3.100.51 \(\int \frac {e^{-x^2 \log (x^2)+2 x \log (x^2) \log (4+20 x+25 x^2)-\log (x^2) \log ^2(4+20 x+25 x^2)} (-2-5 x-4 x^2-10 x^3+(16 x^2-10 x^3) \log (x^2)+(8 x+20 x^2+(-16 x+10 x^2) \log (x^2)) \log (4+20 x+25 x^2)+(-4-10 x) \log ^2(4+20 x+25 x^2))}{2 x^2+5 x^3} \, dx\)

Optimal. Leaf size=24 \[ \frac {\left (x^2\right )^{-\left (-x+\log \left ((2+5 x)^2\right )\right )^2}}{x} \]

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Rubi [B]  time = 1.51, antiderivative size = 247, normalized size of antiderivative = 10.29, number of steps used = 2, number of rules used = 2, integrand size = 148, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {1593, 2288} \begin {gather*} \frac {\left (x^2\right )^{-x^2-\log ^2\left (25 x^2+20 x+4\right )+2 x \log \left (25 x^2+20 x+4\right )-1} \left (5 x^3+2 x^2+(5 x+2) \log ^2\left (25 x^2+20 x+4\right )-\left (10 x^2-\left (8 x-5 x^2\right ) \log \left (x^2\right )+4 x\right ) \log \left (25 x^2+20 x+4\right )-\left (8 x^2-5 x^3\right ) \log \left (x^2\right )\right )}{(5 x+2) \left (\frac {\log ^2\left (25 x^2+20 x+4\right )}{x}+\frac {10 (5 x+2) \log \left (x^2\right ) \log \left (25 x^2+20 x+4\right )}{25 x^2+20 x+4}-\log \left (x^2\right ) \log \left (25 x^2+20 x+4\right )-2 \log \left (25 x^2+20 x+4\right )+x \log \left (x^2\right )-\frac {10 x (5 x+2) \log \left (x^2\right )}{25 x^2+20 x+4}+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-(x^2*Log[x^2]) + 2*x*Log[x^2]*Log[4 + 20*x + 25*x^2] - Log[x^2]*Log[4 + 20*x + 25*x^2]^2)*(-2 - 5*x -
4*x^2 - 10*x^3 + (16*x^2 - 10*x^3)*Log[x^2] + (8*x + 20*x^2 + (-16*x + 10*x^2)*Log[x^2])*Log[4 + 20*x + 25*x^
2] + (-4 - 10*x)*Log[4 + 20*x + 25*x^2]^2))/(2*x^2 + 5*x^3),x]

[Out]

((x^2)^(-1 - x^2 + 2*x*Log[4 + 20*x + 25*x^2] - Log[4 + 20*x + 25*x^2]^2)*(2*x^2 + 5*x^3 - (8*x^2 - 5*x^3)*Log
[x^2] - (4*x + 10*x^2 - (8*x - 5*x^2)*Log[x^2])*Log[4 + 20*x + 25*x^2] + (2 + 5*x)*Log[4 + 20*x + 25*x^2]^2))/
((2 + 5*x)*(x + x*Log[x^2] - (10*x*(2 + 5*x)*Log[x^2])/(4 + 20*x + 25*x^2) - 2*Log[4 + 20*x + 25*x^2] - Log[x^
2]*Log[4 + 20*x + 25*x^2] + (10*(2 + 5*x)*Log[x^2]*Log[4 + 20*x + 25*x^2])/(4 + 20*x + 25*x^2) + Log[4 + 20*x
+ 25*x^2]^2/x))

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-x^2 \log \left (x^2\right )+2 x \log \left (x^2\right ) \log \left (4+20 x+25 x^2\right )-\log \left (x^2\right ) \log ^2\left (4+20 x+25 x^2\right )\right ) \left (-2-5 x-4 x^2-10 x^3+\left (16 x^2-10 x^3\right ) \log \left (x^2\right )+\left (8 x+20 x^2+\left (-16 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (4+20 x+25 x^2\right )+(-4-10 x) \log ^2\left (4+20 x+25 x^2\right )\right )}{x^2 (2+5 x)} \, dx\\ &=\frac {\left (x^2\right )^{-1-x^2+2 x \log \left (4+20 x+25 x^2\right )-\log ^2\left (4+20 x+25 x^2\right )} \left (2 x^2+5 x^3-\left (8 x^2-5 x^3\right ) \log \left (x^2\right )-\left (4 x+10 x^2-\left (8 x-5 x^2\right ) \log \left (x^2\right )\right ) \log \left (4+20 x+25 x^2\right )+(2+5 x) \log ^2\left (4+20 x+25 x^2\right )\right )}{(2+5 x) \left (x+x \log \left (x^2\right )-\frac {10 x (2+5 x) \log \left (x^2\right )}{4+20 x+25 x^2}-2 \log \left (4+20 x+25 x^2\right )-\log \left (x^2\right ) \log \left (4+20 x+25 x^2\right )+\frac {10 (2+5 x) \log \left (x^2\right ) \log \left (4+20 x+25 x^2\right )}{4+20 x+25 x^2}+\frac {\log ^2\left (4+20 x+25 x^2\right )}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 24, normalized size = 1.00 \begin {gather*} \frac {\left (x^2\right )^{-\left (x-\log \left ((2+5 x)^2\right )\right )^2}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-(x^2*Log[x^2]) + 2*x*Log[x^2]*Log[4 + 20*x + 25*x^2] - Log[x^2]*Log[4 + 20*x + 25*x^2]^2)*(-2 -
5*x - 4*x^2 - 10*x^3 + (16*x^2 - 10*x^3)*Log[x^2] + (8*x + 20*x^2 + (-16*x + 10*x^2)*Log[x^2])*Log[4 + 20*x +
25*x^2] + (-4 - 10*x)*Log[4 + 20*x + 25*x^2]^2))/(2*x^2 + 5*x^3),x]

[Out]

1/(x*(x^2)^(x - Log[(2 + 5*x)^2])^2)

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fricas [B]  time = 0.62, size = 52, normalized size = 2.17 \begin {gather*} \frac {e^{\left (-x^{2} \log \left (x^{2}\right ) + 2 \, x \log \left (25 \, x^{2} + 20 \, x + 4\right ) \log \left (x^{2}\right ) - \log \left (25 \, x^{2} + 20 \, x + 4\right )^{2} \log \left (x^{2}\right )\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x-4)*log(25*x^2+20*x+4)^2+((10*x^2-16*x)*log(x^2)+20*x^2+8*x)*log(25*x^2+20*x+4)+(-10*x^3+16*x
^2)*log(x^2)-10*x^3-4*x^2-5*x-2)*exp(-log(x^2)*log(25*x^2+20*x+4)^2+2*x*log(x^2)*log(25*x^2+20*x+4)-x^2*log(x^
2))/(5*x^3+2*x^2),x, algorithm="fricas")

[Out]

e^(-x^2*log(x^2) + 2*x*log(25*x^2 + 20*x + 4)*log(x^2) - log(25*x^2 + 20*x + 4)^2*log(x^2))/x

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giac [B]  time = 3.37, size = 52, normalized size = 2.17 \begin {gather*} \frac {e^{\left (-x^{2} \log \left (x^{2}\right ) + 2 \, x \log \left (25 \, x^{2} + 20 \, x + 4\right ) \log \left (x^{2}\right ) - \log \left (25 \, x^{2} + 20 \, x + 4\right )^{2} \log \left (x^{2}\right )\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x-4)*log(25*x^2+20*x+4)^2+((10*x^2-16*x)*log(x^2)+20*x^2+8*x)*log(25*x^2+20*x+4)+(-10*x^3+16*x
^2)*log(x^2)-10*x^3-4*x^2-5*x-2)*exp(-log(x^2)*log(25*x^2+20*x+4)^2+2*x*log(x^2)*log(25*x^2+20*x+4)-x^2*log(x^
2))/(5*x^3+2*x^2),x, algorithm="giac")

[Out]

e^(-x^2*log(x^2) + 2*x*log(25*x^2 + 20*x + 4)*log(x^2) - log(25*x^2 + 20*x + 4)^2*log(x^2))/x

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maple [C]  time = 0.34, size = 133, normalized size = 5.54

method	result	size
risch	\(\frac {{\mathrm e}^{-\frac {\left (-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+4 \ln \relax (x )\right ) \left (i \pi \mathrm {csgn}\left (i \left (x +\frac {2}{5}\right )^{2}\right )^{3}-2 i \pi \mathrm {csgn}\left (i \left (x +\frac {2}{5}\right )^{2}\right )^{2} \mathrm {csgn}\left (i \left (x +\frac {2}{5}\right )\right )+i \pi \,\mathrm {csgn}\left (i \left (x +\frac {2}{5}\right )^{2}\right ) \mathrm {csgn}\left (i \left (x +\frac {2}{5}\right )\right )^{2}-4 \ln \left (x +\frac {2}{5}\right )+2 x \right )^{2}}{8}}}{x}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-10*x-4)*ln(25*x^2+20*x+4)^2+((10*x^2-16*x)*ln(x^2)+20*x^2+8*x)*ln(25*x^2+20*x+4)+(-10*x^3+16*x^2)*ln(x^
2)-10*x^3-4*x^2-5*x-2)*exp(-ln(x^2)*ln(25*x^2+20*x+4)^2+2*x*ln(x^2)*ln(25*x^2+20*x+4)-x^2*ln(x^2))/(5*x^3+2*x^
2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 36, normalized size = 1.50 \begin {gather*} \frac {e^{\left (-2 \, x^{2} \log \relax (x) + 8 \, x \log \left (5 \, x + 2\right ) \log \relax (x) - 8 \, \log \left (5 \, x + 2\right )^{2} \log \relax (x)\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x-4)*log(25*x^2+20*x+4)^2+((10*x^2-16*x)*log(x^2)+20*x^2+8*x)*log(25*x^2+20*x+4)+(-10*x^3+16*x
^2)*log(x^2)-10*x^3-4*x^2-5*x-2)*exp(-log(x^2)*log(25*x^2+20*x+4)^2+2*x*log(x^2)*log(25*x^2+20*x+4)-x^2*log(x^
2))/(5*x^3+2*x^2),x, algorithm="maxima")

[Out]

e^(-2*x^2*log(x) + 8*x*log(5*x + 2)*log(x) - 8*log(5*x + 2)^2*log(x))/x

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mupad [B]  time = 8.84, size = 50, normalized size = 2.08 \begin {gather*} \frac {{\left (x^2\right )}^{2\,x\,\ln \left (25\,x^2+20\,x+4\right )}}{x\,{\left (x^2\right )}^{x^2}\,{\left (x^2\right )}^{{\ln \left (25\,x^2+20\,x+4\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x*log(x^2)*log(20*x + 25*x^2 + 4) - log(x^2)*log(20*x + 25*x^2 + 4)^2 - x^2*log(x^2))*(5*x - log(x
^2)*(16*x^2 - 10*x^3) - log(20*x + 25*x^2 + 4)*(8*x - log(x^2)*(16*x - 10*x^2) + 20*x^2) + log(20*x + 25*x^2 +
4)^2*(10*x + 4) + 4*x^2 + 10*x^3 + 2))/(2*x^2 + 5*x^3),x)

[Out]

(x^2)^(2*x*log(20*x + 25*x^2 + 4))/(x*(x^2)^(x^2)*(x^2)^(log(20*x + 25*x^2 + 4)^2))

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sympy [B]  time = 0.69, size = 49, normalized size = 2.04 \begin {gather*} \frac {e^{- x^{2} \log {\left (x^{2} \right )} + 2 x \log {\left (x^{2} \right )} \log {\left (25 x^{2} + 20 x + 4 \right )} - \log {\left (x^{2} \right )} \log {\left (25 x^{2} + 20 x + 4 \right )}^{2}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-10*x-4)*ln(25*x**2+20*x+4)**2+((10*x**2-16*x)*ln(x**2)+20*x**2+8*x)*ln(25*x**2+20*x+4)+(-10*x**3+
16*x**2)*ln(x**2)-10*x**3-4*x**2-5*x-2)*exp(-ln(x**2)*ln(25*x**2+20*x+4)**2+2*x*ln(x**2)*ln(25*x**2+20*x+4)-x*
*2*ln(x**2))/(5*x**3+2*x**2),x)

[Out]

exp(-x**2*log(x**2) + 2*x*log(x**2)*log(25*x**2 + 20*x + 4) - log(x**2)*log(25*x**2 + 20*x + 4)**2)/x

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