3.19.61 \(\int \frac {-2+(5 x-5 e^x x-40 x^2+(x-e^x x-8 x^2) \log (\frac {x^2}{16})) \log ^2(5+\log (\frac {x^2}{16}))}{(5 x+x \log (\frac {x^2}{16})) \log ^2(5+\log (\frac {x^2}{16}))} \, dx\)

Optimal. Leaf size=33 \[ -e^x+x-4 x^2+\frac {1}{3} \left (1+\frac {3}{\log \left (5+\log \left (\frac {x^2}{16}\right )\right )}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.78, antiderivative size = 25, normalized size of antiderivative = 0.76, number of steps used = 8, number of rules used = 6, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.072, Rules used = {2561, 6688, 2194, 2390, 2302, 30} \begin {gather*} -4 x^2+\frac {1}{\log \left (\log \left (\frac {x^2}{16}\right )+5\right )}+x-e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + (5*x - 5*E^x*x - 40*x^2 + (x - E^x*x - 8*x^2)*Log[x^2/16])*Log[5 + Log[x^2/16]]^2)/((5*x + x*Log[x^2
/16])*Log[5 + Log[x^2/16]]^2),x]

[Out]

-E^x + x - 4*x^2 + Log[5 + Log[x^2/16]]^(-1)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2194

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

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2561

Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.))^(p_.), x_Symbol] :> Int[u*x^(p*r)*
(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2+\left (5 x-5 e^x x-40 x^2+\left (x-e^x x-8 x^2\right ) \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )}{x \left (5+\log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )} \, dx\\ &=\int \left (1-e^x-8 x-\frac {2}{x \left (5+\log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )}\right ) \, dx\\ &=x-4 x^2-2 \int \frac {1}{x \left (5+\log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )} \, dx-\int e^x \, dx\\ &=-e^x+x-4 x^2-\operatorname {Subst}\left (\int \frac {1}{(5+x) \log ^2(5+x)} \, dx,x,\log \left (\frac {x^2}{16}\right )\right )\\ &=-e^x+x-4 x^2-\operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,5+\log \left (\frac {x^2}{16}\right )\right )\\ &=-e^x+x-4 x^2-\operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (5+\log \left (\frac {x^2}{16}\right )\right )\right )\\ &=-e^x+x-4 x^2+\frac {1}{\log \left (5+\log \left (\frac {x^2}{16}\right )\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 25, normalized size = 0.76 \begin {gather*} -e^x+x-4 x^2+\frac {1}{\log \left (5+\log \left (\frac {x^2}{16}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + (5*x - 5*E^x*x - 40*x^2 + (x - E^x*x - 8*x^2)*Log[x^2/16])*Log[5 + Log[x^2/16]]^2)/((5*x + x*L
og[x^2/16])*Log[5 + Log[x^2/16]]^2),x]

[Out]

-E^x + x - 4*x^2 + Log[5 + Log[x^2/16]]^(-1)

________________________________________________________________________________________

fricas [A]  time = 0.78, size = 36, normalized size = 1.09 \begin {gather*} -\frac {{\left (4 \, x^{2} - x + e^{x}\right )} \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 1}{\log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-exp(x)*x-8*x^2+x)*log(1/16*x^2)-5*exp(x)*x-40*x^2+5*x)*log(log(1/16*x^2)+5)^2-2)/(x*log(1/16*x^2
)+5*x)/log(log(1/16*x^2)+5)^2,x, algorithm="fricas")

[Out]

-((4*x^2 - x + e^x)*log(log(1/16*x^2) + 5) - 1)/log(log(1/16*x^2) + 5)

________________________________________________________________________________________

giac [B]  time = 0.54, size = 186, normalized size = 5.64 \begin {gather*} -\frac {16 \, x^{2} \log \relax (2) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 4 \, x^{2} \log \left (x^{2}\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 20 \, x^{2} \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 4 \, x \log \relax (2) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) + 4 \, e^{x} \log \relax (2) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) + x \log \left (x^{2}\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - e^{x} \log \left (x^{2}\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) + 5 \, x \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 5 \, e^{x} \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) + \log \left (\frac {1}{16} \, x^{2}\right ) + 5}{4 \, \log \relax (2) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - \log \left (x^{2}\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 5 \, \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-exp(x)*x-8*x^2+x)*log(1/16*x^2)-5*exp(x)*x-40*x^2+5*x)*log(log(1/16*x^2)+5)^2-2)/(x*log(1/16*x^2
)+5*x)/log(log(1/16*x^2)+5)^2,x, algorithm="giac")

[Out]

-(16*x^2*log(2)*log(log(1/16*x^2) + 5) - 4*x^2*log(x^2)*log(log(1/16*x^2) + 5) - 20*x^2*log(log(1/16*x^2) + 5)
- 4*x*log(2)*log(log(1/16*x^2) + 5) + 4*e^x*log(2)*log(log(1/16*x^2) + 5) + x*log(x^2)*log(log(1/16*x^2) + 5)
- e^x*log(x^2)*log(log(1/16*x^2) + 5) + 5*x*log(log(1/16*x^2) + 5) - 5*e^x*log(log(1/16*x^2) + 5) + log(1/16*
x^2) + 5)/(4*log(2)*log(log(1/16*x^2) + 5) - log(x^2)*log(log(1/16*x^2) + 5) - 5*log(log(1/16*x^2) + 5))

________________________________________________________________________________________

maple [A]  time = 0.17, size = 23, normalized size = 0.70

method	result	size
default	\(-{\mathrm e}^{x}-4 x^{2}+x +\frac {1}{\ln \left (\ln \left (\frac {x^{2}}{16}\right )+5\right )}\)	\(23\)
risch	\(-4 x^{2}+x -{\mathrm e}^{x}+\frac {1}{\ln \left (-4 \ln \relax (2)+2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}+5\right )}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-exp(x)*x-8*x^2+x)*ln(1/16*x^2)-5*exp(x)*x-40*x^2+5*x)*ln(ln(1/16*x^2)+5)^2-2)/(x*ln(1/16*x^2)+5*x)/ln(
ln(1/16*x^2)+5)^2,x,method=_RETURNVERBOSE)

[Out]

-exp(x)-4*x^2+x+1/ln(ln(1/16*x^2)+5)

________________________________________________________________________________________

maxima [A]  time = 0.58, size = 24, normalized size = 0.73 \begin {gather*} -4 \, x^{2} + x + \frac {1}{\log \left (-4 \, \log \relax (2) + 2 \, \log \relax (x) + 5\right )} - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-exp(x)*x-8*x^2+x)*log(1/16*x^2)-5*exp(x)*x-40*x^2+5*x)*log(log(1/16*x^2)+5)^2-2)/(x*log(1/16*x^2
)+5*x)/log(log(1/16*x^2)+5)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.20, size = 22, normalized size = 0.67 \begin {gather*} x-{\mathrm {e}}^x+\frac {1}{\ln \left (\ln \left (\frac {x^2}{16}\right )+5\right )}-4\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(x^2/16) + 5)^2*(log(x^2/16)*(x*exp(x) - x + 8*x^2) - 5*x + 5*x*exp(x) + 40*x^2) + 2)/(log(log(x^
2/16) + 5)^2*(5*x + x*log(x^2/16))),x)

[Out]

x - exp(x) + 1/log(log(x^2/16) + 5) - 4*x^2

________________________________________________________________________________________

sympy [A]  time = 0.47, size = 20, normalized size = 0.61 \begin {gather*} - 4 x^{2} + x - e^{x} + \frac {1}{\log {\left (\log {\left (\frac {x^{2}}{16} \right )} + 5 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-exp(x)*x-8*x**2+x)*ln(1/16*x**2)-5*exp(x)*x-40*x**2+5*x)*ln(ln(1/16*x**2)+5)**2-2)/(x*ln(1/16*x*
*2)+5*x)/ln(ln(1/16*x**2)+5)**2,x)

[Out]

-4*x**2 + x - exp(x) + 1/log(log(x**2/16) + 5)

________________________________________________________________________________________