3.7.59 \(\int \frac {e^{-1-\log ^2(4)-(-2-2 \log (4)) \log (x^2)-\log ^2(x^2)} (15-21 x+3 x^2+(12-20 x+4 x^2) \log (4)+(4 x+4 x \log (4)) \log (x)+(-12+20 x-4 x^2-4 x \log (x)) \log (x^2))}{16 (18-60 x+62 x^2-20 x^3+2 x^4+(12 x-20 x^2+4 x^3) \log (x)+2 x^2 \log ^2(x))} \, dx\)

Optimal. Leaf size=32 \[ \frac {e^{-\left (1+\log (4)-\log \left (x^2\right )\right )^2}}{2 \left (-5+\frac {3}{x}+x+\log (x)\right )} \]

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Rubi [B]  time = 0.88, antiderivative size = 87, normalized size of antiderivative = 2.72, number of steps used = 4, number of rules used = 4, integrand size = 137, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {12, 2274, 15, 2288} \begin {gather*} \frac {x^5 e^{-\log ^2\left (x^2\right )-1-\log ^2(4)} \left (x^2\right )^{\log (16)} \left (x^2-5 x+x \log (x)+3\right )}{32 \left (x^4-10 x^3+31 x^2+x^2 \log ^2(x)+2 \left (x^3-5 x^2+3 x\right ) \log (x)-30 x+9\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-1 - Log[4]^2 - (-2 - 2*Log[4])*Log[x^2] - Log[x^2]^2)*(15 - 21*x + 3*x^2 + (12 - 20*x + 4*x^2)*Log[4]
+ (4*x + 4*x*Log[4])*Log[x] + (-12 + 20*x - 4*x^2 - 4*x*Log[x])*Log[x^2]))/(16*(18 - 60*x + 62*x^2 - 20*x^3 +
2*x^4 + (12*x - 20*x^2 + 4*x^3)*Log[x] + 2*x^2*Log[x]^2)),x]

[Out]

(E^(-1 - Log[4]^2 - Log[x^2]^2)*x^5*(x^2)^Log[16]*(3 - 5*x + x^2 + x*Log[x]))/(32*(9 - 30*x + 31*x^2 - 10*x^3
+ x^4 + 2*(3*x - 5*x^2 + x^3)*Log[x] + x^2*Log[x]^2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

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} &=\frac {1}{16} \int \frac {\exp \left (-1-\log ^2(4)-(-2-2 \log (4)) \log \left (x^2\right )-\log ^2\left (x^2\right )\right ) \left (15-21 x+3 x^2+\left (12-20 x+4 x^2\right ) \log (4)+(4 x+4 x \log (4)) \log (x)+\left (-12+20 x-4 x^2-4 x \log (x)\right ) \log \left (x^2\right )\right )}{18-60 x+62 x^2-20 x^3+2 x^4+\left (12 x-20 x^2+4 x^3\right ) \log (x)+2 x^2 \log ^2(x)} \, dx\\ &=\frac {1}{16} \int \frac {e^{-1-\log ^2(4)-\log ^2\left (x^2\right )} \left (x^2\right )^{2+2 \log (4)} \left (15-21 x+3 x^2+\left (12-20 x+4 x^2\right ) \log (4)+(4 x+4 x \log (4)) \log (x)+\left (-12+20 x-4 x^2-4 x \log (x)\right ) \log \left (x^2\right )\right )}{18-60 x+62 x^2-20 x^3+2 x^4+\left (12 x-20 x^2+4 x^3\right ) \log (x)+2 x^2 \log ^2(x)} \, dx\\ &=\frac {1}{16} \left (x^{-4 \log (4)} \left (x^2\right )^{2 \log (4)}\right ) \int \frac {e^{-1-\log ^2(4)-\log ^2\left (x^2\right )} x^{2 (2+2 \log (4))} \left (15-21 x+3 x^2+\left (12-20 x+4 x^2\right ) \log (4)+(4 x+4 x \log (4)) \log (x)+\left (-12+20 x-4 x^2-4 x \log (x)\right ) \log \left (x^2\right )\right )}{18-60 x+62 x^2-20 x^3+2 x^4+\left (12 x-20 x^2+4 x^3\right ) \log (x)+2 x^2 \log ^2(x)} \, dx\\ &=\frac {e^{-1-\log ^2(4)-\log ^2\left (x^2\right )} x^5 \left (x^2\right )^{\log (16)} \left (3-5 x+x^2+x \log (x)\right )}{32 \left (9-30 x+31 x^2-10 x^3+x^4+2 \left (3 x-5 x^2+x^3\right ) \log (x)+x^2 \log ^2(x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 2.20, size = 140, normalized size = 4.38 \begin {gather*} \frac {1}{16} \int \frac {e^{-1-\log ^2(4)-(-2-2 \log (4)) \log \left (x^2\right )-\log ^2\left (x^2\right )} \left (15-21 x+3 x^2+\left (12-20 x+4 x^2\right ) \log (4)+(4 x+4 x \log (4)) \log (x)+\left (-12+20 x-4 x^2-4 x \log (x)\right ) \log \left (x^2\right )\right )}{18-60 x+62 x^2-20 x^3+2 x^4+\left (12 x-20 x^2+4 x^3\right ) \log (x)+2 x^2 \log ^2(x)} \, dx \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-1 - Log[4]^2 - (-2 - 2*Log[4])*Log[x^2] - Log[x^2]^2)*(15 - 21*x + 3*x^2 + (12 - 20*x + 4*x^2)*
Log[4] + (4*x + 4*x*Log[4])*Log[x] + (-12 + 20*x - 4*x^2 - 4*x*Log[x])*Log[x^2]))/(16*(18 - 60*x + 62*x^2 - 20
*x^3 + 2*x^4 + (12*x - 20*x^2 + 4*x^3)*Log[x] + 2*x^2*Log[x]^2)),x]

[Out]

Integrate[(E^(-1 - Log[4]^2 - (-2 - 2*Log[4])*Log[x^2] - Log[x^2]^2)*(15 - 21*x + 3*x^2 + (12 - 20*x + 4*x^2)*
Log[4] + (4*x + 4*x*Log[4])*Log[x] + (-12 + 20*x - 4*x^2 - 4*x*Log[x])*Log[x^2]))/(18 - 60*x + 62*x^2 - 20*x^3
+ 2*x^4 + (12*x - 20*x^2 + 4*x^3)*Log[x] + 2*x^2*Log[x]^2), x]/16

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fricas [A]  time = 1.19, size = 46, normalized size = 1.44 \begin {gather*} \frac {x e^{\left (-4 \, \log \relax (2)^{2} + 4 \, {\left (2 \, \log \relax (2) + 1\right )} \log \relax (x) - 4 \, \log \relax (x)^{2} - 4 \, \log \relax (2) - 1\right )}}{2 \, {\left (x^{2} + x \log \relax (x) - 5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(x)-4*x^2+20*x-12)*log(x^2)+(8*x*log(2)+4*x)*log(x)+2*(4*x^2-20*x+12)*log(2)+3*x^2-21*x+15
)/(2*x^2*log(x)^2+(4*x^3-20*x^2+12*x)*log(x)+2*x^4-20*x^3+62*x^2-60*x+18)/exp(log(x^2)^2+(-4*log(2)-2)*log(x^2
)+4*log(2)^2+4*log(2)+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.47, size = 42, normalized size = 1.31 \begin {gather*} \frac {x e^{\left (-4 \, \log \relax (2)^{2} + 8 \, \log \relax (2) \log \relax (x) - 4 \, \log \relax (x)^{2} + 4 \, \log \relax (x) - 1\right )}}{32 \, {\left (x^{2} + x \log \relax (x) - 5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(x)-4*x^2+20*x-12)*log(x^2)+(8*x*log(2)+4*x)*log(x)+2*(4*x^2-20*x+12)*log(2)+3*x^2-21*x+15
)/(2*x^2*log(x)^2+(4*x^3-20*x^2+12*x)*log(x)+2*x^4-20*x^3+62*x^2-60*x+18)/exp(log(x^2)^2+(-4*log(2)-2)*log(x^2
)+4*log(2)^2+4*log(2)+1),x, algorithm="giac")

[Out]

1/32*x*e^(-4*log(2)^2 + 8*log(2)*log(x) - 4*log(x)^2 + 4*log(x) - 1)/(x^2 + x*log(x) - 5*x + 3)

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maple [C]  time = 0.79, size = 244, normalized size = 7.62

method	result	size
risch	\(\frac {x^{5} x^{8 \ln \relax (2)} x^{-4 i \pi \,\mathrm {csgn}\left (i x \right )} 2^{4 i \pi \,\mathrm {csgn}\left (i x \right )} 2^{-4 i \pi \,\mathrm {csgn}\left (i x^{2}\right )} x^{4 i \pi \,\mathrm {csgn}\left (i x^{2}\right )} {\mathrm e}^{-4 \ln \relax (x )^{2}-1+\frac {\pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{6}}{4}-\pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{5} \mathrm {csgn}\left (i x \right )+\frac {3 \pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{4} \mathrm {csgn}\left (i x \right )^{2}}{2}-\pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{3} \mathrm {csgn}\left (i x \right )^{3}+\frac {\pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )^{4}}{4}-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-4 \ln \relax (2)^{2}}}{32 x \ln \relax (x )+32 x^{2}-160 x +96}\)	\(244\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x*ln(x)-4*x^2+20*x-12)*ln(x^2)+(8*x*ln(2)+4*x)*ln(x)+2*(4*x^2-20*x+12)*ln(2)+3*x^2-21*x+15)/(2*x^2*ln
(x)^2+(4*x^3-20*x^2+12*x)*ln(x)+2*x^4-20*x^3+62*x^2-60*x+18)/exp(ln(x^2)^2+(-4*ln(2)-2)*ln(x^2)+4*ln(2)^2+4*ln
(2)+1),x,method=_RETURNVERBOSE)

[Out]

1/32*x^5/(x*ln(x)+x^2-5*x+3)/(x^(-8*ln(2)))/(x^(4*I*Pi*csgn(I*x)))/(2^(-4*I*Pi*csgn(I*x)))/(2^(2*I*Pi*csgn(I*x
^2)))^2/(x^(-2*I*Pi*csgn(I*x^2)))^2*exp(-4*ln(x)^2-1+1/4*Pi^2*csgn(I*x^2)^6-Pi^2*csgn(I*x^2)^5*csgn(I*x)+3/2*P
i^2*csgn(I*x^2)^4*csgn(I*x)^2-Pi^2*csgn(I*x^2)^3*csgn(I*x)^3+1/4*Pi^2*csgn(I*x^2)^2*csgn(I*x)^4-I*Pi*csgn(I*x^
2)*csgn(I*x)^2+2*I*Pi*csgn(I*x^2)^2*csgn(I*x)-I*Pi*csgn(I*x^2)^3-4*ln(2)^2)

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maxima [B]  time = 0.82, size = 71, normalized size = 2.22 \begin {gather*} \frac {x^{5} e^{\left (8 \, \log \relax (2) \log \relax (x) - 4 \, \log \relax (x)^{2}\right )}}{32 \, {\left (x^{2} e^{\left (4 \, \log \relax (2)^{2} + 1\right )} + x e^{\left (4 \, \log \relax (2)^{2} + 1\right )} \log \relax (x) - 5 \, x e^{\left (4 \, \log \relax (2)^{2} + 1\right )} + 3 \, e^{\left (4 \, \log \relax (2)^{2} + 1\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*log(x)-4*x^2+20*x-12)*log(x^2)+(8*x*log(2)+4*x)*log(x)+2*(4*x^2-20*x+12)*log(2)+3*x^2-21*x+15
)/(2*x^2*log(x)^2+(4*x^3-20*x^2+12*x)*log(x)+2*x^4-20*x^3+62*x^2-60*x+18)/exp(log(x^2)^2+(-4*log(2)-2)*log(x^2
)+4*log(2)^2+4*log(2)+1),x, algorithm="maxima")

[Out]

1/32*x^5*e^(8*log(2)*log(x) - 4*log(x)^2)/(x^2*e^(4*log(2)^2 + 1) + x*e^(4*log(2)^2 + 1)*log(x) - 5*x*e^(4*log
(2)^2 + 1) + 3*e^(4*log(2)^2 + 1))

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mupad [B]  time = 1.20, size = 47, normalized size = 1.47 \begin {gather*} \frac {x^5\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{-{\ln \left (x^2\right )}^2}\,{\mathrm {e}}^{-4\,{\ln \relax (2)}^2}\,{\left (x^2\right )}^{4\,\ln \relax (2)}}{32\,\left (x\,\ln \relax (x)-5\,x+x^2+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(x^2)*(4*log(2) + 2) - log(x^2)^2 - 4*log(2)^2 - 4*log(2) - 1)*(2*log(2)*(4*x^2 - 20*x + 12) - 21*
x + log(x)*(4*x + 8*x*log(2)) - log(x^2)*(4*x*log(x) - 20*x + 4*x^2 + 12) + 3*x^2 + 15))/(2*x^2*log(x)^2 - 60*
x + 62*x^2 - 20*x^3 + 2*x^4 + log(x)*(12*x - 20*x^2 + 4*x^3) + 18),x)

[Out]

(x^5*exp(-1)*exp(-log(x^2)^2)*exp(-4*log(2)^2)*(x^2)^(4*log(2)))/(32*(x*log(x) - 5*x + x^2 + 3))

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sympy [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*ln(x)-4*x**2+20*x-12)*ln(x**2)+(8*x*ln(2)+4*x)*ln(x)+2*(4*x**2-20*x+12)*ln(2)+3*x**2-21*x+15)
/(2*x**2*ln(x)**2+(4*x**3-20*x**2+12*x)*ln(x)+2*x**4-20*x**3+62*x**2-60*x+18)/exp(ln(x**2)**2+(-4*ln(2)-2)*ln(
x**2)+4*ln(2)**2+4*ln(2)+1),x)

[Out]

Timed out

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