3.26.21 \(\int \frac {-50+x+\log (\frac {e^{16}}{4}) (x^3-2 x \log (x)+x \log ^2(x))}{x^3 \log (\frac {e^{16}}{4})} \, dx\)

Optimal. Leaf size=30 \[ -2+x-\frac {-25+x}{x^2 \log \left (\frac {e^{16}}{4}\right )}-\frac {\log ^2(x)}{x} \]

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Rubi [A]  time = 0.09, antiderivative size = 53, normalized size of antiderivative = 1.77, number of steps used = 8, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {12, 14, 2304, 2305} \begin {gather*} \frac {25}{x^2 (16-\log (4))}-\frac {\log ^2(x)}{x}-\frac {1}{x (16-\log (4))}+\frac {2 x (8-\log (2))}{16-\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50 + x + Log[E^16/4]*(x^3 - 2*x*Log[x] + x*Log[x]^2))/(x^3*Log[E^16/4]),x]

[Out]

25/(x^2*(16 - Log[4])) - 1/(x*(16 - Log[4])) + (2*x*(8 - Log[2]))/(16 - Log[4]) - Log[x]^2/x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-50+x+\log \left (\frac {e^{16}}{4}\right ) \left (x^3-2 x \log (x)+x \log ^2(x)\right )}{x^3} \, dx}{16-\log (4)}\\ &=\frac {\int \left (\frac {-50+x+x^3 (16-\log (4))}{x^3}+\frac {2 (-16+\log (4)) \log (x)}{x^2}-\frac {(-16+\log (4)) \log ^2(x)}{x^2}\right ) \, dx}{16-\log (4)}\\ &=-\left (2 \int \frac {\log (x)}{x^2} \, dx\right )+\frac {\int \frac {-50+x+x^3 (16-\log (4))}{x^3} \, dx}{16-\log (4)}+\int \frac {\log ^2(x)}{x^2} \, dx\\ &=\frac {2}{x}+\frac {2 \log (x)}{x}-\frac {\log ^2(x)}{x}+2 \int \frac {\log (x)}{x^2} \, dx+\frac {\int \left (-\frac {50}{x^3}+\frac {1}{x^2}+16 \left (1-\frac {\log (2)}{8}\right )\right ) \, dx}{16-\log (4)}\\ &=\frac {25}{x^2 (16-\log (4))}-\frac {1}{x (16-\log (4))}+\frac {2 x (8-\log (2))}{16-\log (4)}-\frac {\log ^2(x)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 48, normalized size = 1.60 \begin {gather*} \frac {\frac {25}{x^2}-\frac {1}{x}+16 x-2 x \log (2)-\frac {16 \log ^2(x)}{x}+\frac {2 \log (2) \log ^2(x)}{x}}{16-\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50 + x + Log[E^16/4]*(x^3 - 2*x*Log[x] + x*Log[x]^2))/(x^3*Log[E^16/4]),x]

[Out]

(25/x^2 - x^(-1) + 16*x - 2*x*Log[2] - (16*Log[x]^2)/x + (2*Log[2]*Log[x]^2)/x)/(16 - Log[4])

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fricas [A]  time = 0.50, size = 45, normalized size = 1.50 \begin {gather*} \frac {2 \, x^{3} \log \relax (2) - 16 \, x^{3} - 2 \, {\left (x \log \relax (2) - 8 \, x\right )} \log \relax (x)^{2} + x - 25}{2 \, {\left (x^{2} \log \relax (2) - 8 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)^2-2*x*log(x)+x^3)*log(1/4*exp(16))+x-50)/x^3/log(1/4*exp(16)),x, algorithm="fricas")

[Out]

1/2*(2*x^3*log(2) - 16*x^3 - 2*(x*log(2) - 8*x)*log(x)^2 + x - 25)/(x^2*log(2) - 8*x^2)

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giac [A]  time = 0.16, size = 37, normalized size = 1.23 \begin {gather*} -\frac {2 \, x {\left (\log \relax (2) - 8\right )} - \frac {2 \, {\left (\log \relax (2) - 8\right )} \log \relax (x)^{2}}{x} + \frac {x - 25}{x^{2}}}{\log \left (\frac {1}{4} \, e^{16}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)^2-2*x*log(x)+x^3)*log(1/4*exp(16))+x-50)/x^3/log(1/4*exp(16)),x, algorithm="giac")

[Out]

-(2*x*(log(2) - 8) - 2*(log(2) - 8)*log(x)^2/x + (x - 25)/x^2)/log(1/4*e^16)

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maple [A]  time = 0.05, size = 33, normalized size = 1.10

method	result	size
norman	\(\frac {x^{3}-\frac {25}{2 \left (\ln \relax (2)-8\right )}-x \ln \relax (x )^{2}+\frac {x}{2 \ln \relax (2)-16}}{x^{2}}\)	\(33\)
risch	\(\frac {2 \left (\ln \relax (2)-8\right ) \ln \relax (x )^{2}}{\left (-2 \ln \relax (2)+16\right ) x}-\frac {2 x^{3} \ln \relax (2)-16 x^{3}+x -25}{\left (-2 \ln \relax (2)+16\right ) x^{2}}\)	\(51\)
default	\(\frac {-2 \ln \relax (2) \left (-\frac {\ln \relax (x )^{2}}{x}-\frac {2 \ln \relax (x )}{x}-\frac {2}{x}\right )-\frac {16 \ln \relax (x )^{2}}{x}-\frac {1}{x}-2 x \ln \relax (2)+16 x +4 \ln \relax (2) \left (-\frac {\ln \relax (x )}{x}-\frac {1}{x}\right )+\frac {25}{x^{2}}}{\ln \left (\frac {{\mathrm e}^{16}}{4}\right )}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*ln(x)^2-2*x*ln(x)+x^3)*ln(1/4*exp(16))+x-50)/x^3/ln(1/4*exp(16)),x,method=_RETURNVERBOSE)

[Out]

(x^3-25/2/(ln(2)-8)-x*ln(x)^2+1/2/(ln(2)-8)*x)/x^2

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maxima [B]  time = 0.41, size = 63, normalized size = 2.10 \begin {gather*} \frac {x \log \left (\frac {1}{4} \, e^{16}\right ) + 2 \, {\left (\frac {\log \relax (x)}{x} + \frac {1}{x}\right )} \log \left (\frac {1}{4} \, e^{16}\right ) - \frac {{\left (\log \relax (x)^{2} + 2 \, \log \relax (x) + 2\right )} \log \left (\frac {1}{4} \, e^{16}\right )}{x} - \frac {1}{x} + \frac {25}{x^{2}}}{\log \left (\frac {1}{4} \, e^{16}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)^2-2*x*log(x)+x^3)*log(1/4*exp(16))+x-50)/x^3/log(1/4*exp(16)),x, algorithm="maxima")

[Out]

(x*log(1/4*e^16) + 2*(log(x)/x + 1/x)*log(1/4*e^16) - (log(x)^2 + 2*log(x) + 2)*log(1/4*e^16)/x - 1/x + 25/x^2
)/log(1/4*e^16)

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mupad [B]  time = 1.50, size = 41, normalized size = 1.37 \begin {gather*} \frac {\left (16-\ln \relax (4)\right )\,x^4+\left (\ln \relax (4)-16\right )\,x^2\,{\ln \relax (x)}^2-x^2+25\,x}{x^3\,\ln \left (\frac {{\mathrm {e}}^{16}}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + log(exp(16)/4)*(x*log(x)^2 - 2*x*log(x) + x^3) - 50)/(x^3*log(exp(16)/4)),x)

[Out]

(25*x - x^4*(log(4) - 16) - x^2 + x^2*log(x)^2*(log(4) - 16))/(x^3*log(exp(16)/4))

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sympy [A]  time = 0.19, size = 29, normalized size = 0.97 \begin {gather*} \frac {- x \left (16 - 2 \log {\relax (2 )}\right ) - \frac {25 - x}{x^{2}}}{-16 + 2 \log {\relax (2 )}} - \frac {\log {\relax (x )}^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*ln(x)**2-2*x*ln(x)+x**3)*ln(1/4*exp(16))+x-50)/x**3/ln(1/4*exp(16)),x)

[Out]

(-x*(16 - 2*log(2)) - (25 - x)/x**2)/(-16 + 2*log(2)) - log(x)**2/x

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