3.90.29 \(\int \frac {-16 x^3+(32 x^2-32 x^3) \log (e (-1+x))+e^{6+2 x+2 \log ^2(x)} (-16 x+\log (e (-1+x)) (32 x-32 x^2+(64-64 x) \log (x)))+e^{3+x+\log ^2(x)} (-32 x^2+\log (e (-1+x)) (32 x-32 x^3+(64 x-64 x^2) \log (x)))}{-x+x^2} \, dx\)

Optimal. Leaf size=23 \[ 5-16 \left (e^{3+x+\log ^2(x)}+x\right )^2 \log (e (-1+x)) \]

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Rubi [F]  time = 7.97, 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 {-16 x^3+\left (32 x^2-32 x^3\right ) \log (e (-1+x))+e^{6+2 x+2 \log ^2(x)} \left (-16 x+\log (e (-1+x)) \left (32 x-32 x^2+(64-64 x) \log (x)\right )\right )+e^{3+x+\log ^2(x)} \left (-32 x^2+\log (e (-1+x)) \left (32 x-32 x^3+\left (64 x-64 x^2\right ) \log (x)\right )\right )}{-x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-16*x^3 + (32*x^2 - 32*x^3)*Log[E*(-1 + x)] + E^(6 + 2*x + 2*Log[x]^2)*(-16*x + Log[E*(-1 + x)]*(32*x - 3
2*x^2 + (64 - 64*x)*Log[x])) + E^(3 + x + Log[x]^2)*(-32*x^2 + Log[E*(-1 + x)]*(32*x - 32*x^3 + (64*x - 64*x^2
)*Log[x])))/(-x + x^2),x]

[Out]

-16*x^2 - 16*x^2*Log[-1 + x] - 64*Defer[Int][E^(3 + x + Log[x]^2), x] - 32*Defer[Int][E^(2*(3 + x + Log[x]^2))
, x] + 16*Defer[Int][E^(2*(3 + x + Log[x]^2))/(1 - x), x] - 32*Defer[Int][E^(3 + x + Log[x]^2)/(-1 + x), x] -
32*Defer[Int][E^(3 + x + Log[x]^2)*x, x] - 32*Defer[Int][E^(3 + x + Log[x]^2)*Log[-1 + x], x] - 32*Defer[Int][
E^(2*(3 + x + Log[x]^2))*Log[-1 + x], x] - 32*Defer[Int][E^(3 + x + Log[x]^2)*x*Log[-1 + x], x] - 64*Defer[Int
][E^(3 + x + Log[x]^2)*Log[x], x] - 64*Defer[Int][(E^(2*(3 + x + Log[x]^2))*Log[x])/x, x] - 64*Defer[Int][E^(3
+ x + Log[x]^2)*Log[-1 + x]*Log[x], x] - 64*Defer[Int][(E^(2*(3 + x + Log[x]^2))*Log[-1 + x]*Log[x])/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-16 x^3+\left (32 x^2-32 x^3\right ) \log (e (-1+x))+e^{6+2 x+2 \log ^2(x)} \left (-16 x+\log (e (-1+x)) \left (32 x-32 x^2+(64-64 x) \log (x)\right )\right )+e^{3+x+\log ^2(x)} \left (-32 x^2+\log (e (-1+x)) \left (32 x-32 x^3+\left (64 x-64 x^2\right ) \log (x)\right )\right )}{(-1+x) x} \, dx\\ &=\int \frac {16 \left (e^{3+x+\log ^2(x)}+x\right ) \left (x \left (e^{3+x+\log ^2(x)}+x\right )+2 (-1+x) (1+\log (-1+x)) \left (x+e^{3+x+\log ^2(x)} x+2 e^{3+x+\log ^2(x)} \log (x)\right )\right )}{(1-x) x} \, dx\\ &=16 \int \frac {\left (e^{3+x+\log ^2(x)}+x\right ) \left (x \left (e^{3+x+\log ^2(x)}+x\right )+2 (-1+x) (1+\log (-1+x)) \left (x+e^{3+x+\log ^2(x)} x+2 e^{3+x+\log ^2(x)} \log (x)\right )\right )}{(1-x) x} \, dx\\ &=16 \int \left (-\frac {x (-2+3 x-2 \log (-1+x)+2 x \log (-1+x))}{-1+x}-\frac {2 e^{3+x+\log ^2(x)} \left (-1+x+x^2-\log (-1+x)+x^2 \log (-1+x)-2 \log (x)+2 x \log (x)-2 \log (-1+x) \log (x)+2 x \log (-1+x) \log (x)\right )}{-1+x}+\frac {e^{2 \left (3+x+\log ^2(x)\right )} \left (-x+2 x^2-2 x \log (-1+x)+2 x^2 \log (-1+x)-4 \log (x)+4 x \log (x)-4 \log (-1+x) \log (x)+4 x \log (-1+x) \log (x)\right )}{(1-x) x}\right ) \, dx\\ &=-\left (16 \int \frac {x (-2+3 x-2 \log (-1+x)+2 x \log (-1+x))}{-1+x} \, dx\right )+16 \int \frac {e^{2 \left (3+x+\log ^2(x)\right )} \left (-x+2 x^2-2 x \log (-1+x)+2 x^2 \log (-1+x)-4 \log (x)+4 x \log (x)-4 \log (-1+x) \log (x)+4 x \log (-1+x) \log (x)\right )}{(1-x) x} \, dx-32 \int \frac {e^{3+x+\log ^2(x)} \left (-1+x+x^2-\log (-1+x)+x^2 \log (-1+x)-2 \log (x)+2 x \log (x)-2 \log (-1+x) \log (x)+2 x \log (-1+x) \log (x)\right )}{-1+x} \, dx\\ &=-\left (16 \int \left (\frac {x (-2+3 x)}{-1+x}+2 x \log (-1+x)\right ) \, dx\right )+16 \int \frac {e^{2 \left (3+x+\log ^2(x)\right )} (x (-1+2 x)+4 (-1+x) \log (x)+2 (-1+x) \log (-1+x) (x+2 \log (x)))}{(1-x) x} \, dx-32 \int \left (\frac {e^{3+x+\log ^2(x)} \left (-1+x+x^2-\log (-1+x)+x^2 \log (-1+x)\right )}{-1+x}+2 e^{3+x+\log ^2(x)} (1+\log (-1+x)) \log (x)\right ) \, dx\\ &=-\left (16 \int \frac {x (-2+3 x)}{-1+x} \, dx\right )+16 \int \left (\frac {e^{2 \left (3+x+\log ^2(x)\right )} (1-2 x+2 \log (-1+x)-2 x \log (-1+x))}{-1+x}-\frac {4 e^{2 \left (3+x+\log ^2(x)\right )} (1+\log (-1+x)) \log (x)}{x}\right ) \, dx-32 \int x \log (-1+x) \, dx-32 \int \frac {e^{3+x+\log ^2(x)} \left (-1+x+x^2-\log (-1+x)+x^2 \log (-1+x)\right )}{-1+x} \, dx-64 \int e^{3+x+\log ^2(x)} (1+\log (-1+x)) \log (x) \, dx\\ &=-16 x^2 \log (-1+x)+16 \int \frac {x^2}{-1+x} \, dx-16 \int \left (1+\frac {1}{-1+x}+3 x\right ) \, dx+16 \int \frac {e^{2 \left (3+x+\log ^2(x)\right )} (1-2 x+2 \log (-1+x)-2 x \log (-1+x))}{-1+x} \, dx-32 \int \left (\frac {e^{3+x+\log ^2(x)} \left (-1+x+x^2\right )}{-1+x}+e^{3+x+\log ^2(x)} (1+x) \log (-1+x)\right ) \, dx-64 \int \frac {e^{2 \left (3+x+\log ^2(x)\right )} (1+\log (-1+x)) \log (x)}{x} \, dx-64 \int \left (e^{3+x+\log ^2(x)} \log (x)+e^{3+x+\log ^2(x)} \log (-1+x) \log (x)\right ) \, dx\\ &=-16 x-24 x^2-16 \log (1-x)-16 x^2 \log (-1+x)+16 \int \left (1+\frac {1}{-1+x}+x\right ) \, dx+16 \int \left (\frac {e^{2 \left (3+x+\log ^2(x)\right )} (1-2 x)}{-1+x}-2 e^{2 \left (3+x+\log ^2(x)\right )} \log (-1+x)\right ) \, dx-32 \int \frac {e^{3+x+\log ^2(x)} \left (-1+x+x^2\right )}{-1+x} \, dx-32 \int e^{3+x+\log ^2(x)} (1+x) \log (-1+x) \, dx-64 \int e^{3+x+\log ^2(x)} \log (x) \, dx-64 \int e^{3+x+\log ^2(x)} \log (-1+x) \log (x) \, dx-64 \int \left (\frac {e^{2 \left (3+x+\log ^2(x)\right )} \log (x)}{x}+\frac {e^{2 \left (3+x+\log ^2(x)\right )} \log (-1+x) \log (x)}{x}\right ) \, dx\\ &=-16 x^2-16 x^2 \log (-1+x)+16 \int \frac {e^{2 \left (3+x+\log ^2(x)\right )} (1-2 x)}{-1+x} \, dx-32 \int \left (2 e^{3+x+\log ^2(x)}+\frac {e^{3+x+\log ^2(x)}}{-1+x}+e^{3+x+\log ^2(x)} x\right ) \, dx-32 \int e^{2 \left (3+x+\log ^2(x)\right )} \log (-1+x) \, dx-32 \int \left (e^{3+x+\log ^2(x)} \log (-1+x)+e^{3+x+\log ^2(x)} x \log (-1+x)\right ) \, dx-64 \int e^{3+x+\log ^2(x)} \log (x) \, dx-64 \int \frac {e^{2 \left (3+x+\log ^2(x)\right )} \log (x)}{x} \, dx-64 \int e^{3+x+\log ^2(x)} \log (-1+x) \log (x) \, dx-64 \int \frac {e^{2 \left (3+x+\log ^2(x)\right )} \log (-1+x) \log (x)}{x} \, dx\\ &=-16 x^2-16 x^2 \log (-1+x)+16 \int \left (-2 e^{2 \left (3+x+\log ^2(x)\right )}+\frac {e^{2 \left (3+x+\log ^2(x)\right )}}{1-x}\right ) \, dx-32 \int \frac {e^{3+x+\log ^2(x)}}{-1+x} \, dx-32 \int e^{3+x+\log ^2(x)} x \, dx-32 \int e^{3+x+\log ^2(x)} \log (-1+x) \, dx-32 \int e^{2 \left (3+x+\log ^2(x)\right )} \log (-1+x) \, dx-32 \int e^{3+x+\log ^2(x)} x \log (-1+x) \, dx-64 \int e^{3+x+\log ^2(x)} \, dx-64 \int e^{3+x+\log ^2(x)} \log (x) \, dx-64 \int \frac {e^{2 \left (3+x+\log ^2(x)\right )} \log (x)}{x} \, dx-64 \int e^{3+x+\log ^2(x)} \log (-1+x) \log (x) \, dx-64 \int \frac {e^{2 \left (3+x+\log ^2(x)\right )} \log (-1+x) \log (x)}{x} \, dx\\ &=-16 x^2-16 x^2 \log (-1+x)+16 \int \frac {e^{2 \left (3+x+\log ^2(x)\right )}}{1-x} \, dx-32 \int e^{2 \left (3+x+\log ^2(x)\right )} \, dx-32 \int \frac {e^{3+x+\log ^2(x)}}{-1+x} \, dx-32 \int e^{3+x+\log ^2(x)} x \, dx-32 \int e^{3+x+\log ^2(x)} \log (-1+x) \, dx-32 \int e^{2 \left (3+x+\log ^2(x)\right )} \log (-1+x) \, dx-32 \int e^{3+x+\log ^2(x)} x \log (-1+x) \, dx-64 \int e^{3+x+\log ^2(x)} \, dx-64 \int e^{3+x+\log ^2(x)} \log (x) \, dx-64 \int \frac {e^{2 \left (3+x+\log ^2(x)\right )} \log (x)}{x} \, dx-64 \int e^{3+x+\log ^2(x)} \log (-1+x) \log (x) \, dx-64 \int \frac {e^{2 \left (3+x+\log ^2(x)\right )} \log (-1+x) \log (x)}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 2.20, size = 21, normalized size = 0.91 \begin {gather*} -16 \left (e^{3+x+\log ^2(x)}+x\right )^2 (1+\log (-1+x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*x^3 + (32*x^2 - 32*x^3)*Log[E*(-1 + x)] + E^(6 + 2*x + 2*Log[x]^2)*(-16*x + Log[E*(-1 + x)]*(32
*x - 32*x^2 + (64 - 64*x)*Log[x])) + E^(3 + x + Log[x]^2)*(-32*x^2 + Log[E*(-1 + x)]*(32*x - 32*x^3 + (64*x -
64*x^2)*Log[x])))/(-x + x^2),x]

[Out]

-16*(E^(3 + x + Log[x]^2) + x)^2*(1 + Log[-1 + x])

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fricas [B]  time = 0.54, size = 52, normalized size = 2.26 \begin {gather*} -16 \, x^{2} \log \left ({\left (x - 1\right )} e\right ) - 32 \, x e^{\left (\log \relax (x)^{2} + x + 3\right )} \log \left ({\left (x - 1\right )} e\right ) - 16 \, e^{\left (2 \, \log \relax (x)^{2} + 2 \, x + 6\right )} \log \left ({\left (x - 1\right )} e\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-64*x+64)*log(x)-32*x^2+32*x)*log((x-1)*exp(1))-16*x)*exp(log(x)^2+3+x)^2+(((-64*x^2+64*x)*log(x
)-32*x^3+32*x)*log((x-1)*exp(1))-32*x^2)*exp(log(x)^2+3+x)+(-32*x^3+32*x^2)*log((x-1)*exp(1))-16*x^3)/(x^2-x),
x, algorithm="fricas")

[Out]

-16*x^2*log((x - 1)*e) - 32*x*e^(log(x)^2 + x + 3)*log((x - 1)*e) - 16*e^(2*log(x)^2 + 2*x + 6)*log((x - 1)*e)

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giac [B]  time = 0.28, size = 73, normalized size = 3.17 \begin {gather*} -16 \, x^{2} \log \left (x - 1\right ) - 32 \, x e^{\left (\log \relax (x)^{2} + x + 3\right )} \log \left (x - 1\right ) - 16 \, x^{2} - 32 \, x e^{\left (\log \relax (x)^{2} + x + 3\right )} - 16 \, e^{\left (2 \, \log \relax (x)^{2} + 2 \, x + 6\right )} \log \left (x - 1\right ) - 16 \, e^{\left (2 \, \log \relax (x)^{2} + 2 \, x + 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-64*x+64)*log(x)-32*x^2+32*x)*log((x-1)*exp(1))-16*x)*exp(log(x)^2+3+x)^2+(((-64*x^2+64*x)*log(x
)-32*x^3+32*x)*log((x-1)*exp(1))-32*x^2)*exp(log(x)^2+3+x)+(-32*x^3+32*x^2)*log((x-1)*exp(1))-16*x^3)/(x^2-x),
x, algorithm="giac")

[Out]

-16*x^2*log(x - 1) - 32*x*e^(log(x)^2 + x + 3)*log(x - 1) - 16*x^2 - 32*x*e^(log(x)^2 + x + 3) - 16*e^(2*log(x
)^2 + 2*x + 6)*log(x - 1) - 16*e^(2*log(x)^2 + 2*x + 6)

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maple [A]  time = 0.44, size = 53, normalized size = 2.30

method	result	size
risch	\(-16 x^{2} \ln \left (\left (x -1\right ) {\mathrm e}\right )-32 \ln \left (\left (x -1\right ) {\mathrm e}\right ) {\mathrm e}^{\ln \relax (x )^{2}+3+x} x -16 \ln \left (\left (x -1\right ) {\mathrm e}\right ) {\mathrm e}^{2 \ln \relax (x )^{2}+6+2 x}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-64*x+64)*ln(x)-32*x^2+32*x)*ln((x-1)*exp(1))-16*x)*exp(ln(x)^2+3+x)^2+(((-64*x^2+64*x)*ln(x)-32*x^3+3
2*x)*ln((x-1)*exp(1))-32*x^2)*exp(ln(x)^2+3+x)+(-32*x^3+32*x^2)*ln((x-1)*exp(1))-16*x^3)/(x^2-x),x,method=_RET
URNVERBOSE)

[Out]

-16*x^2*ln((x-1)*exp(1))-32*ln((x-1)*exp(1))*exp(ln(x)^2+3+x)*x-16*ln((x-1)*exp(1))*exp(2*ln(x)^2+6+2*x)

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maxima [B]  time = 0.44, size = 74, normalized size = 3.22 \begin {gather*} -16 \, x^{2} - 16 \, {\left (e^{\left (2 \, x + 6\right )} \log \left (x - 1\right ) + e^{\left (2 \, x + 6\right )}\right )} e^{\left (2 \, \log \relax (x)^{2}\right )} - 32 \, {\left (x e^{\left (x + 3\right )} \log \left (x - 1\right ) + x e^{\left (x + 3\right )}\right )} e^{\left (\log \relax (x)^{2}\right )} - 16 \, {\left (x^{2} - 1\right )} \log \left (x - 1\right ) - 16 \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-64*x+64)*log(x)-32*x^2+32*x)*log((x-1)*exp(1))-16*x)*exp(log(x)^2+3+x)^2+(((-64*x^2+64*x)*log(x
)-32*x^3+32*x)*log((x-1)*exp(1))-32*x^2)*exp(log(x)^2+3+x)+(-32*x^3+32*x^2)*log((x-1)*exp(1))-16*x^3)/(x^2-x),
x, algorithm="maxima")

[Out]

-16*x^2 - 16*(e^(2*x + 6)*log(x - 1) + e^(2*x + 6))*e^(2*log(x)^2) - 32*(x*e^(x + 3)*log(x - 1) + x*e^(x + 3))
*e^(log(x)^2) - 16*(x^2 - 1)*log(x - 1) - 16*log(x - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{2\,{\ln \relax (x)}^2+2\,x+6}\,\left (16\,x+\ln \left (\mathrm {e}\,\left (x-1\right )\right )\,\left (\ln \relax (x)\,\left (64\,x-64\right )-32\,x+32\,x^2\right )\right )-\ln \left (\mathrm {e}\,\left (x-1\right )\right )\,\left (32\,x^2-32\,x^3\right )+16\,x^3-{\mathrm {e}}^{{\ln \relax (x)}^2+x+3}\,\left (\ln \left (\mathrm {e}\,\left (x-1\right )\right )\,\left (32\,x+\ln \relax (x)\,\left (64\,x-64\,x^2\right )-32\,x^3\right )-32\,x^2\right )}{x-x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x + 2*log(x)^2 + 6)*(16*x + log(exp(1)*(x - 1))*(log(x)*(64*x - 64) - 32*x + 32*x^2)) - log(exp(1)*
(x - 1))*(32*x^2 - 32*x^3) + 16*x^3 - exp(x + log(x)^2 + 3)*(log(exp(1)*(x - 1))*(32*x + log(x)*(64*x - 64*x^2
) - 32*x^3) - 32*x^2))/(x - x^2),x)

[Out]

int((exp(2*x + 2*log(x)^2 + 6)*(16*x + log(exp(1)*(x - 1))*(log(x)*(64*x - 64) - 32*x + 32*x^2)) - log(exp(1)*
(x - 1))*(32*x^2 - 32*x^3) + 16*x^3 - exp(x + log(x)^2 + 3)*(log(exp(1)*(x - 1))*(32*x + log(x)*(64*x - 64*x^2
) - 32*x^3) - 32*x^2))/(x - x^2), x)

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sympy [A]  time = 8.52, size = 71, normalized size = 3.09 \begin {gather*} - 32 x e^{x + \log {\relax (x )}^{2} + 3} \log {\left (e \left (x - 1\right ) \right )} + \left (\frac {16}{3} - 16 x^{2}\right ) \log {\left (e \left (x - 1\right ) \right )} - 16 e^{2 x + 2 \log {\relax (x )}^{2} + 6} \log {\left (e \left (x - 1\right ) \right )} - \frac {16 \log {\left (3 x - 3 \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-64*x+64)*ln(x)-32*x**2+32*x)*ln((x-1)*exp(1))-16*x)*exp(ln(x)**2+3+x)**2+(((-64*x**2+64*x)*ln(x
)-32*x**3+32*x)*ln((x-1)*exp(1))-32*x**2)*exp(ln(x)**2+3+x)+(-32*x**3+32*x**2)*ln((x-1)*exp(1))-16*x**3)/(x**2
-x),x)

[Out]

-32*x*exp(x + log(x)**2 + 3)*log(E*(x - 1)) + (16/3 - 16*x**2)*log(E*(x - 1)) - 16*exp(2*x + 2*log(x)**2 + 6)*
log(E*(x - 1)) - 16*log(3*x - 3)/3

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