3.50.52 \(\int \frac {-60+169 x^2+(-5+26 x^2) \log ^2(4)+x^2 \log ^4(4)+(5-26 x^2-2 x^2 \log ^2(4)) \log (x)+x^2 \log ^2(x)}{169 x^2+26 x^2 \log ^2(4)+x^2 \log ^4(4)+(-26 x^2-2 x^2 \log ^2(4)) \log (x)+x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=19 \[ x+\frac {5}{x \left (13+\log ^2(4)-\log (x)\right )} \]

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Rubi [A]  time = 0.52, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 106, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {6, 6688, 6742, 2306, 2309, 2178} \begin {gather*} x+\frac {5}{x \left (-\log (x)+13+\log ^2(4)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-60 + 169*x^2 + (-5 + 26*x^2)*Log[4]^2 + x^2*Log[4]^4 + (5 - 26*x^2 - 2*x^2*Log[4]^2)*Log[x] + x^2*Log[x]
^2)/(169*x^2 + 26*x^2*Log[4]^2 + x^2*Log[4]^4 + (-26*x^2 - 2*x^2*Log[4]^2)*Log[x] + x^2*Log[x]^2),x]

[Out]

x + 5/(x*(13 + Log[4]^2 - 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 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 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] && LtQ[p, -1]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-60+169 x^2+\left (-5+26 x^2\right ) \log ^2(4)+x^2 \log ^4(4)+\left (5-26 x^2-2 x^2 \log ^2(4)\right ) \log (x)+x^2 \log ^2(x)}{x^2 \log ^4(4)+x^2 \left (169+26 \log ^2(4)\right )+\left (-26 x^2-2 x^2 \log ^2(4)\right ) \log (x)+x^2 \log ^2(x)} \, dx\\ &=\int \frac {-60+169 x^2+\left (-5+26 x^2\right ) \log ^2(4)+x^2 \log ^4(4)+\left (5-26 x^2-2 x^2 \log ^2(4)\right ) \log (x)+x^2 \log ^2(x)}{x^2 \left (169+26 \log ^2(4)+\log ^4(4)\right )+\left (-26 x^2-2 x^2 \log ^2(4)\right ) \log (x)+x^2 \log ^2(x)} \, dx\\ &=\int \frac {-60+\left (-5+26 x^2\right ) \log ^2(4)+x^2 \left (169+\log ^4(4)\right )+\left (5-26 x^2-2 x^2 \log ^2(4)\right ) \log (x)+x^2 \log ^2(x)}{x^2 \left (169+26 \log ^2(4)+\log ^4(4)\right )+\left (-26 x^2-2 x^2 \log ^2(4)\right ) \log (x)+x^2 \log ^2(x)} \, dx\\ &=\int \frac {-5 \left (12+\log ^2(4)\right )+x^2 \left (13+\log ^2(4)\right )^2+\left (5-2 x^2 \left (13+\log ^2(4)\right )\right ) \log (x)+x^2 \log ^2(x)}{x^2 \left (13 \left (1+\frac {\log ^2(4)}{13}\right )-\log (x)\right )^2} \, dx\\ &=\int \left (1+\frac {5}{x^2 \left (13 \left (1+\frac {\log ^2(4)}{13}\right )-\log (x)\right )^2}+\frac {5}{x^2 \left (-13 \left (1+\frac {\log ^2(4)}{13}\right )+\log (x)\right )}\right ) \, dx\\ &=x+5 \int \frac {1}{x^2 \left (13 \left (1+\frac {\log ^2(4)}{13}\right )-\log (x)\right )^2} \, dx+5 \int \frac {1}{x^2 \left (-13 \left (1+\frac {\log ^2(4)}{13}\right )+\log (x)\right )} \, dx\\ &=x+\frac {5}{x \left (13+\log ^2(4)-\log (x)\right )}+5 \int \frac {1}{x^2 \left (13 \left (1+\frac {\log ^2(4)}{13}\right )-\log (x)\right )} \, dx+5 \operatorname {Subst}\left (\int \frac {e^{-x}}{x-13 \left (1+\frac {\log ^2(4)}{13}\right )} \, dx,x,\log (x)\right )\\ &=x+5 e^{-13-\log ^2(4)} \text {Ei}\left (13+\log ^2(4)-\log (x)\right )+\frac {5}{x \left (13+\log ^2(4)-\log (x)\right )}+5 \operatorname {Subst}\left (\int \frac {e^{-x}}{-x+13 \left (1+\frac {\log ^2(4)}{13}\right )} \, dx,x,\log (x)\right )\\ &=x+\frac {5}{x \left (13+\log ^2(4)-\log (x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 19, normalized size = 1.00 \begin {gather*} x-\frac {5}{x \left (-13-\log ^2(4)+\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-60 + 169*x^2 + (-5 + 26*x^2)*Log[4]^2 + x^2*Log[4]^4 + (5 - 26*x^2 - 2*x^2*Log[4]^2)*Log[x] + x^2*
Log[x]^2)/(169*x^2 + 26*x^2*Log[4]^2 + x^2*Log[4]^4 + (-26*x^2 - 2*x^2*Log[4]^2)*Log[x] + x^2*Log[x]^2),x]

[Out]

x - 5/(x*(-13 - Log[4]^2 + Log[x]))

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fricas [A]  time = 0.71, size = 42, normalized size = 2.21 \begin {gather*} \frac {4 \, x^{2} \log \relax (2)^{2} - x^{2} \log \relax (x) + 13 \, x^{2} + 5}{4 \, x \log \relax (2)^{2} - x \log \relax (x) + 13 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2+(-8*x^2*log(2)^2-26*x^2+5)*log(x)+16*x^2*log(2)^4+4*(26*x^2-5)*log(2)^2+169*x^2-60)/(x
^2*log(x)^2+(-8*x^2*log(2)^2-26*x^2)*log(x)+16*x^2*log(2)^4+104*x^2*log(2)^2+169*x^2),x, algorithm="fricas")

[Out]

(4*x^2*log(2)^2 - x^2*log(x) + 13*x^2 + 5)/(4*x*log(2)^2 - x*log(x) + 13*x)

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giac [A]  time = 0.15, size = 22, normalized size = 1.16 \begin {gather*} x + \frac {5}{4 \, x \log \relax (2)^{2} - x \log \relax (x) + 13 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2+(-8*x^2*log(2)^2-26*x^2+5)*log(x)+16*x^2*log(2)^4+4*(26*x^2-5)*log(2)^2+169*x^2-60)/(x
^2*log(x)^2+(-8*x^2*log(2)^2-26*x^2)*log(x)+16*x^2*log(2)^4+104*x^2*log(2)^2+169*x^2),x, algorithm="giac")

[Out]

x + 5/(4*x*log(2)^2 - x*log(x) + 13*x)

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maple [A]  time = 0.13, size = 22, normalized size = 1.16

method	result	size
risch	\(x +\frac {5}{\left (13-\ln \relax (x )+4 \ln \relax (2)^{2}\right ) x}\)	\(22\)
norman	\(\frac {5+\left (4 \ln \relax (2)^{2}+13\right ) x^{2}-x^{2} \ln \relax (x )}{x \left (13-\ln \relax (x )+4 \ln \relax (2)^{2}\right )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*ln(x)^2+(-8*x^2*ln(2)^2-26*x^2+5)*ln(x)+16*x^2*ln(2)^4+4*(26*x^2-5)*ln(2)^2+169*x^2-60)/(x^2*ln(x)^2+
(-8*x^2*ln(2)^2-26*x^2)*ln(x)+16*x^2*ln(2)^4+104*x^2*ln(2)^2+169*x^2),x,method=_RETURNVERBOSE)

[Out]

x+5/(13-ln(x)+4*ln(2)^2)/x

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maxima [A]  time = 0.47, size = 40, normalized size = 2.11 \begin {gather*} \frac {{\left (4 \, \log \relax (2)^{2} + 13\right )} x^{2} - x^{2} \log \relax (x) + 5}{{\left (4 \, \log \relax (2)^{2} + 13\right )} x - x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2+(-8*x^2*log(2)^2-26*x^2+5)*log(x)+16*x^2*log(2)^4+4*(26*x^2-5)*log(2)^2+169*x^2-60)/(x
^2*log(x)^2+(-8*x^2*log(2)^2-26*x^2)*log(x)+16*x^2*log(2)^4+104*x^2*log(2)^2+169*x^2),x, algorithm="maxima")

[Out]

((4*log(2)^2 + 13)*x^2 - x^2*log(x) + 5)/((4*log(2)^2 + 13)*x - x*log(x))

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mupad [B]  time = 3.32, size = 21, normalized size = 1.11 \begin {gather*} x+\frac {5}{x\,\left (4\,{\ln \relax (2)}^2-\ln \relax (x)+13\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x^2*log(2)^4 + x^2*log(x)^2 + 4*log(2)^2*(26*x^2 - 5) + 169*x^2 - log(x)*(8*x^2*log(2)^2 + 26*x^2 - 5)
- 60)/(104*x^2*log(2)^2 + 16*x^2*log(2)^4 + x^2*log(x)^2 + 169*x^2 - log(x)*(8*x^2*log(2)^2 + 26*x^2)),x)

[Out]

x + 5/(x*(4*log(2)^2 - log(x) + 13))

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sympy [A]  time = 0.12, size = 19, normalized size = 1.00 \begin {gather*} x - \frac {5}{x \log {\relax (x )} - 13 x - 4 x \log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*ln(x)**2+(-8*x**2*ln(2)**2-26*x**2+5)*ln(x)+16*x**2*ln(2)**4+4*(26*x**2-5)*ln(2)**2+169*x**2-6
0)/(x**2*ln(x)**2+(-8*x**2*ln(2)**2-26*x**2)*ln(x)+16*x**2*ln(2)**4+104*x**2*ln(2)**2+169*x**2),x)

[Out]

x - 5/(x*log(x) - 13*x - 4*x*log(2)**2)

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