∫ e−5−6x+6 log( 16 x2 ) ________________________ −x+log( 16 x2 ) (−40x−30x2+3x3+x4+ex(10x+5x2−2x3+x4)+(−16x2−2x3+ex(4x2−2x3)) log( 16 x2 )+(8x+x2+ex(−2x+x2)) log2( 16 x2 )) ____________________________________________________________________________________________________________________________________________________________________________________16x2 +e2xx2+8x3+x4+ex(−8x2−2x3)+(−32x−2e2xx−16x2−2x3+ex(16x+4x2)) log( 16 x2 )+(16+e2x+ex(−8−2x)+8x+x2) log2( 16 x2 ) dx

3.84.56 $\int \frac {e^{\frac {-5-6 x+6 \log (\frac {16}{x^2})}{-x+\log (\frac {16}{x^2})}} (-40 x-30 x^2+3 x^3+x^4+e^x (10 x+5 x^2-2 x^3+x^4)+(-16 x^2-2 x^3+e^x (4 x^2-2 x^3)) \log (\frac {16}{x^2})+(8 x+x^2+e^x (-2 x+x^2)) \log ^2(\frac {16}{x^2}))}{16 x^2+e^{2 x} x^2+8 x^3+x^4+e^x (-8 x^2-2 x^3)+(-32 x-2 e^{2 x} x-16 x^2-2 x^3+e^x (16 x+4 x^2)) \log (\frac {16}{x^2})+(16+e^{2 x}+e^x (-8-2 x)+8 x+x^2) \log ^2(\frac {16}{x^2})} \, dx$

Optimal. Leaf size=32 \[ \frac {e^{6-\frac {5}{-x+\log \left (\frac {16}{x^2}\right )}} x^2}{4-e^x+x} \]

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Rubi [F] time = 180.00, 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*} \text {\$Aborted} \end {gather*}

Veriﬁcation is not applicable to the result.

[In]

Int[(E^((-5 - 6*x + 6*Log[16/x^2])/(-x + Log[16/x^2]))*(-40*x - 30*x^2 + 3*x^3 + x^4 + E^x*(10*x + 5*x^2 - 2*x

^3 + x^4) + (-16*x^2 - 2*x^3 + E^x*(4*x^2 - 2*x^3))*Log[16/x^2] + (8*x + x^2 + E^x*(-2*x + x^2))*Log[16/x^2]^2

))/(16*x^2 + E^(2*x)*x^2 + 8*x^3 + x^4 + E^x*(-8*x^2 - 2*x^3) + (-32*x - 2*E^(2*x)*x - 16*x^2 - 2*x^3 + E^x*(1

6*x + 4*x^2))*Log[16/x^2] + (16 + E^(2*x) + E^x*(-8 - 2*x) + 8*x + x^2)*Log[16/x^2]^2),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [F] time = 7.68, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {-5-6 x+6 \log \left (\frac {16}{x^2}\right )}{-x+\log \left (\frac {16}{x^2}\right )}} \left (-40 x-30 x^2+3 x^3+x^4+e^x \left (10 x+5 x^2-2 x^3+x^4\right )+\left (-16 x^2-2 x^3+e^x \left (4 x^2-2 x^3\right )\right ) \log \left (\frac {16}{x^2}\right )+\left (8 x+x^2+e^x \left (-2 x+x^2\right )\right ) \log ^2\left (\frac {16}{x^2}\right )\right )}{16 x^2+e^{2 x} x^2+8 x^3+x^4+e^x \left (-8 x^2-2 x^3\right )+\left (-32 x-2 e^{2 x} x-16 x^2-2 x^3+e^x \left (16 x+4 x^2\right )\right ) \log \left (\frac {16}{x^2}\right )+\left (16+e^{2 x}+e^x (-8-2 x)+8 x+x^2\right ) \log ^2\left (\frac {16}{x^2}\right )} \, dx \end {gather*}

Veriﬁcation is not applicable to the result.

[In]

Integrate[(E^((-5 - 6*x + 6*Log[16/x^2])/(-x + Log[16/x^2]))*(-40*x - 30*x^2 + 3*x^3 + x^4 + E^x*(10*x + 5*x^2

- 2*x^3 + x^4) + (-16*x^2 - 2*x^3 + E^x*(4*x^2 - 2*x^3))*Log[16/x^2] + (8*x + x^2 + E^x*(-2*x + x^2))*Log[16/

x^2]^2))/(16*x^2 + E^(2*x)*x^2 + 8*x^3 + x^4 + E^x*(-8*x^2 - 2*x^3) + (-32*x - 2*E^(2*x)*x - 16*x^2 - 2*x^3 +

E^x*(16*x + 4*x^2))*Log[16/x^2] + (16 + E^(2*x) + E^x*(-8 - 2*x) + 8*x + x^2)*Log[16/x^2]^2),x]

[Out]

Integrate[(E^((-5 - 6*x + 6*Log[16/x^2])/(-x + Log[16/x^2]))*(-40*x - 30*x^2 + 3*x^3 + x^4 + E^x*(10*x + 5*x^2

- 2*x^3 + x^4) + (-16*x^2 - 2*x^3 + E^x*(4*x^2 - 2*x^3))*Log[16/x^2] + (8*x + x^2 + E^x*(-2*x + x^2))*Log[16/

x^2]^2))/(16*x^2 + E^(2*x)*x^2 + 8*x^3 + x^4 + E^x*(-8*x^2 - 2*x^3) + (-32*x - 2*E^(2*x)*x - 16*x^2 - 2*x^3 +

E^x*(16*x + 4*x^2))*Log[16/x^2] + (16 + E^(2*x) + E^x*(-8 - 2*x) + 8*x + x^2)*Log[16/x^2]^2), x]

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fricas [A] time = 0.85, size = 40, normalized size = 1.25 \begin {gather*} \frac {x^{2} e^{\left (\frac {6 \, x - 6 \, \log \left (\frac {16}{x^{2}}\right ) + 5}{x - \log \left (\frac {16}{x^{2}}\right )}\right )}}{x - e^{x} + 4} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-2*x)*exp(x)+x^2+8*x)*log(16/x^2)^2+((-2*x^3+4*x^2)*exp(x)-2*x^3-16*x^2)*log(16/x^2)+(x^4-2*x^

3+5*x^2+10*x)*exp(x)+x^4+3*x^3-30*x^2-40*x)*exp((6*log(16/x^2)-5-6*x)/(log(16/x^2)-x))/((exp(x)^2+(-2*x-8)*exp

(x)+x^2+8*x+16)*log(16/x^2)^2+(-2*x*exp(x)^2+(4*x^2+16*x)*exp(x)-2*x^3-16*x^2-32*x)*log(16/x^2)+exp(x)^2*x^2+(

-2*x^3-8*x^2)*exp(x)+x^4+8*x^3+16*x^2),x, algorithm="fricas")

[Out]

x^2*e^((6*x - 6*log(16/x^2) + 5)/(x - log(16/x^2)))/(x - e^x + 4)

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giac [A] time = 4.23, size = 40, normalized size = 1.25 \begin {gather*} \frac {x^{2} e^{\left (\frac {6 \, x - 6 \, \log \left (\frac {16}{x^{2}}\right ) + 5}{x - \log \left (\frac {16}{x^{2}}\right )}\right )}}{x - e^{x} + 4} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-2*x)*exp(x)+x^2+8*x)*log(16/x^2)^2+((-2*x^3+4*x^2)*exp(x)-2*x^3-16*x^2)*log(16/x^2)+(x^4-2*x^

3+5*x^2+10*x)*exp(x)+x^4+3*x^3-30*x^2-40*x)*exp((6*log(16/x^2)-5-6*x)/(log(16/x^2)-x))/((exp(x)^2+(-2*x-8)*exp

(x)+x^2+8*x+16)*log(16/x^2)^2+(-2*x*exp(x)^2+(4*x^2+16*x)*exp(x)-2*x^3-16*x^2-32*x)*log(16/x^2)+exp(x)^2*x^2+(

-2*x^3-8*x^2)*exp(x)+x^4+8*x^3+16*x^2),x, algorithm="giac")

[Out]

x^2*e^((6*x - 6*log(16/x^2) + 5)/(x - log(16/x^2)))/(x - e^x + 4)

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maple [C] time = 4.56, size = 142, normalized size = 4.44


method	result	size

risch	$\frac {x^{2} {\mathrm e}^{\frac {6 i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-12 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )+6 i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}-24 \ln \relax (x )+48 \ln \relax (2)-12 x -10}{i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )+i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}-4 \ln \relax (x )+8 \ln \relax (2)-2 x}}}{x -{\mathrm e}^{x}+4}$	$142$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2-2*x)*exp(x)+x^2+8*x)*ln(16/x^2)^2+((-2*x^3+4*x^2)*exp(x)-2*x^3-16*x^2)*ln(16/x^2)+(x^4-2*x^3+5*x^2+

10*x)*exp(x)+x^4+3*x^3-30*x^2-40*x)*exp((6*ln(16/x^2)-5-6*x)/(ln(16/x^2)-x))/((exp(x)^2+(-2*x-8)*exp(x)+x^2+8*

x+16)*ln(16/x^2)^2+(-2*x*exp(x)^2+(4*x^2+16*x)*exp(x)-2*x^3-16*x^2-32*x)*ln(16/x^2)+exp(x)^2*x^2+(-2*x^3-8*x^2

)*exp(x)+x^4+8*x^3+16*x^2),x,method=_RETURNVERBOSE)

[Out]

x^2/(x-exp(x)+4)*exp(2*(3*I*Pi*csgn(I*x^2)^3-6*I*Pi*csgn(I*x^2)^2*csgn(I*x)+3*I*Pi*csgn(I*x^2)*csgn(I*x)^2-12*

ln(x)+24*ln(2)-6*x-5)/(I*Pi*csgn(I*x^2)^3-2*I*Pi*csgn(I*x^2)^2*csgn(I*x)+I*Pi*csgn(I*x^2)*csgn(I*x)^2-4*ln(x)+

8*ln(2)-2*x))

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maxima [A] time = 0.69, size = 30, normalized size = 0.94 \begin {gather*} \frac {x^{2} e^{\left (\frac {5}{x - 4 \, \log \relax (2) + 2 \, \log \relax (x)} + 6\right )}}{x - e^{x} + 4} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-2*x)*exp(x)+x^2+8*x)*log(16/x^2)^2+((-2*x^3+4*x^2)*exp(x)-2*x^3-16*x^2)*log(16/x^2)+(x^4-2*x^

3+5*x^2+10*x)*exp(x)+x^4+3*x^3-30*x^2-40*x)*exp((6*log(16/x^2)-5-6*x)/(log(16/x^2)-x))/((exp(x)^2+(-2*x-8)*exp

(x)+x^2+8*x+16)*log(16/x^2)^2+(-2*x*exp(x)^2+(4*x^2+16*x)*exp(x)-2*x^3-16*x^2-32*x)*log(16/x^2)+exp(x)^2*x^2+(

-2*x^3-8*x^2)*exp(x)+x^4+8*x^3+16*x^2),x, algorithm="maxima")

[Out]

x^2*e^(5/(x - 4*log(2) + 2*log(x)) + 6)/(x - e^x + 4)

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mupad [B] time = 5.56, size = 68, normalized size = 2.12 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{-\frac {5}{\ln \left (\frac {1}{x^2}\right )-x+4\,\ln \relax (2)}}\,{\mathrm {e}}^{-\frac {6\,x}{\ln \left (\frac {1}{x^2}\right )-x+4\,\ln \relax (2)}}\,{\left (\frac {16777216}{x^{12}}\right )}^{\frac {1}{\ln \left (\frac {1}{x^2}\right )-x+4\,\ln \relax (2)}}}{x-{\mathrm {e}}^x+4} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((exp((6*x - 6*log(16/x^2) + 5)/(x - log(16/x^2)))*(log(16/x^2)^2*(8*x - exp(x)*(2*x - x^2) + x^2) - 40*x +

exp(x)*(10*x + 5*x^2 - 2*x^3 + x^4) - 30*x^2 + 3*x^3 + x^4 - log(16/x^2)*(16*x^2 - exp(x)*(4*x^2 - 2*x^3) + 2

*x^3)))/(log(16/x^2)^2*(8*x + exp(2*x) - exp(x)*(2*x + 8) + x^2 + 16) - exp(x)*(8*x^2 + 2*x^3) - log(16/x^2)*(

32*x + 2*x*exp(2*x) - exp(x)*(16*x + 4*x^2) + 16*x^2 + 2*x^3) + x^2*exp(2*x) + 16*x^2 + 8*x^3 + x^4),x)

[Out]

(x^2*exp(-5/(log(1/x^2) - x + 4*log(2)))*exp(-(6*x)/(log(1/x^2) - x + 4*log(2)))*(16777216/x^12)^(1/(log(1/x^2

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

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sympy [A] time = 1.73, size = 32, normalized size = 1.00 \begin {gather*} \frac {x^{2} e^{\frac {- 6 x + 6 \log {\left (\frac {16}{x^{2}} \right )} - 5}{- x + \log {\left (\frac {16}{x^{2}} \right )}}}}{x - e^{x} + 4} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2-2*x)*exp(x)+x**2+8*x)*ln(16/x**2)**2+((-2*x**3+4*x**2)*exp(x)-2*x**3-16*x**2)*ln(16/x**2)+(x

**4-2*x**3+5*x**2+10*x)*exp(x)+x**4+3*x**3-30*x**2-40*x)*exp((6*ln(16/x**2)-5-6*x)/(ln(16/x**2)-x))/((exp(x)**

2+(-2*x-8)*exp(x)+x**2+8*x+16)*ln(16/x**2)**2+(-2*x*exp(x)**2+(4*x**2+16*x)*exp(x)-2*x**3-16*x**2-32*x)*ln(16/

x**2)+exp(x)**2*x**2+(-2*x**3-8*x**2)*exp(x)+x**4+8*x**3+16*x**2),x)

[Out]

x**2*exp((-6*x + 6*log(16/x**2) - 5)/(-x + log(16/x**2)))/(x - exp(x) + 4)

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