3.3.77 \(\int (3+(-8+4 x^2) \log ^3(\frac {e^{\frac {x^2}{2}}}{x^2})+\log ^4(\frac {e^{\frac {x^2}{2}}}{x^2})) \, dx\)

Optimal. Leaf size=25 \[ \log \left (2 e^{x \left (3+\log ^4\left (\frac {e^{\frac {x^2}{2}}}{x^2}\right )\right )}\right ) \]

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Rubi [F]  time = 0.15, 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 \left (3+\left (-8+4 x^2\right ) \log ^3\left (\frac {e^{\frac {x^2}{2}}}{x^2}\right )+\log ^4\left (\frac {e^{\frac {x^2}{2}}}{x^2}\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[3 + (-8 + 4*x^2)*Log[E^(x^2/2)/x^2]^3 + Log[E^(x^2/2)/x^2]^4,x]

[Out]

3*x - 8*Defer[Int][Log[E^(x^2/2)/x^2]^3, x] + 4*Defer[Int][x^2*Log[E^(x^2/2)/x^2]^3, x] + Defer[Int][Log[E^(x^
2/2)/x^2]^4, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=3 x+\int \left (-8+4 x^2\right ) \log ^3\left (\frac {e^{\frac {x^2}{2}}}{x^2}\right ) \, dx+\int \log ^4\left (\frac {e^{\frac {x^2}{2}}}{x^2}\right ) \, dx\\ &=3 x+\int \log ^4\left (\frac {e^{\frac {x^2}{2}}}{x^2}\right ) \, dx+\int \left (-8 \log ^3\left (\frac {e^{\frac {x^2}{2}}}{x^2}\right )+4 x^2 \log ^3\left (\frac {e^{\frac {x^2}{2}}}{x^2}\right )\right ) \, dx\\ &=3 x+4 \int x^2 \log ^3\left (\frac {e^{\frac {x^2}{2}}}{x^2}\right ) \, dx-8 \int \log ^3\left (\frac {e^{\frac {x^2}{2}}}{x^2}\right ) \, dx+\int \log ^4\left (\frac {e^{\frac {x^2}{2}}}{x^2}\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 20, normalized size = 0.80 \begin {gather*} x \left (3+\log ^4\left (\frac {e^{\frac {x^2}{2}}}{x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[3 + (-8 + 4*x^2)*Log[E^(x^2/2)/x^2]^3 + Log[E^(x^2/2)/x^2]^4,x]

[Out]

x*(3 + Log[E^(x^2/2)/x^2]^4)

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fricas [A]  time = 0.77, size = 19, normalized size = 0.76 \begin {gather*} x \log \left (\frac {e^{\left (\frac {1}{2} \, x^{2}\right )}}{x^{2}}\right )^{4} + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(e^(1/2*x^2)/x^2)^4 + 3*x

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giac [B]  time = 1.29, size = 231, normalized size = 9.24 \begin {gather*} -\frac {17}{210} \, x^{9} + \frac {8}{35} \, x^{7} \log \left (\frac {e^{\left (\frac {1}{2} \, x^{2}\right )}}{x^{2}}\right ) + \frac {59}{75} \, x^{7} - \frac {4}{5} \, x^{5} \log \left (\frac {e^{\left (\frac {1}{2} \, x^{2}\right )}}{x^{2}}\right )^{2} - \frac {368}{75} \, x^{5} \log \left (\frac {e^{\left (\frac {1}{2} \, x^{2}\right )}}{x^{2}}\right ) + \frac {44}{9} \, x^{5} + \frac {32}{3} \, x^{3} \log \left (\frac {e^{\left (\frac {1}{2} \, x^{2}\right )}}{x^{2}}\right )^{2} + x \log \left (\frac {e^{\left (\frac {1}{2} \, x^{2}\right )}}{x^{2}}\right )^{4} + \frac {4}{3} \, {\left (x^{3} - 6 \, x\right )} \log \left (x^{2}\right )^{3} + \frac {128}{9} \, x^{3} \log \left (\frac {e^{\left (\frac {1}{2} \, x^{2}\right )}}{x^{2}}\right ) + \frac {4}{3} \, {\left (x^{3} - 6 \, x\right )} \log \left (\frac {e^{\left (\frac {1}{2} \, x^{2}\right )}}{x^{2}}\right )^{3} - \frac {2}{15} \, {\left (9 \, x^{5} - 10 \, x^{3} - 360 \, x\right )} \log \left (x^{2}\right )^{2} - 48 \, x \log \left (\frac {e^{\left (\frac {1}{2} \, x^{2}\right )}}{x^{2}}\right )^{2} + \frac {1}{1575} \, {\left (675 \, x^{7} - 378 \, x^{5} - 2800 \, x^{3} - 302400 \, x\right )} \log \left (x^{2}\right ) - 32 \, {\left (x^{3} - 6 \, x\right )} \log \left (x^{2}\right ) + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-17/210*x^9 + 8/35*x^7*log(e^(1/2*x^2)/x^2) + 59/75*x^7 - 4/5*x^5*log(e^(1/2*x^2)/x^2)^2 - 368/75*x^5*log(e^(1
/2*x^2)/x^2) + 44/9*x^5 + 32/3*x^3*log(e^(1/2*x^2)/x^2)^2 + x*log(e^(1/2*x^2)/x^2)^4 + 4/3*(x^3 - 6*x)*log(x^2
)^3 + 128/9*x^3*log(e^(1/2*x^2)/x^2) + 4/3*(x^3 - 6*x)*log(e^(1/2*x^2)/x^2)^3 - 2/15*(9*x^5 - 10*x^3 - 360*x)*
log(x^2)^2 - 48*x*log(e^(1/2*x^2)/x^2)^2 + 1/1575*(675*x^7 - 378*x^5 - 2800*x^3 - 302400*x)*log(x^2) - 32*(x^3
- 6*x)*log(x^2) + 3*x

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maple [B]  time = 1.65, size = 323, normalized size = 12.92

method	result	size
default	\(3 x +\frac {x^{9}}{16}-x^{7} \ln \relax (x )-16 x^{3} \ln \relax (x )^{3}+16 x \ln \relax (x )^{4}+\frac {\left (\ln \left (\frac {{\mathrm e}^{\frac {x^{2}}{2}}}{x^{2}}\right )-\frac {x^{2}}{2}+2 \ln \relax (x )\right ) x^{7}}{2}+\left (\ln \left (\frac {{\mathrm e}^{\frac {x^{2}}{2}}}{x^{2}}\right )-\frac {x^{2}}{2}+2 \ln \relax (x )\right )^{4} x +\frac {3 \left (\ln \left (\frac {{\mathrm e}^{\frac {x^{2}}{2}}}{x^{2}}\right )-\frac {x^{2}}{2}+2 \ln \relax (x )\right )^{2} x^{5}}{2}+2 \left (\ln \left (\frac {{\mathrm e}^{\frac {x^{2}}{2}}}{x^{2}}\right )-\frac {x^{2}}{2}+2 \ln \relax (x )\right )^{3} x^{3}+6 x^{5} \ln \relax (x )^{2}+24 \left (\ln \left (\frac {{\mathrm e}^{\frac {x^{2}}{2}}}{x^{2}}\right )-\frac {x^{2}}{2}+2 \ln \relax (x )\right ) x^{3} \ln \relax (x )^{2}-12 \left (\ln \left (\frac {{\mathrm e}^{\frac {x^{2}}{2}}}{x^{2}}\right )-\frac {x^{2}}{2}+2 \ln \relax (x )\right )^{2} x^{3} \ln \relax (x )+24 \ln \relax (x )^{2} x \left (\ln \left (\frac {{\mathrm e}^{\frac {x^{2}}{2}}}{x^{2}}\right )-\frac {x^{2}}{2}+2 \ln \relax (x )\right )^{2}-6 \left (\ln \left (\frac {{\mathrm e}^{\frac {x^{2}}{2}}}{x^{2}}\right )-\frac {x^{2}}{2}+2 \ln \relax (x )\right ) x^{5} \ln \relax (x )-8 \ln \relax (x ) x \left (\ln \left (\frac {{\mathrm e}^{\frac {x^{2}}{2}}}{x^{2}}\right )-\frac {x^{2}}{2}+2 \ln \relax (x )\right )^{3}-32 \ln \relax (x )^{3} x \left (\ln \left (\frac {{\mathrm e}^{\frac {x^{2}}{2}}}{x^{2}}\right )-\frac {x^{2}}{2}+2 \ln \relax (x )\right )\)	\(323\)
risch	\(\text {Expression too large to display}\)	\(17452\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x+1/16*x^9-x^7*ln(x)-16*x^3*ln(x)^3+16*x*ln(x)^4+1/2*(ln(exp(1/2*x^2)/x^2)-1/2*x^2+2*ln(x))*x^7+(ln(exp(1/2*
x^2)/x^2)-1/2*x^2+2*ln(x))^4*x+3/2*(ln(exp(1/2*x^2)/x^2)-1/2*x^2+2*ln(x))^2*x^5+2*(ln(exp(1/2*x^2)/x^2)-1/2*x^
2+2*ln(x))^3*x^3+6*x^5*ln(x)^2+24*(ln(exp(1/2*x^2)/x^2)-1/2*x^2+2*ln(x))*x^3*ln(x)^2-12*(ln(exp(1/2*x^2)/x^2)-
1/2*x^2+2*ln(x))^2*x^3*ln(x)+24*ln(x)^2*x*(ln(exp(1/2*x^2)/x^2)-1/2*x^2+2*ln(x))^2-6*(ln(exp(1/2*x^2)/x^2)-1/2
*x^2+2*ln(x))*x^5*ln(x)-8*ln(x)*x*(ln(exp(1/2*x^2)/x^2)-1/2*x^2+2*ln(x))^3-32*ln(x)^3*x*(ln(exp(1/2*x^2)/x^2)-
1/2*x^2+2*ln(x))

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maxima [A]  time = 0.39, size = 19, normalized size = 0.76 \begin {gather*} x \log \left (\frac {e^{\left (\frac {1}{2} \, x^{2}\right )}}{x^{2}}\right )^{4} + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(e^(1/2*x^2)/x^2)^4 + 3*x

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mupad [B]  time = 0.40, size = 17, normalized size = 0.68 \begin {gather*} x\,\left ({\ln \left (\frac {{\mathrm {e}}^{\frac {x^2}{2}}}{x^2}\right )}^4+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(exp(x^2/2)/x^2)^3*(4*x^2 - 8) + log(exp(x^2/2)/x^2)^4 + 3,x)

[Out]

x*(log(exp(x^2/2)/x^2)^4 + 3)

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sympy [A]  time = 0.17, size = 17, normalized size = 0.68 \begin {gather*} x \log {\left (\frac {e^{\frac {x^{2}}{2}}}{x^{2}} \right )}^{4} + 3 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(exp(x**2/2)/x**2)**4 + 3*x

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