\(\int \frac {-8 x^2+2 x^3+3 x^4-x^5+(2-x) \log (5)+(-8 x^2+4 x^3+9 x^4-4 x^5-2 \log (5)) \log (x)}{x^2} \, dx\) [1372]

Optimal result

Integrand size = 61, antiderivative size = 22 \[ \int \frac {-8 x^2+2 x^3+3 x^4-x^5+(2-x) \log (5)+\left (-8 x^2+4 x^3+9 x^4-4 x^5-2 \log (5)\right ) \log (x)}{x^2} \, dx=(-2+x) x \left (4+x-x^2-\frac {\log (5)}{x^2}\right ) \log (x) \]

[Out]

x*(-2+x)*ln(x)*(4+x-x^2-ln(5)/x^2)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(54\) vs. \(2(22)=44\).

Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.45, number of steps used = 11, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {14, 1634, 2404, 2332, 2341} \[ \int \frac {-8 x^2+2 x^3+3 x^4-x^5+(2-x) \log (5)+\left (-8 x^2+4 x^3+9 x^4-4 x^5-2 \log (5)\right ) \log (x)}{x^2} \, dx=x^4 (-\log (x))+3 x^3 \log (x)+2 x^2 \log (x)-8 x \log (x)-\log (5) \log (x)+\frac {\log (25) \log (x)}{x}+\frac {\log (25)}{x}-\frac {2 \log (5)}{x} \]

[In]

Int[(-8*x^2 + 2*x^3 + 3*x^4 - x^5 + (2 - x)*Log[5] + (-8*x^2 + 4*x^3 + 9*x^4 - 4*x^5 - 2*Log[5])*Log[x])/x^2,x
]

[Out]

(-2*Log[5])/x + Log[25]/x - 8*x*Log[x] + 2*x^2*Log[x] + 3*x^3*Log[x] - x^4*Log[x] - Log[5]*Log[x] + (Log[25]*L
og[x])/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 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 2332

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

Rule 2341

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 2404

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {(-2+x) \left (-4 x^2-x^3+x^4+\log (5)\right )}{x^2}-\frac {\left (8 x^2-4 x^3-9 x^4+4 x^5+\log (25)\right ) \log (x)}{x^2}\right ) \, dx \\ & = -\int \frac {(-2+x) \left (-4 x^2-x^3+x^4+\log (5)\right )}{x^2} \, dx-\int \frac {\left (8 x^2-4 x^3-9 x^4+4 x^5+\log (25)\right ) \log (x)}{x^2} \, dx \\ & = -\int \left (8-2 x-3 x^2+x^3-\frac {2 \log (5)}{x^2}+\frac {\log (5)}{x}\right ) \, dx-\int \left (8 \log (x)-4 x \log (x)-9 x^2 \log (x)+4 x^3 \log (x)+\frac {\log (25) \log (x)}{x^2}\right ) \, dx \\ & = -8 x+x^2+x^3-\frac {x^4}{4}-\frac {2 \log (5)}{x}-\log (5) \log (x)+4 \int x \log (x) \, dx-4 \int x^3 \log (x) \, dx-8 \int \log (x) \, dx+9 \int x^2 \log (x) \, dx-\log (25) \int \frac {\log (x)}{x^2} \, dx \\ & = -\frac {2 \log (5)}{x}+\frac {\log (25)}{x}-8 x \log (x)+2 x^2 \log (x)+3 x^3 \log (x)-x^4 \log (x)-\log (5) \log (x)+\frac {\log (25) \log (x)}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \[ \int \frac {-8 x^2+2 x^3+3 x^4-x^5+(2-x) \log (5)+\left (-8 x^2+4 x^3+9 x^4-4 x^5-2 \log (5)\right ) \log (x)}{x^2} \, dx=-8 x \log (x)+2 x^2 \log (x)+3 x^3 \log (x)-x^4 \log (x)-\log (5) \log (x)+\frac {2 \log (5) \log (x)}{x} \]

[In]

Integrate[(-8*x^2 + 2*x^3 + 3*x^4 - x^5 + (2 - x)*Log[5] + (-8*x^2 + 4*x^3 + 9*x^4 - 4*x^5 - 2*Log[5])*Log[x])
/x^2,x]

[Out]

-8*x*Log[x] + 2*x^2*Log[x] + 3*x^3*Log[x] - x^4*Log[x] - Log[5]*Log[x] + (2*Log[5]*Log[x])/x

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77

method	result	size
risch	\(\frac {\left (-x^{5}+3 x^{4}+2 x^{3}-8 x^{2}+2 \ln \left (5\right )\right ) \ln \left (x \right )}{x}-\ln \left (5\right ) \ln \left (x \right )\)	\(39\)
parallelrisch	\(-\frac {x^{5} \ln \left (x \right )-3 x^{4} \ln \left (x \right )-2 x^{3} \ln \left (x \right )+x \ln \left (5\right ) \ln \left (x \right )+8 x^{2} \ln \left (x \right )-2 \ln \left (5\right ) \ln \left (x \right )}{x}\)	\(46\)
norman	\(\frac {-x \ln \left (5\right ) \ln \left (x \right )-8 x^{2} \ln \left (x \right )+2 x^{3} \ln \left (x \right )+3 x^{4} \ln \left (x \right )-x^{5} \ln \left (x \right )+2 \ln \left (5\right ) \ln \left (x \right )}{x}\)	\(47\)
default	\(-x^{4} \ln \left (x \right )+3 x^{3} \ln \left (x \right )+2 x^{2} \ln \left (x \right )-8 x \ln \left (x \right )-2 \ln \left (5\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )-\ln \left (5\right ) \ln \left (x \right )-\frac {2 \ln \left (5\right )}{x}\)	\(58\)
parts	\(-x^{4} \ln \left (x \right )+3 x^{3} \ln \left (x \right )+2 x^{2} \ln \left (x \right )-8 x \ln \left (x \right )-2 \ln \left (5\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )-\ln \left (5\right ) \ln \left (x \right )-\frac {2 \ln \left (5\right )}{x}\)	\(58\)

[In]

int(((-2*ln(5)-4*x^5+9*x^4+4*x^3-8*x^2)*ln(x)+(2-x)*ln(5)-x^5+3*x^4+2*x^3-8*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

(-x^5+3*x^4+2*x^3-8*x^2+2*ln(5))/x*ln(x)-ln(5)*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {-8 x^2+2 x^3+3 x^4-x^5+(2-x) \log (5)+\left (-8 x^2+4 x^3+9 x^4-4 x^5-2 \log (5)\right ) \log (x)}{x^2} \, dx=-\frac {{\left (x^{5} - 3 \, x^{4} - 2 \, x^{3} + 8 \, x^{2} + {\left (x - 2\right )} \log \left (5\right )\right )} \log \left (x\right )}{x} \]

[In]

integrate(((-2*log(5)-4*x^5+9*x^4+4*x^3-8*x^2)*log(x)+(2-x)*log(5)-x^5+3*x^4+2*x^3-8*x^2)/x^2,x, algorithm="fr
icas")

[Out]

-(x^5 - 3*x^4 - 2*x^3 + 8*x^2 + (x - 2)*log(5))*log(x)/x

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {-8 x^2+2 x^3+3 x^4-x^5+(2-x) \log (5)+\left (-8 x^2+4 x^3+9 x^4-4 x^5-2 \log (5)\right ) \log (x)}{x^2} \, dx=- \log {\left (5 \right )} \log {\left (x \right )} + \frac {\left (- x^{5} + 3 x^{4} + 2 x^{3} - 8 x^{2} + 2 \log {\left (5 \right )}\right ) \log {\left (x \right )}}{x} \]

[In]

integrate(((-2*ln(5)-4*x**5+9*x**4+4*x**3-8*x**2)*ln(x)+(2-x)*ln(5)-x**5+3*x**4+2*x**3-8*x**2)/x**2,x)

[Out]

-log(5)*log(x) + (-x**5 + 3*x**4 + 2*x**3 - 8*x**2 + 2*log(5))*log(x)/x

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).

Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.45 \[ \int \frac {-8 x^2+2 x^3+3 x^4-x^5+(2-x) \log (5)+\left (-8 x^2+4 x^3+9 x^4-4 x^5-2 \log (5)\right ) \log (x)}{x^2} \, dx=-x^{4} \log \left (x\right ) + 3 \, x^{3} \log \left (x\right ) + 2 \, x^{2} \log \left (x\right ) + 2 \, {\left (\frac {\log \left (x\right )}{x} + \frac {1}{x}\right )} \log \left (5\right ) - 8 \, x \log \left (x\right ) - \log \left (5\right ) \log \left (x\right ) - \frac {2 \, \log \left (5\right )}{x} \]

[In]

integrate(((-2*log(5)-4*x^5+9*x^4+4*x^3-8*x^2)*log(x)+(2-x)*log(5)-x^5+3*x^4+2*x^3-8*x^2)/x^2,x, algorithm="ma
xima")

[Out]

-x^4*log(x) + 3*x^3*log(x) + 2*x^2*log(x) + 2*(log(x)/x + 1/x)*log(5) - 8*x*log(x) - log(5)*log(x) - 2*log(5)/
x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59 \[ \int \frac {-8 x^2+2 x^3+3 x^4-x^5+(2-x) \log (5)+\left (-8 x^2+4 x^3+9 x^4-4 x^5-2 \log (5)\right ) \log (x)}{x^2} \, dx=-{\left (x^{4} - 3 \, x^{3} - 2 \, x^{2} + 8 \, x - \frac {2 \, \log \left (5\right )}{x}\right )} \log \left (x\right ) - \log \left (5\right ) \log \left (x\right ) \]

[In]

integrate(((-2*log(5)-4*x^5+9*x^4+4*x^3-8*x^2)*log(x)+(2-x)*log(5)-x^5+3*x^4+2*x^3-8*x^2)/x^2,x, algorithm="gi
ac")

[Out]

-(x^4 - 3*x^3 - 2*x^2 + 8*x - 2*log(5)/x)*log(x) - log(5)*log(x)

Mupad [B] (verification not implemented)

Time = 9.43 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {-8 x^2+2 x^3+3 x^4-x^5+(2-x) \log (5)+\left (-8 x^2+4 x^3+9 x^4-4 x^5-2 \log (5)\right ) \log (x)}{x^2} \, dx=-\frac {\ln \left (x\right )\,\left (x-2\right )\,\left (x^4-x^3-4\,x^2+\ln \left (5\right )\right )}{x} \]

[In]

int(-(log(x)*(2*log(5) + 8*x^2 - 4*x^3 - 9*x^4 + 4*x^5) + log(5)*(x - 2) + 8*x^2 - 2*x^3 - 3*x^4 + x^5)/x^2,x)

[Out]

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