3.60.69 \(\int \frac {-4 x^2+2 x^3+(8 x-8 x^2+2 x^3) \log (1-x)+(-4+6 x-2 x^2) \log ^2(1-x)+e^{10-2 e^x} (-4 x^2+2 x^3+e^x (-4 x^3+4 x^4)+(4 x-2 x^2+e^x (8 x^2-8 x^3)) \log (1-x)+e^x (-4 x+4 x^2) \log ^2(1-x))+(4 x^2-2 x^3+(-4 x+2 x^2) \log (1-x)) \log (x^2)}{-x^4+x^5+e^{30-6 e^x} (-x+x^2)+(3 x^3-3 x^4) \log (x^2)+(-3 x^2+3 x^3) \log ^2(x^2)+(x-x^2) \log ^3(x^2)+e^{20-4 e^x} (-3 x^2+3 x^3+(3 x-3 x^2) \log (x^2))+e^{10-2 e^x} (-3 x^3+3 x^4+(6 x^2-6 x^3) \log (x^2)+(-3 x+3 x^2) \log ^2(x^2))} \, dx\)

Optimal. Leaf size=32 \[ \frac {(-x+\log (1-x))^2}{\left (e^{10-2 e^x}+x-\log \left (x^2\right )\right )^2} \]

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Rubi [F]  time = 45.65, 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 {-4 x^2+2 x^3+\left (8 x-8 x^2+2 x^3\right ) \log (1-x)+\left (-4+6 x-2 x^2\right ) \log ^2(1-x)+e^{10-2 e^x} \left (-4 x^2+2 x^3+e^x \left (-4 x^3+4 x^4\right )+\left (4 x-2 x^2+e^x \left (8 x^2-8 x^3\right )\right ) \log (1-x)+e^x \left (-4 x+4 x^2\right ) \log ^2(1-x)\right )+\left (4 x^2-2 x^3+\left (-4 x+2 x^2\right ) \log (1-x)\right ) \log \left (x^2\right )}{-x^4+x^5+e^{30-6 e^x} \left (-x+x^2\right )+\left (3 x^3-3 x^4\right ) \log \left (x^2\right )+\left (-3 x^2+3 x^3\right ) \log ^2\left (x^2\right )+\left (x-x^2\right ) \log ^3\left (x^2\right )+e^{20-4 e^x} \left (-3 x^2+3 x^3+\left (3 x-3 x^2\right ) \log \left (x^2\right )\right )+e^{10-2 e^x} \left (-3 x^3+3 x^4+\left (6 x^2-6 x^3\right ) \log \left (x^2\right )+\left (-3 x+3 x^2\right ) \log ^2\left (x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*x^2 + 2*x^3 + (8*x - 8*x^2 + 2*x^3)*Log[1 - x] + (-4 + 6*x - 2*x^2)*Log[1 - x]^2 + E^(10 - 2*E^x)*(-4*
x^2 + 2*x^3 + E^x*(-4*x^3 + 4*x^4) + (4*x - 2*x^2 + E^x*(8*x^2 - 8*x^3))*Log[1 - x] + E^x*(-4*x + 4*x^2)*Log[1
- x]^2) + (4*x^2 - 2*x^3 + (-4*x + 2*x^2)*Log[1 - x])*Log[x^2])/(-x^4 + x^5 + E^(30 - 6*E^x)*(-x + x^2) + (3*
x^3 - 3*x^4)*Log[x^2] + (-3*x^2 + 3*x^3)*Log[x^2]^2 + (x - x^2)*Log[x^2]^3 + E^(20 - 4*E^x)*(-3*x^2 + 3*x^3 +
(3*x - 3*x^2)*Log[x^2]) + E^(10 - 2*E^x)*(-3*x^3 + 3*x^4 + (6*x^2 - 6*x^3)*Log[x^2] + (-3*x + 3*x^2)*Log[x^2]^
2)),x]

[Out]

-2*Defer[Int][E^(6*E^x)/(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3, x] - 2*Defer[Int][E^(2*(5 + 2*E^x))/(E^10
+ E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3, x] - 2*Defer[Int][E^(6*E^x)/((-1 + x)*(E^10 + E^(2*E^x)*x - E^(2*E^x)*
Log[x^2])^3), x] - 2*Defer[Int][E^(2*(5 + 2*E^x))/((-1 + x)*(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3), x] +
2*Defer[Int][(E^(6*E^x)*x)/(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3, x] + 2*Defer[Int][(E^(2*(5 + 2*E^x))*
x)/(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3, x] + 4*Defer[Int][(E^(10 + 4*E^x + x)*x^2)/(E^10 + E^(2*E^x)*x
- E^(2*E^x)*Log[x^2])^3, x] - 6*Defer[Int][(E^(6*E^x)*Log[1 - x])/(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3
, x] - 2*Defer[Int][(E^(2*(5 + 2*E^x))*Log[1 - x])/(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3, x] + 2*Defer[I
nt][(E^(6*E^x)*Log[1 - x])/((-1 + x)*(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3), x] + 2*Defer[Int][(E^(2*(5
+ 2*E^x))*Log[1 - x])/((-1 + x)*(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3), x] + 2*Defer[Int][(E^(6*E^x)*x*L
og[1 - x])/(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3, x] - 8*Defer[Int][(E^(10 + 4*E^x + x)*x*Log[1 - x])/(E
^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3, x] - 2*Defer[Int][(E^(6*E^x)*Log[1 - x]^2)/(E^10 + E^(2*E^x)*x - E^
(2*E^x)*Log[x^2])^3, x] + 4*Defer[Int][(E^(10 + 4*E^x + x)*Log[1 - x]^2)/(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x
^2])^3, x] + 4*Defer[Int][(E^(6*E^x)*Log[1 - x]^2)/(x*(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3), x] + 2*Def
er[Int][(E^(6*E^x)*Log[x^2])/((-1 + x)*(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3), x] - 2*Defer[Int][(E^(6*E
^x)*x*Log[x^2])/(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3, x] - 2*Defer[Int][(E^(6*E^x)*Log[1 - x]*Log[x^2])
/((-1 + x)*(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^3), x] - 2*Defer[Int][(E^(6*E^x)*Log[x^2])/(-E^10 - E^(2*
E^x)*x + E^(2*E^x)*Log[x^2])^3, x] - 2*Defer[Int][(E^(6*E^x)*Log[1 - x]*Log[x^2])/(-E^10 - E^(2*E^x)*x + E^(2*
E^x)*Log[x^2])^3, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{4 e^x} (x-\log (1-x)) \left (-\left ((-1+x) \left (e^{2 e^x} (-2+x)-2 e^{10+x} x\right ) \log (1-x)\right )-x \left (e^{10} (-2+x)+e^{2 e^x} (-2+x)+2 e^{10+x} (-1+x) x-e^{2 e^x} (-2+x) \log \left (x^2\right )\right )\right )}{(1-x) x \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx\\ &=2 \int \frac {e^{4 e^x} (x-\log (1-x)) \left (-\left ((-1+x) \left (e^{2 e^x} (-2+x)-2 e^{10+x} x\right ) \log (1-x)\right )-x \left (e^{10} (-2+x)+e^{2 e^x} (-2+x)+2 e^{10+x} (-1+x) x-e^{2 e^x} (-2+x) \log \left (x^2\right )\right )\right )}{(1-x) x \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx\\ &=2 \int \left (\frac {e^{6 e^x} (-2+x) (x-\log (1-x))}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}+\frac {e^{10+4 e^x} (-2+x) (x-\log (1-x))}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}+\frac {2 e^{10+4 e^x+x} (x-\log (1-x))^2}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}+\frac {e^{6 e^x} (-2+x) (x-\log (1-x)) \log (1-x)}{x \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}-\frac {e^{6 e^x} (-2+x) (x-\log (1-x)) \log \left (x^2\right )}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}\right ) \, dx\\ &=2 \int \frac {e^{6 e^x} (-2+x) (x-\log (1-x))}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx+2 \int \frac {e^{10+4 e^x} (-2+x) (x-\log (1-x))}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx+2 \int \frac {e^{6 e^x} (-2+x) (x-\log (1-x)) \log (1-x)}{x \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx-2 \int \frac {e^{6 e^x} (-2+x) (x-\log (1-x)) \log \left (x^2\right )}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx+4 \int \frac {e^{10+4 e^x+x} (x-\log (1-x))^2}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx\\ &=2 \int \frac {e^{2 \left (5+2 e^x\right )} (2-x) (x-\log (1-x))}{(1-x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx+2 \int \left (\frac {e^{6 e^x} (x-\log (1-x))}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}-\frac {e^{6 e^x} (x-\log (1-x))}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}\right ) \, dx+2 \int \left (\frac {e^{6 e^x} (x-\log (1-x)) \log (1-x)}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}-\frac {2 e^{6 e^x} (x-\log (1-x)) \log (1-x)}{x \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}\right ) \, dx-2 \int \left (\frac {e^{6 e^x} (x-\log (1-x)) \log \left (x^2\right )}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}-\frac {e^{6 e^x} (x-\log (1-x)) \log \left (x^2\right )}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}\right ) \, dx+4 \int \left (\frac {e^{10+4 e^x+x} x^2}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}-\frac {2 e^{10+4 e^x+x} x \log (1-x)}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}+\frac {e^{10+4 e^x+x} \log ^2(1-x)}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}\right ) \, dx\\ &=2 \int \frac {e^{6 e^x} (x-\log (1-x))}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx-2 \int \frac {e^{6 e^x} (x-\log (1-x))}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx+2 \int \frac {e^{6 e^x} (x-\log (1-x)) \log (1-x)}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx-2 \int \frac {e^{6 e^x} (x-\log (1-x)) \log \left (x^2\right )}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx+2 \int \frac {e^{6 e^x} (x-\log (1-x)) \log \left (x^2\right )}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx+2 \int \left (\frac {e^{2 \left (5+2 e^x\right )} (x-\log (1-x))}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}-\frac {e^{2 \left (5+2 e^x\right )} (x-\log (1-x))}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}\right ) \, dx+4 \int \frac {e^{10+4 e^x+x} x^2}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx-4 \int \frac {e^{6 e^x} (x-\log (1-x)) \log (1-x)}{x \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx+4 \int \frac {e^{10+4 e^x+x} \log ^2(1-x)}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx-8 \int \frac {e^{10+4 e^x+x} x \log (1-x)}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx\\ &=2 \int \frac {e^{2 \left (5+2 e^x\right )} (x-\log (1-x))}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx-2 \int \frac {e^{2 \left (5+2 e^x\right )} (x-\log (1-x))}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx+2 \int \left (\frac {e^{6 e^x} x}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}-\frac {e^{6 e^x} \log (1-x)}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}\right ) \, dx-2 \int \left (\frac {e^{6 e^x} x}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}-\frac {e^{6 e^x} \log (1-x)}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}\right ) \, dx+2 \int \left (\frac {e^{6 e^x} x \log (1-x)}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}-\frac {e^{6 e^x} \log ^2(1-x)}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}\right ) \, dx+2 \int \left (\frac {e^{6 e^x} x \log \left (x^2\right )}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}-\frac {e^{6 e^x} \log (1-x) \log \left (x^2\right )}{(-1+x) \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}\right ) \, dx-2 \int \left (\frac {e^{6 e^x} x \log \left (x^2\right )}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}+\frac {e^{6 e^x} \log (1-x) \log \left (x^2\right )}{\left (-e^{10}-e^{2 e^x} x+e^{2 e^x} \log \left (x^2\right )\right )^3}\right ) \, dx+4 \int \frac {e^{10+4 e^x+x} x^2}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx+4 \int \frac {e^{10+4 e^x+x} \log ^2(1-x)}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx-4 \int \left (\frac {e^{6 e^x} \log (1-x)}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}-\frac {e^{6 e^x} \log ^2(1-x)}{x \left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3}\right ) \, dx-8 \int \frac {e^{10+4 e^x+x} x \log (1-x)}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^3} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.43, size = 48, normalized size = 1.50 \begin {gather*} \frac {e^{4 e^x} (x-\log (1-x))^2}{\left (e^{10}+e^{2 e^x} x-e^{2 e^x} \log \left (x^2\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*x^2 + 2*x^3 + (8*x - 8*x^2 + 2*x^3)*Log[1 - x] + (-4 + 6*x - 2*x^2)*Log[1 - x]^2 + E^(10 - 2*E^x
)*(-4*x^2 + 2*x^3 + E^x*(-4*x^3 + 4*x^4) + (4*x - 2*x^2 + E^x*(8*x^2 - 8*x^3))*Log[1 - x] + E^x*(-4*x + 4*x^2)
*Log[1 - x]^2) + (4*x^2 - 2*x^3 + (-4*x + 2*x^2)*Log[1 - x])*Log[x^2])/(-x^4 + x^5 + E^(30 - 6*E^x)*(-x + x^2)
+ (3*x^3 - 3*x^4)*Log[x^2] + (-3*x^2 + 3*x^3)*Log[x^2]^2 + (x - x^2)*Log[x^2]^3 + E^(20 - 4*E^x)*(-3*x^2 + 3*
x^3 + (3*x - 3*x^2)*Log[x^2]) + E^(10 - 2*E^x)*(-3*x^3 + 3*x^4 + (6*x^2 - 6*x^3)*Log[x^2] + (-3*x + 3*x^2)*Log
[x^2]^2)),x]

[Out]

(E^(4*E^x)*(x - Log[1 - x])^2)/(E^10 + E^(2*E^x)*x - E^(2*E^x)*Log[x^2])^2

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fricas [B]  time = 0.95, size = 65, normalized size = 2.03 \begin {gather*} \frac {x^{2} - 2 \, x \log \left (-x + 1\right ) + \log \left (-x + 1\right )^{2}}{x^{2} + 2 \, {\left (x - \log \left (x^{2}\right )\right )} e^{\left (-2 \, e^{x} + 10\right )} - 2 \, x \log \left (x^{2}\right ) + \log \left (x^{2}\right )^{2} + e^{\left (-4 \, e^{x} + 20\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2-4*x)*exp(x)*log(-x+1)^2+((-8*x^3+8*x^2)*exp(x)-2*x^2+4*x)*log(-x+1)+(4*x^4-4*x^3)*exp(x)+2*
x^3-4*x^2)*exp(5-exp(x))^2+((2*x^2-4*x)*log(-x+1)-2*x^3+4*x^2)*log(x^2)+(-2*x^2+6*x-4)*log(-x+1)^2+(2*x^3-8*x^
2+8*x)*log(-x+1)+2*x^3-4*x^2)/((x^2-x)*exp(5-exp(x))^6+((-3*x^2+3*x)*log(x^2)+3*x^3-3*x^2)*exp(5-exp(x))^4+((3
*x^2-3*x)*log(x^2)^2+(-6*x^3+6*x^2)*log(x^2)+3*x^4-3*x^3)*exp(5-exp(x))^2+(-x^2+x)*log(x^2)^3+(3*x^3-3*x^2)*lo
g(x^2)^2+(-3*x^4+3*x^3)*log(x^2)+x^5-x^4),x, algorithm="fricas")

[Out]

(x^2 - 2*x*log(-x + 1) + log(-x + 1)^2)/(x^2 + 2*(x - log(x^2))*e^(-2*e^x + 10) - 2*x*log(x^2) + log(x^2)^2 +
e^(-4*e^x + 20))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2-4*x)*exp(x)*log(-x+1)^2+((-8*x^3+8*x^2)*exp(x)-2*x^2+4*x)*log(-x+1)+(4*x^4-4*x^3)*exp(x)+2*
x^3-4*x^2)*exp(5-exp(x))^2+((2*x^2-4*x)*log(-x+1)-2*x^3+4*x^2)*log(x^2)+(-2*x^2+6*x-4)*log(-x+1)^2+(2*x^3-8*x^
2+8*x)*log(-x+1)+2*x^3-4*x^2)/((x^2-x)*exp(5-exp(x))^6+((-3*x^2+3*x)*log(x^2)+3*x^3-3*x^2)*exp(5-exp(x))^4+((3
*x^2-3*x)*log(x^2)^2+(-6*x^3+6*x^2)*log(x^2)+3*x^4-3*x^3)*exp(5-exp(x))^2+(-x^2+x)*log(x^2)^3+(3*x^3-3*x^2)*lo
g(x^2)^2+(-3*x^4+3*x^3)*log(x^2)+x^5-x^4),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(-e

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maple [C]  time = 11.50, size = 92, normalized size = 2.88

method	result	size
risch	\(\frac {4 x^{2}-8 x \ln \left (1-x \right )+4 \ln \left (1-x \right )^{2}}{\left (2 \,{\mathrm e}^{-2 \,{\mathrm e}^{x}+10}-4 \ln \relax (x )+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 x +i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right )^{2}}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x^2-4*x)*exp(x)*ln(1-x)^2+((-8*x^3+8*x^2)*exp(x)-2*x^2+4*x)*ln(1-x)+(4*x^4-4*x^3)*exp(x)+2*x^3-4*x^2)
*exp(5-exp(x))^2+((2*x^2-4*x)*ln(1-x)-2*x^3+4*x^2)*ln(x^2)+(-2*x^2+6*x-4)*ln(1-x)^2+(2*x^3-8*x^2+8*x)*ln(1-x)+
2*x^3-4*x^2)/((x^2-x)*exp(5-exp(x))^6+((-3*x^2+3*x)*ln(x^2)+3*x^3-3*x^2)*exp(5-exp(x))^4+((3*x^2-3*x)*ln(x^2)^
2+(-6*x^3+6*x^2)*ln(x^2)+3*x^4-3*x^3)*exp(5-exp(x))^2+(-x^2+x)*ln(x^2)^3+(3*x^3-3*x^2)*ln(x^2)^2+(-3*x^4+3*x^3
)*ln(x^2)+x^5-x^4),x,method=_RETURNVERBOSE)

[Out]

4*(x^2-2*x*ln(1-x)+ln(1-x)^2)/(2*exp(-2*exp(x)+10)-4*ln(x)+I*Pi*csgn(I*x^2)^3+2*x+I*Pi*csgn(I*x)^2*csgn(I*x^2)
-2*I*Pi*csgn(I*x)*csgn(I*x^2)^2)^2

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maxima [B]  time = 1.18, size = 83, normalized size = 2.59 \begin {gather*} \frac {x^{2} e^{\left (4 \, e^{x}\right )} - 2 \, x e^{\left (4 \, e^{x}\right )} \log \left (-x + 1\right ) + e^{\left (4 \, e^{x}\right )} \log \left (-x + 1\right )^{2}}{{\left (x^{2} - 4 \, x \log \relax (x) + 4 \, \log \relax (x)^{2}\right )} e^{\left (4 \, e^{x}\right )} + 2 \, {\left (x e^{10} - 2 \, e^{10} \log \relax (x)\right )} e^{\left (2 \, e^{x}\right )} + e^{20}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x^2-4*x)*exp(x)*log(-x+1)^2+((-8*x^3+8*x^2)*exp(x)-2*x^2+4*x)*log(-x+1)+(4*x^4-4*x^3)*exp(x)+2*
x^3-4*x^2)*exp(5-exp(x))^2+((2*x^2-4*x)*log(-x+1)-2*x^3+4*x^2)*log(x^2)+(-2*x^2+6*x-4)*log(-x+1)^2+(2*x^3-8*x^
2+8*x)*log(-x+1)+2*x^3-4*x^2)/((x^2-x)*exp(5-exp(x))^6+((-3*x^2+3*x)*log(x^2)+3*x^3-3*x^2)*exp(5-exp(x))^4+((3
*x^2-3*x)*log(x^2)^2+(-6*x^3+6*x^2)*log(x^2)+3*x^4-3*x^3)*exp(5-exp(x))^2+(-x^2+x)*log(x^2)^3+(3*x^3-3*x^2)*lo
g(x^2)^2+(-3*x^4+3*x^3)*log(x^2)+x^5-x^4),x, algorithm="maxima")

[Out]

(x^2*e^(4*e^x) - 2*x*e^(4*e^x)*log(-x + 1) + e^(4*e^x)*log(-x + 1)^2)/((x^2 - 4*x*log(x) + 4*log(x)^2)*e^(4*e^
x) + 2*(x*e^10 - 2*e^10*log(x))*e^(2*e^x) + e^20)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{10-2\,{\mathrm {e}}^x}\,\left ({\mathrm {e}}^x\,\left (4\,x^3-4\,x^4\right )-\ln \left (1-x\right )\,\left (4\,x+{\mathrm {e}}^x\,\left (8\,x^2-8\,x^3\right )-2\,x^2\right )+4\,x^2-2\,x^3+{\mathrm {e}}^x\,{\ln \left (1-x\right )}^2\,\left (4\,x-4\,x^2\right )\right )-\ln \left (1-x\right )\,\left (2\,x^3-8\,x^2+8\,x\right )+{\ln \left (1-x\right )}^2\,\left (2\,x^2-6\,x+4\right )+4\,x^2-2\,x^3+\ln \left (x^2\right )\,\left (\ln \left (1-x\right )\,\left (4\,x-2\,x^2\right )-4\,x^2+2\,x^3\right )}{{\mathrm {e}}^{10-2\,{\mathrm {e}}^x}\,\left ({\ln \left (x^2\right )}^2\,\left (3\,x-3\,x^2\right )-\ln \left (x^2\right )\,\left (6\,x^2-6\,x^3\right )+3\,x^3-3\,x^4\right )-{\ln \left (x^2\right )}^3\,\left (x-x^2\right )-{\mathrm {e}}^{20-4\,{\mathrm {e}}^x}\,\left (\ln \left (x^2\right )\,\left (3\,x-3\,x^2\right )-3\,x^2+3\,x^3\right )+{\mathrm {e}}^{30-6\,{\mathrm {e}}^x}\,\left (x-x^2\right )-\ln \left (x^2\right )\,\left (3\,x^3-3\,x^4\right )+{\ln \left (x^2\right )}^2\,\left (3\,x^2-3\,x^3\right )+x^4-x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(10 - 2*exp(x))*(exp(x)*(4*x^3 - 4*x^4) - log(1 - x)*(4*x + exp(x)*(8*x^2 - 8*x^3) - 2*x^2) + 4*x^2 -
2*x^3 + exp(x)*log(1 - x)^2*(4*x - 4*x^2)) - log(1 - x)*(8*x - 8*x^2 + 2*x^3) + log(1 - x)^2*(2*x^2 - 6*x + 4)
+ 4*x^2 - 2*x^3 + log(x^2)*(log(1 - x)*(4*x - 2*x^2) - 4*x^2 + 2*x^3))/(exp(10 - 2*exp(x))*(log(x^2)^2*(3*x -
3*x^2) - log(x^2)*(6*x^2 - 6*x^3) + 3*x^3 - 3*x^4) - log(x^2)^3*(x - x^2) - exp(20 - 4*exp(x))*(log(x^2)*(3*x
- 3*x^2) - 3*x^2 + 3*x^3) + exp(30 - 6*exp(x))*(x - x^2) - log(x^2)*(3*x^3 - 3*x^4) + log(x^2)^2*(3*x^2 - 3*x
^3) + x^4 - x^5),x)

[Out]

int((exp(10 - 2*exp(x))*(exp(x)*(4*x^3 - 4*x^4) - log(1 - x)*(4*x + exp(x)*(8*x^2 - 8*x^3) - 2*x^2) + 4*x^2 -
2*x^3 + exp(x)*log(1 - x)^2*(4*x - 4*x^2)) - log(1 - x)*(8*x - 8*x^2 + 2*x^3) + log(1 - x)^2*(2*x^2 - 6*x + 4)
+ 4*x^2 - 2*x^3 + log(x^2)*(log(1 - x)*(4*x - 2*x^2) - 4*x^2 + 2*x^3))/(exp(10 - 2*exp(x))*(log(x^2)^2*(3*x -
3*x^2) - log(x^2)*(6*x^2 - 6*x^3) + 3*x^3 - 3*x^4) - log(x^2)^3*(x - x^2) - exp(20 - 4*exp(x))*(log(x^2)*(3*x
- 3*x^2) - 3*x^2 + 3*x^3) + exp(30 - 6*exp(x))*(x - x^2) - log(x^2)*(3*x^3 - 3*x^4) + log(x^2)^2*(3*x^2 - 3*x
^3) + x^4 - x^5), x)

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sympy [B]  time = 0.87, size = 63, normalized size = 1.97 \begin {gather*} \frac {x^{2} - 2 x \log {\left (1 - x \right )} + \log {\left (1 - x \right )}^{2}}{x^{2} - 2 x \log {\left (x^{2} \right )} + \left (2 x - 2 \log {\left (x^{2} \right )}\right ) e^{10 - 2 e^{x}} + e^{20 - 4 e^{x}} + \log {\left (x^{2} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x**2-4*x)*exp(x)*ln(-x+1)**2+((-8*x**3+8*x**2)*exp(x)-2*x**2+4*x)*ln(-x+1)+(4*x**4-4*x**3)*exp(
x)+2*x**3-4*x**2)*exp(5-exp(x))**2+((2*x**2-4*x)*ln(-x+1)-2*x**3+4*x**2)*ln(x**2)+(-2*x**2+6*x-4)*ln(-x+1)**2+
(2*x**3-8*x**2+8*x)*ln(-x+1)+2*x**3-4*x**2)/((x**2-x)*exp(5-exp(x))**6+((-3*x**2+3*x)*ln(x**2)+3*x**3-3*x**2)*
exp(5-exp(x))**4+((3*x**2-3*x)*ln(x**2)**2+(-6*x**3+6*x**2)*ln(x**2)+3*x**4-3*x**3)*exp(5-exp(x))**2+(-x**2+x)
*ln(x**2)**3+(3*x**3-3*x**2)*ln(x**2)**2+(-3*x**4+3*x**3)*ln(x**2)+x**5-x**4),x)

[Out]

(x**2 - 2*x*log(1 - x) + log(1 - x)**2)/(x**2 - 2*x*log(x**2) + (2*x - 2*log(x**2))*exp(10 - 2*exp(x)) + exp(2
0 - 4*exp(x)) + log(x**2)**2)

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