3.81.40 \(\int \frac {-5 x^2+e^x (5 x-x^2+x^3)+(5 x^2-5 e^x x^2) \log (\frac {x}{3})+(-12 x+4 x^2+(60-20 x) \log (\frac {x}{3})) \log (3 x^2)+(-15+11 x-x^2-15 \log (\frac {x}{3})) \log ^2(3 x^2)}{e^{2 x} x^2-2 e^x x^3+x^4+(6 x^2-2 x^3+e^x (-6 x+2 x^2)) \log ^2(3 x^2)+(9-6 x+x^2) \log ^4(3 x^2)} \, dx\)

Optimal. Leaf size=40 \[ \frac {-x+5 \log \left (\frac {x}{3}\right )}{e^x-x-\frac {(3-x) \log ^2\left (3 x^2\right )}{x}} \]

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Rubi [F]  time = 36.33, 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 {-5 x^2+e^x \left (5 x-x^2+x^3\right )+\left (5 x^2-5 e^x x^2\right ) \log \left (\frac {x}{3}\right )+\left (-12 x+4 x^2+(60-20 x) \log \left (\frac {x}{3}\right )\right ) \log \left (3 x^2\right )+\left (-15+11 x-x^2-15 \log \left (\frac {x}{3}\right )\right ) \log ^2\left (3 x^2\right )}{e^{2 x} x^2-2 e^x x^3+x^4+\left (6 x^2-2 x^3+e^x \left (-6 x+2 x^2\right )\right ) \log ^2\left (3 x^2\right )+\left (9-6 x+x^2\right ) \log ^4\left (3 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-5*x^2 + E^x*(5*x - x^2 + x^3) + (5*x^2 - 5*E^x*x^2)*Log[x/3] + (-12*x + 4*x^2 + (60 - 20*x)*Log[x/3])*Lo
g[3*x^2] + (-15 + 11*x - x^2 - 15*Log[x/3])*Log[3*x^2]^2)/(E^(2*x)*x^2 - 2*E^x*x^3 + x^4 + (6*x^2 - 2*x^3 + E^
x*(-6*x + 2*x^2))*Log[3*x^2]^2 + (9 - 6*x + x^2)*Log[3*x^2]^4),x]

[Out]

-Defer[Int][x^3/((E^x - x)*x + (-3 + x)*Log[3*x^2]^2)^2, x] + Defer[Int][x^4/((E^x - x)*x + (-3 + x)*Log[3*x^2
]^2)^2, x] + 5*Defer[Int][(x^2*Log[x/3])/((E^x - x)*x + (-3 + x)*Log[3*x^2]^2)^2, x] - 5*Defer[Int][(x^3*Log[x
/3])/((E^x - x)*x + (-3 + x)*Log[3*x^2]^2)^2, x] - 12*Defer[Int][(x*Log[3*x^2])/((E^x - x)*x + (-3 + x)*Log[3*
x^2]^2)^2, x] + 4*Defer[Int][(x^2*Log[3*x^2])/((E^x - x)*x + (-3 + x)*Log[3*x^2]^2)^2, x] + 60*Defer[Int][(Log
[x/3]*Log[3*x^2])/((E^x - x)*x + (-3 + x)*Log[3*x^2]^2)^2, x] - 20*Defer[Int][(x*Log[x/3]*Log[3*x^2])/((E^x -
x)*x + (-3 + x)*Log[3*x^2]^2)^2, x] + 3*Defer[Int][(x*Log[3*x^2]^2)/((E^x - x)*x + (-3 + x)*Log[3*x^2]^2)^2, x
] + 3*Defer[Int][(x^2*Log[3*x^2]^2)/((E^x - x)*x + (-3 + x)*Log[3*x^2]^2)^2, x] - Defer[Int][(x^3*Log[3*x^2]^2
)/((E^x - x)*x + (-3 + x)*Log[3*x^2]^2)^2, x] - 15*Defer[Int][(Log[x/3]*Log[3*x^2]^2)/((E^x - x)*x + (-3 + x)*
Log[3*x^2]^2)^2, x] - 15*Defer[Int][(x*Log[x/3]*Log[3*x^2]^2)/((E^x - x)*x + (-3 + x)*Log[3*x^2]^2)^2, x] + 5*
Defer[Int][(x^2*Log[x/3]*Log[3*x^2]^2)/((E^x - x)*x + (-3 + x)*Log[3*x^2]^2)^2, x] + 5*Defer[Int][((E^x - x)*x
+ (-3 + x)*Log[3*x^2]^2)^(-1), x] + Defer[Int][x/(-(E^x*x) + x^2 + 3*Log[3*x^2]^2 - x*Log[3*x^2]^2), x] - Def
er[Int][x^2/(-(E^x*x) + x^2 + 3*Log[3*x^2]^2 - x*Log[3*x^2]^2), x] + 5*Defer[Int][(x*Log[x/3])/(-(E^x*x) + x^2
+ 3*Log[3*x^2]^2 - x*Log[3*x^2]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x \left (-5 x+e^x \left (5-x+x^2\right )\right )+4 (-3+x) x \log \left (3 x^2\right )-\left (15-11 x+x^2\right ) \log ^2\left (3 x^2\right )-5 \log \left (\frac {x}{3}\right ) \left (\left (-1+e^x\right ) x^2+4 (-3+x) \log \left (3 x^2\right )+3 \log ^2\left (3 x^2\right )\right )}{\left (\left (e^x-x\right ) x+(-3+x) \log ^2\left (3 x^2\right )\right )^2} \, dx\\ &=\int \left (-\frac {5-x+x^2-5 x \log \left (\frac {x}{3}\right )}{-e^x x+x^2+3 \log ^2\left (3 x^2\right )-x \log ^2\left (3 x^2\right )}+\frac {\left (x-5 \log \left (\frac {x}{3}\right )\right ) \left (-x^2+x^3-12 \log \left (3 x^2\right )+4 x \log \left (3 x^2\right )+3 \log ^2\left (3 x^2\right )+3 x \log ^2\left (3 x^2\right )-x^2 \log ^2\left (3 x^2\right )\right )}{\left (-e^x x+x^2+3 \log ^2\left (3 x^2\right )-x \log ^2\left (3 x^2\right )\right )^2}\right ) \, dx\\ &=-\int \frac {5-x+x^2-5 x \log \left (\frac {x}{3}\right )}{-e^x x+x^2+3 \log ^2\left (3 x^2\right )-x \log ^2\left (3 x^2\right )} \, dx+\int \frac {\left (x-5 \log \left (\frac {x}{3}\right )\right ) \left (-x^2+x^3-12 \log \left (3 x^2\right )+4 x \log \left (3 x^2\right )+3 \log ^2\left (3 x^2\right )+3 x \log ^2\left (3 x^2\right )-x^2 \log ^2\left (3 x^2\right )\right )}{\left (-e^x x+x^2+3 \log ^2\left (3 x^2\right )-x \log ^2\left (3 x^2\right )\right )^2} \, dx\\ &=\int \frac {\left (x-5 \log \left (\frac {x}{3}\right )\right ) \left ((-1+x) x^2+4 (-3+x) \log \left (3 x^2\right )+\left (3+3 x-x^2\right ) \log ^2\left (3 x^2\right )\right )}{\left (\left (e^x-x\right ) x+(-3+x) \log ^2\left (3 x^2\right )\right )^2} \, dx-\int \left (-\frac {x}{-e^x x+x^2+3 \log ^2\left (3 x^2\right )-x \log ^2\left (3 x^2\right )}+\frac {x^2}{-e^x x+x^2+3 \log ^2\left (3 x^2\right )-x \log ^2\left (3 x^2\right )}-\frac {5 x \log \left (\frac {x}{3}\right )}{-e^x x+x^2+3 \log ^2\left (3 x^2\right )-x \log ^2\left (3 x^2\right )}-\frac {5}{e^x x-x^2-3 \log ^2\left (3 x^2\right )+x \log ^2\left (3 x^2\right )}\right ) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 37, normalized size = 0.92 \begin {gather*} \frac {x \left (x-5 \log \left (\frac {x}{3}\right )\right )}{x \left (-e^x+x\right )-(-3+x) \log ^2\left (3 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*x^2 + E^x*(5*x - x^2 + x^3) + (5*x^2 - 5*E^x*x^2)*Log[x/3] + (-12*x + 4*x^2 + (60 - 20*x)*Log[x/
3])*Log[3*x^2] + (-15 + 11*x - x^2 - 15*Log[x/3])*Log[3*x^2]^2)/(E^(2*x)*x^2 - 2*E^x*x^3 + x^4 + (6*x^2 - 2*x^
3 + E^x*(-6*x + 2*x^2))*Log[3*x^2]^2 + (9 - 6*x + x^2)*Log[3*x^2]^4),x]

[Out]

(x*(x - 5*Log[x/3]))/(x*(-E^x + x) - (-3 + x)*Log[3*x^2]^2)

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fricas [A]  time = 1.32, size = 56, normalized size = 1.40 \begin {gather*} -\frac {x^{2} - 5 \, x \log \left (\frac {1}{3} \, x\right )}{9 \, {\left (x - 3\right )} \log \relax (3)^{2} + 12 \, {\left (x - 3\right )} \log \relax (3) \log \left (\frac {1}{3} \, x\right ) + 4 \, {\left (x - 3\right )} \log \left (\frac {1}{3} \, x\right )^{2} - x^{2} + x e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*log(1/3*x)-x^2+11*x-15)*log(3*x^2)^2+((-20*x+60)*log(1/3*x)+4*x^2-12*x)*log(3*x^2)+(-5*exp(x)*
x^2+5*x^2)*log(1/3*x)+(x^3-x^2+5*x)*exp(x)-5*x^2)/((x^2-6*x+9)*log(3*x^2)^4+((2*x^2-6*x)*exp(x)-2*x^3+6*x^2)*l
og(3*x^2)^2+exp(x)^2*x^2-2*exp(x)*x^3+x^4),x, algorithm="fricas")

[Out]

-(x^2 - 5*x*log(1/3*x))/(9*(x - 3)*log(3)^2 + 12*(x - 3)*log(3)*log(1/3*x) + 4*(x - 3)*log(1/3*x)^2 - x^2 + x*
e^x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*log(1/3*x)-x^2+11*x-15)*log(3*x^2)^2+((-20*x+60)*log(1/3*x)+4*x^2-12*x)*log(3*x^2)+(-5*exp(x)*
x^2+5*x^2)*log(1/3*x)+(x^3-x^2+5*x)*exp(x)-5*x^2)/((x^2-6*x+9)*log(3*x^2)^4+((2*x^2-6*x)*exp(x)-2*x^3+6*x^2)*l
og(3*x^2)^2+exp(x)^2*x^2-2*exp(x)*x^3+x^4),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 67.54, size = 492, normalized size = 12.30

method	result	size
risch	\(\frac {-20 x \ln \relax (x )+20 x \ln \relax (3)+4 x^{2}}{-16 x \ln \relax (3) \ln \relax (x )+12 \ln \relax (3)^{2}+48 \ln \relax (x )^{2}+4 x^{2}-4 x \ln \relax (3)^{2}+48 \ln \relax (3) \ln \relax (x )-16 x \ln \relax (x )^{2}-4 \,{\mathrm e}^{x} x -3 \pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{6}-12 i \ln \relax (3) \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-24 i \ln \relax (x ) \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+x \,\pi ^{2} \mathrm {csgn}\left (i x \right )^{4} \mathrm {csgn}\left (i x^{2}\right )^{2}-4 x \,\pi ^{2} \mathrm {csgn}\left (i x \right )^{3} \mathrm {csgn}\left (i x^{2}\right )^{3}+6 x \,\pi ^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )^{4}-4 x \,\pi ^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{5}+8 i x \ln \relax (x ) \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+48 i \ln \relax (x ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+4 i x \ln \relax (3) \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-12 i \ln \relax (3) \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+24 i \ln \relax (3) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+4 i x \ln \relax (3) \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-8 i x \ln \relax (3) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+8 i x \ln \relax (x ) \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-16 i x \ln \relax (x ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+12 \pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{5} \mathrm {csgn}\left (i x \right )-18 \pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{4} \mathrm {csgn}\left (i x \right )^{2}+12 \pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{3} \mathrm {csgn}\left (i x \right )^{3}-3 \pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )^{4}+x \,\pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{6}-24 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}\)	\(492\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-15*ln(1/3*x)-x^2+11*x-15)*ln(3*x^2)^2+((-20*x+60)*ln(1/3*x)+4*x^2-12*x)*ln(3*x^2)+(-5*exp(x)*x^2+5*x^2)
*ln(1/3*x)+(x^3-x^2+5*x)*exp(x)-5*x^2)/((x^2-6*x+9)*ln(3*x^2)^4+((2*x^2-6*x)*exp(x)-2*x^3+6*x^2)*ln(3*x^2)^2+e
xp(x)^2*x^2-2*exp(x)*x^3+x^4),x,method=_RETURNVERBOSE)

[Out]

2*(-10*x*ln(x)+10*x*ln(3)+2*x^2)/(-16*x*ln(3)*ln(x)-12*I*ln(3)*Pi*csgn(I*x^2)^3-24*I*ln(x)*Pi*csgn(I*x^2)^3+12
*ln(3)^2+48*ln(x)^2+4*x^2-4*x*ln(3)^2+48*ln(3)*ln(x)-16*x*ln(x)^2-4*exp(x)*x+x*Pi^2*csgn(I*x)^4*csgn(I*x^2)^2-
4*x*Pi^2*csgn(I*x)^3*csgn(I*x^2)^3+6*x*Pi^2*csgn(I*x)^2*csgn(I*x^2)^4-4*x*Pi^2*csgn(I*x)*csgn(I*x^2)^5+8*I*x*l
n(x)*Pi*csgn(I*x^2)^3-24*I*ln(x)*Pi*csgn(I*x)^2*csgn(I*x^2)+48*I*ln(x)*Pi*csgn(I*x)*csgn(I*x^2)^2+4*I*x*ln(3)*
Pi*csgn(I*x^2)^3-12*I*ln(3)*Pi*csgn(I*x)^2*csgn(I*x^2)+24*I*ln(3)*Pi*csgn(I*x)*csgn(I*x^2)^2+4*I*x*ln(3)*Pi*cs
gn(I*x)^2*csgn(I*x^2)-8*I*x*ln(3)*Pi*csgn(I*x)*csgn(I*x^2)^2+8*I*x*ln(x)*Pi*csgn(I*x)^2*csgn(I*x^2)-16*I*x*ln(
x)*Pi*csgn(I*x)*csgn(I*x^2)^2+12*Pi^2*csgn(I*x^2)^5*csgn(I*x)-18*Pi^2*csgn(I*x^2)^4*csgn(I*x)^2+12*Pi^2*csgn(I
*x^2)^3*csgn(I*x)^3-3*Pi^2*csgn(I*x^2)^2*csgn(I*x)^4-3*Pi^2*csgn(I*x^2)^6+x*Pi^2*csgn(I*x^2)^6)

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maxima [A]  time = 0.53, size = 62, normalized size = 1.55 \begin {gather*} -\frac {x^{2} + 5 \, x \log \relax (3) - 5 \, x \log \relax (x)}{x \log \relax (3)^{2} + 4 \, {\left (x - 3\right )} \log \relax (x)^{2} - x^{2} + x e^{x} - 3 \, \log \relax (3)^{2} + 4 \, {\left (x \log \relax (3) - 3 \, \log \relax (3)\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*log(1/3*x)-x^2+11*x-15)*log(3*x^2)^2+((-20*x+60)*log(1/3*x)+4*x^2-12*x)*log(3*x^2)+(-5*exp(x)*
x^2+5*x^2)*log(1/3*x)+(x^3-x^2+5*x)*exp(x)-5*x^2)/((x^2-6*x+9)*log(3*x^2)^4+((2*x^2-6*x)*exp(x)-2*x^3+6*x^2)*l
og(3*x^2)^2+exp(x)^2*x^2-2*exp(x)*x^3+x^4),x, algorithm="maxima")

[Out]

-(x^2 + 5*x*log(3) - 5*x*log(x))/(x*log(3)^2 + 4*(x - 3)*log(x)^2 - x^2 + x*e^x - 3*log(3)^2 + 4*(x*log(3) - 3
*log(3))*log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int -\frac {{\ln \left (3\,x^2\right )}^2\,\left (15\,\ln \left (\frac {x}{3}\right )-11\,x+x^2+15\right )+\ln \left (3\,x^2\right )\,\left (12\,x-4\,x^2+\ln \left (\frac {x}{3}\right )\,\left (20\,x-60\right )\right )+\ln \left (\frac {x}{3}\right )\,\left (5\,x^2\,{\mathrm {e}}^x-5\,x^2\right )-{\mathrm {e}}^x\,\left (x^3-x^2+5\,x\right )+5\,x^2}{{\ln \left (3\,x^2\right )}^4\,\left (x^2-6\,x+9\right )-2\,x^3\,{\mathrm {e}}^x+x^2\,{\mathrm {e}}^{2\,x}+x^4-{\ln \left (3\,x^2\right )}^2\,\left ({\mathrm {e}}^x\,\left (6\,x-2\,x^2\right )-6\,x^2+2\,x^3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(3*x^2)^2*(15*log(x/3) - 11*x + x^2 + 15) + log(3*x^2)*(12*x - 4*x^2 + log(x/3)*(20*x - 60)) + log(x/
3)*(5*x^2*exp(x) - 5*x^2) - exp(x)*(5*x - x^2 + x^3) + 5*x^2)/(log(3*x^2)^4*(x^2 - 6*x + 9) - 2*x^3*exp(x) + x
^2*exp(2*x) + x^4 - log(3*x^2)^2*(exp(x)*(6*x - 2*x^2) - 6*x^2 + 2*x^3)),x)

[Out]

int(-(log(3*x^2)^2*(15*log(x/3) - 11*x + x^2 + 15) + log(3*x^2)*(12*x - 4*x^2 + log(x/3)*(20*x - 60)) + log(x/
3)*(5*x^2*exp(x) - 5*x^2) - exp(x)*(5*x - x^2 + x^3) + 5*x^2)/(log(3*x^2)^4*(x^2 - 6*x + 9) - 2*x^3*exp(x) + x
^2*exp(2*x) + x^4 - log(3*x^2)^2*(exp(x)*(6*x - 2*x^2) - 6*x^2 + 2*x^3)), x)

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sympy [B]  time = 0.73, size = 75, normalized size = 1.88 \begin {gather*} \frac {- x^{2} + 5 x \log {\left (\frac {x}{3} \right )}}{- x^{2} + x e^{x} + 4 x \log {\left (\frac {x}{3} \right )}^{2} + 12 x \log {\relax (3 )} \log {\left (\frac {x}{3} \right )} + 9 x \log {\relax (3 )}^{2} - 12 \log {\left (\frac {x}{3} \right )}^{2} - 36 \log {\relax (3 )} \log {\left (\frac {x}{3} \right )} - 27 \log {\relax (3 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-15*ln(1/3*x)-x**2+11*x-15)*ln(3*x**2)**2+((-20*x+60)*ln(1/3*x)+4*x**2-12*x)*ln(3*x**2)+(-5*exp(x)
*x**2+5*x**2)*ln(1/3*x)+(x**3-x**2+5*x)*exp(x)-5*x**2)/((x**2-6*x+9)*ln(3*x**2)**4+((2*x**2-6*x)*exp(x)-2*x**3
+6*x**2)*ln(3*x**2)**2+exp(x)**2*x**2-2*exp(x)*x**3+x**4),x)

[Out]

(-x**2 + 5*x*log(x/3))/(-x**2 + x*exp(x) + 4*x*log(x/3)**2 + 12*x*log(3)*log(x/3) + 9*x*log(3)**2 - 12*log(x/3
)**2 - 36*log(3)*log(x/3) - 27*log(3)**2)

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