 \log ^2(2-x)+6 x \log ^2(2-x)-2 x^2 \log ^2(2-x)\right )}{2 (2-x) \log ^2(2-x)}\right ) \, dx\\ &=\frac {x^2}{4}+\frac {1}{2} \int \frac {\exp \left (-2 x+\frac {9+3 x+2 x \log (2-x)}{2 \log (2-x)}\right ) \left (-9 x-3 x^2-6 x \log (2-x)+3 x^2 \log (2-x)-4 \log ^2(2-x)+6 x \log ^2(2-x)-2 x^2 \log ^2(2-x)\right )}{(2-x) \log ^2(2-x)} \, dx+\int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \left (9+3 x+6 \log (2-x)-3 x \log (2-x)-4 \log ^2(2-x)+2 x \log ^2(2-x)\right )}{(2-x) \log ^2(2-x)} \, dx\\ &=\frac {x^2}{4}+\frac {1}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} \left (-3 x (3+x)+3 (-2+x) x \log (2-x)-2 \left (2-3 x+x^2\right ) \log ^2(2-x)\right )}{(2-x) \log ^2(2-x)} \, dx+\int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \left (3 (3+x)-3 (-2+x) \log (2-x)+2 (-2+x) \log ^2(2-x)\right )}{(2-x) \log ^2(2-x)} \, dx\\ &=\frac {x^2}{4}+\frac {1}{2} \int \left (2 e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (-1+x)+\frac {3 e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x (3+x)}{(-2+x) \log ^2(2-x)}-\frac {3 e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x}{\log (2-x)}\right ) \, dx+\int \left (-2 e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}-\frac {3 e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} (3+x)}{(-2+x) \log ^2(2-x)}+\frac {3 e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log (2-x)}\right ) \, dx\\ &=\frac {x^2}{4}+\frac {3}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x (3+x)}{(-2+x) \log ^2(2-x)} \, dx-\frac {3}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x}{\log (2-x)} \, dx-2 \int e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \, dx-3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} (3+x)}{(-2+x) \log ^2(2-x)} \, dx+3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log (2-x)} \, dx+\int e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (-1+x) \, dx\\ &=\frac {x^2}{4}+\frac {3}{2} \int \left (\frac {5 e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log ^2(2-x)}+\frac {10 e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{(-2+x) \log ^2(2-x)}+\frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x}{\log ^2(2-x)}\right ) \, dx-\frac {3}{2} \int \left (\frac {2 e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log (2-x)}-\frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (2-x)}{\log (2-x)}\right ) \, dx-2 \int e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \, dx-3 \int \left (\frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log ^2(2-x)}+\frac {5 e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{(-2+x) \log ^2(2-x)}\right ) \, dx+3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log (2-x)} \, dx+\int \left (-e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}+e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x\right ) \, dx\\ &=\frac {x^2}{4}+\frac {3}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x}{\log ^2(2-x)} \, dx+\frac {3}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (2-x)}{\log (2-x)} \, dx-2 \int e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \, dx-3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log ^2(2-x)} \, dx+3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log (2-x)} \, dx-3 \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log (2-x)} \, dx+\frac {15}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log ^2(2-x)} \, dx-15 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{(-2+x) \log ^2(2-x)} \, dx+15 \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{(-2+x) \log ^2(2-x)} \, dx-\int e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} \, dx+\int e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x \, dx\\ &=\frac {x^2}{4}+\frac {3}{2} \int \left (\frac {2 e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log ^2(2-x)}-\frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (2-x)}{\log ^2(2-x)}\right ) \, dx+\frac {3}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (2-x)}{\log (2-x)} \, dx-2 \int e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \, dx-3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log ^2(2-x)} \, dx+3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log (2-x)} \, dx-3 \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log (2-x)} \, dx+\frac {15}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log ^2(2-x)} \, dx-15 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{(-2+x) \log ^2(2-x)} \, dx+15 \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{(-2+x) \log ^2(2-x)} \, dx-\int e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} \, dx+\int e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x \, dx\\ &=\frac {x^2}{4}-\frac {3}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (2-x)}{\log ^2(2-x)} \, dx+\frac {3}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} (2-x)}{\log (2-x)} \, dx-2 \int e^{-2 x+\frac {3 (3+x)}{\log (2-x)}} \, dx-3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log ^2(2-x)} \, dx+3 \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log ^2(2-x)} \, dx+3 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{\log (2-x)} \, dx-3 \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log (2-x)} \, dx+\frac {15}{2} \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{\log ^2(2-x)} \, dx-15 \int \frac {e^{-2 x+\frac {3 (3+x)}{\log (2-x)}}}{(-2+x) \log ^2(2-x)} \, dx+15 \int \frac {e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}}}{(-2+x) \log ^2(2-x)} \, dx-\int e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} \, dx+\int e^{\frac {9+3 x-2 x \log (2-x)}{2 \log (2-x)}} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 36, normalized size = 1.03 \begin {gather*} \frac {1}{4} e^{-2 x} \left (-2 e^{\frac {3 (3+x)}{2 \log (2-x)}}+e^x x\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(-2*x + x^2)*Log[2 - x]^2 + E^((9 + 3*x)/Log[2 - x])*(-18 - 6*x + (-12 + 6*x)*Log[2 - x] +
(8 - 4*x)*Log[2 - x]^2) + E^((9 + 3*x)/(2*Log[2 - x]))*(E^x*(9*x + 3*x^2) + E^x*(6*x - 3*x^2)*Log[2 - x] + E^x
*(4 - 6*x + 2*x^2)*Log[2 - x]^2))/(E^(2*x)*(-4 + 2*x)*Log[2 - x]^2),x]

[Out]

(-2*E^((3*(3 + x))/(2*Log[2 - x])) + E^x*x)^2/(4*E^(2*x))

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fricas [A]  time = 0.73, size = 50, normalized size = 1.43 \begin {gather*} \frac {1}{4} \, {\left (x^{2} e^{\left (2 \, x\right )} - 4 \, x e^{\left (x + \frac {3 \, {\left (x + 3\right )}}{2 \, \log \left (-x + 2\right )}\right )} + 4 \, e^{\left (\frac {3 \, {\left (x + 3\right )}}{\log \left (-x + 2\right )}\right )}\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+8)*log(2-x)^2+(6*x-12)*log(2-x)-6*x-18)*exp(1/2*(3*x+9)/log(2-x))^2+((2*x^2-6*x+4)*exp(x)*lo
g(2-x)^2+(-3*x^2+6*x)*exp(x)*log(2-x)+(3*x^2+9*x)*exp(x))*exp(1/2*(3*x+9)/log(2-x))+(x^2-2*x)*exp(x)^2*log(2-x
)^2)/(2*x-4)/exp(x)^2/log(2-x)^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.61, size = 59, normalized size = 1.69 \begin {gather*} \frac {1}{4} \, x^{2} - x e^{\left (-\frac {2 \, x \log \left (-x + 2\right ) - 3 \, x - 9}{2 \, \log \left (-x + 2\right )}\right )} + e^{\left (-\frac {2 \, x \log \left (-x + 2\right ) - 3 \, x - 9}{\log \left (-x + 2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+8)*log(2-x)^2+(6*x-12)*log(2-x)-6*x-18)*exp(1/2*(3*x+9)/log(2-x))^2+((2*x^2-6*x+4)*exp(x)*lo
g(2-x)^2+(-3*x^2+6*x)*exp(x)*log(2-x)+(3*x^2+9*x)*exp(x))*exp(1/2*(3*x+9)/log(2-x))+(x^2-2*x)*exp(x)^2*log(2-x
)^2)/(2*x-4)/exp(x)^2/log(2-x)^2,x, algorithm="giac")

[Out]

1/4*x^2 - x*e^(-1/2*(2*x*log(-x + 2) - 3*x - 9)/log(-x + 2)) + e^(-(2*x*log(-x + 2) - 3*x - 9)/log(-x + 2))

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maple [B]  time = 0.56, size = 60, normalized size = 1.71

method	result	size
risch	\(\frac {x^{2}}{4}+{\mathrm e}^{-\frac {2 x \ln \left (2-x \right )-3 x -9}{\ln \left (2-x \right )}}-x \,{\mathrm e}^{-\frac {2 x \ln \left (2-x \right )-3 x -9}{2 \ln \left (2-x \right )}}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x+8)*ln(2-x)^2+(6*x-12)*ln(2-x)-6*x-18)*exp(1/2*(3*x+9)/ln(2-x))^2+((2*x^2-6*x+4)*exp(x)*ln(2-x)^2+(
-3*x^2+6*x)*exp(x)*ln(2-x)+(3*x^2+9*x)*exp(x))*exp(1/2*(3*x+9)/ln(2-x))+(x^2-2*x)*exp(x)^2*ln(2-x)^2)/(2*x-4)/
exp(x)^2/ln(2-x)^2,x,method=_RETURNVERBOSE)

[Out]

1/4*x^2+exp(-(2*x*ln(2-x)-3*x-9)/ln(2-x))-x*exp(-1/2*(2*x*ln(2-x)-3*x-9)/ln(2-x))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+8)*log(2-x)^2+(6*x-12)*log(2-x)-6*x-18)*exp(1/2*(3*x+9)/log(2-x))^2+((2*x^2-6*x+4)*exp(x)*lo
g(2-x)^2+(-3*x^2+6*x)*exp(x)*log(2-x)+(3*x^2+9*x)*exp(x))*exp(1/2*(3*x+9)/log(2-x))+(x^2-2*x)*exp(x)^2*log(2-x
)^2)/(2*x-4)/exp(x)^2/log(2-x)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {{\mathrm {e}}^{-2\,x}\,\left ({\mathrm {e}}^{\frac {2\,\left (\frac {3\,x}{2}+\frac {9}{2}\right )}{\ln \left (2-x\right )}}\,\left (\left (4\,x-8\right )\,{\ln \left (2-x\right )}^2+\left (12-6\,x\right )\,\ln \left (2-x\right )+6\,x+18\right )-{\mathrm {e}}^{\frac {\frac {3\,x}{2}+\frac {9}{2}}{\ln \left (2-x\right )}}\,\left ({\mathrm {e}}^x\,\left (2\,x^2-6\,x+4\right )\,{\ln \left (2-x\right )}^2+{\mathrm {e}}^x\,\left (6\,x-3\,x^2\right )\,\ln \left (2-x\right )+{\mathrm {e}}^x\,\left (3\,x^2+9\,x\right )\right )+{\mathrm {e}}^{2\,x}\,{\ln \left (2-x\right )}^2\,\left (2\,x-x^2\right )\right )}{{\ln \left (2-x\right )}^2\,\left (2\,x-4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2*x)*(exp((2*((3*x)/2 + 9/2))/log(2 - x))*(6*x - log(2 - x)*(6*x - 12) + log(2 - x)^2*(4*x - 8) + 1
8) - exp(((3*x)/2 + 9/2)/log(2 - x))*(exp(x)*(9*x + 3*x^2) + exp(x)*log(2 - x)*(6*x - 3*x^2) + exp(x)*log(2 -
x)^2*(2*x^2 - 6*x + 4)) + exp(2*x)*log(2 - x)^2*(2*x - x^2)))/(log(2 - x)^2*(2*x - 4)),x)

[Out]

-int((exp(-2*x)*(exp((2*((3*x)/2 + 9/2))/log(2 - x))*(6*x - log(2 - x)*(6*x - 12) + log(2 - x)^2*(4*x - 8) + 1
8) - exp(((3*x)/2 + 9/2)/log(2 - x))*(exp(x)*(9*x + 3*x^2) + exp(x)*log(2 - x)*(6*x - 3*x^2) + exp(x)*log(2 -
x)^2*(2*x^2 - 6*x + 4)) + exp(2*x)*log(2 - x)^2*(2*x - x^2)))/(log(2 - x)^2*(2*x - 4)), x)

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sympy [B]  time = 0.84, size = 46, normalized size = 1.31 \begin {gather*} \frac {x^{2}}{4} + \left (- x e^{2 x} e^{\frac {\frac {3 x}{2} + \frac {9}{2}}{\log {\left (2 - x \right )}}} + e^{x} e^{\frac {2 \left (\frac {3 x}{2} + \frac {9}{2}\right )}{\log {\left (2 - x \right )}}}\right ) e^{- 3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+8)*ln(2-x)**2+(6*x-12)*ln(2-x)-6*x-18)*exp(1/2*(3*x+9)/ln(2-x))**2+((2*x**2-6*x+4)*exp(x)*ln
(2-x)**2+(-3*x**2+6*x)*exp(x)*ln(2-x)+(3*x**2+9*x)*exp(x))*exp(1/2*(3*x+9)/ln(2-x))+(x**2-2*x)*exp(x)**2*ln(2-
x)**2)/(2*x-4)/exp(x)**2/ln(2-x)**2,x)

[Out]

x**2/4 + (-x*exp(2*x)*exp((3*x/2 + 9/2)/log(2 - x)) + exp(x)*exp(2*(3*x/2 + 9/2)/log(2 - x)))*exp(-3*x)

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