3.69.2 \(\int \frac {e^x (10 x+14 x^2+4 x^3+(16 x+8 x^2) \log (3)+4 x \log ^2(3))+(e^x (2 x+5 x^2+5 x^3+2 x^4+(4 x+8 x^2+4 x^3) \log (3)+(2 x+2 x^2) \log ^2(3))+e^x (2+5 x+5 x^2+2 x^3+(4+8 x+4 x^2) \log (3)+(2+2 x) \log ^2(3)) \log (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3))) \log (x+\log (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3))) \log (\log ^2(x+\log (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3))))}{(2 x+3 x^2+2 x^3+(4 x+4 x^2) \log (3)+2 x \log ^2(3)+(2+3 x+2 x^2+(4+4 x) \log (3)+2 \log ^2(3)) \log (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3))) \log (x+\log (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)))} \, dx\)

Optimal. Leaf size=25 \[ -3+e^x x \log \left (\log ^2\left (x+\log \left (x-2 (1+x+\log (3))^2\right )\right )\right ) \]

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Rubi [F]  time = 15.49, 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 {e^x \left (10 x+14 x^2+4 x^3+\left (16 x+8 x^2\right ) \log (3)+4 x \log ^2(3)\right )+\left (e^x \left (2 x+5 x^2+5 x^3+2 x^4+\left (4 x+8 x^2+4 x^3\right ) \log (3)+\left (2 x+2 x^2\right ) \log ^2(3)\right )+e^x \left (2+5 x+5 x^2+2 x^3+\left (4+8 x+4 x^2\right ) \log (3)+(2+2 x) \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (\log ^2\left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )\right )}{\left (2 x+3 x^2+2 x^3+\left (4 x+4 x^2\right ) \log (3)+2 x \log ^2(3)+\left (2+3 x+2 x^2+(4+4 x) \log (3)+2 \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(10*x + 14*x^2 + 4*x^3 + (16*x + 8*x^2)*Log[3] + 4*x*Log[3]^2) + (E^x*(2*x + 5*x^2 + 5*x^3 + 2*x^4 +
(4*x + 8*x^2 + 4*x^3)*Log[3] + (2*x + 2*x^2)*Log[3]^2) + E^x*(2 + 5*x + 5*x^2 + 2*x^3 + (4 + 8*x + 4*x^2)*Log[
3] + (2 + 2*x)*Log[3]^2)*Log[-2 - 3*x - 2*x^2 + (-4 - 4*x)*Log[3] - 2*Log[3]^2])*Log[x + Log[-2 - 3*x - 2*x^2
+ (-4 - 4*x)*Log[3] - 2*Log[3]^2]]*Log[Log[x + Log[-2 - 3*x - 2*x^2 + (-4 - 4*x)*Log[3] - 2*Log[3]^2]]^2])/((2
*x + 3*x^2 + 2*x^3 + (4*x + 4*x^2)*Log[3] + 2*x*Log[3]^2 + (2 + 3*x + 2*x^2 + (4 + 4*x)*Log[3] + 2*Log[3]^2)*L
og[-2 - 3*x - 2*x^2 + (-4 - 4*x)*Log[3] - 2*Log[3]^2])*Log[x + Log[-2 - 3*x - 2*x^2 + (-4 - 4*x)*Log[3] - 2*Lo
g[3]^2]]),x]

[Out]

4*Defer[Int][E^x/((x + Log[-2*x^2 - 2*(1 + Log[3])^2 - x*(3 + Log[81])])*Log[x + Log[-2*x^2 - 2*(1 + Log[3])^2
- x*(3 + Log[81])]]), x] + 2*Defer[Int][(E^x*x)/((x + Log[-2*x^2 - 2*(1 + Log[3])^2 - x*(3 + Log[81])])*Log[x
+ Log[-2*x^2 - 2*(1 + Log[3])^2 - x*(3 + Log[81])]]), x] - 2*(3 + Log[81])*(1 + (I*(3 + Log[81]))/Sqrt[7 + 16
*Log[3]^2 - Log[81]^2 + Log[6561]])*Defer[Int][E^x/((3 + 4*x + Log[81] - I*Sqrt[7 + 16*Log[3]^2 + 2*Log[81] -
Log[81]^2])*(x + Log[-2*x^2 - 2*(1 + Log[3])^2 - x*(3 + Log[81])])*Log[x + Log[-2*x^2 - 2*(1 + Log[3])^2 - x*(
3 + Log[81])]]), x] - ((32*I)*(1 + Log[3]^2 + Log[9])*Defer[Int][E^x/((-3 - 4*x - Log[81] + I*Sqrt[7 + 16*Log[
3]^2 + 2*Log[81] - Log[81]^2])*(x + Log[-2*x^2 - 2*(1 + Log[3])^2 - x*(3 + Log[81])])*Log[x + Log[-2*x^2 - 2*(
1 + Log[3])^2 - x*(3 + Log[81])]]), x])/Sqrt[7 + 16*Log[3]^2 - Log[81]^2 + Log[6561]] - ((32*I)*(1 + Log[3]^2
+ Log[9])*Defer[Int][E^x/((3 + 4*x + Log[81] + I*Sqrt[7 + 16*Log[3]^2 + 2*Log[81] - Log[81]^2])*(x + Log[-2*x^
2 - 2*(1 + Log[3])^2 - x*(3 + Log[81])])*Log[x + Log[-2*x^2 - 2*(1 + Log[3])^2 - x*(3 + Log[81])]]), x])/Sqrt[
7 + 16*Log[3]^2 - Log[81]^2 + Log[6561]] - 2*(3 + Log[81])*(1 - (I*(3 + Log[81]))/Sqrt[7 + 16*Log[3]^2 - Log[8
1]^2 + Log[6561]])*Defer[Int][E^x/((3 + 4*x + Log[81] + I*Sqrt[7 + 16*Log[3]^2 + 2*Log[81] - Log[81]^2])*(x +
Log[-2*x^2 - 2*(1 + Log[3])^2 - x*(3 + Log[81])])*Log[x + Log[-2*x^2 - 2*(1 + Log[3])^2 - x*(3 + Log[81])]]),
x] + Defer[Int][E^x*Log[Log[x + Log[-2*x^2 - 2*(1 + Log[3])^2 - x*(3 + Log[81])]]^2], x] + Defer[Int][E^x*x*Lo
g[Log[x + Log[-2*x^2 - 2*(1 + Log[3])^2 - x*(3 + Log[81])]]^2], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (10 x+14 x^2+4 x^3+\left (16 x+8 x^2\right ) \log (3)+4 x \log ^2(3)\right )+\left (e^x \left (2 x+5 x^2+5 x^3+2 x^4+\left (4 x+8 x^2+4 x^3\right ) \log (3)+\left (2 x+2 x^2\right ) \log ^2(3)\right )+e^x \left (2+5 x+5 x^2+2 x^3+\left (4+8 x+4 x^2\right ) \log (3)+(2+2 x) \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (\log ^2\left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )\right )}{\left (3 x^2+2 x^3+\left (4 x+4 x^2\right ) \log (3)+x \left (2+2 \log ^2(3)\right )+\left (2+3 x+2 x^2+(4+4 x) \log (3)+2 \log ^2(3)\right ) \log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right ) \log \left (x+\log \left (-2-3 x-2 x^2+(-4-4 x) \log (3)-2 \log ^2(3)\right )\right )} \, dx\\ &=\int e^x \left (\frac {2 x \left (5+2 x^2+8 \log (3)+2 \log ^2(3)+x (7+\log (81))\right )}{\left (2+2 x^2+2 \log ^2(3)+\log (81)+x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}+(1+x) \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right )\right ) \, dx\\ &=\int \left (\frac {2 e^x x \left (5+2 x^2+8 \log (3)+2 \log ^2(3)+x (7+\log (81))\right )}{\left (2+2 x^2+2 \log ^2(3)+\log (81)+x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}+e^x (1+x) \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right )\right ) \, dx\\ &=2 \int \frac {e^x x \left (5+2 x^2+8 \log (3)+2 \log ^2(3)+x (7+\log (81))\right )}{\left (2+2 x^2+2 \log ^2(3)+\log (81)+x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+\int e^x (1+x) \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx\\ &=2 \int \left (\frac {2 e^x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}+\frac {e^x x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}+\frac {e^x \left (-4-4 \log ^2(3)-x (3+\log (81))-\log (6561)\right )}{\left (2+2 x^2+2 \log ^2(3)+\log (81)+x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}\right ) \, dx+\int \left (e^x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right )+e^x x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right )\right ) \, dx\\ &=2 \int \frac {e^x x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+2 \int \frac {e^x \left (-4-4 \log ^2(3)-x (3+\log (81))-\log (6561)\right )}{\left (2+2 x^2+2 \log ^2(3)+\log (81)+x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+4 \int \frac {e^x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+\int e^x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx+\int e^x x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx\\ &=2 \int \left (\frac {4 e^x \left (1+\log ^2(3)+\log (9)\right )}{\left (-2-2 x^2-2 \log ^2(3)-\log (81)-x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}+\frac {e^x x (-3-\log (81))}{\left (2+2 x^2+2 \log ^2(3)+\log (81)+x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}\right ) \, dx+2 \int \frac {e^x x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+4 \int \frac {e^x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+\int e^x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx+\int e^x x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx\\ &=2 \int \frac {e^x x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+4 \int \frac {e^x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+\left (8 \left (1+\log ^2(3)+\log (9)\right )\right ) \int \frac {e^x}{\left (-2-2 x^2-2 \log ^2(3)-\log (81)-x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx-(2 (3+\log (81))) \int \frac {e^x x}{\left (2+2 x^2+2 \log ^2(3)+\log (81)+x (3+\log (81))\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+\int e^x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx+\int e^x x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx\\ &=2 \int \frac {e^x x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+4 \int \frac {e^x}{\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )} \, dx+\left (8 \left (1+\log ^2(3)+\log (9)\right )\right ) \int \left (-\frac {4 i e^x}{\sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)} \left (-3-4 x-\log (81)+i \sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)}\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}-\frac {4 i e^x}{\sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)} \left (3+4 x+\log (81)+i \sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)}\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}\right ) \, dx-(2 (3+\log (81))) \int \left (\frac {e^x \left (1-\frac {i (3+\log (81))}{\sqrt {7+16 \log ^2(3)-\log ^2(81)+\log (6561)}}\right )}{\left (3+4 x+\log (81)+i \sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)}\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}+\frac {e^x \left (1+\frac {i (3+\log (81))}{\sqrt {7+16 \log ^2(3)-\log ^2(81)+\log (6561)}}\right )}{\left (3+4 x+\log (81)-i \sqrt {7+16 \log ^2(3)+2 \log (81)-\log ^2(81)}\right ) \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right ) \log \left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )}\right ) \, dx+\int e^x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx+\int e^x x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 33, normalized size = 1.32 \begin {gather*} e^x x \log \left (\log ^2\left (x+\log \left (-2 x^2-2 (1+\log (3))^2-x (3+\log (81))\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(10*x + 14*x^2 + 4*x^3 + (16*x + 8*x^2)*Log[3] + 4*x*Log[3]^2) + (E^x*(2*x + 5*x^2 + 5*x^3 + 2*
x^4 + (4*x + 8*x^2 + 4*x^3)*Log[3] + (2*x + 2*x^2)*Log[3]^2) + E^x*(2 + 5*x + 5*x^2 + 2*x^3 + (4 + 8*x + 4*x^2
)*Log[3] + (2 + 2*x)*Log[3]^2)*Log[-2 - 3*x - 2*x^2 + (-4 - 4*x)*Log[3] - 2*Log[3]^2])*Log[x + Log[-2 - 3*x -
2*x^2 + (-4 - 4*x)*Log[3] - 2*Log[3]^2]]*Log[Log[x + Log[-2 - 3*x - 2*x^2 + (-4 - 4*x)*Log[3] - 2*Log[3]^2]]^2
])/((2*x + 3*x^2 + 2*x^3 + (4*x + 4*x^2)*Log[3] + 2*x*Log[3]^2 + (2 + 3*x + 2*x^2 + (4 + 4*x)*Log[3] + 2*Log[3
]^2)*Log[-2 - 3*x - 2*x^2 + (-4 - 4*x)*Log[3] - 2*Log[3]^2])*Log[x + Log[-2 - 3*x - 2*x^2 + (-4 - 4*x)*Log[3]
- 2*Log[3]^2]]),x]

[Out]

E^x*x*Log[Log[x + Log[-2*x^2 - 2*(1 + Log[3])^2 - x*(3 + Log[81])]]^2]

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fricas [A]  time = 0.67, size = 34, normalized size = 1.36 \begin {gather*} x e^{x} \log \left (\log \left (x + \log \left (-2 \, x^{2} - 4 \, {\left (x + 1\right )} \log \relax (3) - 2 \, \log \relax (3)^{2} - 3 \, x - 2\right )\right )^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((2*x+2)*log(3)^2+(4*x^2+8*x+4)*log(3)+2*x^3+5*x^2+5*x+2)*exp(x)*log(-2*log(3)^2+(-4*x-4)*log(3)-2
*x^2-3*x-2)+((2*x^2+2*x)*log(3)^2+(4*x^3+8*x^2+4*x)*log(3)+2*x^4+5*x^3+5*x^2+2*x)*exp(x))*log(log(-2*log(3)^2+
(-4*x-4)*log(3)-2*x^2-3*x-2)+x)*log(log(log(-2*log(3)^2+(-4*x-4)*log(3)-2*x^2-3*x-2)+x)^2)+(4*x*log(3)^2+(8*x^
2+16*x)*log(3)+4*x^3+14*x^2+10*x)*exp(x))/((2*log(3)^2+(4*x+4)*log(3)+2*x^2+3*x+2)*log(-2*log(3)^2+(-4*x-4)*lo
g(3)-2*x^2-3*x-2)+2*x*log(3)^2+(4*x^2+4*x)*log(3)+2*x^3+3*x^2+2*x)/log(log(-2*log(3)^2+(-4*x-4)*log(3)-2*x^2-3
*x-2)+x),x, algorithm="fricas")

[Out]

x*e^x*log(log(x + log(-2*x^2 - 4*(x + 1)*log(3) - 2*log(3)^2 - 3*x - 2))^2)

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giac [A]  time = 67.41, size = 36, normalized size = 1.44 \begin {gather*} x e^{x} \log \left (\log \left (x + \log \left (-2 \, x^{2} - 4 \, x \log \relax (3) - 2 \, \log \relax (3)^{2} - 3 \, x - 4 \, \log \relax (3) - 2\right )\right )^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((2*x+2)*log(3)^2+(4*x^2+8*x+4)*log(3)+2*x^3+5*x^2+5*x+2)*exp(x)*log(-2*log(3)^2+(-4*x-4)*log(3)-2
*x^2-3*x-2)+((2*x^2+2*x)*log(3)^2+(4*x^3+8*x^2+4*x)*log(3)+2*x^4+5*x^3+5*x^2+2*x)*exp(x))*log(log(-2*log(3)^2+
(-4*x-4)*log(3)-2*x^2-3*x-2)+x)*log(log(log(-2*log(3)^2+(-4*x-4)*log(3)-2*x^2-3*x-2)+x)^2)+(4*x*log(3)^2+(8*x^
2+16*x)*log(3)+4*x^3+14*x^2+10*x)*exp(x))/((2*log(3)^2+(4*x+4)*log(3)+2*x^2+3*x+2)*log(-2*log(3)^2+(-4*x-4)*lo
g(3)-2*x^2-3*x-2)+2*x*log(3)^2+(4*x^2+4*x)*log(3)+2*x^3+3*x^2+2*x)/log(log(-2*log(3)^2+(-4*x-4)*log(3)-2*x^2-3
*x-2)+x),x, algorithm="giac")

[Out]

x*e^x*log(log(x + log(-2*x^2 - 4*x*log(3) - 2*log(3)^2 - 3*x - 4*log(3) - 2))^2)

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maple [C]  time = 0.33, size = 216, normalized size = 8.64

method	result	size
risch	\(2 x \,{\mathrm e}^{x} \ln \left (\ln \left (\ln \left (-2 \ln \relax (3)^{2}+\left (-4 x -4\right ) \ln \relax (3)-2 x^{2}-3 x -2\right )+x \right )\right )-\frac {i \pi \,\mathrm {csgn}\left (i \ln \left (\ln \left (-2 \ln \relax (3)^{2}+\left (-4 x -4\right ) \ln \relax (3)-2 x^{2}-3 x -2\right )+x \right )^{2}\right ) \left (\mathrm {csgn}\left (i \ln \left (\ln \left (-2 \ln \relax (3)^{2}+\left (-4 x -4\right ) \ln \relax (3)-2 x^{2}-3 x -2\right )+x \right )\right )^{2}-2 \,\mathrm {csgn}\left (i \ln \left (\ln \left (-2 \ln \relax (3)^{2}+\left (-4 x -4\right ) \ln \relax (3)-2 x^{2}-3 x -2\right )+x \right )^{2}\right ) \mathrm {csgn}\left (i \ln \left (\ln \left (-2 \ln \relax (3)^{2}+\left (-4 x -4\right ) \ln \relax (3)-2 x^{2}-3 x -2\right )+x \right )\right )+\mathrm {csgn}\left (i \ln \left (\ln \left (-2 \ln \relax (3)^{2}+\left (-4 x -4\right ) \ln \relax (3)-2 x^{2}-3 x -2\right )+x \right )^{2}\right )^{2}\right ) x \,{\mathrm e}^{x}}{2}\)	\(216\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((2*x+2)*ln(3)^2+(4*x^2+8*x+4)*ln(3)+2*x^3+5*x^2+5*x+2)*exp(x)*ln(-2*ln(3)^2+(-4*x-4)*ln(3)-2*x^2-3*x-2)
+((2*x^2+2*x)*ln(3)^2+(4*x^3+8*x^2+4*x)*ln(3)+2*x^4+5*x^3+5*x^2+2*x)*exp(x))*ln(ln(-2*ln(3)^2+(-4*x-4)*ln(3)-2
*x^2-3*x-2)+x)*ln(ln(ln(-2*ln(3)^2+(-4*x-4)*ln(3)-2*x^2-3*x-2)+x)^2)+(4*x*ln(3)^2+(8*x^2+16*x)*ln(3)+4*x^3+14*
x^2+10*x)*exp(x))/((2*ln(3)^2+(4*x+4)*ln(3)+2*x^2+3*x+2)*ln(-2*ln(3)^2+(-4*x-4)*ln(3)-2*x^2-3*x-2)+2*x*ln(3)^2
+(4*x^2+4*x)*ln(3)+2*x^3+3*x^2+2*x)/ln(ln(-2*ln(3)^2+(-4*x-4)*ln(3)-2*x^2-3*x-2)+x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.62, size = 36, normalized size = 1.44 \begin {gather*} 2 \, x e^{x} \log \left (\log \left (x + \log \left (-2 \, x^{2} - x {\left (4 \, \log \relax (3) + 3\right )} - 2 \, \log \relax (3)^{2} - 4 \, \log \relax (3) - 2\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((2*x+2)*log(3)^2+(4*x^2+8*x+4)*log(3)+2*x^3+5*x^2+5*x+2)*exp(x)*log(-2*log(3)^2+(-4*x-4)*log(3)-2
*x^2-3*x-2)+((2*x^2+2*x)*log(3)^2+(4*x^3+8*x^2+4*x)*log(3)+2*x^4+5*x^3+5*x^2+2*x)*exp(x))*log(log(-2*log(3)^2+
(-4*x-4)*log(3)-2*x^2-3*x-2)+x)*log(log(log(-2*log(3)^2+(-4*x-4)*log(3)-2*x^2-3*x-2)+x)^2)+(4*x*log(3)^2+(8*x^
2+16*x)*log(3)+4*x^3+14*x^2+10*x)*exp(x))/((2*log(3)^2+(4*x+4)*log(3)+2*x^2+3*x+2)*log(-2*log(3)^2+(-4*x-4)*lo
g(3)-2*x^2-3*x-2)+2*x*log(3)^2+(4*x^2+4*x)*log(3)+2*x^3+3*x^2+2*x)/log(log(-2*log(3)^2+(-4*x-4)*log(3)-2*x^2-3
*x-2)+x),x, algorithm="maxima")

[Out]

2*x*e^x*log(log(x + log(-2*x^2 - x*(4*log(3) + 3) - 2*log(3)^2 - 4*log(3) - 2)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^x\,\left (10\,x+\ln \relax (3)\,\left (8\,x^2+16\,x\right )+4\,x\,{\ln \relax (3)}^2+14\,x^2+4\,x^3\right )+\ln \left ({\ln \left (x+\ln \left (-3\,x-\ln \relax (3)\,\left (4\,x+4\right )-2\,{\ln \relax (3)}^2-2\,x^2-2\right )\right )}^2\right )\,\ln \left (x+\ln \left (-3\,x-\ln \relax (3)\,\left (4\,x+4\right )-2\,{\ln \relax (3)}^2-2\,x^2-2\right )\right )\,\left ({\mathrm {e}}^x\,\left (2\,x+\ln \relax (3)\,\left (4\,x^3+8\,x^2+4\,x\right )+{\ln \relax (3)}^2\,\left (2\,x^2+2\,x\right )+5\,x^2+5\,x^3+2\,x^4\right )+\ln \left (-3\,x-\ln \relax (3)\,\left (4\,x+4\right )-2\,{\ln \relax (3)}^2-2\,x^2-2\right )\,{\mathrm {e}}^x\,\left (5\,x+\ln \relax (3)\,\left (4\,x^2+8\,x+4\right )+{\ln \relax (3)}^2\,\left (2\,x+2\right )+5\,x^2+2\,x^3+2\right )\right )}{\ln \left (x+\ln \left (-3\,x-\ln \relax (3)\,\left (4\,x+4\right )-2\,{\ln \relax (3)}^2-2\,x^2-2\right )\right )\,\left (2\,x+\ln \left (-3\,x-\ln \relax (3)\,\left (4\,x+4\right )-2\,{\ln \relax (3)}^2-2\,x^2-2\right )\,\left (3\,x+\ln \relax (3)\,\left (4\,x+4\right )+2\,{\ln \relax (3)}^2+2\,x^2+2\right )+\ln \relax (3)\,\left (4\,x^2+4\,x\right )+2\,x\,{\ln \relax (3)}^2+3\,x^2+2\,x^3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(10*x + log(3)*(16*x + 8*x^2) + 4*x*log(3)^2 + 14*x^2 + 4*x^3) + log(log(x + log(- 3*x - log(3)*(4
*x + 4) - 2*log(3)^2 - 2*x^2 - 2))^2)*log(x + log(- 3*x - log(3)*(4*x + 4) - 2*log(3)^2 - 2*x^2 - 2))*(exp(x)*
(2*x + log(3)*(4*x + 8*x^2 + 4*x^3) + log(3)^2*(2*x + 2*x^2) + 5*x^2 + 5*x^3 + 2*x^4) + log(- 3*x - log(3)*(4*
x + 4) - 2*log(3)^2 - 2*x^2 - 2)*exp(x)*(5*x + log(3)*(8*x + 4*x^2 + 4) + log(3)^2*(2*x + 2) + 5*x^2 + 2*x^3 +
2)))/(log(x + log(- 3*x - log(3)*(4*x + 4) - 2*log(3)^2 - 2*x^2 - 2))*(2*x + log(- 3*x - log(3)*(4*x + 4) - 2
*log(3)^2 - 2*x^2 - 2)*(3*x + log(3)*(4*x + 4) + 2*log(3)^2 + 2*x^2 + 2) + log(3)*(4*x + 4*x^2) + 2*x*log(3)^2
+ 3*x^2 + 2*x^3)),x)

[Out]

int((exp(x)*(10*x + log(3)*(16*x + 8*x^2) + 4*x*log(3)^2 + 14*x^2 + 4*x^3) + log(log(x + log(- 3*x - log(3)*(4
*x + 4) - 2*log(3)^2 - 2*x^2 - 2))^2)*log(x + log(- 3*x - log(3)*(4*x + 4) - 2*log(3)^2 - 2*x^2 - 2))*(exp(x)*
(2*x + log(3)*(4*x + 8*x^2 + 4*x^3) + log(3)^2*(2*x + 2*x^2) + 5*x^2 + 5*x^3 + 2*x^4) + log(- 3*x - log(3)*(4*
x + 4) - 2*log(3)^2 - 2*x^2 - 2)*exp(x)*(5*x + log(3)*(8*x + 4*x^2 + 4) + log(3)^2*(2*x + 2) + 5*x^2 + 2*x^3 +
2)))/(log(x + log(- 3*x - log(3)*(4*x + 4) - 2*log(3)^2 - 2*x^2 - 2))*(2*x + log(- 3*x - log(3)*(4*x + 4) - 2
*log(3)^2 - 2*x^2 - 2)*(3*x + log(3)*(4*x + 4) + 2*log(3)^2 + 2*x^2 + 2) + log(3)*(4*x + 4*x^2) + 2*x*log(3)^2
+ 3*x^2 + 2*x^3)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((2*x+2)*ln(3)**2+(4*x**2+8*x+4)*ln(3)+2*x**3+5*x**2+5*x+2)*exp(x)*ln(-2*ln(3)**2+(-4*x-4)*ln(3)-2
*x**2-3*x-2)+((2*x**2+2*x)*ln(3)**2+(4*x**3+8*x**2+4*x)*ln(3)+2*x**4+5*x**3+5*x**2+2*x)*exp(x))*ln(ln(-2*ln(3)
**2+(-4*x-4)*ln(3)-2*x**2-3*x-2)+x)*ln(ln(ln(-2*ln(3)**2+(-4*x-4)*ln(3)-2*x**2-3*x-2)+x)**2)+(4*x*ln(3)**2+(8*
x**2+16*x)*ln(3)+4*x**3+14*x**2+10*x)*exp(x))/((2*ln(3)**2+(4*x+4)*ln(3)+2*x**2+3*x+2)*ln(-2*ln(3)**2+(-4*x-4)
*ln(3)-2*x**2-3*x-2)+2*x*ln(3)**2+(4*x**2+4*x)*ln(3)+2*x**3+3*x**2+2*x)/ln(ln(-2*ln(3)**2+(-4*x-4)*ln(3)-2*x**
2-3*x-2)+x),x)

[Out]

Timed out

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