3.83.100 \(\int \frac {e^4 (14-14 x)-28 x-42 x^2+14 x^4+(56 x+28 x^2-28 x^3+e^4 (-14+7 x)) \log (2-x)+(-28 x+14 x^2) \log ^2(2-x)}{-2-3 x+x^3+(4+2 x-2 x^2) \log (2-x)+(-2+x) \log ^2(2-x)} \, dx\)

Optimal. Leaf size=27 \[ e^4+7 x \left (x-\frac {e^4}{1+x-\log (2-x)}\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 0.66, 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^4 (14-14 x)-28 x-42 x^2+14 x^4+\left (56 x+28 x^2-28 x^3+e^4 (-14+7 x)\right ) \log (2-x)+\left (-28 x+14 x^2\right ) \log ^2(2-x)}{-2-3 x+x^3+\left (4+2 x-2 x^2\right ) \log (2-x)+(-2+x) \log ^2(2-x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^4*(14 - 14*x) - 28*x - 42*x^2 + 14*x^4 + (56*x + 28*x^2 - 28*x^3 + E^4*(-14 + 7*x))*Log[2 - x] + (-28*x
+ 14*x^2)*Log[2 - x]^2)/(-2 - 3*x + x^3 + (4 + 2*x - 2*x^2)*Log[2 - x] + (-2 + x)*Log[2 - x]^2),x]

[Out]

7*x^2 - 7*E^4*Defer[Int][(1 + x - Log[2 - x])^(-2), x] - 14*E^4*Defer[Int][1/((-2 + x)*(1 + x - Log[2 - x])^2)
, x] + 7*E^4*Defer[Int][x/(1 + x - Log[2 - x])^2, x] - 7*E^4*Defer[Int][(1 + x - Log[2 - x])^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {7 \left (2 e^4 (-1+x)-2 (-2+x) x (1+x)^2-\left (e^4 (-2+x)+4 x \left (2+x-x^2\right )\right ) \log (2-x)-2 (-2+x) x \log ^2(2-x)\right )}{(2-x) (1+x-\log (2-x))^2} \, dx\\ &=7 \int \frac {2 e^4 (-1+x)-2 (-2+x) x (1+x)^2-\left (e^4 (-2+x)+4 x \left (2+x-x^2\right )\right ) \log (2-x)-2 (-2+x) x \log ^2(2-x)}{(2-x) (1+x-\log (2-x))^2} \, dx\\ &=7 \int \left (2 x+\frac {e^4 (-3+x) x}{(-2+x) (1+x-\log (2-x))^2}-\frac {e^4}{1+x-\log (2-x)}\right ) \, dx\\ &=7 x^2+\left (7 e^4\right ) \int \frac {(-3+x) x}{(-2+x) (1+x-\log (2-x))^2} \, dx-\left (7 e^4\right ) \int \frac {1}{1+x-\log (2-x)} \, dx\\ &=7 x^2+\left (7 e^4\right ) \int \left (-\frac {1}{(1+x-\log (2-x))^2}-\frac {2}{(-2+x) (1+x-\log (2-x))^2}+\frac {x}{(1+x-\log (2-x))^2}\right ) \, dx-\left (7 e^4\right ) \int \frac {1}{1+x-\log (2-x)} \, dx\\ &=7 x^2-\left (7 e^4\right ) \int \frac {1}{(1+x-\log (2-x))^2} \, dx+\left (7 e^4\right ) \int \frac {x}{(1+x-\log (2-x))^2} \, dx-\left (7 e^4\right ) \int \frac {1}{1+x-\log (2-x)} \, dx-\left (14 e^4\right ) \int \frac {1}{(-2+x) (1+x-\log (2-x))^2} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.25, size = 24, normalized size = 0.89 \begin {gather*} 7 \left (x^2+\frac {e^4 x}{-1-x+\log (2-x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^4*(14 - 14*x) - 28*x - 42*x^2 + 14*x^4 + (56*x + 28*x^2 - 28*x^3 + E^4*(-14 + 7*x))*Log[2 - x] +
(-28*x + 14*x^2)*Log[2 - x]^2)/(-2 - 3*x + x^3 + (4 + 2*x - 2*x^2)*Log[2 - x] + (-2 + x)*Log[2 - x]^2),x]

[Out]

7*(x^2 + (E^4*x)/(-1 - x + Log[2 - x]))

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 38, normalized size = 1.41 \begin {gather*} \frac {7 \, {\left (x^{3} - x^{2} \log \left (-x + 2\right ) + x^{2} - x e^{4}\right )}}{x - \log \left (-x + 2\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((14*x^2-28*x)*log(2-x)^2+((7*x-14)*exp(2)^2-28*x^3+28*x^2+56*x)*log(2-x)+(-14*x+14)*exp(2)^2+14*x^4
-42*x^2-28*x)/((x-2)*log(2-x)^2+(-2*x^2+2*x+4)*log(2-x)+x^3-3*x-2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.22, size = 38, normalized size = 1.41 \begin {gather*} \frac {7 \, {\left (x^{3} - x^{2} \log \left (-x + 2\right ) + x^{2} - x e^{4}\right )}}{x - \log \left (-x + 2\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((14*x^2-28*x)*log(2-x)^2+((7*x-14)*exp(2)^2-28*x^3+28*x^2+56*x)*log(2-x)+(-14*x+14)*exp(2)^2+14*x^4
-42*x^2-28*x)/((x-2)*log(2-x)^2+(-2*x^2+2*x+4)*log(2-x)+x^3-3*x-2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.11, size = 25, normalized size = 0.93

method	result	size
risch	\(7 x^{2}-\frac {7 \,{\mathrm e}^{4} x}{1+x -\ln \left (2-x \right )}\)	\(25\)
norman	\(\frac {-7 \,{\mathrm e}^{4} \ln \left (2-x \right )+7 x^{2}+7 x^{3}-7 x^{2} \ln \left (2-x \right )+7 \,{\mathrm e}^{4}}{1+x -\ln \left (2-x \right )}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((14*x^2-28*x)*ln(2-x)^2+((7*x-14)*exp(2)^2-28*x^3+28*x^2+56*x)*ln(2-x)+(-14*x+14)*exp(2)^2+14*x^4-42*x^2-
28*x)/((x-2)*ln(2-x)^2+(-2*x^2+2*x+4)*ln(2-x)+x^3-3*x-2),x,method=_RETURNVERBOSE)

[Out]

7*x^2-7*exp(4)*x/(1+x-ln(2-x))

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 38, normalized size = 1.41 \begin {gather*} \frac {7 \, {\left (x^{3} - x^{2} \log \left (-x + 2\right ) + x^{2} - x e^{4}\right )}}{x - \log \left (-x + 2\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((14*x^2-28*x)*log(2-x)^2+((7*x-14)*exp(2)^2-28*x^3+28*x^2+56*x)*log(2-x)+(-14*x+14)*exp(2)^2+14*x^4
-42*x^2-28*x)/((x-2)*log(2-x)^2+(-2*x^2+2*x+4)*log(2-x)+x^3-3*x-2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.31, size = 45, normalized size = 1.67 \begin {gather*} \frac {7\,\left ({\mathrm {e}}^4-{\mathrm {e}}^4\,\ln \left (2-x\right )+x^2+x^3-x^2\,\ln \left (2-x\right )\right )}{x-\ln \left (2-x\right )+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(28*x - log(2 - x)*(56*x + 28*x^2 - 28*x^3 + exp(4)*(7*x - 14)) + log(2 - x)^2*(28*x - 14*x^2) + 42*x^2 -
14*x^4 + exp(4)*(14*x - 14))/(log(2 - x)^2*(x - 2) - 3*x + log(2 - x)*(2*x - 2*x^2 + 4) + x^3 - 2),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.21, size = 19, normalized size = 0.70 \begin {gather*} 7 x^{2} + \frac {7 x e^{4}}{- x + \log {\left (2 - x \right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((14*x**2-28*x)*ln(2-x)**2+((7*x-14)*exp(2)**2-28*x**3+28*x**2+56*x)*ln(2-x)+(-14*x+14)*exp(2)**2+14
*x**4-42*x**2-28*x)/((x-2)*ln(2-x)**2+(-2*x**2+2*x+4)*ln(2-x)+x**3-3*x-2),x)

[Out]

7*x**2 + 7*x*exp(4)/(-x + log(2 - x) - 1)

________________________________________________________________________________________