3.2.51 \(\int \frac {e^{-\log ^2(x)} (25-48 x+42 x^2-16 x^3+5 x^4+(-50+48 x-28 x^2+8 x^3-2 x^4) \log (x))}{4 \log (3)} \, dx\)

Optimal. Leaf size=33 \[ \frac {e^{-\log ^2(x)} x \left (-x+\left (x+\frac {1}{2} \left (-5-x^2\right )\right )^2\right )}{\log (3)} \]

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Rubi [A]  time = 0.04, antiderivative size = 35, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {12, 2288} \begin {gather*} \frac {x \left (x^4-4 x^3+14 x^2-24 x+25\right ) e^{-\log ^2(x)}}{4 \log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 - 48*x + 42*x^2 - 16*x^3 + 5*x^4 + (-50 + 48*x - 28*x^2 + 8*x^3 - 2*x^4)*Log[x])/(4*E^Log[x]^2*Log[3])
,x]

[Out]

(x*(25 - 24*x + 14*x^2 - 4*x^3 + x^4))/(4*E^Log[x]^2*Log[3])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[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 {\int e^{-\log ^2(x)} \left (25-48 x+42 x^2-16 x^3+5 x^4+\left (-50+48 x-28 x^2+8 x^3-2 x^4\right ) \log (x)\right ) \, dx}{4 \log (3)}\\ &=\frac {e^{-\log ^2(x)} x \left (25-24 x+14 x^2-4 x^3+x^4\right )}{4 \log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 32, normalized size = 0.97 \begin {gather*} \frac {e^{-\log ^2(x)} x \left (25-24 x+14 x^2-4 x^3+x^4\right )}{\log (81)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 - 48*x + 42*x^2 - 16*x^3 + 5*x^4 + (-50 + 48*x - 28*x^2 + 8*x^3 - 2*x^4)*Log[x])/(4*E^Log[x]^2*L
og[3]),x]

[Out]

(x*(25 - 24*x + 14*x^2 - 4*x^3 + x^4))/(E^Log[x]^2*Log[81])

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fricas [A]  time = 0.96, size = 35, normalized size = 1.06 \begin {gather*} \frac {{\left (x^{5} - 4 \, x^{4} + 14 \, x^{3} - 24 \, x^{2} + 25 \, x\right )} e^{\left (-\log \relax (x)^{2}\right )}}{4 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*x^4+8*x^3-28*x^2+48*x-50)*log(x)+5*x^4-16*x^3+42*x^2-48*x+25)/log(3)/exp(log(x)^2),x, algor
ithm="fricas")

[Out]

1/4*(x^5 - 4*x^4 + 14*x^3 - 24*x^2 + 25*x)*e^(-log(x)^2)/log(3)

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giac [B]  time = 0.24, size = 64, normalized size = 1.94 \begin {gather*} \frac {x^{5} e^{\left (-\log \relax (x)^{2}\right )} - 4 \, x^{4} e^{\left (-\log \relax (x)^{2}\right )} + 14 \, x^{3} e^{\left (-\log \relax (x)^{2}\right )} - 24 \, x^{2} e^{\left (-\log \relax (x)^{2}\right )} + 25 \, x e^{\left (-\log \relax (x)^{2}\right )}}{4 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*x^4+8*x^3-28*x^2+48*x-50)*log(x)+5*x^4-16*x^3+42*x^2-48*x+25)/log(3)/exp(log(x)^2),x, algor
ithm="giac")

[Out]

1/4*(x^5*e^(-log(x)^2) - 4*x^4*e^(-log(x)^2) + 14*x^3*e^(-log(x)^2) - 24*x^2*e^(-log(x)^2) + 25*x*e^(-log(x)^2
))/log(3)

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maple [A]  time = 0.07, size = 33, normalized size = 1.00

method	result	size
risch	\(\frac {\left (x^{4}-4 x^{3}+14 x^{2}-24 x +25\right ) x \,{\mathrm e}^{-\ln \relax (x )^{2}}}{4 \ln \relax (3)}\)	\(33\)
norman	\(\left (\frac {25 x}{4 \ln \relax (3)}-\frac {6 x^{2}}{\ln \relax (3)}+\frac {7 x^{3}}{2 \ln \relax (3)}-\frac {x^{4}}{\ln \relax (3)}+\frac {x^{5}}{4 \ln \relax (3)}\right ) {\mathrm e}^{-\ln \relax (x )^{2}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-2*x^4+8*x^3-28*x^2+48*x-50)*ln(x)+5*x^4-16*x^3+42*x^2-48*x+25)/ln(3)/exp(ln(x)^2),x,method=_RETURNV
ERBOSE)

[Out]

1/4/ln(3)*(x^4-4*x^3+14*x^2-24*x+25)*x*exp(-ln(x)^2)

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maxima [C]  time = 0.55, size = 317, normalized size = 9.61 \begin {gather*} \frac {5 \, \sqrt {\pi } \operatorname {erf}\left (\log \relax (x) - \frac {5}{2}\right ) e^{\frac {25}{4}} - 16 \, \sqrt {\pi } \operatorname {erf}\left (\log \relax (x) - 2\right ) e^{4} + 42 \, \sqrt {\pi } \operatorname {erf}\left (\log \relax (x) - \frac {3}{2}\right ) e^{\frac {9}{4}} - 48 \, \sqrt {\pi } \operatorname {erf}\left (\log \relax (x) - 1\right ) e + 25 \, \sqrt {\pi } \operatorname {erf}\left (\log \relax (x) - \frac {1}{2}\right ) e^{\frac {1}{4}} + i \, {\left (\frac {5 i \, \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {{\left (2 \, \log \relax (x) - 5\right )}^{2}}\right ) - 1\right )} {\left (2 \, \log \relax (x) - 5\right )}}{\sqrt {{\left (2 \, \log \relax (x) - 5\right )}^{2}}} - 2 i \, e^{\left (-\frac {1}{4} \, {\left (2 \, \log \relax (x) - 5\right )}^{2}\right )}\right )} e^{\frac {25}{4}} + 8 i \, {\left (-\frac {2 i \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {{\left (\log \relax (x) - 2\right )}^{2}}\right ) - 1\right )} {\left (\log \relax (x) - 2\right )}}{\sqrt {{\left (\log \relax (x) - 2\right )}^{2}}} + i \, e^{\left (-{\left (\log \relax (x) - 2\right )}^{2}\right )}\right )} e^{4} + 14 i \, {\left (\frac {3 i \, \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {{\left (2 \, \log \relax (x) - 3\right )}^{2}}\right ) - 1\right )} {\left (2 \, \log \relax (x) - 3\right )}}{\sqrt {{\left (2 \, \log \relax (x) - 3\right )}^{2}}} - 2 i \, e^{\left (-\frac {1}{4} \, {\left (2 \, \log \relax (x) - 3\right )}^{2}\right )}\right )} e^{\frac {9}{4}} + 48 i \, {\left (-\frac {i \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {{\left (\log \relax (x) - 1\right )}^{2}}\right ) - 1\right )} {\left (\log \relax (x) - 1\right )}}{\sqrt {{\left (\log \relax (x) - 1\right )}^{2}}} + i \, e^{\left (-{\left (\log \relax (x) - 1\right )}^{2}\right )}\right )} e + 25 i \, {\left (\frac {i \, \sqrt {\pi } {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {{\left (2 \, \log \relax (x) - 1\right )}^{2}}\right ) - 1\right )} {\left (2 \, \log \relax (x) - 1\right )}}{\sqrt {{\left (2 \, \log \relax (x) - 1\right )}^{2}}} - 2 i \, e^{\left (-\frac {1}{4} \, {\left (2 \, \log \relax (x) - 1\right )}^{2}\right )}\right )} e^{\frac {1}{4}}}{8 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*x^4+8*x^3-28*x^2+48*x-50)*log(x)+5*x^4-16*x^3+42*x^2-48*x+25)/log(3)/exp(log(x)^2),x, algor
ithm="maxima")

[Out]

1/8*(5*sqrt(pi)*erf(log(x) - 5/2)*e^(25/4) - 16*sqrt(pi)*erf(log(x) - 2)*e^4 + 42*sqrt(pi)*erf(log(x) - 3/2)*e
^(9/4) - 48*sqrt(pi)*erf(log(x) - 1)*e + 25*sqrt(pi)*erf(log(x) - 1/2)*e^(1/4) + I*(5*I*sqrt(pi)*(erf(1/2*sqrt
((2*log(x) - 5)^2)) - 1)*(2*log(x) - 5)/sqrt((2*log(x) - 5)^2) - 2*I*e^(-1/4*(2*log(x) - 5)^2))*e^(25/4) + 8*I
*(-2*I*sqrt(pi)*(erf(sqrt((log(x) - 2)^2)) - 1)*(log(x) - 2)/sqrt((log(x) - 2)^2) + I*e^(-(log(x) - 2)^2))*e^4
+ 14*I*(3*I*sqrt(pi)*(erf(1/2*sqrt((2*log(x) - 3)^2)) - 1)*(2*log(x) - 3)/sqrt((2*log(x) - 3)^2) - 2*I*e^(-1/
4*(2*log(x) - 3)^2))*e^(9/4) + 48*I*(-I*sqrt(pi)*(erf(sqrt((log(x) - 1)^2)) - 1)*(log(x) - 1)/sqrt((log(x) - 1
)^2) + I*e^(-(log(x) - 1)^2))*e + 25*I*(I*sqrt(pi)*(erf(1/2*sqrt((2*log(x) - 1)^2)) - 1)*(2*log(x) - 1)/sqrt((
2*log(x) - 1)^2) - 2*I*e^(-1/4*(2*log(x) - 1)^2))*e^(1/4))/log(3)

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mupad [B]  time = 0.37, size = 69, normalized size = 2.09 \begin {gather*} \frac {25\,x^2\,{\mathrm {e}}^{-{\ln \relax (x)}^2}-24\,x^3\,{\mathrm {e}}^{-{\ln \relax (x)}^2}+14\,x^4\,{\mathrm {e}}^{-{\ln \relax (x)}^2}-4\,x^5\,{\mathrm {e}}^{-{\ln \relax (x)}^2}+x^6\,{\mathrm {e}}^{-{\ln \relax (x)}^2}}{4\,x\,\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-log(x)^2)*(12*x + (log(x)*(28*x^2 - 48*x - 8*x^3 + 2*x^4 + 50))/4 - (21*x^2)/2 + 4*x^3 - (5*x^4)/4
- 25/4))/log(3),x)

[Out]

(25*x^2*exp(-log(x)^2) - 24*x^3*exp(-log(x)^2) + 14*x^4*exp(-log(x)^2) - 4*x^5*exp(-log(x)^2) + x^6*exp(-log(x
)^2))/(4*x*log(3))

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sympy [A]  time = 0.40, size = 32, normalized size = 0.97 \begin {gather*} \frac {\left (x^{5} - 4 x^{4} + 14 x^{3} - 24 x^{2} + 25 x\right ) e^{- \log {\relax (x )}^{2}}}{4 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*x**4+8*x**3-28*x**2+48*x-50)*ln(x)+5*x**4-16*x**3+42*x**2-48*x+25)/ln(3)/exp(ln(x)**2),x)

[Out]

(x**5 - 4*x**4 + 14*x**3 - 24*x**2 + 25*x)*exp(-log(x)**2)/(4*log(3))

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