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

Optimal. Leaf size=26 \[ \frac {4 x}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \]

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Rubi [F]  time = 6.30, 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 {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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.60, size = 26, normalized size = 1.00 \begin {gather*} \frac {4 x}{e^{5 x^2}+\log \left (\frac {(1+x)^2}{9 x^4}\right )} \end {gather*}

Antiderivative was successfully verified.

[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)])

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fricas [A]  time = 0.57, size = 26, normalized size = 1.00 \begin {gather*} \frac {4 \, x}{e^{\left (5 \, x^{2}\right )} + \log \left (\frac {x^{2} + 2 \, x + 1}{9 \, x^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 0.83, size = 26, normalized size = 1.00 \begin {gather*} \frac {4 \, x}{e^{\left (5 \, x^{2}\right )} + \log \left (\frac {x^{2} + 2 \, x + 1}{9 \, x^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [C]  time = 0.26, size = 366, normalized size = 14.08

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x+4)*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)/((x+1)*ln(1/9*(x^2+2*x+1)/x^4)^
2+(2*x+2)*exp(5*x^2)*ln(1/9*(x^2+2*x+1)/x^4)+(x+1)*exp(5*x^2)^2),x,method=_RETURNVERBOSE)

[Out]

-8*I*x/(-Pi*csgn(I/x^4*(x+1)^2)^3+Pi*csgn(I*x^4)^3+Pi*csgn(I*x^3)^3-Pi*csgn(I*(x+1)^2)^3-Pi*csgn(I/x^4)*csgn(I
*(x+1)^2)*csgn(I/x^4*(x+1)^2)+Pi*csgn(I*x)*csgn(I*x^3)*csgn(I*x^4)+Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)+8*I*ln
(x)+4*I*ln(3)-2*I*exp(5*x^2)-4*I*ln(x+1)-Pi*csgn(I*x)*csgn(I*x^3)^2-Pi*csgn(I*x^2)*csgn(I*x^3)^2-Pi*csgn(I*x)*
csgn(I*x^4)^2-Pi*csgn(I*x^3)*csgn(I*x^4)^2+Pi*csgn(I/x^4)*csgn(I/x^4*(x+1)^2)^2+Pi*csgn(I*(x+1)^2)*csgn(I/x^4*
(x+1)^2)^2-Pi*csgn(I*(x+1))^2*csgn(I*(x+1)^2)+2*Pi*csgn(I*(x+1))*csgn(I*(x+1)^2)^2+Pi*csgn(I*x^2)^3+Pi*csgn(I*
x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2)

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maxima [A]  time = 0.60, size = 26, normalized size = 1.00 \begin {gather*} \frac {4 \, x}{e^{\left (5 \, x^{2}\right )} - 2 \, \log \relax (3) + 2 \, \log \left (x + 1\right ) - 4 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.39, size = 27, normalized size = 1.04 \begin {gather*} \frac {4 x}{e^{5 x^{2}} + \log {\left (\frac {\frac {x^{2}}{9} + \frac {2 x}{9} + \frac {1}{9}}{x^{4}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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