3.14.67 \(\int \frac {-96 x+24 x^2+e^{x^2} (-96+24 x)+(-32 x+16 x^2+e^{x^2} (8 x-64 x^2+16 x^3)) \log (\frac {1}{x^3})}{16 x^3-8 x^4+x^5+e^{2 x^2} (16 x-8 x^2+x^3)+e^{x^2} (32 x^2-16 x^3+2 x^4)} \, dx\)

Optimal. Leaf size=28 \[ \frac {4 \log \left (\frac {1}{x^3}\right )}{\left (e^{x^2}+x\right ) \left (-x+\frac {4+x}{2}\right )} \]

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Rubi [F]  time = 7.25, 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 {-96 x+24 x^2+e^{x^2} (-96+24 x)+\left (-32 x+16 x^2+e^{x^2} \left (8 x-64 x^2+16 x^3\right )\right ) \log \left (\frac {1}{x^3}\right )}{16 x^3-8 x^4+x^5+e^{2 x^2} \left (16 x-8 x^2+x^3\right )+e^{x^2} \left (32 x^2-16 x^3+2 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-96*x + 24*x^2 + E^x^2*(-96 + 24*x) + (-32*x + 16*x^2 + E^x^2*(8*x - 64*x^2 + 16*x^3))*Log[x^(-3)])/(16*x
^3 - 8*x^4 + x^5 + E^(2*x^2)*(16*x - 8*x^2 + x^3) + E^x^2*(32*x^2 - 16*x^3 + 2*x^4)),x]

[Out]

-64*Log[x^(-3)]*Defer[Int][(E^x^2 + x)^(-2), x] - 248*Log[x^(-3)]*Defer[Int][1/((-4 + x)*(E^x^2 + x)^2), x] -
16*Log[x^(-3)]*Defer[Int][x/(E^x^2 + x)^2, x] + 16*Log[x^(-3)]*Defer[Int][(E^x^2 + x)^(-1), x] + 8*Log[x^(-3)]
*Defer[Int][1/((-4 + x)^2*(E^x^2 + x)), x] + 6*Defer[Int][1/((-4 + x)*(E^x^2 + x)), x] + 64*Log[x^(-3)]*Defer[
Int][1/((-4 + x)*(E^x^2 + x)), x] - 6*Defer[Int][1/(x*(E^x^2 + x)), x] - 192*Defer[Int][Defer[Int][(E^x^2 + x)
^(-2), x]/x, x] - 744*Defer[Int][Defer[Int][1/((-4 + x)*(E^x^2 + x)^2), x]/x, x] - 48*Defer[Int][Defer[Int][x/
(E^x^2 + x)^2, x]/x, x] + 48*Defer[Int][Defer[Int][(E^x^2 + x)^(-1), x]/x, x] + 24*Defer[Int][Defer[Int][1/((-
4 + x)^2*(E^x^2 + x)), x]/x, x] + 192*Defer[Int][Defer[Int][1/((-4 + x)*(E^x^2 + x)), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8 \left (3 (-4+x) \left (e^{x^2}+x\right )+x \left (2 (-2+x)+e^{x^2} \left (1-8 x+2 x^2\right )\right ) \log \left (\frac {1}{x^3}\right )\right )}{(4-x)^2 x \left (e^{x^2}+x\right )^2} \, dx\\ &=8 \int \frac {3 (-4+x) \left (e^{x^2}+x\right )+x \left (2 (-2+x)+e^{x^2} \left (1-8 x+2 x^2\right )\right ) \log \left (\frac {1}{x^3}\right )}{(4-x)^2 x \left (e^{x^2}+x\right )^2} \, dx\\ &=8 \int \left (-\frac {\left (-1+2 x^2\right ) \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )^2}+\frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 x \left (e^{x^2}+x\right )}\right ) \, dx\\ &=-\left (8 \int \frac {\left (-1+2 x^2\right ) \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx\right )+8 \int \frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 x \left (e^{x^2}+x\right )} \, dx\\ &=8 \int \left (\frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{4 (-4+x)^2 \left (e^{x^2}+x\right )}-\frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{16 (-4+x) \left (e^{x^2}+x\right )}+\frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{16 x \left (e^{x^2}+x\right )}\right ) \, dx+8 \int -\frac {3 \left (8 \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx+31 \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx+2 \int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx\right )}{x} \, dx-\left (16 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx-\left (64 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx-\left (248 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx\\ &=-\left (\frac {1}{2} \int \frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )} \, dx\right )+\frac {1}{2} \int \frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{x \left (e^{x^2}+x\right )} \, dx+2 \int \frac {-12+3 x+x \log \left (\frac {1}{x^3}\right )-8 x^2 \log \left (\frac {1}{x^3}\right )+2 x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 \left (e^{x^2}+x\right )} \, dx-24 \int \frac {8 \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx+31 \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx+2 \int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx}{x} \, dx-\left (16 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx-\left (64 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx-\left (248 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {3}{e^{x^2}+x}-\frac {12}{x \left (e^{x^2}+x\right )}+\frac {\log \left (\frac {1}{x^3}\right )}{e^{x^2}+x}-\frac {8 x \log \left (\frac {1}{x^3}\right )}{e^{x^2}+x}+\frac {2 x^2 \log \left (\frac {1}{x^3}\right )}{e^{x^2}+x}\right ) \, dx-\frac {1}{2} \int \left (-\frac {12}{(-4+x) \left (e^{x^2}+x\right )}+\frac {3 x}{(-4+x) \left (e^{x^2}+x\right )}+\frac {x \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )}-\frac {8 x^2 \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )}+\frac {2 x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )}\right ) \, dx+2 \int \left (-\frac {12}{(-4+x)^2 \left (e^{x^2}+x\right )}+\frac {3 x}{(-4+x)^2 \left (e^{x^2}+x\right )}+\frac {x \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 \left (e^{x^2}+x\right )}-\frac {8 x^2 \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 \left (e^{x^2}+x\right )}+\frac {2 x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 \left (e^{x^2}+x\right )}\right ) \, dx-24 \int \left (\frac {8 \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx+31 \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx}{x}+\frac {2 \int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx}{x}\right ) \, dx-\left (16 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx-\left (64 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx-\left (248 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {\log \left (\frac {1}{x^3}\right )}{e^{x^2}+x} \, dx-\frac {1}{2} \int \frac {x \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )} \, dx+\frac {3}{2} \int \frac {1}{e^{x^2}+x} \, dx-\frac {3}{2} \int \frac {x}{(-4+x) \left (e^{x^2}+x\right )} \, dx+2 \int \frac {x \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 \left (e^{x^2}+x\right )} \, dx-4 \int \frac {x \log \left (\frac {1}{x^3}\right )}{e^{x^2}+x} \, dx+4 \int \frac {x^2 \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )} \, dx+4 \int \frac {x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 \left (e^{x^2}+x\right )} \, dx+6 \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )} \, dx-6 \int \frac {1}{x \left (e^{x^2}+x\right )} \, dx+6 \int \frac {x}{(-4+x)^2 \left (e^{x^2}+x\right )} \, dx-16 \int \frac {x^2 \log \left (\frac {1}{x^3}\right )}{(-4+x)^2 \left (e^{x^2}+x\right )} \, dx-24 \int \frac {1}{(-4+x)^2 \left (e^{x^2}+x\right )} \, dx-24 \int \frac {8 \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx+31 \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx}{x} \, dx-48 \int \frac {\int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx}{x} \, dx-\left (16 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {x}{\left (e^{x^2}+x\right )^2} \, dx-\left (64 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{\left (e^{x^2}+x\right )^2} \, dx-\left (248 \log \left (\frac {1}{x^3}\right )\right ) \int \frac {1}{(-4+x) \left (e^{x^2}+x\right )^2} \, dx+\int \frac {x^2 \log \left (\frac {1}{x^3}\right )}{e^{x^2}+x} \, dx-\int \frac {x^3 \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.87, size = 20, normalized size = 0.71 \begin {gather*} -\frac {8 \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-96*x + 24*x^2 + E^x^2*(-96 + 24*x) + (-32*x + 16*x^2 + E^x^2*(8*x - 64*x^2 + 16*x^3))*Log[x^(-3)])
/(16*x^3 - 8*x^4 + x^5 + E^(2*x^2)*(16*x - 8*x^2 + x^3) + E^x^2*(32*x^2 - 16*x^3 + 2*x^4)),x]

[Out]

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

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fricas [A]  time = 0.80, size = 23, normalized size = 0.82 \begin {gather*} -\frac {8 \, \log \left (\frac {1}{x^{3}}\right )}{x^{2} + {\left (x - 4\right )} e^{\left (x^{2}\right )} - 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3-64*x^2+8*x)*exp(x^2)+16*x^2-32*x)*log(1/x^3)+(24*x-96)*exp(x^2)+24*x^2-96*x)/((x^3-8*x^2+1
6*x)*exp(x^2)^2+(2*x^4-16*x^3+32*x^2)*exp(x^2)+x^5-8*x^4+16*x^3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.37, size = 25, normalized size = 0.89 \begin {gather*} \frac {24 \, \log \relax (x)}{x^{2} + x e^{\left (x^{2}\right )} - 4 \, x - 4 \, e^{\left (x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3-64*x^2+8*x)*exp(x^2)+16*x^2-32*x)*log(1/x^3)+(24*x-96)*exp(x^2)+24*x^2-96*x)/((x^3-8*x^2+1
6*x)*exp(x^2)^2+(2*x^4-16*x^3+32*x^2)*exp(x^2)+x^5-8*x^4+16*x^3),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.16, size = 140, normalized size = 5.00




method result size



risch \(\frac {24 \ln \relax (x )}{\left (x -4\right ) \left ({\mathrm e}^{x^{2}}+x \right )}-\frac {4 i \pi \left (\mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\mathrm {csgn}\left (i x^{2}\right )^{3}-\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\mathrm {csgn}\left (i x^{3}\right )^{3}\right )}{\left (x -4\right ) \left ({\mathrm e}^{x^{2}}+x \right )}\) \(140\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x^3-64*x^2+8*x)*exp(x^2)+16*x^2-32*x)*ln(1/x^3)+(24*x-96)*exp(x^2)+24*x^2-96*x)/((x^3-8*x^2+16*x)*ex
p(x^2)^2+(2*x^4-16*x^3+32*x^2)*exp(x^2)+x^5-8*x^4+16*x^3),x,method=_RETURNVERBOSE)

[Out]

24/(x-4)/(exp(x^2)+x)*ln(x)-4*I*Pi*(csgn(I*x)^2*csgn(I*x^2)-2*csgn(I*x)*csgn(I*x^2)^2+csgn(I*x)*csgn(I*x^2)*cs
gn(I*x^3)-csgn(I*x)*csgn(I*x^3)^2+csgn(I*x^2)^3-csgn(I*x^2)*csgn(I*x^3)^2+csgn(I*x^3)^3)/(x-4)/(exp(x^2)+x)

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maxima [A]  time = 1.00, size = 21, normalized size = 0.75 \begin {gather*} \frac {24 \, \log \relax (x)}{x^{2} + {\left (x - 4\right )} e^{\left (x^{2}\right )} - 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3-64*x^2+8*x)*exp(x^2)+16*x^2-32*x)*log(1/x^3)+(24*x-96)*exp(x^2)+24*x^2-96*x)/((x^3-8*x^2+1
6*x)*exp(x^2)^2+(2*x^4-16*x^3+32*x^2)*exp(x^2)+x^5-8*x^4+16*x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.10, size = 19, normalized size = 0.68 \begin {gather*} -\frac {8\,\ln \left (\frac {1}{x^3}\right )}{\left (x+{\mathrm {e}}^{x^2}\right )\,\left (x-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(1/x^3)*(exp(x^2)*(8*x - 64*x^2 + 16*x^3) - 32*x + 16*x^2) - 96*x + exp(x^2)*(24*x - 96) + 24*x^2)/(ex
p(2*x^2)*(16*x - 8*x^2 + x^3) + exp(x^2)*(32*x^2 - 16*x^3 + 2*x^4) + 16*x^3 - 8*x^4 + x^5),x)

[Out]

-(8*log(1/x^3))/((x + exp(x^2))*(x - 4))

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sympy [A]  time = 0.30, size = 24, normalized size = 0.86 \begin {gather*} - \frac {8 \log {\left (\frac {1}{x^{3}} \right )}}{x^{2} - 4 x + \left (x - 4\right ) e^{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x**3-64*x**2+8*x)*exp(x**2)+16*x**2-32*x)*ln(1/x**3)+(24*x-96)*exp(x**2)+24*x**2-96*x)/((x**3-
8*x**2+16*x)*exp(x**2)**2+(2*x**4-16*x**3+32*x**2)*exp(x**2)+x**5-8*x**4+16*x**3),x)

[Out]

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

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