\(\int \frac {16+8 x+e^{5 x^2} (4+4 x-40 x^2-40 x^3)+(4+4 x) \log (\frac {1+2 x+x^2}{9 x^4})}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log (\frac {1+2 x+x^2}{9 x^4})+(1+x) \log ^2(\frac {1+2 x+x^2}{9 x^4})} \, dx\) [3202]

Optimal result

Integrand size = 116, antiderivative size = 26 \[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\frac {4 x}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \]

[Out]

4/(ln(1/9*(1+x)^2/x^4)+exp(5*x^2))*x

Rubi [F]

\[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx \]

[In]

Int[(16 + 8*x + E^(5*x^2)*(4 + 4*x - 40*x^2 - 40*x^3) + (4 + 4*x)*Log[(1 + 2*x + x^2)/(9*x^4)])/(E^(10*x^2)*(1
+ x) + E^(5*x^2)*(2 + 2*x)*Log[(1 + 2*x + x^2)/(9*x^4)] + (1 + x)*Log[(1 + 2*x + x^2)/(9*x^4)]^2),x]

[Out]

8*Defer[Int][(E^(5*x^2) + Log[(1 + x)^2/(9*x^4)])^(-2), x] + 8*Defer[Int][1/((1 + x)*(E^(5*x^2) + Log[(1 + x)^
2/(9*x^4)])^2), x] + 40*Defer[Int][(x^2*Log[(1 + x)^2/(9*x^4)])/(E^(5*x^2) + Log[(1 + x)^2/(9*x^4)])^2, x] + 4
*Defer[Int][(E^(5*x^2) + Log[(1 + x)^2/(9*x^4)])^(-1), x] - 40*Defer[Int][x^2/(E^(5*x^2) + Log[(1 + x)^2/(9*x^
4)]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {4 \left (2 (2+x)+e^{5 x^2} \left (1+x-10 x^2-10 x^3\right )+(1+x) \log \left (\frac {(1+x)^2}{9 x^4}\right )\right )}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2} \, dx \\ & = 4 \int \frac {2 (2+x)+e^{5 x^2} \left (1+x-10 x^2-10 x^3\right )+(1+x) \log \left (\frac {(1+x)^2}{9 x^4}\right )}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2} \, dx \\ & = 4 \int \left (-\frac {-1+10 x^2}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )}+\frac {2 \left (2+x+5 x^2 \log \left (\frac {(1+x)^2}{9 x^4}\right )+5 x^3 \log \left (\frac {(1+x)^2}{9 x^4}\right )\right )}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2}\right ) \, dx \\ & = -\left (4 \int \frac {-1+10 x^2}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \, dx\right )+8 \int \frac {2+x+5 x^2 \log \left (\frac {(1+x)^2}{9 x^4}\right )+5 x^3 \log \left (\frac {(1+x)^2}{9 x^4}\right )}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2} \, dx \\ & = -\left (4 \int \left (-\frac {1}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )}+\frac {10 x^2}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )}\right ) \, dx\right )+8 \int \frac {2+x+5 x^2 (1+x) \log \left (\frac {(1+x)^2}{9 x^4}\right )}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2} \, dx \\ & = 4 \int \frac {1}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \, dx+8 \int \left (\frac {2}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2}+\frac {x}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2}+\frac {5 x^2 \log \left (\frac {(1+x)^2}{9 x^4}\right )}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2}+\frac {5 x^3 \log \left (\frac {(1+x)^2}{9 x^4}\right )}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2}\right ) \, dx-40 \int \frac {x^2}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \, dx \\ & = 4 \int \frac {1}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \, dx+8 \int \frac {x}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2} \, dx+16 \int \frac {1}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2} \, dx+40 \int \frac {x^2 \log \left (\frac {(1+x)^2}{9 x^4}\right )}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2} \, dx+40 \int \frac {x^3 \log \left (\frac {(1+x)^2}{9 x^4}\right )}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2} \, dx-40 \int \frac {x^2}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \, dx \\ & = 4 \int \frac {1}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \, dx+8 \int \left (\frac {1}{\left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2}-\frac {1}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2}\right ) \, dx+16 \int \frac {1}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2} \, dx-40 \int \frac {x^2}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \, dx+40 \int \left (\frac {\log \left (\frac {(1+x)^2}{9 x^4}\right )}{\left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2}-\frac {x \log \left (\frac {(1+x)^2}{9 x^4}\right )}{\left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2}+\frac {x^2 \log \left (\frac {(1+x)^2}{9 x^4}\right )}{\left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2}-\frac {\log \left (\frac {(1+x)^2}{9 x^4}\right )}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2}\right ) \, dx+40 \int \left (-\frac {\log \left (\frac {(1+x)^2}{9 x^4}\right )}{\left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2}+\frac {x \log \left (\frac {(1+x)^2}{9 x^4}\right )}{\left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2}+\frac {\log \left (\frac {(1+x)^2}{9 x^4}\right )}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2}\right ) \, dx \\ & = 4 \int \frac {1}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \, dx+8 \int \frac {1}{\left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2} \, dx-8 \int \frac {1}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2} \, dx+16 \int \frac {1}{(1+x) \left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2} \, dx+40 \int \frac {x^2 \log \left (\frac {(1+x)^2}{9 x^4}\right )}{\left (e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )\right )^2} \, dx-40 \int \frac {x^2}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\frac {4 x}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \]

[In]

Integrate[(16 + 8*x + E^(5*x^2)*(4 + 4*x - 40*x^2 - 40*x^3) + (4 + 4*x)*Log[(1 + 2*x + x^2)/(9*x^4)])/(E^(10*x
^2)*(1 + x) + E^(5*x^2)*(2 + 2*x)*Log[(1 + 2*x + x^2)/(9*x^4)] + (1 + x)*Log[(1 + 2*x + x^2)/(9*x^4)]^2),x]

[Out]

(4*x)/(E^(5*x^2) + Log[(1 + x)^2/(9*x^4)])

Maple [A] (verified)

Time = 2.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\(\frac {4 x}{{\mathrm e}^{5 x^{2}}+\ln \left (\frac {x^{2}+2 x +1}{9 x^{4}}\right )}\)	\(27\)
risch	\(-\frac {8 i x}{8 i \ln \left (x \right )+4 i \ln \left (3\right )-4 i \ln \left (1+x \right )-2 i {\mathrm e}^{5 x^{2}}+\pi \,\operatorname {csgn}\left (\frac {i}{x^{4}}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )^{2}}{x^{4}}\right )^{2}+\pi \,\operatorname {csgn}\left (i \left (1+x \right )^{2}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )^{2}}{x^{4}}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i \left (1+x \right )^{2}}{x^{4}}\right )^{3}-\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )^{2}-\pi \operatorname {csgn}\left (i x^{4}\right )^{2} \operatorname {csgn}\left (i x \right )-\pi \operatorname {csgn}\left (i \left (1+x \right )\right )^{2} \operatorname {csgn}\left (i \left (1+x \right )^{2}\right )+2 \pi \,\operatorname {csgn}\left (i \left (1+x \right )\right ) \operatorname {csgn}\left (i \left (1+x \right )^{2}\right )^{2}+\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )-\pi \operatorname {csgn}\left (i \left (1+x \right )^{2}\right )^{3}+\pi \operatorname {csgn}\left (i x^{4}\right )^{3}+\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+\pi \operatorname {csgn}\left (i x^{3}\right )^{3}-\pi \operatorname {csgn}\left (i x^{3}\right )^{2} \operatorname {csgn}\left (i x \right )-\pi \operatorname {csgn}\left (i x^{3}\right )^{2} \operatorname {csgn}\left (i x^{2}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{x^{4}}\right ) \operatorname {csgn}\left (i \left (1+x \right )^{2}\right ) \operatorname {csgn}\left (\frac {i \left (1+x \right )^{2}}{x^{4}}\right )+\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right ) \operatorname {csgn}\left (i x \right )}\)	\(366\)

[In]

int(((4+4*x)*ln(1/9*(x^2+2*x+1)/x^4)+(-40*x^3-40*x^2+4*x+4)*exp(5*x^2)+8*x+16)/((1+x)*ln(1/9*(x^2+2*x+1)/x^4)^
2+(2+2*x)*exp(5*x^2)*ln(1/9*(x^2+2*x+1)/x^4)+(1+x)*exp(5*x^2)^2),x,method=_RETURNVERBOSE)

[Out]

4*x/(exp(5*x^2)+ln(1/9*(x^2+2*x+1)/x^4))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\frac {4 \, x}{e^{\left (5 \, x^{2}\right )} + \log \left (\frac {x^{2} + 2 \, x + 1}{9 \, x^{4}}\right )} \]

[In]

integrate(((4+4*x)*log(1/9*(x^2+2*x+1)/x^4)+(-40*x^3-40*x^2+4*x+4)*exp(5*x^2)+8*x+16)/((1+x)*log(1/9*(x^2+2*x+
1)/x^4)^2+(2+2*x)*exp(5*x^2)*log(1/9*(x^2+2*x+1)/x^4)+(1+x)*exp(5*x^2)^2),x, algorithm="fricas")

[Out]

4*x/(e^(5*x^2) + log(1/9*(x^2 + 2*x + 1)/x^4))

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\frac {4 x}{e^{5 x^{2}} + \log {\left (\frac {\frac {x^{2}}{9} + \frac {2 x}{9} + \frac {1}{9}}{x^{4}} \right )}} \]

[In]

integrate(((4+4*x)*ln(1/9*(x**2+2*x+1)/x**4)+(-40*x**3-40*x**2+4*x+4)*exp(5*x**2)+8*x+16)/((1+x)*ln(1/9*(x**2+
2*x+1)/x**4)**2+(2+2*x)*exp(5*x**2)*ln(1/9*(x**2+2*x+1)/x**4)+(1+x)*exp(5*x**2)**2),x)

[Out]

4*x/(exp(5*x**2) + log((x**2/9 + 2*x/9 + 1/9)/x**4))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\frac {4 \, x}{e^{\left (5 \, x^{2}\right )} - 2 \, \log \left (3\right ) + 2 \, \log \left (x + 1\right ) - 4 \, \log \left (x\right )} \]

[In]

integrate(((4+4*x)*log(1/9*(x^2+2*x+1)/x^4)+(-40*x^3-40*x^2+4*x+4)*exp(5*x^2)+8*x+16)/((1+x)*log(1/9*(x^2+2*x+
1)/x^4)^2+(2+2*x)*exp(5*x^2)*log(1/9*(x^2+2*x+1)/x^4)+(1+x)*exp(5*x^2)^2),x, algorithm="maxima")

[Out]

4*x/(e^(5*x^2) - 2*log(3) + 2*log(x + 1) - 4*log(x))

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\frac {4 \, x}{e^{\left (5 \, x^{2}\right )} + \log \left (\frac {x^{2} + 2 \, x + 1}{9 \, x^{4}}\right )} \]

[In]

integrate(((4+4*x)*log(1/9*(x^2+2*x+1)/x^4)+(-40*x^3-40*x^2+4*x+4)*exp(5*x^2)+8*x+16)/((1+x)*log(1/9*(x^2+2*x+
1)/x^4)^2+(2+2*x)*exp(5*x^2)*log(1/9*(x^2+2*x+1)/x^4)+(1+x)*exp(5*x^2)^2),x, algorithm="giac")

[Out]

4*x/(e^(5*x^2) + log(1/9*(x^2 + 2*x + 1)/x^4))

Mupad [F(-1)]

Timed out. \[ \int \frac {16+8 x+e^{5 x^2} \left (4+4 x-40 x^2-40 x^3\right )+(4+4 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )}{e^{10 x^2} (1+x)+e^{5 x^2} (2+2 x) \log \left (\frac {1+2 x+x^2}{9 x^4}\right )+(1+x) \log ^2\left (\frac {1+2 x+x^2}{9 x^4}\right )} \, dx=\int \frac {8\,x+{\mathrm {e}}^{5\,x^2}\,\left (-40\,x^3-40\,x^2+4\,x+4\right )+\ln \left (\frac {\frac {x^2}{9}+\frac {2\,x}{9}+\frac {1}{9}}{x^4}\right )\,\left (4\,x+4\right )+16}{\left (x+1\right )\,{\ln \left (\frac {\frac {x^2}{9}+\frac {2\,x}{9}+\frac {1}{9}}{x^4}\right )}^2+{\mathrm {e}}^{5\,x^2}\,\left (2\,x+2\right )\,\ln \left (\frac {\frac {x^2}{9}+\frac {2\,x}{9}+\frac {1}{9}}{x^4}\right )+{\mathrm {e}}^{10\,x^2}\,\left (x+1\right )} \,d x \]

[In]

int((8*x + exp(5*x^2)*(4*x - 40*x^2 - 40*x^3 + 4) + log(((2*x)/9 + x^2/9 + 1/9)/x^4)*(4*x + 4) + 16)/(log(((2*
x)/9 + x^2/9 + 1/9)/x^4)^2*(x + 1) + exp(10*x^2)*(x + 1) + log(((2*x)/9 + x^2/9 + 1/9)/x^4)*exp(5*x^2)*(2*x +
2)),x)

[Out]

int((8*x + exp(5*x^2)*(4*x - 40*x^2 - 40*x^3 + 4) + log(((2*x)/9 + x^2/9 + 1/9)/x^4)*(4*x + 4) + 16)/(log(((2*
x)/9 + x^2/9 + 1/9)/x^4)^2*(x + 1) + exp(10*x^2)*(x + 1) + log(((2*x)/9 + x^2/9 + 1/9)/x^4)*exp(5*x^2)*(2*x +
2)), x)