3.57.52 \(\int (16+20 x+80 x^2-40 x^2 \log (5)+(320-40 x-120 x^2+(-120+60 x^2) \log (5)) \log (x)) \, dx\)

Optimal. Leaf size=25 \[ 4 x \left (4-5 \left (-4+x-\left (-6+x^2\right ) (-2+\log (5))\right ) (-1+\log (x))\right ) \]

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Rubi [B]  time = 0.06, antiderivative size = 60, normalized size of antiderivative = 2.40, number of steps used = 7, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6, 2356, 2304, 2295} \begin {gather*} -20 x^3 (2-\log (5)) \log (x)+20 x^3 (2-\log (5))+20 x^2-20 x^2 \log (x)+16 x+40 x (8-\log (125)) \log (x)-40 x (8-\log (125)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[16 + 20*x + 80*x^2 - 40*x^2*Log[5] + (320 - 40*x - 120*x^2 + (-120 + 60*x^2)*Log[5])*Log[x],x]

[Out]

16*x + 20*x^2 + 20*x^3*(2 - Log[5]) - 40*x*(8 - Log[125]) - 20*x^2*Log[x] - 20*x^3*(2 - Log[5])*Log[x] + 40*x*
(8 - Log[125])*Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Polyx*(a + b*Log[c*
x^n])^p, x], x] /; FreeQ[{a, b, c, n, p}, x] && PolynomialQ[Polyx, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (16+20 x+x^2 (80-40 \log (5))+\left (320-40 x-120 x^2+\left (-120+60 x^2\right ) \log (5)\right ) \log (x)\right ) \, dx\\ &=16 x+10 x^2+\frac {40}{3} x^3 (2-\log (5))+\int \left (320-40 x-120 x^2+\left (-120+60 x^2\right ) \log (5)\right ) \log (x) \, dx\\ &=16 x+10 x^2+\frac {40}{3} x^3 (2-\log (5))+\int \left (-40 x \log (x)+60 x^2 (-2+\log (5)) \log (x)-40 (-8+\log (125)) \log (x)\right ) \, dx\\ &=16 x+10 x^2+\frac {40}{3} x^3 (2-\log (5))-40 \int x \log (x) \, dx-(60 (2-\log (5))) \int x^2 \log (x) \, dx+(40 (8-\log (125))) \int \log (x) \, dx\\ &=16 x+20 x^2+20 x^3 (2-\log (5))-40 x (8-\log (125))-20 x^2 \log (x)-20 x^3 (2-\log (5)) \log (x)+40 x (8-\log (125)) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.02, size = 61, normalized size = 2.44 \begin {gather*} -304 x+20 x^2+40 x^3+120 x \log (5)-20 x^3 \log (5)+320 x \log (x)-20 x^2 \log (x)-40 x^3 \log (x)-120 x \log (5) \log (x)+20 x^3 \log (5) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[16 + 20*x + 80*x^2 - 40*x^2*Log[5] + (320 - 40*x - 120*x^2 + (-120 + 60*x^2)*Log[5])*Log[x],x]

[Out]

-304*x + 20*x^2 + 40*x^3 + 120*x*Log[5] - 20*x^3*Log[5] + 320*x*Log[x] - 20*x^2*Log[x] - 40*x^3*Log[x] - 120*x
*Log[5]*Log[x] + 20*x^3*Log[5]*Log[x]

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fricas [A]  time = 0.55, size = 52, normalized size = 2.08 \begin {gather*} 40 \, x^{3} + 20 \, x^{2} - 20 \, {\left (x^{3} - 6 \, x\right )} \log \relax (5) - 20 \, {\left (2 \, x^{3} + x^{2} - {\left (x^{3} - 6 \, x\right )} \log \relax (5) - 16 \, x\right )} \log \relax (x) - 304 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((60*x^2-120)*log(5)-120*x^2-40*x+320)*log(x)-40*x^2*log(5)+80*x^2+20*x+16,x, algorithm="fricas")

[Out]

40*x^3 + 20*x^2 - 20*(x^3 - 6*x)*log(5) - 20*(2*x^3 + x^2 - (x^3 - 6*x)*log(5) - 16*x)*log(x) - 304*x

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giac [B]  time = 0.13, size = 61, normalized size = 2.44 \begin {gather*} 20 \, x^{3} \log \relax (5) \log \relax (x) - 20 \, x^{3} \log \relax (5) - 40 \, x^{3} \log \relax (x) + 40 \, x^{3} - 20 \, x^{2} \log \relax (x) - 120 \, x \log \relax (5) \log \relax (x) + 20 \, x^{2} + 120 \, x \log \relax (5) + 320 \, x \log \relax (x) - 304 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((60*x^2-120)*log(5)-120*x^2-40*x+320)*log(x)-40*x^2*log(5)+80*x^2+20*x+16,x, algorithm="giac")

[Out]

20*x^3*log(5)*log(x) - 20*x^3*log(5) - 40*x^3*log(x) + 40*x^3 - 20*x^2*log(x) - 120*x*log(5)*log(x) + 20*x^2 +
120*x*log(5) + 320*x*log(x) - 304*x

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maple [B]  time = 0.04, size = 54, normalized size = 2.16

method	result	size
norman	\(\left (-304+120 \ln \relax (5)\right ) x +\left (-20 \ln \relax (5)+40\right ) x^{3}+\left (-120 \ln \relax (5)+320\right ) x \ln \relax (x )+\left (20 \ln \relax (5)-40\right ) x^{3} \ln \relax (x )+20 x^{2}-20 x^{2} \ln \relax (x )\)	\(54\)
risch	\(\left (20 x^{3} \ln \relax (5)-40 x^{3}-120 x \ln \relax (5)-20 x^{2}+320 x \right ) \ln \relax (x )-20 x^{3} \ln \relax (5)+40 x^{3}+120 x \ln \relax (5)+20 x^{2}-304 x\)	\(56\)
default	\(-304 x +20 \ln \relax (5) \ln \relax (x ) x^{3}-20 x^{3} \ln \relax (5)-40 x^{3} \ln \relax (x )+40 x^{3}-120 x \ln \relax (5) \ln \relax (x )+120 x \ln \relax (5)-20 x^{2} \ln \relax (x )+20 x^{2}+320 x \ln \relax (x )\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((60*x^2-120)*ln(5)-120*x^2-40*x+320)*ln(x)-40*x^2*ln(5)+80*x^2+20*x+16,x,method=_RETURNVERBOSE)

[Out]

(-304+120*ln(5))*x+(-20*ln(5)+40)*x^3+(-120*ln(5)+320)*x*ln(x)+(20*ln(5)-40)*x^3*ln(x)+20*x^2-20*x^2*ln(x)

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maxima [B]  time = 0.38, size = 66, normalized size = 2.64 \begin {gather*} -\frac {20}{3} \, x^{3} {\left (\log \relax (5) - 2\right )} - \frac {40}{3} \, x^{3} \log \relax (5) + \frac {80}{3} \, x^{3} + 20 \, x^{2} + 40 \, x {\left (3 \, \log \relax (5) - 8\right )} - 20 \, {\left (2 \, x^{3} + x^{2} - {\left (x^{3} - 6 \, x\right )} \log \relax (5) - 16 \, x\right )} \log \relax (x) + 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((60*x^2-120)*log(5)-120*x^2-40*x+320)*log(x)-40*x^2*log(5)+80*x^2+20*x+16,x, algorithm="maxima")

[Out]

-20/3*x^3*(log(5) - 2) - 40/3*x^3*log(5) + 80/3*x^3 + 20*x^2 + 40*x*(3*log(5) - 8) - 20*(2*x^3 + x^2 - (x^3 -
6*x)*log(5) - 16*x)*log(x) + 16*x

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mupad [B]  time = 3.57, size = 49, normalized size = 1.96 \begin {gather*} x^3\,\left (\ln \relax (x)\,\left (20\,\ln \relax (5)-40\right )-20\,\ln \relax (5)+40\right )-x\,\left (\ln \relax (x)\,\left (120\,\ln \relax (5)-320\right )-120\,\ln \relax (5)+304\right )-x^2\,\left (20\,\ln \relax (x)-20\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(20*x - log(x)*(40*x - log(5)*(60*x^2 - 120) + 120*x^2 - 320) - 40*x^2*log(5) + 80*x^2 + 16,x)

[Out]

x^3*(log(x)*(20*log(5) - 40) - 20*log(5) + 40) - x*(log(x)*(120*log(5) - 320) - 120*log(5) + 304) - x^2*(20*lo
g(x) - 20)

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sympy [B]  time = 0.14, size = 54, normalized size = 2.16 \begin {gather*} x^{3} \left (40 - 20 \log {\relax (5 )}\right ) + 20 x^{2} + x \left (-304 + 120 \log {\relax (5 )}\right ) + \left (- 40 x^{3} + 20 x^{3} \log {\relax (5 )} - 20 x^{2} - 120 x \log {\relax (5 )} + 320 x\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((60*x**2-120)*ln(5)-120*x**2-40*x+320)*ln(x)-40*x**2*ln(5)+80*x**2+20*x+16,x)

[Out]

x**3*(40 - 20*log(5)) + 20*x**2 + x*(-304 + 120*log(5)) + (-40*x**3 + 20*x**3*log(5) - 20*x**2 - 120*x*log(5)
+ 320*x)*log(x)

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