3.8.63 \(\int \frac {-2 x^2+(-2-4 x+e^x (-1-x+2 x^2)) \log (-3-6 x)}{(4-24 x+128 x^3+e^{2 x} (1+2 x)+e^x (4-8 x-32 x^2)) \log (-3-6 x)+(-4 x+8 x^2+32 x^3+e^x (-2 x-4 x^2)) \log (-3-6 x) \log (\log (-3-6 x))+(x^2+2 x^3) \log (-3-6 x) \log ^2(\log (-3-6 x))} \, dx\)

Optimal. Leaf size=24 \[ \frac {x}{-2-e^x+x (8+\log (\log (-3 (1+2 x))))} \]

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Rubi [F]  time = 11.32, 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 {-2 x^2+\left (-2-4 x+e^x \left (-1-x+2 x^2\right )\right ) \log (-3-6 x)}{\left (4-24 x+128 x^3+e^{2 x} (1+2 x)+e^x \left (4-8 x-32 x^2\right )\right ) \log (-3-6 x)+\left (-4 x+8 x^2+32 x^3+e^x \left (-2 x-4 x^2\right )\right ) \log (-3-6 x) \log (\log (-3-6 x))+\left (x^2+2 x^3\right ) \log (-3-6 x) \log ^2(\log (-3-6 x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*x^2 + (-2 - 4*x + E^x*(-1 - x + 2*x^2))*Log[-3 - 6*x])/((4 - 24*x + 128*x^3 + E^(2*x)*(1 + 2*x) + E^x*
(4 - 8*x - 32*x^2))*Log[-3 - 6*x] + (-4*x + 8*x^2 + 32*x^3 + E^x*(-2*x - 4*x^2))*Log[-3 - 6*x]*Log[Log[-3 - 6*
x]] + (x^2 + 2*x^3)*Log[-3 - 6*x]*Log[Log[-3 - 6*x]]^2),x]

[Out]

Defer[Int][1/(Log[-3 - 6*x]*(2 + E^x - 8*x - x*Log[Log[-3 - 6*x]])^2), x]/2 - Defer[Int][(2 + E^x - 8*x - x*Lo
g[Log[-3 - 6*x]])^(-1), x] - 10*Defer[Int][x/(-2 - E^x + 8*x + x*Log[Log[-3 - 6*x]])^2, x] + 8*Defer[Int][x^2/
(-2 - E^x + 8*x + x*Log[Log[-3 - 6*x]])^2, x] - Defer[Int][x/(Log[-3 - 6*x]*(-2 - E^x + 8*x + x*Log[Log[-3 - 6
*x]])^2), x] - Defer[Int][1/((1 + 2*x)*Log[-3 - 6*x]*(-2 - E^x + 8*x + x*Log[Log[-3 - 6*x]])^2), x]/2 - Defer[
Int][(x*Log[Log[-3 - 6*x]])/(-2 - E^x + 8*x + x*Log[Log[-3 - 6*x]])^2, x] + Defer[Int][(x^2*Log[Log[-3 - 6*x]]
)/(-2 - E^x + 8*x + x*Log[Log[-3 - 6*x]])^2, x] - Defer[Int][x/(-2 - E^x + 8*x + x*Log[Log[-3 - 6*x]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 x^2+\left (-2+e^x (-1+x)\right ) (1+2 x) \log (-3-6 x)}{(1+2 x) \log (-3-6 x) \left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2} \, dx\\ &=\int \left (-\frac {-1+x}{-2-e^x+8 x+x \log (\log (-3-6 x))}+\frac {x \left (-2 x-10 \log (-3-6 x)-12 x \log (-3-6 x)+16 x^2 \log (-3-6 x)-\log (-3-6 x) \log (\log (-3-6 x))-x \log (-3-6 x) \log (\log (-3-6 x))+2 x^2 \log (-3-6 x) \log (\log (-3-6 x))\right )}{(1+2 x) \log (-3-6 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}\right ) \, dx\\ &=-\int \frac {-1+x}{-2-e^x+8 x+x \log (\log (-3-6 x))} \, dx+\int \frac {x \left (-2 x-10 \log (-3-6 x)-12 x \log (-3-6 x)+16 x^2 \log (-3-6 x)-\log (-3-6 x) \log (\log (-3-6 x))-x \log (-3-6 x) \log (\log (-3-6 x))+2 x^2 \log (-3-6 x) \log (\log (-3-6 x))\right )}{(1+2 x) \log (-3-6 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx\\ &=\int \frac {x (-2 x+(1+2 x) \log (-3-6 x) (-10+8 x+(-1+x) \log (\log (-3-6 x))))}{(1+2 x) \log (-3-6 x) \left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2} \, dx-\int \left (\frac {1}{2+e^x-8 x-x \log (\log (-3-6 x))}+\frac {x}{-2-e^x+8 x+x \log (\log (-3-6 x))}\right ) \, dx\\ &=-\int \frac {1}{2+e^x-8 x-x \log (\log (-3-6 x))} \, dx-\int \frac {x}{-2-e^x+8 x+x \log (\log (-3-6 x))} \, dx+\int \left (\frac {-2 x-10 \log (-3-6 x)-12 x \log (-3-6 x)+16 x^2 \log (-3-6 x)-\log (-3-6 x) \log (\log (-3-6 x))-x \log (-3-6 x) \log (\log (-3-6 x))+2 x^2 \log (-3-6 x) \log (\log (-3-6 x))}{2 \log (-3-6 x) \left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2}-\frac {-2 x-10 \log (-3-6 x)-12 x \log (-3-6 x)+16 x^2 \log (-3-6 x)-\log (-3-6 x) \log (\log (-3-6 x))-x \log (-3-6 x) \log (\log (-3-6 x))+2 x^2 \log (-3-6 x) \log (\log (-3-6 x))}{2 (1+2 x) \log (-3-6 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}\right ) \, dx\\ &=\frac {1}{2} \int \frac {-2 x-10 \log (-3-6 x)-12 x \log (-3-6 x)+16 x^2 \log (-3-6 x)-\log (-3-6 x) \log (\log (-3-6 x))-x \log (-3-6 x) \log (\log (-3-6 x))+2 x^2 \log (-3-6 x) \log (\log (-3-6 x))}{\log (-3-6 x) \left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2} \, dx-\frac {1}{2} \int \frac {-2 x-10 \log (-3-6 x)-12 x \log (-3-6 x)+16 x^2 \log (-3-6 x)-\log (-3-6 x) \log (\log (-3-6 x))-x \log (-3-6 x) \log (\log (-3-6 x))+2 x^2 \log (-3-6 x) \log (\log (-3-6 x))}{(1+2 x) \log (-3-6 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-\int \frac {1}{2+e^x-8 x-x \log (\log (-3-6 x))} \, dx-\int \frac {x}{-2-e^x+8 x+x \log (\log (-3-6 x))} \, dx\\ &=\frac {1}{2} \int \frac {-2 x+(1+2 x) \log (-3-6 x) (-10+8 x+(-1+x) \log (\log (-3-6 x)))}{\log (-3-6 x) \left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2} \, dx-\frac {1}{2} \int \frac {-2 x+(1+2 x) \log (-3-6 x) (-10+8 x+(-1+x) \log (\log (-3-6 x)))}{(1+2 x) \log (-3-6 x) \left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2} \, dx-\int \frac {1}{2+e^x-8 x-x \log (\log (-3-6 x))} \, dx-\int \frac {x}{-2-e^x+8 x+x \log (\log (-3-6 x))} \, dx\\ &=\frac {1}{2} \int \left (-\frac {10}{\left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2}-\frac {12 x}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}+\frac {16 x^2}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}-\frac {2 x}{\log (-3-6 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}-\frac {\log (\log (-3-6 x))}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}-\frac {x \log (\log (-3-6 x))}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}+\frac {2 x^2 \log (\log (-3-6 x))}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}\right ) \, dx-\frac {1}{2} \int \left (-\frac {10}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}-\frac {12 x}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}+\frac {16 x^2}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}-\frac {2 x}{(1+2 x) \log (-3-6 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}-\frac {\log (\log (-3-6 x))}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}-\frac {x \log (\log (-3-6 x))}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}+\frac {2 x^2 \log (\log (-3-6 x))}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}\right ) \, dx-\int \frac {1}{2+e^x-8 x-x \log (\log (-3-6 x))} \, dx-\int \frac {x}{-2-e^x+8 x+x \log (\log (-3-6 x))} \, dx\\ &=-\left (\frac {1}{2} \int \frac {\log (\log (-3-6 x))}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx\right )-\frac {1}{2} \int \frac {x \log (\log (-3-6 x))}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx+\frac {1}{2} \int \frac {\log (\log (-3-6 x))}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx+\frac {1}{2} \int \frac {x \log (\log (-3-6 x))}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-5 \int \frac {1}{\left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2} \, dx+5 \int \frac {1}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-6 \int \frac {x}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx+6 \int \frac {x}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx+8 \int \frac {x^2}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-8 \int \frac {x^2}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-\int \frac {1}{2+e^x-8 x-x \log (\log (-3-6 x))} \, dx-\int \frac {x}{\log (-3-6 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx+\int \frac {x}{(1+2 x) \log (-3-6 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx+\int \frac {x^2 \log (\log (-3-6 x))}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-\int \frac {x^2 \log (\log (-3-6 x))}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-\int \frac {x}{-2-e^x+8 x+x \log (\log (-3-6 x))} \, dx\\ &=-\left (\frac {1}{2} \int \frac {\log (\log (-3-6 x))}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx\right )-\frac {1}{2} \int \frac {x \log (\log (-3-6 x))}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx+\frac {1}{2} \int \frac {\log (\log (-3-6 x))}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx+\frac {1}{2} \int \left (\frac {\log (\log (-3-6 x))}{2 \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}-\frac {\log (\log (-3-6 x))}{2 (1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}\right ) \, dx-5 \int \frac {1}{\left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2} \, dx+5 \int \frac {1}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-6 \int \frac {x}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx+6 \int \left (\frac {1}{2 \left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2}-\frac {1}{2 (1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}\right ) \, dx+8 \int \frac {x^2}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-8 \int \left (-\frac {1}{4 \left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2}+\frac {x}{2 \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}+\frac {1}{4 (1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}\right ) \, dx-\int \frac {1}{2+e^x-8 x-x \log (\log (-3-6 x))} \, dx-\int \frac {x}{\log (-3-6 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx+\int \frac {x^2 \log (\log (-3-6 x))}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-\int \frac {x}{-2-e^x+8 x+x \log (\log (-3-6 x))} \, dx+\int \left (\frac {1}{2 \log (-3-6 x) \left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2}-\frac {1}{2 (1+2 x) \log (-3-6 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}\right ) \, dx-\int \left (-\frac {\log (\log (-3-6 x))}{4 \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}+\frac {x \log (\log (-3-6 x))}{2 \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}+\frac {\log (\log (-3-6 x))}{4 (1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2}\right ) \, dx\\ &=2 \left (\frac {1}{4} \int \frac {\log (\log (-3-6 x))}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx\right )-2 \left (\frac {1}{4} \int \frac {\log (\log (-3-6 x))}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx\right )+\frac {1}{2} \int \frac {1}{\log (-3-6 x) \left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2} \, dx-\frac {1}{2} \int \frac {1}{(1+2 x) \log (-3-6 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-\frac {1}{2} \int \frac {\log (\log (-3-6 x))}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-2 \left (\frac {1}{2} \int \frac {x \log (\log (-3-6 x))}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx\right )+\frac {1}{2} \int \frac {\log (\log (-3-6 x))}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx+2 \int \frac {1}{\left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2} \, dx-2 \int \frac {1}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx+3 \int \frac {1}{\left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2} \, dx-3 \int \frac {1}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-4 \int \frac {x}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-5 \int \frac {1}{\left (2+e^x-8 x-x \log (\log (-3-6 x))\right )^2} \, dx+5 \int \frac {1}{(1+2 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-6 \int \frac {x}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx+8 \int \frac {x^2}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-\int \frac {1}{2+e^x-8 x-x \log (\log (-3-6 x))} \, dx-\int \frac {x}{\log (-3-6 x) \left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx+\int \frac {x^2 \log (\log (-3-6 x))}{\left (-2-e^x+8 x+x \log (\log (-3-6 x))\right )^2} \, dx-\int \frac {x}{-2-e^x+8 x+x \log (\log (-3-6 x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.69, size = 23, normalized size = 0.96 \begin {gather*} \frac {x}{-2-e^x+8 x+x \log (\log (-3-6 x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x^2 + (-2 - 4*x + E^x*(-1 - x + 2*x^2))*Log[-3 - 6*x])/((4 - 24*x + 128*x^3 + E^(2*x)*(1 + 2*x)
+ E^x*(4 - 8*x - 32*x^2))*Log[-3 - 6*x] + (-4*x + 8*x^2 + 32*x^3 + E^x*(-2*x - 4*x^2))*Log[-3 - 6*x]*Log[Log[-
3 - 6*x]] + (x^2 + 2*x^3)*Log[-3 - 6*x]*Log[Log[-3 - 6*x]]^2),x]

[Out]

x/(-2 - E^x + 8*x + x*Log[Log[-3 - 6*x]])

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fricas [A]  time = 1.01, size = 22, normalized size = 0.92 \begin {gather*} \frac {x}{x \log \left (\log \left (-6 \, x - 3\right )\right ) + 8 \, x - e^{x} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-x-1)*exp(x)-4*x-2)*log(-6*x-3)-2*x^2)/((2*x^3+x^2)*log(-6*x-3)*log(log(-6*x-3))^2+((-4*x^2-
2*x)*exp(x)+32*x^3+8*x^2-4*x)*log(-6*x-3)*log(log(-6*x-3))+((2*x+1)*exp(x)^2+(-32*x^2-8*x+4)*exp(x)+128*x^3-24
*x+4)*log(-6*x-3)),x, algorithm="fricas")

[Out]

x/(x*log(log(-6*x - 3)) + 8*x - e^x - 2)

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giac [A]  time = 0.68, size = 22, normalized size = 0.92 \begin {gather*} \frac {x}{x \log \left (\log \left (-6 \, x - 3\right )\right ) + 8 \, x - e^{x} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-x-1)*exp(x)-4*x-2)*log(-6*x-3)-2*x^2)/((2*x^3+x^2)*log(-6*x-3)*log(log(-6*x-3))^2+((-4*x^2-
2*x)*exp(x)+32*x^3+8*x^2-4*x)*log(-6*x-3)*log(log(-6*x-3))+((2*x+1)*exp(x)^2+(-32*x^2-8*x+4)*exp(x)+128*x^3-24
*x+4)*log(-6*x-3)),x, algorithm="giac")

[Out]

x/(x*log(log(-6*x - 3)) + 8*x - e^x - 2)

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maple [A]  time = 0.05, size = 23, normalized size = 0.96

method	result	size
risch	\(\frac {x}{x \ln \left (\ln \left (-6 x -3\right )\right )+8 x -{\mathrm e}^{x}-2}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2-x-1)*exp(x)-4*x-2)*ln(-6*x-3)-2*x^2)/((2*x^3+x^2)*ln(-6*x-3)*ln(ln(-6*x-3))^2+((-4*x^2-2*x)*exp(x
)+32*x^3+8*x^2-4*x)*ln(-6*x-3)*ln(ln(-6*x-3))+((2*x+1)*exp(x)^2+(-32*x^2-8*x+4)*exp(x)+128*x^3-24*x+4)*ln(-6*x
-3)),x,method=_RETURNVERBOSE)

[Out]

x/(x*ln(ln(-6*x-3))+8*x-exp(x)-2)

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maxima [C]  time = 0.83, size = 28, normalized size = 1.17 \begin {gather*} \frac {x}{x \log \left (i \, \pi + \log \relax (3) + \log \left (2 \, x + 1\right )\right ) + 8 \, x - e^{x} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2-x-1)*exp(x)-4*x-2)*log(-6*x-3)-2*x^2)/((2*x^3+x^2)*log(-6*x-3)*log(log(-6*x-3))^2+((-4*x^2-
2*x)*exp(x)+32*x^3+8*x^2-4*x)*log(-6*x-3)*log(log(-6*x-3))+((2*x+1)*exp(x)^2+(-32*x^2-8*x+4)*exp(x)+128*x^3-24
*x+4)*log(-6*x-3)),x, algorithm="maxima")

[Out]

x/(x*log(I*pi + log(3) + log(2*x + 1)) + 8*x - e^x - 2)

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mupad [B]  time = 1.09, size = 226, normalized size = 9.42 \begin {gather*} \frac {x\,{\left (\ln \left (-6\,x-3\right )+2\,x\,\ln \left (-6\,x-3\right )\right )}^2\,\left (2\,\ln \left (-6\,x-3\right )+4\,x\,\ln \left (-6\,x-3\right )+2\,x^2\right )+x\,{\mathrm {e}}^x\,{\left (\ln \left (-6\,x-3\right )+2\,x\,\ln \left (-6\,x-3\right )\right )}^2\,\left (\ln \left (-6\,x-3\right )+x\,\ln \left (-6\,x-3\right )-2\,x^2\,\ln \left (-6\,x-3\right )\right )}{\ln \left (-6\,x-3\right )\,\left (2\,x+1\right )\,\left (8\,x-{\mathrm {e}}^x+x\,\ln \left (\ln \left (-6\,x-3\right )\right )-2\right )\,\left (2\,{\ln \left (-6\,x-3\right )}^2+8\,x^2\,{\ln \left (-6\,x-3\right )}^2+{\mathrm {e}}^x\,{\ln \left (-6\,x-3\right )}^2+8\,x\,{\ln \left (-6\,x-3\right )}^2+2\,x^2\,\ln \left (-6\,x-3\right )+4\,x^3\,\ln \left (-6\,x-3\right )-4\,x^3\,{\mathrm {e}}^x\,{\ln \left (-6\,x-3\right )}^2+3\,x\,{\mathrm {e}}^x\,{\ln \left (-6\,x-3\right )}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(- 6*x - 3)*(4*x + exp(x)*(x - 2*x^2 + 1) + 2) + 2*x^2)/(log(- 6*x - 3)*(exp(2*x)*(2*x + 1) - exp(x)*
(8*x + 32*x^2 - 4) - 24*x + 128*x^3 + 4) - log(log(- 6*x - 3))*log(- 6*x - 3)*(4*x + exp(x)*(2*x + 4*x^2) - 8*
x^2 - 32*x^3) + log(log(- 6*x - 3))^2*log(- 6*x - 3)*(x^2 + 2*x^3)),x)

[Out]

(x*(log(- 6*x - 3) + 2*x*log(- 6*x - 3))^2*(2*log(- 6*x - 3) + 4*x*log(- 6*x - 3) + 2*x^2) + x*exp(x)*(log(- 6
*x - 3) + 2*x*log(- 6*x - 3))^2*(log(- 6*x - 3) + x*log(- 6*x - 3) - 2*x^2*log(- 6*x - 3)))/(log(- 6*x - 3)*(2
*x + 1)*(8*x - exp(x) + x*log(log(- 6*x - 3)) - 2)*(2*log(- 6*x - 3)^2 + 8*x^2*log(- 6*x - 3)^2 + exp(x)*log(-
6*x - 3)^2 + 8*x*log(- 6*x - 3)^2 + 2*x^2*log(- 6*x - 3) + 4*x^3*log(- 6*x - 3) - 4*x^3*exp(x)*log(- 6*x - 3)
^2 + 3*x*exp(x)*log(- 6*x - 3)^2))

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sympy [A]  time = 0.48, size = 22, normalized size = 0.92 \begin {gather*} - \frac {x}{- x \log {\left (\log {\left (- 6 x - 3 \right )} \right )} - 8 x + e^{x} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2-x-1)*exp(x)-4*x-2)*ln(-6*x-3)-2*x**2)/((2*x**3+x**2)*ln(-6*x-3)*ln(ln(-6*x-3))**2+((-4*x**
2-2*x)*exp(x)+32*x**3+8*x**2-4*x)*ln(-6*x-3)*ln(ln(-6*x-3))+((2*x+1)*exp(x)**2+(-32*x**2-8*x+4)*exp(x)+128*x**
3-24*x+4)*ln(-6*x-3)),x)

[Out]

-x/(-x*log(log(-6*x - 3)) - 8*x + exp(x) + 2)

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