3.15.91 \(\int \frac {48+12 x-48 x^2-9 x^3-9 x^4+e^{-4+x} (-24-24 x+6 x^2-12 x^3-6 x^4)+(-48+24 x+60 x^2+18 x^3+e^{-4+x} (24+48 x+30 x^2+6 x^3)) \log (x)}{64 x^2-32 x^3-44 x^4+12 x^5+9 x^6+e^{-8+2 x} (16 x^2+16 x^3+4 x^4)+e^{-4+x} (-64 x^2-16 x^3+32 x^4+12 x^5)} \, dx\)

Optimal. Leaf size=33 \[ \frac {3 \left (\frac {x^2}{2+x}-\log (x)\right )}{x \left (-4+2 e^{-4+x}+3 x\right )} \]

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Rubi [F]  time = 18.45, 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 {48+12 x-48 x^2-9 x^3-9 x^4+e^{-4+x} \left (-24-24 x+6 x^2-12 x^3-6 x^4\right )+\left (-48+24 x+60 x^2+18 x^3+e^{-4+x} \left (24+48 x+30 x^2+6 x^3\right )\right ) \log (x)}{64 x^2-32 x^3-44 x^4+12 x^5+9 x^6+e^{-8+2 x} \left (16 x^2+16 x^3+4 x^4\right )+e^{-4+x} \left (-64 x^2-16 x^3+32 x^4+12 x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(48 + 12*x - 48*x^2 - 9*x^3 - 9*x^4 + E^(-4 + x)*(-24 - 24*x + 6*x^2 - 12*x^3 - 6*x^4) + (-48 + 24*x + 60*
x^2 + 18*x^3 + E^(-4 + x)*(24 + 48*x + 30*x^2 + 6*x^3))*Log[x])/(64*x^2 - 32*x^3 - 44*x^4 + 12*x^5 + 9*x^6 + E
^(-8 + 2*x)*(16*x^2 + 16*x^3 + 4*x^4) + E^(-4 + x)*(-64*x^2 - 16*x^3 + 32*x^4 + 12*x^5)),x]

[Out]

-39*E^8*Defer[Int][(-4*E^4 + 2*E^x + 3*E^4*x)^(-2), x] - 9*E^8*Log[x]*Defer[Int][(-4*E^4 + 2*E^x + 3*E^4*x)^(-
2), x] + 21*E^8*Log[x]*Defer[Int][1/(x*(-4*E^4 + 2*E^x + 3*E^4*x)^2), x] + 9*E^8*Defer[Int][x/(-4*E^4 + 2*E^x
+ 3*E^4*x)^2, x] + 78*E^8*Defer[Int][1/((2 + x)*(-4*E^4 + 2*E^x + 3*E^4*x)^2), x] - 3*E^4*Defer[Int][(-4*E^4 +
2*E^x + 3*E^4*x)^(-1), x] - 3*E^4*Defer[Int][1/(x^2*(-4*E^4 + 2*E^x + 3*E^4*x)), x] + 3*E^4*Log[x]*Defer[Int]
[1/(x^2*(-4*E^4 + 2*E^x + 3*E^4*x)), x] + 3*E^4*Log[x]*Defer[Int][1/(x*(-4*E^4 + 2*E^x + 3*E^4*x)), x] + 6*E^4
*Defer[Int][1/((2 + x)^2*(-4*E^4 + 2*E^x + 3*E^4*x)), x] + 6*E^4*Defer[Int][1/((2 + x)*(-4*E^4 + 2*E^x + 3*E^4
*x)), x] + 9*E^8*Defer[Int][Defer[Int][(2*E^x + E^4*(-4 + 3*x))^(-2), x]/x, x] - 21*E^8*Defer[Int][Defer[Int][
1/(x*(2*E^x + E^4*(-4 + 3*x))^2), x]/x, x] - 3*E^4*Defer[Int][Defer[Int][1/(x^2*(2*E^x + E^4*(-4 + 3*x))), x]/
x, x] - 3*E^4*Defer[Int][Defer[Int][1/(x*(2*E^x + E^4*(-4 + 3*x))), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 e^4 \left (-2 e^x \left (4+4 x-x^2+2 x^3+x^4\right )-e^4 \left (-16-4 x+16 x^2+3 x^3+3 x^4\right )+2 (2+x)^2 \left (e^x (1+x)+e^4 (-2+3 x)\right ) \log (x)\right )}{x^2 (2+x)^2 \left (2 e^x+e^4 (-4+3 x)\right )^2} \, dx\\ &=\left (3 e^4\right ) \int \frac {-2 e^x \left (4+4 x-x^2+2 x^3+x^4\right )-e^4 \left (-16-4 x+16 x^2+3 x^3+3 x^4\right )+2 (2+x)^2 \left (e^x (1+x)+e^4 (-2+3 x)\right ) \log (x)}{x^2 (2+x)^2 \left (2 e^x+e^4 (-4+3 x)\right )^2} \, dx\\ &=\left (3 e^4\right ) \int \left (\frac {e^4 (-7+3 x) \left (x^2-2 \log (x)-x \log (x)\right )}{x (2+x) \left (-4 e^4+2 e^x+3 e^4 x\right )^2}-\frac {4+4 x-x^2+2 x^3+x^4-4 \log (x)-8 x \log (x)-5 x^2 \log (x)-x^3 \log (x)}{x^2 (2+x)^2 \left (-4 e^4+2 e^x+3 e^4 x\right )}\right ) \, dx\\ &=-\left (\left (3 e^4\right ) \int \frac {4+4 x-x^2+2 x^3+x^4-4 \log (x)-8 x \log (x)-5 x^2 \log (x)-x^3 \log (x)}{x^2 (2+x)^2 \left (-4 e^4+2 e^x+3 e^4 x\right )} \, dx\right )+\left (3 e^8\right ) \int \frac {(-7+3 x) \left (x^2-2 \log (x)-x \log (x)\right )}{x (2+x) \left (-4 e^4+2 e^x+3 e^4 x\right )^2} \, dx\\ &=-\left (\left (3 e^4\right ) \int \left (\frac {4+4 x-x^2+2 x^3+x^4-4 \log (x)-8 x \log (x)-5 x^2 \log (x)-x^3 \log (x)}{4 x^2 \left (-4 e^4+2 e^x+3 e^4 x\right )}-\frac {4+4 x-x^2+2 x^3+x^4-4 \log (x)-8 x \log (x)-5 x^2 \log (x)-x^3 \log (x)}{4 x \left (-4 e^4+2 e^x+3 e^4 x\right )}+\frac {4+4 x-x^2+2 x^3+x^4-4 \log (x)-8 x \log (x)-5 x^2 \log (x)-x^3 \log (x)}{4 (2+x)^2 \left (-4 e^4+2 e^x+3 e^4 x\right )}+\frac {4+4 x-x^2+2 x^3+x^4-4 \log (x)-8 x \log (x)-5 x^2 \log (x)-x^3 \log (x)}{4 (2+x) \left (-4 e^4+2 e^x+3 e^4 x\right )}\right ) \, dx\right )+\left (3 e^8\right ) \int \left (-\frac {7 \left (x^2-2 \log (x)-x \log (x)\right )}{2 x \left (-4 e^4+2 e^x+3 e^4 x\right )^2}+\frac {13 \left (x^2-2 \log (x)-x \log (x)\right )}{2 (2+x) \left (-4 e^4+2 e^x+3 e^4 x\right )^2}\right ) \, dx\\ &=-\left (\frac {1}{4} \left (3 e^4\right ) \int \frac {4+4 x-x^2+2 x^3+x^4-4 \log (x)-8 x \log (x)-5 x^2 \log (x)-x^3 \log (x)}{x^2 \left (-4 e^4+2 e^x+3 e^4 x\right )} \, dx\right )+\frac {1}{4} \left (3 e^4\right ) \int \frac {4+4 x-x^2+2 x^3+x^4-4 \log (x)-8 x \log (x)-5 x^2 \log (x)-x^3 \log (x)}{x \left (-4 e^4+2 e^x+3 e^4 x\right )} \, dx-\frac {1}{4} \left (3 e^4\right ) \int \frac {4+4 x-x^2+2 x^3+x^4-4 \log (x)-8 x \log (x)-5 x^2 \log (x)-x^3 \log (x)}{(2+x)^2 \left (-4 e^4+2 e^x+3 e^4 x\right )} \, dx-\frac {1}{4} \left (3 e^4\right ) \int \frac {4+4 x-x^2+2 x^3+x^4-4 \log (x)-8 x \log (x)-5 x^2 \log (x)-x^3 \log (x)}{(2+x) \left (-4 e^4+2 e^x+3 e^4 x\right )} \, dx-\frac {1}{2} \left (21 e^8\right ) \int \frac {x^2-2 \log (x)-x \log (x)}{x \left (-4 e^4+2 e^x+3 e^4 x\right )^2} \, dx+\frac {1}{2} \left (39 e^8\right ) \int \frac {x^2-2 \log (x)-x \log (x)}{(2+x) \left (-4 e^4+2 e^x+3 e^4 x\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 42, normalized size = 1.27 \begin {gather*} -\frac {3 e^4 \left (-x^2+(2+x) \log (x)\right )}{x (2+x) \left (2 e^x+e^4 (-4+3 x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(48 + 12*x - 48*x^2 - 9*x^3 - 9*x^4 + E^(-4 + x)*(-24 - 24*x + 6*x^2 - 12*x^3 - 6*x^4) + (-48 + 24*x
+ 60*x^2 + 18*x^3 + E^(-4 + x)*(24 + 48*x + 30*x^2 + 6*x^3))*Log[x])/(64*x^2 - 32*x^3 - 44*x^4 + 12*x^5 + 9*x
^6 + E^(-8 + 2*x)*(16*x^2 + 16*x^3 + 4*x^4) + E^(-4 + x)*(-64*x^2 - 16*x^3 + 32*x^4 + 12*x^5)),x]

[Out]

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

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fricas [A]  time = 0.75, size = 42, normalized size = 1.27 \begin {gather*} \frac {3 \, {\left (x^{2} - {\left (x + 2\right )} \log \relax (x)\right )}}{3 \, x^{3} + 2 \, x^{2} + 2 \, {\left (x^{2} + 2 \, x\right )} e^{\left (x - 4\right )} - 8 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^3+30*x^2+48*x+24)*exp(x-4)+18*x^3+60*x^2+24*x-48)*log(x)+(-6*x^4-12*x^3+6*x^2-24*x-24)*exp(x-
4)-9*x^4-9*x^3-48*x^2+12*x+48)/((4*x^4+16*x^3+16*x^2)*exp(x-4)^2+(12*x^5+32*x^4-16*x^3-64*x^2)*exp(x-4)+9*x^6+
12*x^5-44*x^4-32*x^3+64*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.34, size = 56, normalized size = 1.70 \begin {gather*} \frac {3 \, {\left (x^{2} e^{4} - x e^{4} \log \relax (x) - 2 \, e^{4} \log \relax (x)\right )}}{3 \, x^{3} e^{4} + 2 \, x^{2} e^{4} + 2 \, x^{2} e^{x} - 8 \, x e^{4} + 4 \, x e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^3+30*x^2+48*x+24)*exp(x-4)+18*x^3+60*x^2+24*x-48)*log(x)+(-6*x^4-12*x^3+6*x^2-24*x-24)*exp(x-
4)-9*x^4-9*x^3-48*x^2+12*x+48)/((4*x^4+16*x^3+16*x^2)*exp(x-4)^2+(12*x^5+32*x^4-16*x^3-64*x^2)*exp(x-4)+9*x^6+
12*x^5-44*x^4-32*x^3+64*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 43, normalized size = 1.30

method	result	size
risch	\(-\frac {3 \ln \relax (x )}{x \left (3 x -4+2 \,{\mathrm e}^{x -4}\right )}+\frac {3 x}{\left (2+x \right ) \left (3 x -4+2 \,{\mathrm e}^{x -4}\right )}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((6*x^3+30*x^2+48*x+24)*exp(x-4)+18*x^3+60*x^2+24*x-48)*ln(x)+(-6*x^4-12*x^3+6*x^2-24*x-24)*exp(x-4)-9*x^
4-9*x^3-48*x^2+12*x+48)/((4*x^4+16*x^3+16*x^2)*exp(x-4)^2+(12*x^5+32*x^4-16*x^3-64*x^2)*exp(x-4)+9*x^6+12*x^5-
44*x^4-32*x^3+64*x^2),x,method=_RETURNVERBOSE)

[Out]

-3/x/(3*x-4+2*exp(x-4))*ln(x)+3*x/(2+x)/(3*x-4+2*exp(x-4))

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maxima [A]  time = 0.53, size = 55, normalized size = 1.67 \begin {gather*} \frac {3 \, {\left (x^{2} e^{4} - {\left (x e^{4} + 2 \, e^{4}\right )} \log \relax (x)\right )}}{3 \, x^{3} e^{4} + 2 \, x^{2} e^{4} - 8 \, x e^{4} + 2 \, {\left (x^{2} + 2 \, x\right )} e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x^3+30*x^2+48*x+24)*exp(x-4)+18*x^3+60*x^2+24*x-48)*log(x)+(-6*x^4-12*x^3+6*x^2-24*x-24)*exp(x-
4)-9*x^4-9*x^3-48*x^2+12*x+48)/((4*x^4+16*x^3+16*x^2)*exp(x-4)^2+(12*x^5+32*x^4-16*x^3-64*x^2)*exp(x-4)+9*x^6+
12*x^5-44*x^4-32*x^3+64*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.29, size = 37, normalized size = 1.12 \begin {gather*} -\frac {3\,\left (2\,\ln \relax (x)+x\,\ln \relax (x)-x^2\right )}{x\,\left (x+2\right )\,\left (3\,x+2\,{\mathrm {e}}^{x-4}-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x - 4)*(24*x - 6*x^2 + 12*x^3 + 6*x^4 + 24) - log(x)*(24*x + exp(x - 4)*(48*x + 30*x^2 + 6*x^3 + 24)
+ 60*x^2 + 18*x^3 - 48) - 12*x + 48*x^2 + 9*x^3 + 9*x^4 - 48)/(64*x^2 - exp(x - 4)*(64*x^2 + 16*x^3 - 32*x^4
- 12*x^5) - 32*x^3 - 44*x^4 + 12*x^5 + 9*x^6 + exp(2*x - 8)*(16*x^2 + 16*x^3 + 4*x^4)),x)

[Out]

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

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sympy [A]  time = 0.41, size = 42, normalized size = 1.27 \begin {gather*} \frac {3 x^{2} - 3 x \log {\relax (x )} - 6 \log {\relax (x )}}{3 x^{3} + 2 x^{2} - 8 x + \left (2 x^{2} + 4 x\right ) e^{x - 4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((6*x**3+30*x**2+48*x+24)*exp(x-4)+18*x**3+60*x**2+24*x-48)*ln(x)+(-6*x**4-12*x**3+6*x**2-24*x-24)*
exp(x-4)-9*x**4-9*x**3-48*x**2+12*x+48)/((4*x**4+16*x**3+16*x**2)*exp(x-4)**2+(12*x**5+32*x**4-16*x**3-64*x**2
)*exp(x-4)+9*x**6+12*x**5-44*x**4-32*x**3+64*x**2),x)

[Out]

(3*x**2 - 3*x*log(x) - 6*log(x))/(3*x**3 + 2*x**2 - 8*x + (2*x**2 + 4*x)*exp(x - 4))

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