3.100.36 \(\int \frac {(-5+10 x-x^2-2 x^3) \log (x)+(5-x^2) \log (\frac {5-x^2}{x})}{-10 x+2 x^3} \, dx\)

Optimal. Leaf size=29 \[ -4+x-\log (x)+\left (1-x-\frac {1}{2} \log \left (\frac {5}{x}-x\right )\right ) \log (x) \]

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Rubi [A]  time = 0.45, antiderivative size = 25, normalized size of antiderivative = 0.86, number of steps used = 15, number of rules used = 8, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.157, Rules used = {1593, 6725, 2357, 2295, 2301, 2337, 2391, 2524} \begin {gather*} -\frac {1}{2} \log (x) \log \left (\frac {5-x^2}{x}\right )+x+x (-\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-5 + 10*x - x^2 - 2*x^3)*Log[x] + (5 - x^2)*Log[(5 - x^2)/x])/(-10*x + 2*x^3),x]

[Out]

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2337

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Si
mp[(f^m*Log[1 + (e*x^r)/d]*(a + b*Log[c*x^n])^p)/(e*r), x] - Dist[(b*f^m*n*p)/(e*r), Int[(Log[1 + (e*x^r)/d]*(
a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] &
& (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]

Rule 2357

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]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2524

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[d + e*x]*(a + b
*Log[c*RFx^p])^n)/e, x] - Dist[(b*n*p)/e, Int[(Log[d + e*x]*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x
] /; FreeQ[{a, b, c, d, e, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-5+10 x-x^2-2 x^3\right ) \log (x)+\left (5-x^2\right ) \log \left (\frac {5-x^2}{x}\right )}{x \left (-10+2 x^2\right )} \, dx\\ &=\int \left (-\frac {\left (5-10 x+x^2+2 x^3\right ) \log (x)}{2 x \left (-5+x^2\right )}-\frac {\log \left (\frac {5-x^2}{x}\right )}{2 x}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {\left (5-10 x+x^2+2 x^3\right ) \log (x)}{x \left (-5+x^2\right )} \, dx\right )-\frac {1}{2} \int \frac {\log \left (\frac {5-x^2}{x}\right )}{x} \, dx\\ &=-\frac {1}{2} \log (x) \log \left (\frac {5-x^2}{x}\right )+\frac {1}{2} \int \frac {x \left (-2-\frac {5-x^2}{x^2}\right ) \log (x)}{5-x^2} \, dx-\frac {1}{2} \int \left (2 \log (x)-\frac {\log (x)}{x}+\frac {2 x \log (x)}{-5+x^2}\right ) \, dx\\ &=-\frac {1}{2} \log (x) \log \left (\frac {5-x^2}{x}\right )+\frac {1}{2} \int \frac {\log (x)}{x} \, dx+\frac {1}{2} \int \left (-\frac {\log (x)}{x}+\frac {2 x \log (x)}{-5+x^2}\right ) \, dx-\int \log (x) \, dx-\int \frac {x \log (x)}{-5+x^2} \, dx\\ &=x-x \log (x)+\frac {\log ^2(x)}{4}-\frac {1}{2} \log (x) \log \left (\frac {5-x^2}{x}\right )-\frac {1}{2} \log (x) \log \left (1-\frac {x^2}{5}\right )-\frac {1}{2} \int \frac {\log (x)}{x} \, dx+\frac {1}{2} \int \frac {\log \left (1-\frac {x^2}{5}\right )}{x} \, dx+\int \frac {x \log (x)}{-5+x^2} \, dx\\ &=x-x \log (x)-\frac {1}{2} \log (x) \log \left (\frac {5-x^2}{x}\right )-\frac {\text {Li}_2\left (\frac {x^2}{5}\right )}{4}-\frac {1}{2} \int \frac {\log \left (1-\frac {x^2}{5}\right )}{x} \, dx\\ &=x-x \log (x)-\frac {1}{2} \log (x) \log \left (\frac {5-x^2}{x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 29, normalized size = 1.00 \begin {gather*} \frac {1}{2} \left (2 x-2 x \log (x)-\log (x) \log \left (\frac {5-x^2}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-5 + 10*x - x^2 - 2*x^3)*Log[x] + (5 - x^2)*Log[(5 - x^2)/x])/(-10*x + 2*x^3),x]

[Out]

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

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fricas [A]  time = 0.57, size = 21, normalized size = 0.72 \begin {gather*} -\frac {1}{2} \, {\left (2 \, x + \log \left (-\frac {x^{2} - 5}{x}\right )\right )} \log \relax (x) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-x^2+10*x-5)*log(x)+(-x^2+5)*log((-x^2+5)/x))/(2*x^3-10*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 25, normalized size = 0.86 \begin {gather*} -x \log \relax (x) - \frac {1}{2} \, \log \left (-x^{2} + 5\right ) \log \relax (x) + \frac {1}{2} \, \log \relax (x)^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-x^2+10*x-5)*log(x)+(-x^2+5)*log((-x^2+5)/x))/(2*x^3-10*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.22, size = 24, normalized size = 0.83

method	result	size
default	\(-\frac {\ln \relax (x ) \ln \left (\frac {-x^{2}+5}{x}\right )}{2}-x \ln \relax (x )+x\)	\(24\)
risch	\(-\frac {\ln \relax (x ) \ln \left (x^{2}-5\right )}{2}+\frac {\ln \relax (x )^{2}}{2}-x \ln \relax (x )+x +\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i \left (x^{2}-5\right )}{x}\right )^{2}}{2}+\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x^{2}-5\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-5\right )}{x}\right )}{4}-\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-5\right )}{x}\right )^{2}}{4}-\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (i \left (x^{2}-5\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}-5\right )}{x}\right )^{2}}{4}-\frac {i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i \left (x^{2}-5\right )}{x}\right )^{3}}{4}-\frac {i \pi \ln \relax (x )}{2}\)	\(160\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3-x^2+10*x-5)*ln(x)+(-x^2+5)*ln((-x^2+5)/x))/(2*x^3-10*x),x,method=_RETURNVERBOSE)

[Out]

-1/2*ln(x)*ln((-x^2+5)/x)-x*ln(x)+x

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maxima [A]  time = 0.40, size = 25, normalized size = 0.86 \begin {gather*} -x \log \relax (x) - \frac {1}{2} \, \log \left (-x^{2} + 5\right ) \log \relax (x) + \frac {1}{2} \, \log \relax (x)^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-x^2+10*x-5)*log(x)+(-x^2+5)*log((-x^2+5)/x))/(2*x^3-10*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 7.86, size = 22, normalized size = 0.76 \begin {gather*} x-\frac {\ln \left (-\frac {x^2-5}{x}\right )\,\ln \relax (x)}{2}-x\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(-(x^2 - 5)/x)*(x^2 - 5) + log(x)*(x^2 - 10*x + 2*x^3 + 5))/(10*x - 2*x^3),x)

[Out]

x - (log(-(x^2 - 5)/x)*log(x))/2 - x*log(x)

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sympy [A]  time = 0.36, size = 19, normalized size = 0.66 \begin {gather*} - x \log {\relax (x )} + x - \frac {\log {\relax (x )} \log {\left (\frac {5 - x^{2}}{x} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**3-x**2+10*x-5)*ln(x)+(-x**2+5)*ln((-x**2+5)/x))/(2*x**3-10*x),x)

[Out]

-x*log(x) + x - log(x)*log((5 - x**2)/x)/2

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