3.9.76 \(\int \frac {490 x^2+35630 x^3+629520 x^4-665640 x^5+e^{2 x} (-56-4072 x-262148 x^2+266256 x^3)+(140 x^3+5020 x^4-5160 x^5+e^{2 x} (-16 x-2048 x^2+2064 x^3)) \log (1-x)+(10 x^4-10 x^5+e^{2 x} (-4 x^2+4 x^3)) \log ^2(1-x)}{e^{2 x} (-49 x^2-3563 x^3-62952 x^4+66564 x^5)+e^{2 x} (-14 x^3-502 x^4+516 x^5) \log (1-x)+e^{2 x} (-x^4+x^5) \log ^2(1-x)} \, dx\)

Optimal. Leaf size=31 \[ 5 e^{-2 x}-\frac {4}{x+\frac {5}{258+\frac {2}{x}+\log (1-x)}} \]

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Rubi [F]  time = 6.78, 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 {490 x^2+35630 x^3+629520 x^4-665640 x^5+e^{2 x} \left (-56-4072 x-262148 x^2+266256 x^3\right )+\left (140 x^3+5020 x^4-5160 x^5+e^{2 x} \left (-16 x-2048 x^2+2064 x^3\right )\right ) \log (1-x)+\left (10 x^4-10 x^5+e^{2 x} \left (-4 x^2+4 x^3\right )\right ) \log ^2(1-x)}{e^{2 x} \left (-49 x^2-3563 x^3-62952 x^4+66564 x^5\right )+e^{2 x} \left (-14 x^3-502 x^4+516 x^5\right ) \log (1-x)+e^{2 x} \left (-x^4+x^5\right ) \log ^2(1-x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(490*x^2 + 35630*x^3 + 629520*x^4 - 665640*x^5 + E^(2*x)*(-56 - 4072*x - 262148*x^2 + 266256*x^3) + (140*x
^3 + 5020*x^4 - 5160*x^5 + E^(2*x)*(-16*x - 2048*x^2 + 2064*x^3))*Log[1 - x] + (10*x^4 - 10*x^5 + E^(2*x)*(-4*
x^2 + 4*x^3))*Log[1 - x]^2)/(E^(2*x)*(-49*x^2 - 3563*x^3 - 62952*x^4 + 66564*x^5) + E^(2*x)*(-14*x^3 - 502*x^4
+ 516*x^5)*Log[1 - x] + E^(2*x)*(-x^4 + x^5)*Log[1 - x]^2),x]

[Out]

5/E^(2*x) - 4/x - 20*Defer[Int][1/((-1 + x)*(7 + 258*x + x*Log[1 - x])^2), x] + 140*Defer[Int][1/(x^2*(7 + 258
*x + x*Log[1 - x])^2), x] - 40*Defer[Int][1/(x^2*(7 + 258*x + x*Log[1 - x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{-2 x} \left (5 (-1+x) x^2 (7+258 x)^2-2 e^{2 x} \left (-14-1018 x-65537 x^2+66564 x^3\right )-2 (-1+x) x \left (-5 x^2 (7+258 x)+e^{2 x} (4+516 x)\right ) \log (1-x)-(-1+x) x^2 \left (2 e^{2 x}-5 x^2\right ) \log ^2(1-x)\right )}{(1-x) x^2 (7+258 x+x \log (1-x))^2} \, dx\\ &=2 \int \frac {e^{-2 x} \left (5 (-1+x) x^2 (7+258 x)^2-2 e^{2 x} \left (-14-1018 x-65537 x^2+66564 x^3\right )-2 (-1+x) x \left (-5 x^2 (7+258 x)+e^{2 x} (4+516 x)\right ) \log (1-x)-(-1+x) x^2 \left (2 e^{2 x}-5 x^2\right ) \log ^2(1-x)\right )}{(1-x) x^2 (7+258 x+x \log (1-x))^2} \, dx\\ &=2 \int \left (-5 e^{-2 x}+\frac {2 \left (-14-1018 x-65537 x^2+66564 x^3-4 x \log (1-x)-512 x^2 \log (1-x)+516 x^3 \log (1-x)-x^2 \log ^2(1-x)+x^3 \log ^2(1-x)\right )}{(-1+x) x^2 (7+258 x+x \log (1-x))^2}\right ) \, dx\\ &=4 \int \frac {-14-1018 x-65537 x^2+66564 x^3-4 x \log (1-x)-512 x^2 \log (1-x)+516 x^3 \log (1-x)-x^2 \log ^2(1-x)+x^3 \log ^2(1-x)}{(-1+x) x^2 (7+258 x+x \log (1-x))^2} \, dx-10 \int e^{-2 x} \, dx\\ &=5 e^{-2 x}+4 \int \left (\frac {1}{x^2}-\frac {5 \left (7-7 x+x^2\right )}{(-1+x) x^2 (7+258 x+x \log (1-x))^2}-\frac {10}{x^2 (7+258 x+x \log (1-x))}\right ) \, dx\\ &=5 e^{-2 x}-\frac {4}{x}-20 \int \frac {7-7 x+x^2}{(-1+x) x^2 (7+258 x+x \log (1-x))^2} \, dx-40 \int \frac {1}{x^2 (7+258 x+x \log (1-x))} \, dx\\ &=5 e^{-2 x}-\frac {4}{x}-20 \int \left (\frac {1}{(-1+x) (7+258 x+x \log (1-x))^2}-\frac {7}{x^2 (7+258 x+x \log (1-x))^2}\right ) \, dx-40 \int \frac {1}{x^2 (7+258 x+x \log (1-x))} \, dx\\ &=5 e^{-2 x}-\frac {4}{x}-20 \int \frac {1}{(-1+x) (7+258 x+x \log (1-x))^2} \, dx-40 \int \frac {1}{x^2 (7+258 x+x \log (1-x))} \, dx+140 \int \frac {1}{x^2 (7+258 x+x \log (1-x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 37, normalized size = 1.19 \begin {gather*} 2 \left (\frac {5 e^{-2 x}}{2}-\frac {2}{x}+\frac {10}{x (7+258 x+x \log (1-x))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(490*x^2 + 35630*x^3 + 629520*x^4 - 665640*x^5 + E^(2*x)*(-56 - 4072*x - 262148*x^2 + 266256*x^3) +
(140*x^3 + 5020*x^4 - 5160*x^5 + E^(2*x)*(-16*x - 2048*x^2 + 2064*x^3))*Log[1 - x] + (10*x^4 - 10*x^5 + E^(2*x
)*(-4*x^2 + 4*x^3))*Log[1 - x]^2)/(E^(2*x)*(-49*x^2 - 3563*x^3 - 62952*x^4 + 66564*x^5) + E^(2*x)*(-14*x^3 - 5
02*x^4 + 516*x^5)*Log[1 - x] + E^(2*x)*(-x^4 + x^5)*Log[1 - x]^2),x]

[Out]

2*(5/(2*E^(2*x)) - 2/x + 10/(x*(7 + 258*x + x*Log[1 - x])))

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fricas [B]  time = 0.85, size = 72, normalized size = 2.32 \begin {gather*} \frac {1290 \, x^{2} - 8 \, {\left (129 \, x + 1\right )} e^{\left (2 \, x\right )} + {\left (5 \, x^{2} - 4 \, x e^{\left (2 \, x\right )}\right )} \log \left (-x + 1\right ) + 35 \, x}{x^{2} e^{\left (2 \, x\right )} \log \left (-x + 1\right ) + {\left (258 \, x^{2} + 7 \, x\right )} e^{\left (2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^3-4*x^2)*exp(x)^2-10*x^5+10*x^4)*log(-x+1)^2+((2064*x^3-2048*x^2-16*x)*exp(x)^2-5160*x^5+5020
*x^4+140*x^3)*log(-x+1)+(266256*x^3-262148*x^2-4072*x-56)*exp(x)^2-665640*x^5+629520*x^4+35630*x^3+490*x^2)/((
x^5-x^4)*exp(x)^2*log(-x+1)^2+(516*x^5-502*x^4-14*x^3)*exp(x)^2*log(-x+1)+(66564*x^5-62952*x^4-3563*x^3-49*x^2
)*exp(x)^2),x, algorithm="fricas")

[Out]

(1290*x^2 - 8*(129*x + 1)*e^(2*x) + (5*x^2 - 4*x*e^(2*x))*log(-x + 1) + 35*x)/(x^2*e^(2*x)*log(-x + 1) + (258*
x^2 + 7*x)*e^(2*x))

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giac [B]  time = 0.66, size = 67, normalized size = 2.16 \begin {gather*} \frac {5 \, x^{2} e^{\left (-2 \, x\right )} \log \left (-x + 1\right ) + 1290 \, x^{2} e^{\left (-2 \, x\right )} + 35 \, x e^{\left (-2 \, x\right )} - 4 \, x \log \left (-x + 1\right ) - 1032 \, x - 8}{x^{2} \log \left (-x + 1\right ) + 258 \, x^{2} + 7 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^3-4*x^2)*exp(x)^2-10*x^5+10*x^4)*log(-x+1)^2+((2064*x^3-2048*x^2-16*x)*exp(x)^2-5160*x^5+5020
*x^4+140*x^3)*log(-x+1)+(266256*x^3-262148*x^2-4072*x-56)*exp(x)^2-665640*x^5+629520*x^4+35630*x^3+490*x^2)/((
x^5-x^4)*exp(x)^2*log(-x+1)^2+(516*x^5-502*x^4-14*x^3)*exp(x)^2*log(-x+1)+(66564*x^5-62952*x^4-3563*x^3-49*x^2
)*exp(x)^2),x, algorithm="giac")

[Out]

(5*x^2*e^(-2*x)*log(-x + 1) + 1290*x^2*e^(-2*x) + 35*x*e^(-2*x) - 4*x*log(-x + 1) - 1032*x - 8)/(x^2*log(-x +
1) + 258*x^2 + 7*x)

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maple [A]  time = 0.11, size = 40, normalized size = 1.29

method	result	size
risch	\(\frac {\left (-4 \,{\mathrm e}^{2 x}+5 x \right ) {\mathrm e}^{-2 x}}{x}+\frac {20}{x \left (x \ln \left (1-x \right )+258 x +7\right )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x^3-4*x^2)*exp(x)^2-10*x^5+10*x^4)*ln(1-x)^2+((2064*x^3-2048*x^2-16*x)*exp(x)^2-5160*x^5+5020*x^4+140
*x^3)*ln(1-x)+(266256*x^3-262148*x^2-4072*x-56)*exp(x)^2-665640*x^5+629520*x^4+35630*x^3+490*x^2)/((x^5-x^4)*e
xp(x)^2*ln(1-x)^2+(516*x^5-502*x^4-14*x^3)*exp(x)^2*ln(1-x)+(66564*x^5-62952*x^4-3563*x^3-49*x^2)*exp(x)^2),x,
method=_RETURNVERBOSE)

[Out]

(-4*exp(2*x)+5*x)/x*exp(-2*x)+20/x/(x*ln(1-x)+258*x+7)

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maxima [B]  time = 0.77, size = 72, normalized size = 2.32 \begin {gather*} \frac {1290 \, x^{2} - 8 \, {\left (129 \, x + 1\right )} e^{\left (2 \, x\right )} + {\left (5 \, x^{2} - 4 \, x e^{\left (2 \, x\right )}\right )} \log \left (-x + 1\right ) + 35 \, x}{x^{2} e^{\left (2 \, x\right )} \log \left (-x + 1\right ) + {\left (258 \, x^{2} + 7 \, x\right )} e^{\left (2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^3-4*x^2)*exp(x)^2-10*x^5+10*x^4)*log(-x+1)^2+((2064*x^3-2048*x^2-16*x)*exp(x)^2-5160*x^5+5020
*x^4+140*x^3)*log(-x+1)+(266256*x^3-262148*x^2-4072*x-56)*exp(x)^2-665640*x^5+629520*x^4+35630*x^3+490*x^2)/((
x^5-x^4)*exp(x)^2*log(-x+1)^2+(516*x^5-502*x^4-14*x^3)*exp(x)^2*log(-x+1)+(66564*x^5-62952*x^4-3563*x^3-49*x^2
)*exp(x)^2),x, algorithm="maxima")

[Out]

(1290*x^2 - 8*(129*x + 1)*e^(2*x) + (5*x^2 - 4*x*e^(2*x))*log(-x + 1) + 35*x)/(x^2*e^(2*x)*log(-x + 1) + (258*
x^2 + 7*x)*e^(2*x))

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mupad [B]  time = 0.84, size = 32, normalized size = 1.03 \begin {gather*} 5\,{\mathrm {e}}^{-2\,x}+\frac {20}{x\,\left (258\,x+x\,\ln \left (1-x\right )+7\right )}-\frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*(4072*x + 262148*x^2 - 266256*x^3 + 56) + log(1 - x)*(exp(2*x)*(16*x + 2048*x^2 - 2064*x^3) - 14
0*x^3 - 5020*x^4 + 5160*x^5) + log(1 - x)^2*(exp(2*x)*(4*x^2 - 4*x^3) - 10*x^4 + 10*x^5) - 490*x^2 - 35630*x^3
- 629520*x^4 + 665640*x^5)/(exp(2*x)*(49*x^2 + 3563*x^3 + 62952*x^4 - 66564*x^5) + exp(2*x)*log(1 - x)^2*(x^4
- x^5) + exp(2*x)*log(1 - x)*(14*x^3 + 502*x^4 - 516*x^5)),x)

[Out]

5*exp(-2*x) + 20/(x*(258*x + x*log(1 - x) + 7)) - 4/x

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sympy [A]  time = 0.55, size = 27, normalized size = 0.87 \begin {gather*} 5 e^{- 2 x} + \frac {20}{x^{2} \log {\left (1 - x \right )} + 258 x^{2} + 7 x} - \frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x**3-4*x**2)*exp(x)**2-10*x**5+10*x**4)*ln(-x+1)**2+((2064*x**3-2048*x**2-16*x)*exp(x)**2-5160*
x**5+5020*x**4+140*x**3)*ln(-x+1)+(266256*x**3-262148*x**2-4072*x-56)*exp(x)**2-665640*x**5+629520*x**4+35630*
x**3+490*x**2)/((x**5-x**4)*exp(x)**2*ln(-x+1)**2+(516*x**5-502*x**4-14*x**3)*exp(x)**2*ln(-x+1)+(66564*x**5-6
2952*x**4-3563*x**3-49*x**2)*exp(x)**2),x)

[Out]

5*exp(-2*x) + 20/(x**2*log(1 - x) + 258*x**2 + 7*x) - 4/x

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