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

Optimal. Leaf size=25 $$-3+e^{x}x\log ({\log }^2(x+\log (x-2(1+x+\log (3){)}^2)))$$

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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 = {} $$\begin{array}{c}\int \frac{e^x(10x+14x^2+4x^3+(16x+8x^2)\log (3)+4x{\log }^2(3))+(e^x(2x+5x^2+5x^3+2x^4+(4x+8x^2+4x^3)\log (3)+(2x+2x^2){\log }^2(3))+e^x(2+5x+5x^2+2x^3+(4+8x+4x^2)\log (3)+(2+2x){\log }^2(3))\log (-2-3x-2x^2+(-4-4x)\log (3)-2{\log }^2(3)))\log (x+\log (-2-3x-2x^2+(-4-4x)\log (3)-2{\log }^2(3)))\log ({\log }^2(x+\log (-2-3x-2x^2+(-4-4x)\log (3)-2{\log }^2(3))))}{(2x+3x^2+2x^3+(4x+4x^2)\log (3)+2x{\log }^2(3)+(2+3x+2x^2+(4+4x)\log (3)+2{\log }^2(3))\log (-2-3x-2x^2+(-4-4x)\log (3)-2{\log }^2(3)))\log (x+\log (-2-3x-2x^2+(-4-4x)\log (3)-2{\log }^2(3)))}dx\end{array}$$

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

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

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 $$\begin{array}{c}x{e}^x\log (\log {(x+\log (-2x^2-4(x+1)\log (3)-2\log (3{)}^2-3x-2))}^2)\end{array}$$

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 $$\begin{array}{c}x{e}^x\log (\log {(x+\log (-2x^2-4x\log (3)-2\log (3{)}^2-3x-4\log (3)-2))}^2)\end{array}$$

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	$2x{\mathrm{e}}^x\ln (\ln (\ln (-2\ln (3{)}^2+(-4x-4)\ln (3)-2x^2-3x-2)+x))-\frac{i\pi \mathrm{csgn}(i\ln {(\ln (-2\ln (3{)}^2+(-4x-4)\ln (3)-2x^2-3x-2)+x)}^2)(\mathrm{csgn}{(i\ln (\ln (-2\ln (3{)}^2+(-4x-4)\ln (3)-2x^2-3x-2)+x))}^2-2\mathrm{csgn}(i\ln {(\ln (-2\ln (3{)}^2+(-4x-4)\ln (3)-2x^2-3x-2)+x)}^2)\mathrm{csgn}(i\ln (\ln (-2\ln (3{)}^2+(-4x-4)\ln (3)-2x^2-3x-2)+x))+\mathrm{csgn}{(i\ln {(\ln (-2\ln (3{)}^2+(-4x-4)\ln (3)-2x^2-3x-2)+x)}^2)}^2)x{\mathrm{e}}^x}{2}$	$216$

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 $$\begin{array}{c}2x{e}^x\log (\log (x+\log (-2x^2-x(4\log (3)+3)-2\log (3{)}^2-4\log (3)-2)))\end{array}$$

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 $$\begin{array}{c}\int \frac{{\mathrm{e}}^x(10x+\ln (3)(8x^2+16x)+4x{\ln (3)}^2+14x^2+4x^3)+\ln \left({\ln (x+\ln (-3x-\ln (3)(4x+4)-2{\ln (3)}^2-2x^2-2))}^2\right)\ln (x+\ln (-3x-\ln (3)(4x+4)-2{\ln (3)}^2-2x^2-2))({\mathrm{e}}^x(2x+\ln (3)(4x^3+8x^2+4x)+{\ln (3)}^2(2x^2+2x)+5x^2+5x^3+2x^4)+\ln (-3x-\ln (3)(4x+4)-2{\ln (3)}^2-2x^2-2){\mathrm{e}}^x(5x+\ln (3)(4x^2+8x+4)+{\ln (3)}^2(2x+2)+5x^2+2x^3+2))}{\ln (x+\ln (-3x-\ln (3)(4x+4)-2{\ln (3)}^2-2x^2-2))(2x+\ln (-3x-\ln (3)(4x+4)-2{\ln (3)}^2-2x^2-2)(3x+\ln (3)(4x+4)+2{\ln (3)}^2+2x^2+2)+\ln (3)(4x^2+4x)+2x{\ln (3)}^2+3x^2+2x^3)}dx\end{array}$$

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 $$\begin{array}{c}\text{Timed out}\end{array}$$

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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