3.82.81 \(\int \frac {-9 x+6 x^2-x^3+(6 x-2 x^2) \log (\frac {4}{x^3})-x \log ^2(\frac {4}{x^3})+e^{\frac {e^{9+12 x+4 x^2}}{-3+x+\log (\frac {4}{x^3})}} (e^{9+12 x+4 x^2} (3-37 x-12 x^2+8 x^3)+e^{9+12 x+4 x^2} (12 x+8 x^2) \log (\frac {4}{x^3}))}{9 x-6 x^2+x^3+(-6 x+2 x^2) \log (\frac {4}{x^3})+x \log ^2(\frac {4}{x^3})} \, dx\)

Optimal. Leaf size=27 \[ e^{\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}}-x \]

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Rubi [F]  time = 31.61, 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 {-9 x+6 x^2-x^3+\left (6 x-2 x^2\right ) \log \left (\frac {4}{x^3}\right )-x \log ^2\left (\frac {4}{x^3}\right )+e^{\frac {e^{9+12 x+4 x^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}} \left (e^{9+12 x+4 x^2} \left (3-37 x-12 x^2+8 x^3\right )+e^{9+12 x+4 x^2} \left (12 x+8 x^2\right ) \log \left (\frac {4}{x^3}\right )\right )}{9 x-6 x^2+x^3+\left (-6 x+2 x^2\right ) \log \left (\frac {4}{x^3}\right )+x \log ^2\left (\frac {4}{x^3}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-9*x + 6*x^2 - x^3 + (6*x - 2*x^2)*Log[4/x^3] - x*Log[4/x^3]^2 + E^(E^(9 + 12*x + 4*x^2)/(-3 + x + Log[4/
x^3]))*(E^(9 + 12*x + 4*x^2)*(3 - 37*x - 12*x^2 + 8*x^3) + E^(9 + 12*x + 4*x^2)*(12*x + 8*x^2)*Log[4/x^3]))/(9
*x - 6*x^2 + x^3 + (-6*x + 2*x^2)*Log[4/x^3] + x*Log[4/x^3]^2),x]

[Out]

-x - Defer[Int][E^(9 + 12*x + 4*x^2 + E^(3 + 2*x)^2/(-3 + x + Log[4/x^3]))/(-3 + x + Log[4/x^3])^2, x] + 3*Def
er[Int][E^(9 + 12*x + 4*x^2 + E^(3 + 2*x)^2/(-3 + x + Log[4/x^3]))/(x*(-3 + x + Log[4/x^3])^2), x] + 12*Defer[
Int][E^(9 + 12*x + 4*x^2 + E^(3 + 2*x)^2/(-3 + x + Log[4/x^3]))/(-3 + x + Log[4/x^3]), x] + 8*Defer[Int][(E^(9
+ 12*x + 4*x^2 + E^(3 + 2*x)^2/(-3 + x + Log[4/x^3]))*x)/(-3 + x + Log[4/x^3]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-9 x+6 x^2-x^3+\left (6 x-2 x^2\right ) \log \left (\frac {4}{x^3}\right )-x \log ^2\left (\frac {4}{x^3}\right )+e^{\frac {e^{9+12 x+4 x^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}} \left (e^{9+12 x+4 x^2} \left (3-37 x-12 x^2+8 x^3\right )+e^{9+12 x+4 x^2} \left (12 x+8 x^2\right ) \log \left (\frac {4}{x^3}\right )\right )}{x \left (3-x-\log \left (\frac {4}{x^3}\right )\right )^2} \, dx\\ &=\int \left (-\frac {9}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}+\frac {6 x}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}-\frac {x^2}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}-\frac {2 (-3+x) \log \left (\frac {4}{x^3}\right )}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}-\frac {\log ^2\left (\frac {4}{x^3}\right )}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}+\frac {\exp \left (9+12 x+4 x^2+\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}\right ) \left (3-37 x-12 x^2+8 x^3+12 x \log \left (\frac {4}{x^3}\right )+8 x^2 \log \left (\frac {4}{x^3}\right )\right )}{x \left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}\right ) \, dx\\ &=-\left (2 \int \frac {(-3+x) \log \left (\frac {4}{x^3}\right )}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx\right )+6 \int \frac {x}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx-9 \int \frac {1}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx-\int \frac {x^2}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx-\int \frac {\log ^2\left (\frac {4}{x^3}\right )}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx+\int \frac {\exp \left (9+12 x+4 x^2+\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}\right ) \left (3-37 x-12 x^2+8 x^3+12 x \log \left (\frac {4}{x^3}\right )+8 x^2 \log \left (\frac {4}{x^3}\right )\right )}{x \left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx\\ &=-\left (2 \int \left (-\frac {(-3+x)^2}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}+\frac {-3+x}{-3+x+\log \left (\frac {4}{x^3}\right )}\right ) \, dx\right )+6 \int \frac {x}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx-9 \int \frac {1}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx-\int \frac {x^2}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx-\int \left (1+\frac {(-3+x)^2}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}-\frac {2 (-3+x)}{-3+x+\log \left (\frac {4}{x^3}\right )}\right ) \, dx+\int \left (\frac {\exp \left (9+12 x+4 x^2+\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}\right ) (3-x)}{x \left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}+\frac {4 \exp \left (9+12 x+4 x^2+\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}\right ) (3+2 x)}{-3+x+\log \left (\frac {4}{x^3}\right )}\right ) \, dx\\ &=-x+2 \int \frac {(-3+x)^2}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx+4 \int \frac {\exp \left (9+12 x+4 x^2+\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}\right ) (3+2 x)}{-3+x+\log \left (\frac {4}{x^3}\right )} \, dx+6 \int \frac {x}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx-9 \int \frac {1}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx-\int \frac {(-3+x)^2}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx+\int \frac {\exp \left (9+12 x+4 x^2+\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}\right ) (3-x)}{x \left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx-\int \frac {x^2}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx\\ &=-x+2 \int \left (\frac {9}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}-\frac {6 x}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}+\frac {x^2}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}\right ) \, dx+4 \int \left (\frac {3 \exp \left (9+12 x+4 x^2+\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}\right )}{-3+x+\log \left (\frac {4}{x^3}\right )}+\frac {2 \exp \left (9+12 x+4 x^2+\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}\right ) x}{-3+x+\log \left (\frac {4}{x^3}\right )}\right ) \, dx+6 \int \frac {x}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx-9 \int \frac {1}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx-\int \frac {x^2}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx+\int \left (-\frac {\exp \left (9+12 x+4 x^2+\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}\right )}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}+\frac {3 \exp \left (9+12 x+4 x^2+\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}\right )}{x \left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}\right ) \, dx-\int \left (\frac {9}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}-\frac {6 x}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}+\frac {x^2}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2}\right ) \, dx\\ &=-x+2 \int \frac {x^2}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx+3 \int \frac {\exp \left (9+12 x+4 x^2+\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}\right )}{x \left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx+2 \left (6 \int \frac {x}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx\right )+8 \int \frac {\exp \left (9+12 x+4 x^2+\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}\right ) x}{-3+x+\log \left (\frac {4}{x^3}\right )} \, dx-2 \left (9 \int \frac {1}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx\right )-12 \int \frac {x}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx+12 \int \frac {\exp \left (9+12 x+4 x^2+\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}\right )}{-3+x+\log \left (\frac {4}{x^3}\right )} \, dx+18 \int \frac {1}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx-\int \frac {\exp \left (9+12 x+4 x^2+\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}\right )}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx-2 \int \frac {x^2}{\left (-3+x+\log \left (\frac {4}{x^3}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.92, size = 27, normalized size = 1.00 \begin {gather*} e^{\frac {e^{(3+2 x)^2}}{-3+x+\log \left (\frac {4}{x^3}\right )}}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9*x + 6*x^2 - x^3 + (6*x - 2*x^2)*Log[4/x^3] - x*Log[4/x^3]^2 + E^(E^(9 + 12*x + 4*x^2)/(-3 + x +
Log[4/x^3]))*(E^(9 + 12*x + 4*x^2)*(3 - 37*x - 12*x^2 + 8*x^3) + E^(9 + 12*x + 4*x^2)*(12*x + 8*x^2)*Log[4/x^3
]))/(9*x - 6*x^2 + x^3 + (-6*x + 2*x^2)*Log[4/x^3] + x*Log[4/x^3]^2),x]

[Out]

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

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fricas [A]  time = 0.58, size = 28, normalized size = 1.04 \begin {gather*} -x + e^{\left (\frac {e^{\left (4 \, x^{2} + 12 \, x + 9\right )}}{x + \log \left (\frac {4}{x^{3}}\right ) - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+12*x)*exp(4*x^2+12*x+9)*log(4/x^3)+(8*x^3-12*x^2-37*x+3)*exp(4*x^2+12*x+9))*exp(exp(4*x^2+1
2*x+9)/(log(4/x^3)+x-3))-x*log(4/x^3)^2+(-2*x^2+6*x)*log(4/x^3)-x^3+6*x^2-9*x)/(x*log(4/x^3)^2+(2*x^2-6*x)*log
(4/x^3)+x^3-6*x^2+9*x),x, algorithm="fricas")

[Out]

-x + e^(e^(4*x^2 + 12*x + 9)/(x + log(4/x^3) - 3))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {x^{3} + x \log \left (\frac {4}{x^{3}}\right )^{2} - 6 \, x^{2} - {\left (4 \, {\left (2 \, x^{2} + 3 \, x\right )} e^{\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (\frac {4}{x^{3}}\right ) + {\left (8 \, x^{3} - 12 \, x^{2} - 37 \, x + 3\right )} e^{\left (4 \, x^{2} + 12 \, x + 9\right )}\right )} e^{\left (\frac {e^{\left (4 \, x^{2} + 12 \, x + 9\right )}}{x + \log \left (\frac {4}{x^{3}}\right ) - 3}\right )} + 2 \, {\left (x^{2} - 3 \, x\right )} \log \left (\frac {4}{x^{3}}\right ) + 9 \, x}{x^{3} + x \log \left (\frac {4}{x^{3}}\right )^{2} - 6 \, x^{2} + 2 \, {\left (x^{2} - 3 \, x\right )} \log \left (\frac {4}{x^{3}}\right ) + 9 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+12*x)*exp(4*x^2+12*x+9)*log(4/x^3)+(8*x^3-12*x^2-37*x+3)*exp(4*x^2+12*x+9))*exp(exp(4*x^2+1
2*x+9)/(log(4/x^3)+x-3))-x*log(4/x^3)^2+(-2*x^2+6*x)*log(4/x^3)-x^3+6*x^2-9*x)/(x*log(4/x^3)^2+(2*x^2-6*x)*log
(4/x^3)+x^3-6*x^2+9*x),x, algorithm="giac")

[Out]

integrate(-(x^3 + x*log(4/x^3)^2 - 6*x^2 - (4*(2*x^2 + 3*x)*e^(4*x^2 + 12*x + 9)*log(4/x^3) + (8*x^3 - 12*x^2
- 37*x + 3)*e^(4*x^2 + 12*x + 9))*e^(e^(4*x^2 + 12*x + 9)/(x + log(4/x^3) - 3)) + 2*(x^2 - 3*x)*log(4/x^3) + 9
*x)/(x^3 + x*log(4/x^3)^2 - 6*x^2 + 2*(x^2 - 3*x)*log(4/x^3) + 9*x), x)

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maple [C]  time = 44.92, size = 154, normalized size = 5.70

method	result	size
risch	\(-x +{\mathrm e}^{\frac {2 \,{\mathrm e}^{\left (2 x +3\right )^{2}}}{i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )+i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{3}\right )^{3}+4 \ln \relax (2)-6 \ln \relax (x )+2 x -6}}\)	\(154\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x^2+12*x)*exp(4*x^2+12*x+9)*ln(4/x^3)+(8*x^3-12*x^2-37*x+3)*exp(4*x^2+12*x+9))*exp(exp(4*x^2+12*x+9)/
(ln(4/x^3)+x-3))-x*ln(4/x^3)^2+(-2*x^2+6*x)*ln(4/x^3)-x^3+6*x^2-9*x)/(x*ln(4/x^3)^2+(2*x^2-6*x)*ln(4/x^3)+x^3-
6*x^2+9*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.51, size = 59, normalized size = 2.19 \begin {gather*} -{\left (x e^{\left (-\frac {e^{\left (4 \, x^{2} + 12 \, x + 9\right )}}{x + 2 \, \log \relax (2) - 3 \, \log \relax (x) - 3}\right )} - 1\right )} e^{\left (\frac {e^{\left (4 \, x^{2} + 12 \, x + 9\right )}}{x + 2 \, \log \relax (2) - 3 \, \log \relax (x) - 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^2+12*x)*exp(4*x^2+12*x+9)*log(4/x^3)+(8*x^3-12*x^2-37*x+3)*exp(4*x^2+12*x+9))*exp(exp(4*x^2+1
2*x+9)/(log(4/x^3)+x-3))-x*log(4/x^3)^2+(-2*x^2+6*x)*log(4/x^3)-x^3+6*x^2-9*x)/(x*log(4/x^3)^2+(2*x^2-6*x)*log
(4/x^3)+x^3-6*x^2+9*x),x, algorithm="maxima")

[Out]

-(x*e^(-e^(4*x^2 + 12*x + 9)/(x + 2*log(2) - 3*log(x) - 3)) - 1)*e^(e^(4*x^2 + 12*x + 9)/(x + 2*log(2) - 3*log
(x) - 3))

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mupad [B]  time = 6.01, size = 29, normalized size = 1.07 \begin {gather*} {\mathrm {e}}^{\frac {{\mathrm {e}}^{12\,x}\,{\mathrm {e}}^9\,{\mathrm {e}}^{4\,x^2}}{x+\ln \left (\frac {4}{x^3}\right )-3}}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(9*x + exp(exp(12*x + 4*x^2 + 9)/(x + log(4/x^3) - 3))*(exp(12*x + 4*x^2 + 9)*(37*x + 12*x^2 - 8*x^3 - 3)
- exp(12*x + 4*x^2 + 9)*log(4/x^3)*(12*x + 8*x^2)) - log(4/x^3)*(6*x - 2*x^2) - 6*x^2 + x^3 + x*log(4/x^3)^2)
/(9*x - log(4/x^3)*(6*x - 2*x^2) - 6*x^2 + x^3 + x*log(4/x^3)^2),x)

[Out]

exp((exp(12*x)*exp(9)*exp(4*x^2))/(x + log(4/x^3) - 3)) - x

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sympy [A]  time = 1.45, size = 24, normalized size = 0.89 \begin {gather*} - x + e^{\frac {e^{4 x^{2} + 12 x + 9}}{x + \log {\left (\frac {4}{x^{3}} \right )} - 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x**2+12*x)*exp(4*x**2+12*x+9)*ln(4/x**3)+(8*x**3-12*x**2-37*x+3)*exp(4*x**2+12*x+9))*exp(exp(4*
x**2+12*x+9)/(ln(4/x**3)+x-3))-x*ln(4/x**3)**2+(-2*x**2+6*x)*ln(4/x**3)-x**3+6*x**2-9*x)/(x*ln(4/x**3)**2+(2*x
**2-6*x)*ln(4/x**3)+x**3-6*x**2+9*x),x)

[Out]

-x + exp(exp(4*x**2 + 12*x + 9)/(x + log(4/x**3) - 3))

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