3.63.50 \(\int \frac {e^{2 x} (4 x+16 x^2)+e^x (-8 x-32 x^2) \log (3)+(4 x+16 x^2) \log ^2(3)+(e^{2 x} (-8 x-16 x^2)-8 x^3 \log (3)+(-8 x-16 x^2) \log ^2(3)+e^x (8 x^3+(16 x+32 x^2) \log (3))) \log (x)-4 e^x x^4 \log ^2(x)}{(e^{2 x} (1+8 x+16 x^2)+e^x (-2-16 x-32 x^2) \log (3)+(1+8 x+16 x^2) \log ^2(3)) \log ^2(x)+(e^x (2 x^2+8 x^3)+(-2 x^2-8 x^3) \log (3)) \log ^3(x)+x^4 \log ^4(x)} \, dx\)

Optimal. Leaf size=34 \[ \frac {4 x}{\log (x) \left (-\frac {1+4 x}{x}+\frac {x \log (x)}{-e^x+\log (3)}\right )} \]

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Rubi [F]  time = 76.64, 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^{2 x} \left (4 x+16 x^2\right )+e^x \left (-8 x-32 x^2\right ) \log (3)+\left (4 x+16 x^2\right ) \log ^2(3)+\left (e^{2 x} \left (-8 x-16 x^2\right )-8 x^3 \log (3)+\left (-8 x-16 x^2\right ) \log ^2(3)+e^x \left (8 x^3+\left (16 x+32 x^2\right ) \log (3)\right )\right ) \log (x)-4 e^x x^4 \log ^2(x)}{\left (e^{2 x} \left (1+8 x+16 x^2\right )+e^x \left (-2-16 x-32 x^2\right ) \log (3)+\left (1+8 x+16 x^2\right ) \log ^2(3)\right ) \log ^2(x)+\left (e^x \left (2 x^2+8 x^3\right )+\left (-2 x^2-8 x^3\right ) \log (3)\right ) \log ^3(x)+x^4 \log ^4(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*x)*(4*x + 16*x^2) + E^x*(-8*x - 32*x^2)*Log[3] + (4*x + 16*x^2)*Log[3]^2 + (E^(2*x)*(-8*x - 16*x^2)
- 8*x^3*Log[3] + (-8*x - 16*x^2)*Log[3]^2 + E^x*(8*x^3 + (16*x + 32*x^2)*Log[3]))*Log[x] - 4*E^x*x^4*Log[x]^2)
/((E^(2*x)*(1 + 8*x + 16*x^2) + E^x*(-2 - 16*x - 32*x^2)*Log[3] + (1 + 8*x + 16*x^2)*Log[3]^2)*Log[x]^2 + (E^x
*(2*x^2 + 8*x^3) + (-2*x^2 - 8*x^3)*Log[3])*Log[x]^3 + x^4*Log[x]^4),x]

[Out]

4*Defer[Int][x/((1 + 4*x)*Log[x]^2), x] - 8*Defer[Int][(x*(1 + 2*x))/((1 + 4*x)^2*Log[x]), x] - (3*Log[3]*Defe
r[Int][(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^(-2), x])/64 - (5*(1 + Log[81])*Defer[Int][(E^x + 4*
E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^(-2), x])/256 + ((1 + Log[6561])*Defer[Int][(E^x + 4*E^x*x - Log[3]
- 4*x*Log[3] + x^2*Log[x])^(-2), x])/64 + (Log[3]*Defer[Int][x/(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[
x])^2, x])/8 + ((1 + Log[81])*Defer[Int][x/(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2, x])/16 - (3*(
1 + Log[6561])*Defer[Int][x/(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2, x])/64 - (Log[3]*Defer[Int][
x^2/(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2, x])/4 - (3*(1 + Log[81])*Defer[Int][x^2/(E^x + 4*E^x
*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2, x])/16 + ((1 + Log[6561])*Defer[Int][x^2/(E^x + 4*E^x*x - Log[3] - 4
*x*Log[3] + x^2*Log[x])^2, x])/8 + ((1 + Log[81])*Defer[Int][x^3/(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Lo
g[x])^2, x])/2 - ((1 + Log[6561])*Defer[Int][x^3/(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2, x])/4 -
(1 + Log[81])*Defer[Int][x^4/(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2, x] - (Log[3]*Defer[Int][1/
((1 + 4*x)^2*(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2), x])/64 - ((1 + Log[81])*Defer[Int][1/((1 +
4*x)^2*(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2), x])/256 + ((1 + Log[6561])*Defer[Int][1/((1 + 4
*x)^2*(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2), x])/256 + (Log[3]*Defer[Int][1/((1 + 4*x)*(E^x +
4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2), x])/16 + (3*(1 + Log[81])*Defer[Int][1/((1 + 4*x)*(E^x + 4*E^x
*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2), x])/128 - (5*(1 + Log[6561])*Defer[Int][1/((1 + 4*x)*(E^x + 4*E^x*x
- Log[3] - 4*x*Log[3] + x^2*Log[x])^2), x])/256 - ((40*Log[3]^2 - 16*Log[3]*Log[9] - Log[9]*Log[81])*Defer[In
t][1/(Log[x]*(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2), x])/4 + (4*Log[3]^2 - Log[9]^2)*Defer[Int]
[1/((1 + 4*x)^2*Log[x]*(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2), x] - ((40*Log[3]^2 - 16*Log[3]*L
og[9] - Log[9]*Log[81])*Defer[Int][1/((1 + 4*x)^2*Log[x]*(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2)
, x])/4 - (4*Log[3]^2 - Log[9]^2)*Defer[Int][1/((1 + 4*x)*Log[x]*(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Lo
g[x])^2), x] + ((40*Log[3]^2 - 16*Log[3]*Log[9] - Log[9]*Log[81])*Defer[Int][1/((1 + 4*x)*Log[x]*(E^x + 4*E^x*
x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2), x])/2 + (11*Defer[Int][Log[x]/(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] +
x^2*Log[x])^2, x])/1024 - (7*Defer[Int][(x*Log[x])/(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2, x])/
256 + (3*Defer[Int][(x^2*Log[x])/(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2, x])/64 + Defer[Int][(x^
3*Log[x])/(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2, x]/16 - (5*Defer[Int][(x^4*Log[x])/(E^x + 4*E^
x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2, x])/4 + Defer[Int][(x^5*Log[x])/(E^x + 4*E^x*x - Log[3] - 4*x*Log[3
] + x^2*Log[x])^2, x] + Defer[Int][Log[x]/((1 + 4*x)^2*(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2),
x]/256 - (15*Defer[Int][Log[x]/((1 + 4*x)*(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^2), x])/1024 - (7
*Defer[Int][(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])^(-1), x])/64 - Defer[Int][x/(E^x + 4*E^x*x - Lo
g[3] - 4*x*Log[3] + x^2*Log[x]), x]/16 + (9*Defer[Int][x^2/(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x]),
x])/4 - Defer[Int][x^3/(E^x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x]), x] - Defer[Int][1/((1 + 4*x)^2*(E^
x + 4*E^x*x - Log[3] - 4*x*Log[3] + x^2*Log[x])), x]/8 + (15*Defer[Int][1/((1 + 4*x)*(E^x + 4*E^x*x - Log[3] -
4*x*Log[3] + x^2*Log[x])), x])/64

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x \left (e^{2 x} (1+4 x)+(1+4 x) \log ^2(3)-e^x (8 x \log (3)+\log (9))-\left (e^{2 x} (2+4 x)+2 \log ^2(3)+4 x \log ^2(3)+x^2 \log (9)-2 e^x \left (x^2+\log (9)+x \log (81)\right )\right ) \log (x)-e^x x^3 \log ^2(x)\right )}{\log ^2(x) \left ((1+4 x) \left (e^x-\log (3)\right )+x^2 \log (x)\right )^2} \, dx\\ &=4 \int \frac {x \left (e^{2 x} (1+4 x)+(1+4 x) \log ^2(3)-e^x (8 x \log (3)+\log (9))-\left (e^{2 x} (2+4 x)+2 \log ^2(3)+4 x \log ^2(3)+x^2 \log (9)-2 e^x \left (x^2+\log (9)+x \log (81)\right )\right ) \log (x)-e^x x^3 \log ^2(x)\right )}{\log ^2(x) \left ((1+4 x) \left (e^x-\log (3)\right )+x^2 \log (x)\right )^2} \, dx\\ &=4 \int \left (-\frac {x (-1-4 x+2 \log (x)+4 x \log (x))}{(1+4 x)^2 \log ^2(x)}-\frac {x^3 \left (-4-7 x+4 x^2\right )}{(1+4 x)^2 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )}+\frac {x \left (-4 \log ^2(3) \left (1-\frac {\log ^2(9)}{4 \log ^2(3)}\right )-40 x \log ^2(3) \left (1-\frac {\log (9) (16 \log (3)+\log (81))}{40 \log ^2(3)}\right )-x^3 \log (3) \log (x)-4 x^5 (1+\log (81)) \log (x)-x^4 (1+\log (6561)) \log (x)-2 x^4 \log ^2(x)-3 x^5 \log ^2(x)+4 x^6 \log ^2(x)\right )}{(1+4 x)^2 \log (x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {x (-1-4 x+2 \log (x)+4 x \log (x))}{(1+4 x)^2 \log ^2(x)} \, dx\right )-4 \int \frac {x^3 \left (-4-7 x+4 x^2\right )}{(1+4 x)^2 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )} \, dx+4 \int \frac {x \left (-4 \log ^2(3) \left (1-\frac {\log ^2(9)}{4 \log ^2(3)}\right )-40 x \log ^2(3) \left (1-\frac {\log (9) (16 \log (3)+\log (81))}{40 \log ^2(3)}\right )-x^3 \log (3) \log (x)-4 x^5 (1+\log (81)) \log (x)-x^4 (1+\log (6561)) \log (x)-2 x^4 \log ^2(x)-3 x^5 \log ^2(x)+4 x^6 \log ^2(x)\right )}{(1+4 x)^2 \log (x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )^2} \, dx\\ &=-\left (4 \int \left (-\frac {x}{(1+4 x) \log ^2(x)}+\frac {2 x (1+2 x)}{(1+4 x)^2 \log (x)}\right ) \, dx\right )-4 \int \left (\frac {7}{256 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )}+\frac {x}{64 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )}-\frac {9 x^2}{16 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )}+\frac {x^3}{4 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )}+\frac {1}{32 (1+4 x)^2 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )}-\frac {15}{256 (1+4 x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )}\right ) \, dx+4 \int \left (\frac {4 \log ^2(3) \left (1-\frac {\log ^2(9)}{4 \log ^2(3)}\right )+40 x \log ^2(3) \left (1-\frac {\log (9) (16 \log (3)+\log (81))}{40 \log ^2(3)}\right )+x^3 \log (3) \log (x)+4 x^5 (1+\log (81)) \log (x)+x^4 (1+\log (6561)) \log (x)+2 x^4 \log ^2(x)+3 x^5 \log ^2(x)-4 x^6 \log ^2(x)}{4 (1+4 x)^2 \log (x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )^2}+\frac {-4 \log ^2(3) \left (1-\frac {\log ^2(9)}{4 \log ^2(3)}\right )-40 x \log ^2(3) \left (1-\frac {\log (9) (16 \log (3)+\log (81))}{40 \log ^2(3)}\right )-x^3 \log (3) \log (x)-4 x^5 (1+\log (81)) \log (x)-x^4 (1+\log (6561)) \log (x)-2 x^4 \log ^2(x)-3 x^5 \log ^2(x)+4 x^6 \log ^2(x)}{4 (1+4 x) \log (x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )^2}\right ) \, dx\\ &=-\left (\frac {1}{16} \int \frac {x}{e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)} \, dx\right )-\frac {7}{64} \int \frac {1}{e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)} \, dx-\frac {1}{8} \int \frac {1}{(1+4 x)^2 \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )} \, dx+\frac {15}{64} \int \frac {1}{(1+4 x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )} \, dx+\frac {9}{4} \int \frac {x^2}{e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)} \, dx+4 \int \frac {x}{(1+4 x) \log ^2(x)} \, dx-8 \int \frac {x (1+2 x)}{(1+4 x)^2 \log (x)} \, dx-\int \frac {x^3}{e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)} \, dx+\int \frac {4 \log ^2(3) \left (1-\frac {\log ^2(9)}{4 \log ^2(3)}\right )+40 x \log ^2(3) \left (1-\frac {\log (9) (16 \log (3)+\log (81))}{40 \log ^2(3)}\right )+x^3 \log (3) \log (x)+4 x^5 (1+\log (81)) \log (x)+x^4 (1+\log (6561)) \log (x)+2 x^4 \log ^2(x)+3 x^5 \log ^2(x)-4 x^6 \log ^2(x)}{(1+4 x)^2 \log (x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )^2} \, dx+\int \frac {-4 \log ^2(3) \left (1-\frac {\log ^2(9)}{4 \log ^2(3)}\right )-40 x \log ^2(3) \left (1-\frac {\log (9) (16 \log (3)+\log (81))}{40 \log ^2(3)}\right )-x^3 \log (3) \log (x)-4 x^5 (1+\log (81)) \log (x)-x^4 (1+\log (6561)) \log (x)-2 x^4 \log ^2(x)-3 x^5 \log ^2(x)+4 x^6 \log ^2(x)}{(1+4 x) \log (x) \left (e^x+4 e^x x-\log (3)-4 x \log (3)+x^2 \log (x)\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.67, size = 290, normalized size = 8.53 \begin {gather*} -\frac {4 x^2 \left (\frac {e^{2 x} (1+4 x)+(1+4 x) \log ^2(3)-e^x (8 x \log (3)+\log (9))}{\log (x)}-\frac {x^2 \left (e^{3 x} \left (-2-11 x-8 x^2+16 x^3\right )+2 \log ^3(3)+12 x \log ^3(3)+x^2 \log ^2(3) (1+16 \log (3))+x^3 \left (-4 \log ^2(3)+\log (9) \log (81)\right )+e^x \left (-2 \log ^2(3)+4 x^3 \left (4 \log ^2(3)-\log (9)\right )-\log ^2(9)-x^2 \left (8 \log ^2(3)+\log (9)+8 \log (3) \log (81)\right )-x \left (11 \log ^2(3)+8 \log (3) \log (9)+\log (9) \log (81)\right )\right )+e^{2 x} \left (x^3 (4-32 \log (3))+x^2 (1+8 \log (81))+\log (729)+2 x \log (129140163)\right )\right )}{\left (x^2+e^x \left (-2-3 x+4 x^2\right )+\log (9)+x \log (81)\right ) \left ((1+4 x) \left (e^x-\log (3)\right )+x^2 \log (x)\right )}\right )}{(1+4 x)^2 \left (e^x-\log (3)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(4*x + 16*x^2) + E^x*(-8*x - 32*x^2)*Log[3] + (4*x + 16*x^2)*Log[3]^2 + (E^(2*x)*(-8*x - 16
*x^2) - 8*x^3*Log[3] + (-8*x - 16*x^2)*Log[3]^2 + E^x*(8*x^3 + (16*x + 32*x^2)*Log[3]))*Log[x] - 4*E^x*x^4*Log
[x]^2)/((E^(2*x)*(1 + 8*x + 16*x^2) + E^x*(-2 - 16*x - 32*x^2)*Log[3] + (1 + 8*x + 16*x^2)*Log[3]^2)*Log[x]^2
+ (E^x*(2*x^2 + 8*x^3) + (-2*x^2 - 8*x^3)*Log[3])*Log[x]^3 + x^4*Log[x]^4),x]

[Out]

(-4*x^2*((E^(2*x)*(1 + 4*x) + (1 + 4*x)*Log[3]^2 - E^x*(8*x*Log[3] + Log[9]))/Log[x] - (x^2*(E^(3*x)*(-2 - 11*
x - 8*x^2 + 16*x^3) + 2*Log[3]^3 + 12*x*Log[3]^3 + x^2*Log[3]^2*(1 + 16*Log[3]) + x^3*(-4*Log[3]^2 + Log[9]*Lo
g[81]) + E^x*(-2*Log[3]^2 + 4*x^3*(4*Log[3]^2 - Log[9]) - Log[9]^2 - x^2*(8*Log[3]^2 + Log[9] + 8*Log[3]*Log[8
1]) - x*(11*Log[3]^2 + 8*Log[3]*Log[9] + Log[9]*Log[81])) + E^(2*x)*(x^3*(4 - 32*Log[3]) + x^2*(1 + 8*Log[81])
+ Log[729] + 2*x*Log[129140163])))/((x^2 + E^x*(-2 - 3*x + 4*x^2) + Log[9] + x*Log[81])*((1 + 4*x)*(E^x - Log
[3]) + x^2*Log[x]))))/((1 + 4*x)^2*(E^x - Log[3])^2)

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fricas [A]  time = 0.84, size = 48, normalized size = 1.41 \begin {gather*} -\frac {4 \, {\left (x^{2} e^{x} - x^{2} \log \relax (3)\right )}}{x^{2} \log \relax (x)^{2} + {\left ({\left (4 \, x + 1\right )} e^{x} - {\left (4 \, x + 1\right )} \log \relax (3)\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^4*exp(x)*log(x)^2+((-16*x^2-8*x)*exp(x)^2+((32*x^2+16*x)*log(3)+8*x^3)*exp(x)+(-16*x^2-8*x)*lo
g(3)^2-8*x^3*log(3))*log(x)+(16*x^2+4*x)*exp(x)^2+(-32*x^2-8*x)*log(3)*exp(x)+(16*x^2+4*x)*log(3)^2)/(x^4*log(
x)^4+((8*x^3+2*x^2)*exp(x)+(-8*x^3-2*x^2)*log(3))*log(x)^3+((16*x^2+8*x+1)*exp(x)^2+(-32*x^2-16*x-2)*log(3)*ex
p(x)+(16*x^2+8*x+1)*log(3)^2)*log(x)^2),x, algorithm="fricas")

[Out]

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

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giac [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((-4*x^4*exp(x)*log(x)^2+((-16*x^2-8*x)*exp(x)^2+((32*x^2+16*x)*log(3)+8*x^3)*exp(x)+(-16*x^2-8*x)*lo
g(3)^2-8*x^3*log(3))*log(x)+(16*x^2+4*x)*exp(x)^2+(-32*x^2-8*x)*log(3)*exp(x)+(16*x^2+4*x)*log(3)^2)/(x^4*log(
x)^4+((8*x^3+2*x^2)*exp(x)+(-8*x^3-2*x^2)*log(3))*log(x)^3+((16*x^2+8*x+1)*exp(x)^2+(-32*x^2-16*x-2)*log(3)*ex
p(x)+(16*x^2+8*x+1)*log(3)^2)*log(x)^2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.08, size = 43, normalized size = 1.26

method	result	size
risch	\(-\frac {4 \left (\ln \relax (3)-{\mathrm e}^{x}\right ) x^{2}}{\left (-x^{2} \ln \relax (x )+4 x \ln \relax (3)-4 \,{\mathrm e}^{x} x +\ln \relax (3)-{\mathrm e}^{x}\right ) \ln \relax (x )}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x^4*exp(x)*ln(x)^2+((-16*x^2-8*x)*exp(x)^2+((32*x^2+16*x)*ln(3)+8*x^3)*exp(x)+(-16*x^2-8*x)*ln(3)^2-8*
x^3*ln(3))*ln(x)+(16*x^2+4*x)*exp(x)^2+(-32*x^2-8*x)*ln(3)*exp(x)+(16*x^2+4*x)*ln(3)^2)/(x^4*ln(x)^4+((8*x^3+2
*x^2)*exp(x)+(-8*x^3-2*x^2)*ln(3))*ln(x)^3+((16*x^2+8*x+1)*exp(x)^2+(-32*x^2-16*x-2)*ln(3)*exp(x)+(16*x^2+8*x+
1)*ln(3)^2)*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

-4*(ln(3)-exp(x))*x^2/(-x^2*ln(x)+4*x*ln(3)-4*exp(x)*x+ln(3)-exp(x))/ln(x)

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maxima [A]  time = 0.71, size = 49, normalized size = 1.44 \begin {gather*} -\frac {4 \, {\left (x^{2} e^{x} - x^{2} \log \relax (3)\right )}}{x^{2} \log \relax (x)^{2} + {\left (4 \, x + 1\right )} e^{x} \log \relax (x) - {\left (4 \, x \log \relax (3) + \log \relax (3)\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^4*exp(x)*log(x)^2+((-16*x^2-8*x)*exp(x)^2+((32*x^2+16*x)*log(3)+8*x^3)*exp(x)+(-16*x^2-8*x)*lo
g(3)^2-8*x^3*log(3))*log(x)+(16*x^2+4*x)*exp(x)^2+(-32*x^2-8*x)*log(3)*exp(x)+(16*x^2+4*x)*log(3)^2)/(x^4*log(
x)^4+((8*x^3+2*x^2)*exp(x)+(-8*x^3-2*x^2)*log(3))*log(x)^3+((16*x^2+8*x+1)*exp(x)^2+(-32*x^2-16*x-2)*log(3)*ex
p(x)+(16*x^2+8*x+1)*log(3)^2)*log(x)^2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {\ln \relax (x)\,\left ({\mathrm {e}}^{2\,x}\,\left (16\,x^2+8\,x\right )-{\mathrm {e}}^x\,\left (\ln \relax (3)\,\left (32\,x^2+16\,x\right )+8\,x^3\right )+{\ln \relax (3)}^2\,\left (16\,x^2+8\,x\right )+8\,x^3\,\ln \relax (3)\right )-{\ln \relax (3)}^2\,\left (16\,x^2+4\,x\right )-{\mathrm {e}}^{2\,x}\,\left (16\,x^2+4\,x\right )+{\mathrm {e}}^x\,\ln \relax (3)\,\left (32\,x^2+8\,x\right )+4\,x^4\,{\mathrm {e}}^x\,{\ln \relax (x)}^2}{x^4\,{\ln \relax (x)}^4+{\ln \relax (x)}^3\,\left ({\mathrm {e}}^x\,\left (8\,x^3+2\,x^2\right )-\ln \relax (3)\,\left (8\,x^3+2\,x^2\right )\right )+{\ln \relax (x)}^2\,\left ({\mathrm {e}}^{2\,x}\,\left (16\,x^2+8\,x+1\right )+{\ln \relax (3)}^2\,\left (16\,x^2+8\,x+1\right )-{\mathrm {e}}^x\,\ln \relax (3)\,\left (32\,x^2+16\,x+2\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(exp(2*x)*(8*x + 16*x^2) - exp(x)*(log(3)*(16*x + 32*x^2) + 8*x^3) + log(3)^2*(8*x + 16*x^2) + 8*
x^3*log(3)) - log(3)^2*(4*x + 16*x^2) - exp(2*x)*(4*x + 16*x^2) + exp(x)*log(3)*(8*x + 32*x^2) + 4*x^4*exp(x)*
log(x)^2)/(x^4*log(x)^4 + log(x)^3*(exp(x)*(2*x^2 + 8*x^3) - log(3)*(2*x^2 + 8*x^3)) + log(x)^2*(exp(2*x)*(8*x
+ 16*x^2 + 1) + log(3)^2*(8*x + 16*x^2 + 1) - exp(x)*log(3)*(16*x + 32*x^2 + 2))),x)

[Out]

int(-(log(x)*(exp(2*x)*(8*x + 16*x^2) - exp(x)*(log(3)*(16*x + 32*x^2) + 8*x^3) + log(3)^2*(8*x + 16*x^2) + 8*
x^3*log(3)) - log(3)^2*(4*x + 16*x^2) - exp(2*x)*(4*x + 16*x^2) + exp(x)*log(3)*(8*x + 32*x^2) + 4*x^4*exp(x)*
log(x)^2)/(x^4*log(x)^4 + log(x)^3*(exp(x)*(2*x^2 + 8*x^3) - log(3)*(2*x^2 + 8*x^3)) + log(x)^2*(exp(2*x)*(8*x
+ 16*x^2 + 1) + log(3)^2*(8*x + 16*x^2 + 1) - exp(x)*log(3)*(16*x + 32*x^2 + 2))), x)

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sympy [B]  time = 0.57, size = 65, normalized size = 1.91 \begin {gather*} \frac {4 x^{4}}{4 x^{3} \log {\relax (x )} + x^{2} \log {\relax (x )} - 16 x^{2} \log {\relax (3 )} - 8 x \log {\relax (3 )} + \left (16 x^{2} + 8 x + 1\right ) e^{x} - \log {\relax (3 )}} - \frac {4 x^{2}}{\left (4 x + 1\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x**4*exp(x)*ln(x)**2+((-16*x**2-8*x)*exp(x)**2+((32*x**2+16*x)*ln(3)+8*x**3)*exp(x)+(-16*x**2-8*
x)*ln(3)**2-8*x**3*ln(3))*ln(x)+(16*x**2+4*x)*exp(x)**2+(-32*x**2-8*x)*ln(3)*exp(x)+(16*x**2+4*x)*ln(3)**2)/(x
**4*ln(x)**4+((8*x**3+2*x**2)*exp(x)+(-8*x**3-2*x**2)*ln(3))*ln(x)**3+((16*x**2+8*x+1)*exp(x)**2+(-32*x**2-16*
x-2)*ln(3)*exp(x)+(16*x**2+8*x+1)*ln(3)**2)*ln(x)**2),x)

[Out]

4*x**4/(4*x**3*log(x) + x**2*log(x) - 16*x**2*log(3) - 8*x*log(3) + (16*x**2 + 8*x + 1)*exp(x) - log(3)) - 4*x
**2/((4*x + 1)*log(x))

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