3.75.84 \(\int \frac {-8+9 x-8 e^x x+8 x^2+(-18-32 x+e^x (16+16 x)) \log (\frac {1}{2} (8-9 x+8 e^x x-8 x^2))}{8-9 x+8 e^x x-8 x^2} \, dx\)

Optimal. Leaf size=25 \[ -x+\log ^2\left (4-\frac {x}{2}-4 x \left (1-e^x+x\right )\right ) \]

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Rubi [F]  time = 5.90, 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 {-8+9 x-8 e^x x+8 x^2+\left (-18-32 x+e^x (16+16 x)\right ) \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{8-9 x+8 e^x x-8 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8 + 9*x - 8*E^x*x + 8*x^2 + (-18 - 32*x + E^x*(16 + 16*x))*Log[(8 - 9*x + 8*E^x*x - 8*x^2)/2])/(8 - 9*x
+ 8*E^x*x - 8*x^2),x]

[Out]

-3*x - x^2 + 2*x*Log[(8 - 9*x + 8*E^x*x - 8*x^2)/2] - 16*Defer[Int][(-8 + 9*x - 8*E^x*x + 8*x^2)^(-1), x] + 16
*Log[(8 - 9*x + 8*E^x*x - 8*x^2)/2]*Defer[Int][(-8 + 9*x - 8*E^x*x + 8*x^2)^(-1), x] + 16*Log[(8 - 9*x + 8*E^x
*x - 8*x^2)/2]*Defer[Int][1/(x*(-8 + 9*x - 8*E^x*x + 8*x^2)), x] - 16*Defer[Int][x/(-8 + 9*x - 8*E^x*x + 8*x^2
), x] - 2*Log[(8 - 9*x + 8*E^x*x - 8*x^2)/2]*Defer[Int][x/(-8 + 9*x - 8*E^x*x + 8*x^2), x] + 2*Defer[Int][x^2/
(-8 + 9*x - 8*E^x*x + 8*x^2), x] - 16*Log[(8 - 9*x + 8*E^x*x - 8*x^2)/2]*Defer[Int][x^2/(-8 + 9*x - 8*E^x*x +
8*x^2), x] + 16*Defer[Int][x^3/(-8 + 9*x - 8*E^x*x + 8*x^2), x] + 2*Defer[Int][Log[(8 - 9*x + 8*E^x*x - 8*x^2)
/2]/x, x] - 16*Defer[Int][Defer[Int][(-8 + (9 - 8*E^x)*x + 8*x^2)^(-1), x], x] - 16*Defer[Int][Defer[Int][(-8
+ (9 - 8*E^x)*x + 8*x^2)^(-1), x]/x, x] - 128*Defer[Int][Defer[Int][(-8 + (9 - 8*E^x)*x + 8*x^2)^(-1), x]/(-8
+ 9*x - 8*E^x*x + 8*x^2), x] - 128*Defer[Int][Defer[Int][(-8 + (9 - 8*E^x)*x + 8*x^2)^(-1), x]/(x*(-8 + 9*x -
8*E^x*x + 8*x^2)), x] + 16*Defer[Int][(x*Defer[Int][(-8 + (9 - 8*E^x)*x + 8*x^2)^(-1), x])/(-8 + 9*x - 8*E^x*x
+ 8*x^2), x] + 128*Defer[Int][(x^2*Defer[Int][(-8 + (9 - 8*E^x)*x + 8*x^2)^(-1), x])/(-8 + 9*x - 8*E^x*x + 8*
x^2), x] - 16*Defer[Int][Defer[Int][1/(x*(-8 + (9 - 8*E^x)*x + 8*x^2)), x], x] - 16*Defer[Int][Defer[Int][1/(x
*(-8 + (9 - 8*E^x)*x + 8*x^2)), x]/x, x] - 128*Defer[Int][Defer[Int][1/(x*(-8 + (9 - 8*E^x)*x + 8*x^2)), x]/(-
8 + 9*x - 8*E^x*x + 8*x^2), x] - 128*Defer[Int][Defer[Int][1/(x*(-8 + (9 - 8*E^x)*x + 8*x^2)), x]/(x*(-8 + 9*x
- 8*E^x*x + 8*x^2)), x] + 16*Defer[Int][(x*Defer[Int][1/(x*(-8 + (9 - 8*E^x)*x + 8*x^2)), x])/(-8 + 9*x - 8*E
^x*x + 8*x^2), x] + 128*Defer[Int][(x^2*Defer[Int][1/(x*(-8 + (9 - 8*E^x)*x + 8*x^2)), x])/(-8 + 9*x - 8*E^x*x
+ 8*x^2), x] + 2*Defer[Int][Defer[Int][x/(-8 + (9 - 8*E^x)*x + 8*x^2), x], x] + 2*Defer[Int][Defer[Int][x/(-8
+ (9 - 8*E^x)*x + 8*x^2), x]/x, x] + 16*Defer[Int][Defer[Int][x/(-8 + (9 - 8*E^x)*x + 8*x^2), x]/(-8 + 9*x -
8*E^x*x + 8*x^2), x] + 16*Defer[Int][Defer[Int][x/(-8 + (9 - 8*E^x)*x + 8*x^2), x]/(x*(-8 + 9*x - 8*E^x*x + 8*
x^2)), x] - 2*Defer[Int][(x*Defer[Int][x/(-8 + (9 - 8*E^x)*x + 8*x^2), x])/(-8 + 9*x - 8*E^x*x + 8*x^2), x] -
16*Defer[Int][(x^2*Defer[Int][x/(-8 + (9 - 8*E^x)*x + 8*x^2), x])/(-8 + 9*x - 8*E^x*x + 8*x^2), x] + 16*Defer[
Int][Defer[Int][x^2/(-8 + (9 - 8*E^x)*x + 8*x^2), x], x] + 16*Defer[Int][Defer[Int][x^2/(-8 + (9 - 8*E^x)*x +
8*x^2), x]/x, x] + 128*Defer[Int][Defer[Int][x^2/(-8 + (9 - 8*E^x)*x + 8*x^2), x]/(-8 + 9*x - 8*E^x*x + 8*x^2)
, x] + 128*Defer[Int][Defer[Int][x^2/(-8 + (9 - 8*E^x)*x + 8*x^2), x]/(x*(-8 + 9*x - 8*E^x*x + 8*x^2)), x] - 1
6*Defer[Int][(x*Defer[Int][x^2/(-8 + (9 - 8*E^x)*x + 8*x^2), x])/(-8 + 9*x - 8*E^x*x + 8*x^2), x] - 128*Defer[
Int][(x^2*Defer[Int][x^2/(-8 + (9 - 8*E^x)*x + 8*x^2), x])/(-8 + 9*x - 8*E^x*x + 8*x^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 \left (-8-8 x+x^2+8 x^3\right ) \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{x \left (8-9 x+8 e^x x-8 x^2\right )}+\frac {-x+2 \log \left (4+\left (-\frac {9}{2}+4 e^x\right ) x-4 x^2\right )+2 x \log \left (4+\left (-\frac {9}{2}+4 e^x\right ) x-4 x^2\right )}{x}\right ) \, dx\\ &=2 \int \frac {\left (-8-8 x+x^2+8 x^3\right ) \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{x \left (8-9 x+8 e^x x-8 x^2\right )} \, dx+\int \frac {-x+2 \log \left (4+\left (-\frac {9}{2}+4 e^x\right ) x-4 x^2\right )+2 x \log \left (4+\left (-\frac {9}{2}+4 e^x\right ) x-4 x^2\right )}{x} \, dx\\ &=-\left (2 \int \frac {\left (9+16 x-8 e^x (1+x)\right ) \left (-8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx+\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{8+\left (-9+8 e^x\right ) x-8 x^2} \, dx\right )-\left (2 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx+\int \left (-1+\frac {2 (1+x) \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{x}\right ) \, dx\\ &=-x+2 \int \frac {(1+x) \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{x} \, dx-2 \int \left (\frac {(1+x) \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{x}-\frac {\left (-8-8 x+x^2+8 x^3\right ) \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{x \left (-8+9 x-8 e^x x+8 x^2\right )}\right ) \, dx-\left (2 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\\ &=-x+2 \int \left (\log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )+\frac {\log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{x}\right ) \, dx-2 \int \frac {(1+x) \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{x} \, dx+2 \int \frac {\left (-8-8 x+x^2+8 x^3\right ) \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\left (2 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\\ &=-x+2 \int \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right ) \, dx+2 \int \frac {\log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{x} \, dx+2 \int \left (-\frac {8 \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{-8+9 x-8 e^x x+8 x^2}-\frac {8 \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{x \left (-8+9 x-8 e^x x+8 x^2\right )}+\frac {x \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{-8+9 x-8 e^x x+8 x^2}+\frac {8 x^2 \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{-8+9 x-8 e^x x+8 x^2}\right ) \, dx-2 \int \left (\frac {(1+x) \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{x}-\frac {8 (1+x) \int \frac {x^2}{-8+\left (9-8 e^x\right ) x+8 x^2} \, dx}{x}\right ) \, dx-\left (2 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\\ &=-x+2 x \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )-2 \int \frac {x \left (-9-16 x+8 e^x (1+x)\right )}{8-\left (9-8 e^x\right ) x-8 x^2} \, dx+2 \int \frac {\log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )}{x} \, dx-2 \int \frac {(1+x) \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{x} \, dx+2 \int \frac {x \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{-8+9 x-8 e^x x+8 x^2} \, dx-16 \int \frac {8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx}{-8+9 x-8 e^x x+8 x^2} \, dx-16 \int \frac {8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx+16 \int \frac {x^2 \left (8 \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+8 \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx-8 \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\right )}{-8+9 x-8 e^x x+8 x^2} \, dx+16 \int \frac {(1+x) \int \frac {x^2}{-8+\left (9-8 e^x\right ) x+8 x^2} \, dx}{x} \, dx-\left (2 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{-8+9 x-8 e^x x+8 x^2} \, dx+\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {1}{x \left (-8+9 x-8 e^x x+8 x^2\right )} \, dx-\left (16 \log \left (\frac {1}{2} \left (8-9 x+8 e^x x-8 x^2\right )\right )\right ) \int \frac {x^2}{-8+9 x-8 e^x x+8 x^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.55, size = 25, normalized size = 1.00 \begin {gather*} -x+\log ^2\left (4+\left (-\frac {9}{2}+4 e^x\right ) x-4 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 + 9*x - 8*E^x*x + 8*x^2 + (-18 - 32*x + E^x*(16 + 16*x))*Log[(8 - 9*x + 8*E^x*x - 8*x^2)/2])/(8
- 9*x + 8*E^x*x - 8*x^2),x]

[Out]

-x + Log[4 + (-9/2 + 4*E^x)*x - 4*x^2]^2

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fricas [A]  time = 0.66, size = 22, normalized size = 0.88 \begin {gather*} \log \left (-4 \, x^{2} + 4 \, x e^{x} - \frac {9}{2} \, x + 4\right )^{2} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+16)*exp(x)-32*x-18)*log(4*exp(x)*x-4*x^2-9/2*x+4)-8*exp(x)*x+8*x^2+9*x-8)/(8*exp(x)*x-8*x^2-
9*x+8),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {8 \, x^{2} - 8 \, x e^{x} + 2 \, {\left (8 \, {\left (x + 1\right )} e^{x} - 16 \, x - 9\right )} \log \left (-4 \, x^{2} + 4 \, x e^{x} - \frac {9}{2} \, x + 4\right ) + 9 \, x - 8}{8 \, x^{2} - 8 \, x e^{x} + 9 \, x - 8}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+16)*exp(x)-32*x-18)*log(4*exp(x)*x-4*x^2-9/2*x+4)-8*exp(x)*x+8*x^2+9*x-8)/(8*exp(x)*x-8*x^2-
9*x+8),x, algorithm="giac")

[Out]

integrate(-(8*x^2 - 8*x*e^x + 2*(8*(x + 1)*e^x - 16*x - 9)*log(-4*x^2 + 4*x*e^x - 9/2*x + 4) + 9*x - 8)/(8*x^2
- 8*x*e^x + 9*x - 8), x)

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maple [A]  time = 0.10, size = 23, normalized size = 0.92

method	result	size
norman	\(\ln \left (4 \,{\mathrm e}^{x} x -4 x^{2}-\frac {9 x}{2}+4\right )^{2}-x\)	\(23\)
risch	\(\ln \left (4 \,{\mathrm e}^{x} x -4 x^{2}-\frac {9 x}{2}+4\right )^{2}-x\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x+16)*exp(x)-32*x-18)*ln(4*exp(x)*x-4*x^2-9/2*x+4)-8*exp(x)*x+8*x^2+9*x-8)/(8*exp(x)*x-8*x^2-9*x+8),
x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {8 \, x^{2} - 8 \, x e^{x} + 2 \, {\left (8 \, {\left (x + 1\right )} e^{x} - 16 \, x - 9\right )} \log \left (-4 \, x^{2} + 4 \, x e^{x} - \frac {9}{2} \, x + 4\right ) + 9 \, x - 8}{8 \, x^{2} - 8 \, x e^{x} + 9 \, x - 8}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+16)*exp(x)-32*x-18)*log(4*exp(x)*x-4*x^2-9/2*x+4)-8*exp(x)*x+8*x^2+9*x-8)/(8*exp(x)*x-8*x^2-
9*x+8),x, algorithm="maxima")

[Out]

-integrate((8*x^2 - 8*x*e^x + 2*(8*(x + 1)*e^x - 16*x - 9)*log(-4*x^2 + 4*x*e^x - 9/2*x + 4) + 9*x - 8)/(8*x^2
- 8*x*e^x + 9*x - 8), x)

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mupad [B]  time = 5.78, size = 22, normalized size = 0.88 \begin {gather*} {\ln \left (4\,x\,{\mathrm {e}}^x-\frac {9\,x}{2}-4\,x^2+4\right )}^2-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(4*x*exp(x) - (9*x)/2 - 4*x^2 + 4)*(32*x - exp(x)*(16*x + 16) + 18) - 9*x + 8*x*exp(x) - 8*x^2 + 8)/(9
*x - 8*x*exp(x) + 8*x^2 - 8),x)

[Out]

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

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sympy [A]  time = 0.44, size = 22, normalized size = 0.88 \begin {gather*} - x + \log {\left (- 4 x^{2} + 4 x e^{x} - \frac {9 x}{2} + 4 \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+16)*exp(x)-32*x-18)*ln(4*exp(x)*x-4*x**2-9/2*x+4)-8*exp(x)*x+8*x**2+9*x-8)/(8*exp(x)*x-8*x**
2-9*x+8),x)

[Out]

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

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