3.62.82 \(\int \frac {1}{16} (23 x^2+60 x^3-200 x^4+120 x^5+(10 x+60 x^2-80 x^3) \log (x)+10 x \log ^2(x)+x^{2 x} (40 x+40 x^2+40 x^2 \log (x))+x^x (-20 x-120 x^2+120 x^3+40 x^4+(-40 x-20 x^2-40 x^3+40 x^4) \log (x)-20 x^2 \log ^2(x))) \, dx\)

Optimal. Leaf size=30 \[ \frac {1}{16} x^2 \left (x+5 \left (2 \left (x-x^2-x^x\right )+\log (x)\right )^2\right ) \]

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Rubi [F]  time = 0.52, 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 {1}{16} \left (23 x^2+60 x^3-200 x^4+120 x^5+\left (10 x+60 x^2-80 x^3\right ) \log (x)+10 x \log ^2(x)+x^{2 x} \left (40 x+40 x^2+40 x^2 \log (x)\right )+x^x \left (-20 x-120 x^2+120 x^3+40 x^4+\left (-40 x-20 x^2-40 x^3+40 x^4\right ) \log (x)-20 x^2 \log ^2(x)\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(23*x^2 + 60*x^3 - 200*x^4 + 120*x^5 + (10*x + 60*x^2 - 80*x^3)*Log[x] + 10*x*Log[x]^2 + x^(2*x)*(40*x + 4
0*x^2 + 40*x^2*Log[x]) + x^x*(-20*x - 120*x^2 + 120*x^3 + 40*x^4 + (-40*x - 20*x^2 - 40*x^3 + 40*x^4)*Log[x] -
20*x^2*Log[x]^2))/16,x]

[Out]

x^3/16 + (5*x^4)/4 - (5*x^5)/2 + (5*x^6)/4 + (5*x^3*Log[x])/4 - (5*x^4*Log[x])/4 + (5*x^2*Log[x]^2)/16 - (5*De
fer[Int][x^(1 + x), x])/4 - (5*Log[x]*Defer[Int][x^(1 + x), x])/2 - (15*Defer[Int][x^(2 + x), x])/2 - (5*Log[x
]*Defer[Int][x^(2 + x), x])/4 + (15*Defer[Int][x^(3 + x), x])/2 - (5*Log[x]*Defer[Int][x^(3 + x), x])/2 + (5*D
efer[Int][x^(4 + x), x])/2 + (5*Log[x]*Defer[Int][x^(4 + x), x])/2 + (5*Defer[Int][x^(1 + 2*x), x])/2 + (5*Def
er[Int][x^(2 + 2*x), x])/2 + (5*Log[x]*Defer[Int][x^(2 + 2*x), x])/2 - (5*Defer[Int][x^(2 + x)*Log[x]^2, x])/4
+ (5*Defer[Int][Defer[Int][x^(1 + x), x]/x, x])/2 + (5*Defer[Int][Defer[Int][x^(2 + x), x]/x, x])/4 + (5*Defe
r[Int][Defer[Int][x^(3 + x), x]/x, x])/2 - (5*Defer[Int][Defer[Int][x^(4 + x), x]/x, x])/2 - (5*Defer[Int][Def
er[Int][x^(2 + 2*x), x]/x, x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{16} \int \left (23 x^2+60 x^3-200 x^4+120 x^5+\left (10 x+60 x^2-80 x^3\right ) \log (x)+10 x \log ^2(x)+x^{2 x} \left (40 x+40 x^2+40 x^2 \log (x)\right )+x^x \left (-20 x-120 x^2+120 x^3+40 x^4+\left (-40 x-20 x^2-40 x^3+40 x^4\right ) \log (x)-20 x^2 \log ^2(x)\right )\right ) \, dx\\ &=\frac {23 x^3}{48}+\frac {15 x^4}{16}-\frac {5 x^5}{2}+\frac {5 x^6}{4}+\frac {1}{16} \int \left (10 x+60 x^2-80 x^3\right ) \log (x) \, dx+\frac {1}{16} \int x^{2 x} \left (40 x+40 x^2+40 x^2 \log (x)\right ) \, dx+\frac {1}{16} \int x^x \left (-20 x-120 x^2+120 x^3+40 x^4+\left (-40 x-20 x^2-40 x^3+40 x^4\right ) \log (x)-20 x^2 \log ^2(x)\right ) \, dx+\frac {5}{8} \int x \log ^2(x) \, dx\\ &=\frac {23 x^3}{48}+\frac {15 x^4}{16}-\frac {5 x^5}{2}+\frac {5 x^6}{4}+\frac {5}{16} x^2 \log ^2(x)+\frac {1}{16} \int x \left (10+60 x-80 x^2\right ) \log (x) \, dx+\frac {1}{16} \int 40 x^{1+2 x} (1+x+x \log (x)) \, dx+\frac {1}{16} \int 20 x^{1+x} \left (-1-6 x+6 x^2+2 x^3-2 \log (x)-x \log (x)-2 x^2 \log (x)+2 x^3 \log (x)-x \log ^2(x)\right ) \, dx-\frac {5}{8} \int x \log (x) \, dx\\ &=\frac {5 x^2}{32}+\frac {23 x^3}{48}+\frac {15 x^4}{16}-\frac {5 x^5}{2}+\frac {5 x^6}{4}-\frac {5}{16} x^2 \log (x)+\frac {5}{16} x^2 \log ^2(x)+\frac {1}{16} \int \left (10 x \log (x)+60 x^2 \log (x)-80 x^3 \log (x)\right ) \, dx+\frac {5}{4} \int x^{1+x} \left (-1-6 x+6 x^2+2 x^3-2 \log (x)-x \log (x)-2 x^2 \log (x)+2 x^3 \log (x)-x \log ^2(x)\right ) \, dx+\frac {5}{2} \int x^{1+2 x} (1+x+x \log (x)) \, dx\\ &=\frac {5 x^2}{32}+\frac {23 x^3}{48}+\frac {15 x^4}{16}-\frac {5 x^5}{2}+\frac {5 x^6}{4}-\frac {5}{16} x^2 \log (x)+\frac {5}{16} x^2 \log ^2(x)+\frac {5}{8} \int x \log (x) \, dx+\frac {5}{4} \int \left (-x^{1+x}-6 x^{2+x}+6 x^{3+x}+2 x^{4+x}-2 x^{1+x} \log (x)-x^{2+x} \log (x)-2 x^{3+x} \log (x)+2 x^{4+x} \log (x)-x^{2+x} \log ^2(x)\right ) \, dx+\frac {5}{2} \int \left (x^{1+2 x}+x^{2+2 x}+x^{2+2 x} \log (x)\right ) \, dx+\frac {15}{4} \int x^2 \log (x) \, dx-5 \int x^3 \log (x) \, dx\\ &=\frac {x^3}{16}+\frac {5 x^4}{4}-\frac {5 x^5}{2}+\frac {5 x^6}{4}+\frac {5}{4} x^3 \log (x)-\frac {5}{4} x^4 \log (x)+\frac {5}{16} x^2 \log ^2(x)-\frac {5}{4} \int x^{1+x} \, dx-\frac {5}{4} \int x^{2+x} \log (x) \, dx-\frac {5}{4} \int x^{2+x} \log ^2(x) \, dx+\frac {5}{2} \int x^{4+x} \, dx+\frac {5}{2} \int x^{1+2 x} \, dx+\frac {5}{2} \int x^{2+2 x} \, dx-\frac {5}{2} \int x^{1+x} \log (x) \, dx-\frac {5}{2} \int x^{3+x} \log (x) \, dx+\frac {5}{2} \int x^{4+x} \log (x) \, dx+\frac {5}{2} \int x^{2+2 x} \log (x) \, dx-\frac {15}{2} \int x^{2+x} \, dx+\frac {15}{2} \int x^{3+x} \, dx\\ &=\frac {x^3}{16}+\frac {5 x^4}{4}-\frac {5 x^5}{2}+\frac {5 x^6}{4}+\frac {5}{4} x^3 \log (x)-\frac {5}{4} x^4 \log (x)+\frac {5}{16} x^2 \log ^2(x)-\frac {5}{4} \int x^{1+x} \, dx-\frac {5}{4} \int x^{2+x} \log ^2(x) \, dx+\frac {5}{4} \int \frac {\int x^{2+x} \, dx}{x} \, dx+\frac {5}{2} \int x^{4+x} \, dx+\frac {5}{2} \int x^{1+2 x} \, dx+\frac {5}{2} \int x^{2+2 x} \, dx+\frac {5}{2} \int \frac {\int x^{1+x} \, dx}{x} \, dx+\frac {5}{2} \int \frac {\int x^{3+x} \, dx}{x} \, dx-\frac {5}{2} \int \frac {\int x^{4+x} \, dx}{x} \, dx-\frac {5}{2} \int \frac {\int x^{2+2 x} \, dx}{x} \, dx-\frac {15}{2} \int x^{2+x} \, dx+\frac {15}{2} \int x^{3+x} \, dx-\frac {1}{4} (5 \log (x)) \int x^{2+x} \, dx-\frac {1}{2} (5 \log (x)) \int x^{1+x} \, dx-\frac {1}{2} (5 \log (x)) \int x^{3+x} \, dx+\frac {1}{2} (5 \log (x)) \int x^{4+x} \, dx+\frac {1}{2} (5 \log (x)) \int x^{2+2 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.08, size = 65, normalized size = 2.17 \begin {gather*} \frac {1}{16} x^2 \left (x+20 x^2-40 x^3+20 x^4+20 x^{2 x}-40 x^{1+x}+40 x^{2+x}-20 \left (-x+x^2+x^x\right ) \log (x)+5 \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(23*x^2 + 60*x^3 - 200*x^4 + 120*x^5 + (10*x + 60*x^2 - 80*x^3)*Log[x] + 10*x*Log[x]^2 + x^(2*x)*(40
*x + 40*x^2 + 40*x^2*Log[x]) + x^x*(-20*x - 120*x^2 + 120*x^3 + 40*x^4 + (-40*x - 20*x^2 - 40*x^3 + 40*x^4)*Lo
g[x] - 20*x^2*Log[x]^2))/16,x]

[Out]

(x^2*(x + 20*x^2 - 40*x^3 + 20*x^4 + 20*x^(2*x) - 40*x^(1 + x) + 40*x^(2 + x) - 20*(-x + x^2 + x^x)*Log[x] + 5
*Log[x]^2))/16

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fricas [B]  time = 0.71, size = 76, normalized size = 2.53 \begin {gather*} \frac {5}{4} \, x^{6} - \frac {5}{2} \, x^{5} + \frac {5}{4} \, x^{4} + \frac {5}{16} \, x^{2} \log \relax (x)^{2} + \frac {5}{4} \, x^{2} x^{2 \, x} + \frac {1}{16} \, x^{3} + \frac {5}{4} \, {\left (2 \, x^{4} - 2 \, x^{3} - x^{2} \log \relax (x)\right )} x^{x} - \frac {5}{4} \, {\left (x^{4} - x^{3}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(40*x^2*log(x)+40*x^2+40*x)*exp(x*log(x))^2+1/16*(-20*x^2*log(x)^2+(40*x^4-40*x^3-20*x^2-40*x)*
log(x)+40*x^4+120*x^3-120*x^2-20*x)*exp(x*log(x))+5/8*x*log(x)^2+1/16*(-80*x^3+60*x^2+10*x)*log(x)+15/2*x^5-25
/2*x^4+15/4*x^3+23/16*x^2,x, algorithm="fricas")

[Out]

5/4*x^6 - 5/2*x^5 + 5/4*x^4 + 5/16*x^2*log(x)^2 + 5/4*x^2*x^(2*x) + 1/16*x^3 + 5/4*(2*x^4 - 2*x^3 - x^2*log(x)
)*x^x - 5/4*(x^4 - x^3)*log(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(1/16*(40*x^2*log(x)+40*x^2+40*x)*exp(x*log(x))^2+1/16*(-20*x^2*log(x)^2+(40*x^4-40*x^3-20*x^2-40*x)*
log(x)+40*x^4+120*x^3-120*x^2-20*x)*exp(x*log(x))+5/8*x*log(x)^2+1/16*(-80*x^3+60*x^2+10*x)*log(x)+15/2*x^5-25
/2*x^4+15/4*x^3+23/16*x^2,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.15, size = 91, normalized size = 3.03

method	result	size
risch	\(\frac {5 x^{2 x} x^{2}}{4}+\frac {\left (40 x^{4}-40 x^{3}-20 x^{2} \ln \relax (x )\right ) x^{x}}{16}+\frac {5 x^{2} \ln \relax (x )^{2}}{16}-\frac {5 x^{2} \ln \relax (x )}{16}+\frac {\left (-20 x^{4}+20 x^{3}+5 x^{2}\right ) \ln \relax (x )}{16}+\frac {5 x^{4}}{4}+\frac {x^{3}}{16}+\frac {5 x^{6}}{4}-\frac {5 x^{5}}{2}\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*(40*x^2*ln(x)+40*x^2+40*x)*exp(x*ln(x))^2+1/16*(-20*x^2*ln(x)^2+(40*x^4-40*x^3-20*x^2-40*x)*ln(x)+40*
x^4+120*x^3-120*x^2-20*x)*exp(x*ln(x))+5/8*x*ln(x)^2+1/16*(-80*x^3+60*x^2+10*x)*ln(x)+15/2*x^5-25/2*x^4+15/4*x
^3+23/16*x^2,x,method=_RETURNVERBOSE)

[Out]

5/4*(x^x)^2*x^2+1/16*(40*x^4-40*x^3-20*x^2*ln(x))*x^x+5/16*x^2*ln(x)^2-5/16*x^2*ln(x)+1/16*(-20*x^4+20*x^3+5*x
^2)*ln(x)+5/4*x^4+1/16*x^3+5/4*x^6-5/2*x^5

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maxima [B]  time = 0.35, size = 96, normalized size = 3.20 \begin {gather*} \frac {5}{4} \, x^{6} - \frac {5}{2} \, x^{5} + \frac {5}{4} \, x^{4} + \frac {5}{4} \, x^{2} x^{2 \, x} + \frac {5}{32} \, {\left (2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) + 1\right )} x^{2} + \frac {1}{16} \, x^{3} + \frac {5}{4} \, {\left (2 \, x^{4} - 2 \, x^{3} - x^{2} \log \relax (x)\right )} x^{x} - \frac {5}{32} \, x^{2} - \frac {5}{16} \, {\left (4 \, x^{4} - 4 \, x^{3} - x^{2}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(40*x^2*log(x)+40*x^2+40*x)*exp(x*log(x))^2+1/16*(-20*x^2*log(x)^2+(40*x^4-40*x^3-20*x^2-40*x)*
log(x)+40*x^4+120*x^3-120*x^2-20*x)*exp(x*log(x))+5/8*x*log(x)^2+1/16*(-80*x^3+60*x^2+10*x)*log(x)+15/2*x^5-25
/2*x^4+15/4*x^3+23/16*x^2,x, algorithm="maxima")

[Out]

5/4*x^6 - 5/2*x^5 + 5/4*x^4 + 5/4*x^2*x^(2*x) + 5/32*(2*log(x)^2 - 2*log(x) + 1)*x^2 + 1/16*x^3 + 5/4*(2*x^4 -
2*x^3 - x^2*log(x))*x^x - 5/32*x^2 - 5/16*(4*x^4 - 4*x^3 - x^2)*log(x)

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mupad [B]  time = 4.46, size = 68, normalized size = 2.27 \begin {gather*} \frac {x^2\,\left (x-20\,x^2\,\ln \relax (x)-40\,x\,x^x+5\,{\ln \relax (x)}^2+40\,x^x\,x^2-20\,x^x\,\ln \relax (x)+20\,x^{2\,x}+20\,x\,\ln \relax (x)+20\,x^2-40\,x^3+20\,x^4\right )}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x*log(x))*(40*x + 40*x^2*log(x) + 40*x^2))/16 + (5*x*log(x)^2)/8 + (23*x^2)/16 + (15*x^3)/4 - (25*x
^4)/2 + (15*x^5)/2 + (log(x)*(10*x + 60*x^2 - 80*x^3))/16 - (exp(x*log(x))*(20*x + log(x)*(40*x + 20*x^2 + 40*
x^3 - 40*x^4) + 20*x^2*log(x)^2 + 120*x^2 - 120*x^3 - 40*x^4))/16,x)

[Out]

(x^2*(x - 20*x^2*log(x) - 40*x*x^x + 5*log(x)^2 + 40*x^x*x^2 - 20*x^x*log(x) + 20*x^(2*x) + 20*x*log(x) + 20*x
^2 - 40*x^3 + 20*x^4))/16

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sympy [B]  time = 0.53, size = 95, normalized size = 3.17 \begin {gather*} \frac {5 x^{6}}{4} - \frac {5 x^{5}}{2} + \frac {5 x^{4}}{4} + \frac {x^{3}}{16} + \frac {5 x^{2} e^{2 x \log {\relax (x )}}}{4} + \frac {5 x^{2} \log {\relax (x )}^{2}}{16} + \left (- \frac {5 x^{4}}{4} + \frac {5 x^{3}}{4}\right ) \log {\relax (x )} + \frac {\left (40 x^{4} - 40 x^{3} - 20 x^{2} \log {\relax (x )}\right ) e^{x \log {\relax (x )}}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(40*x**2*ln(x)+40*x**2+40*x)*exp(x*ln(x))**2+1/16*(-20*x**2*ln(x)**2+(40*x**4-40*x**3-20*x**2-4
0*x)*ln(x)+40*x**4+120*x**3-120*x**2-20*x)*exp(x*ln(x))+5/8*x*ln(x)**2+1/16*(-80*x**3+60*x**2+10*x)*ln(x)+15/2
*x**5-25/2*x**4+15/4*x**3+23/16*x**2,x)

[Out]

5*x**6/4 - 5*x**5/2 + 5*x**4/4 + x**3/16 + 5*x**2*exp(2*x*log(x))/4 + 5*x**2*log(x)**2/16 + (-5*x**4/4 + 5*x**
3/4)*log(x) + (40*x**4 - 40*x**3 - 20*x**2*log(x))*exp(x*log(x))/16

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