3.62.39 \(\int \frac {x-24 x^2+24 x^3+(-5 e^{24 x-12 x^2}-x) \log (e^{-24 x+12 x^2} (5 e^{24 x-12 x^2}+x))}{5 e^{24 x-12 x^2} x^2+x^3} \, dx\)

Optimal. Leaf size=29 \[ 3+x-\frac {x^2-\log \left (5+e^{-6 (4-2 x) x} x\right )}{x} \]

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Rubi [F]  time = 2.10, 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 {x-24 x^2+24 x^3+\left (-5 e^{24 x-12 x^2}-x\right ) \log \left (e^{-24 x+12 x^2} \left (5 e^{24 x-12 x^2}+x\right )\right )}{5 e^{24 x-12 x^2} x^2+x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x - 24*x^2 + 24*x^3 + (-5*E^(24*x - 12*x^2) - x)*Log[E^(-24*x + 12*x^2)*(5*E^(24*x - 12*x^2) + x)])/(5*E^
(24*x - 12*x^2)*x^2 + x^3),x]

[Out]

-x^(-1) - 24*Log[x] + Log[5 + x/E^(12*(2 - x)*x)]/x + 120*Defer[Int][(5 + E^(12*(-2 + x)*x)*x)^(-1), x] + 24*D
efer[Int][E^(12*(-2 + x)*x)/(5 + E^(12*(-2 + x)*x)*x), x] - Defer[Int][E^(12*(-2 + x)*x)/(x*(5 + E^(12*(-2 + x
)*x)*x)), x] - 120*Defer[Int][E^(24*x)/(5*E^(24*x) + E^(12*x^2)*x), x] - 5*Defer[Int][E^(24*x)/(x^2*(5*E^(24*x
) + E^(12*x^2)*x)), x] + 120*Defer[Int][E^(24*x)/(x*(5*E^(24*x) + E^(12*x^2)*x)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {e^{12 x^2} x \left (1-24 x+24 x^2\right )}{5 e^{24 x}+e^{12 x^2} x}-\log \left (5+e^{12 (-2+x) x} x\right )}{x^2} \, dx\\ &=\int \left (-\frac {5 e^{24 x} \left (1-24 x+24 x^2\right )}{x^2 \left (5 e^{24 x}+e^{12 x^2} x\right )}+\frac {1-24 x+24 x^2-\log \left (5+e^{12 (-2+x) x} x\right )}{x^2}\right ) \, dx\\ &=-\left (5 \int \frac {e^{24 x} \left (1-24 x+24 x^2\right )}{x^2 \left (5 e^{24 x}+e^{12 x^2} x\right )} \, dx\right )+\int \frac {1-24 x+24 x^2-\log \left (5+e^{12 (-2+x) x} x\right )}{x^2} \, dx\\ &=-\left (5 \int \left (\frac {24 e^{24 x}}{5 e^{24 x}+e^{12 x^2} x}+\frac {e^{24 x}}{x^2 \left (5 e^{24 x}+e^{12 x^2} x\right )}-\frac {24 e^{24 x}}{x \left (5 e^{24 x}+e^{12 x^2} x\right )}\right ) \, dx\right )+\int \left (\frac {1-24 x+24 x^2}{x^2}-\frac {\log \left (5+e^{12 (-2+x) x} x\right )}{x^2}\right ) \, dx\\ &=-\left (5 \int \frac {e^{24 x}}{x^2 \left (5 e^{24 x}+e^{12 x^2} x\right )} \, dx\right )-120 \int \frac {e^{24 x}}{5 e^{24 x}+e^{12 x^2} x} \, dx+120 \int \frac {e^{24 x}}{x \left (5 e^{24 x}+e^{12 x^2} x\right )} \, dx+\int \frac {1-24 x+24 x^2}{x^2} \, dx-\int \frac {\log \left (5+e^{12 (-2+x) x} x\right )}{x^2} \, dx\\ &=\frac {\log \left (5+e^{-12 (2-x) x} x\right )}{x}-5 \int \frac {e^{24 x}}{x^2 \left (5 e^{24 x}+e^{12 x^2} x\right )} \, dx-120 \int \frac {e^{24 x}}{5 e^{24 x}+e^{12 x^2} x} \, dx+120 \int \frac {e^{24 x}}{x \left (5 e^{24 x}+e^{12 x^2} x\right )} \, dx+\int \left (24+\frac {1}{x^2}-\frac {24}{x}\right ) \, dx-\int \frac {e^{12 (-2+x) x} \left (1-24 x+24 x^2\right )}{x \left (5+e^{12 (-2+x) x} x\right )} \, dx\\ &=-\frac {1}{x}+24 x-24 \log (x)+\frac {\log \left (5+e^{-12 (2-x) x} x\right )}{x}-5 \int \frac {e^{24 x}}{x^2 \left (5 e^{24 x}+e^{12 x^2} x\right )} \, dx-120 \int \frac {e^{24 x}}{5 e^{24 x}+e^{12 x^2} x} \, dx+120 \int \frac {e^{24 x}}{x \left (5 e^{24 x}+e^{12 x^2} x\right )} \, dx-\int \left (-\frac {24 e^{12 (-2+x) x}}{5+e^{12 (-2+x) x} x}+\frac {e^{12 (-2+x) x}}{x \left (5+e^{12 (-2+x) x} x\right )}+\frac {24 e^{12 (-2+x) x} x}{5+e^{12 (-2+x) x} x}\right ) \, dx\\ &=-\frac {1}{x}+24 x-24 \log (x)+\frac {\log \left (5+e^{-12 (2-x) x} x\right )}{x}-5 \int \frac {e^{24 x}}{x^2 \left (5 e^{24 x}+e^{12 x^2} x\right )} \, dx+24 \int \frac {e^{12 (-2+x) x}}{5+e^{12 (-2+x) x} x} \, dx-24 \int \frac {e^{12 (-2+x) x} x}{5+e^{12 (-2+x) x} x} \, dx-120 \int \frac {e^{24 x}}{5 e^{24 x}+e^{12 x^2} x} \, dx+120 \int \frac {e^{24 x}}{x \left (5 e^{24 x}+e^{12 x^2} x\right )} \, dx-\int \frac {e^{12 (-2+x) x}}{x \left (5+e^{12 (-2+x) x} x\right )} \, dx\\ &=-\frac {1}{x}+24 x-24 \log (x)+\frac {\log \left (5+e^{-12 (2-x) x} x\right )}{x}-5 \int \frac {e^{24 x}}{x^2 \left (5 e^{24 x}+e^{12 x^2} x\right )} \, dx+24 \int \frac {e^{12 (-2+x) x}}{5+e^{12 (-2+x) x} x} \, dx-24 \int \left (1-\frac {5}{5+e^{12 (-2+x) x} x}\right ) \, dx-120 \int \frac {e^{24 x}}{5 e^{24 x}+e^{12 x^2} x} \, dx+120 \int \frac {e^{24 x}}{x \left (5 e^{24 x}+e^{12 x^2} x\right )} \, dx-\int \frac {e^{12 (-2+x) x}}{x \left (5+e^{12 (-2+x) x} x\right )} \, dx\\ &=-\frac {1}{x}-24 \log (x)+\frac {\log \left (5+e^{-12 (2-x) x} x\right )}{x}-5 \int \frac {e^{24 x}}{x^2 \left (5 e^{24 x}+e^{12 x^2} x\right )} \, dx+24 \int \frac {e^{12 (-2+x) x}}{5+e^{12 (-2+x) x} x} \, dx+120 \int \frac {1}{5+e^{12 (-2+x) x} x} \, dx-120 \int \frac {e^{24 x}}{5 e^{24 x}+e^{12 x^2} x} \, dx+120 \int \frac {e^{24 x}}{x \left (5 e^{24 x}+e^{12 x^2} x\right )} \, dx-\int \frac {e^{12 (-2+x) x}}{x \left (5+e^{12 (-2+x) x} x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.78, size = 20, normalized size = 0.69 \begin {gather*} \frac {\log \left (5+e^{-24 x+12 x^2} x\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x - 24*x^2 + 24*x^3 + (-5*E^(24*x - 12*x^2) - x)*Log[E^(-24*x + 12*x^2)*(5*E^(24*x - 12*x^2) + x)])
/(5*E^(24*x - 12*x^2)*x^2 + x^3),x]

[Out]

Log[5 + E^(-24*x + 12*x^2)*x]/x

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fricas [A]  time = 0.89, size = 19, normalized size = 0.66 \begin {gather*} \frac {\log \left (x e^{\left (12 \, x^{2} - 24 \, x\right )} + 5\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(-12*x^2+24*x)-x)*log((5*exp(-12*x^2+24*x)+x)/exp(-12*x^2+24*x))+24*x^3-24*x^2+x)/(5*x^2*exp
(-12*x^2+24*x)+x^3),x, algorithm="fricas")

[Out]

log(x*e^(12*x^2 - 24*x) + 5)/x

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giac [A]  time = 0.24, size = 19, normalized size = 0.66 \begin {gather*} \frac {\log \left (x e^{\left (12 \, x^{2} - 24 \, x\right )} + 5\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(-12*x^2+24*x)-x)*log((5*exp(-12*x^2+24*x)+x)/exp(-12*x^2+24*x))+24*x^3-24*x^2+x)/(5*x^2*exp
(-12*x^2+24*x)+x^3),x, algorithm="giac")

[Out]

log(x*e^(12*x^2 - 24*x) + 5)/x

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maple [A]  time = 0.18, size = 33, normalized size = 1.14

method	result	size
norman	\(\frac {\ln \left (\left (5 \,{\mathrm e}^{-12 x^{2}+24 x}+x \right ) {\mathrm e}^{12 x^{2}-24 x}\right )}{x}\)	\(33\)
risch	\(-\frac {\ln \left ({\mathrm e}^{-12 \left (x -2\right ) x}\right )}{x}+\frac {-i \pi \,\mathrm {csgn}\left (i \left (5 \,{\mathrm e}^{-12 \left (x -2\right ) x}+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{12 \left (x -2\right ) x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{12 \left (x -2\right ) x} \left (5 \,{\mathrm e}^{-12 \left (x -2\right ) x}+x \right )\right )+i \pi \,\mathrm {csgn}\left (i \left (5 \,{\mathrm e}^{-12 \left (x -2\right ) x}+x \right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{12 \left (x -2\right ) x} \left (5 \,{\mathrm e}^{-12 \left (x -2\right ) x}+x \right )\right )^{2}+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{12 \left (x -2\right ) x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{12 \left (x -2\right ) x} \left (5 \,{\mathrm e}^{-12 \left (x -2\right ) x}+x \right )\right )^{2}-i \pi \mathrm {csgn}\left (i {\mathrm e}^{12 \left (x -2\right ) x} \left (5 \,{\mathrm e}^{-12 \left (x -2\right ) x}+x \right )\right )^{3}+2 \ln \left (5 \,{\mathrm e}^{-12 \left (x -2\right ) x}+x \right )}{2 x}\)	\(197\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*exp(-12*x^2+24*x)-x)*ln((5*exp(-12*x^2+24*x)+x)/exp(-12*x^2+24*x))+24*x^3-24*x^2+x)/(5*x^2*exp(-12*x^
2+24*x)+x^3),x,method=_RETURNVERBOSE)

[Out]

ln((5*exp(-12*x^2+24*x)+x)/exp(-12*x^2+24*x))/x

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maxima [A]  time = 0.39, size = 20, normalized size = 0.69 \begin {gather*} \frac {\log \left (x e^{\left (12 \, x^{2}\right )} + 5 \, e^{\left (24 \, x\right )}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(-12*x^2+24*x)-x)*log((5*exp(-12*x^2+24*x)+x)/exp(-12*x^2+24*x))+24*x^3-24*x^2+x)/(5*x^2*exp
(-12*x^2+24*x)+x^3),x, algorithm="maxima")

[Out]

log(x*e^(12*x^2) + 5*e^(24*x))/x

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mupad [B]  time = 4.64, size = 19, normalized size = 0.66 \begin {gather*} \frac {\ln \left (x\,{\mathrm {e}}^{-24\,x}\,{\mathrm {e}}^{12\,x^2}+5\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - log(exp(12*x^2 - 24*x)*(x + 5*exp(24*x - 12*x^2)))*(x + 5*exp(24*x - 12*x^2)) - 24*x^2 + 24*x^3)/(5*x
^2*exp(24*x - 12*x^2) + x^3),x)

[Out]

log(x*exp(-24*x)*exp(12*x^2) + 5)/x

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sympy [A]  time = 0.44, size = 26, normalized size = 0.90 \begin {gather*} \frac {\log {\left (\left (x + 5 e^{- 12 x^{2} + 24 x}\right ) e^{12 x^{2} - 24 x} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(-12*x**2+24*x)-x)*ln((5*exp(-12*x**2+24*x)+x)/exp(-12*x**2+24*x))+24*x**3-24*x**2+x)/(5*x**
2*exp(-12*x**2+24*x)+x**3),x)

[Out]

log((x + 5*exp(-12*x**2 + 24*x))*exp(12*x**2 - 24*x))/x

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