\(\int \frac {-36+12 x-2 x^3-16 e^{2 \log ^2(\frac {16}{x^2})} x^2 \log (\frac {16}{x^2})+e^{\log ^2(\frac {16}{x^2})} (12 x+(48 x-16 x^2) \log (\frac {16}{x^2}))}{x^3} \, dx\) [639]

Optimal result

Integrand size = 68, antiderivative size = 26 \[ \int \frac {-36+12 x-2 x^3-16 e^{2 \log ^2\left (\frac {16}{x^2}\right )} x^2 \log \left (\frac {16}{x^2}\right )+e^{\log ^2\left (\frac {16}{x^2}\right )} \left (12 x+\left (48 x-16 x^2\right ) \log \left (\frac {16}{x^2}\right )\right )}{x^3} \, dx=2 \left (-1+\left (1+e^{\log ^2\left (\frac {16}{x^2}\right )}-\frac {3}{x}\right )^2-x\right ) \]

[Out]

2*(1-3/x+exp(ln(16/x^2)^2))^2-2-2*x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(69\) vs. \(2(26)=52\).

Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.65, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {14, 2308, 2235, 2240, 2326} \[ \int \frac {-36+12 x-2 x^3-16 e^{2 \log ^2\left (\frac {16}{x^2}\right )} x^2 \log \left (\frac {16}{x^2}\right )+e^{\log ^2\left (\frac {16}{x^2}\right )} \left (12 x+\left (48 x-16 x^2\right ) \log \left (\frac {16}{x^2}\right )\right )}{x^3} \, dx=\frac {18}{x^2}+2 e^{2 \log ^2\left (\frac {16}{x^2}\right )}-\frac {4 e^{\log ^2\left (\frac {16}{x^2}\right )} \left (3 \log \left (\frac {16}{x^2}\right )-x \log \left (\frac {16}{x^2}\right )\right )}{x \log \left (\frac {16}{x^2}\right )}-2 x-\frac {12}{x} \]

[In]

Int[(-36 + 12*x - 2*x^3 - 16*E^(2*Log[16/x^2]^2)*x^2*Log[16/x^2] + E^Log[16/x^2]^2*(12*x + (48*x - 16*x^2)*Log
[16/x^2]))/x^3,x]

[Out]

2*E^(2*Log[16/x^2]^2) + 18/x^2 - 12/x - 2*x - (4*E^Log[16/x^2]^2*(3*Log[16/x^2] - x*Log[16/x^2]))/(x*Log[16/x^
2])

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 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2308

Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]^2*(b_.))*(f_.))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol]
:> Dist[(g + h*x)^(m + 1)/(h*n*(c*(d + e*x)^n)^((m + 1)/n)), Subst[Int[E^(a*f*Log[F] + ((m + 1)*x)/n + b*f*Lo
g[F]*x^2), x], x, Log[c*(d + e*x)^n]], x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m, n}, x] && EqQ[e*g - d*h, 0]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 \left (18-6 x+x^3\right )}{x^3}-\frac {16 e^{2 \log ^2\left (\frac {16}{x^2}\right )} \log \left (\frac {16}{x^2}\right )}{x}-\frac {4 e^{\log ^2\left (\frac {16}{x^2}\right )} \left (-3-12 \log \left (\frac {16}{x^2}\right )+4 x \log \left (\frac {16}{x^2}\right )\right )}{x^2}\right ) \, dx \\ & = -\left (2 \int \frac {18-6 x+x^3}{x^3} \, dx\right )-4 \int \frac {e^{\log ^2\left (\frac {16}{x^2}\right )} \left (-3-12 \log \left (\frac {16}{x^2}\right )+4 x \log \left (\frac {16}{x^2}\right )\right )}{x^2} \, dx-16 \int \frac {e^{2 \log ^2\left (\frac {16}{x^2}\right )} \log \left (\frac {16}{x^2}\right )}{x} \, dx \\ & = -\frac {4 e^{\log ^2\left (\frac {16}{x^2}\right )} \left (3 \log \left (\frac {16}{x^2}\right )-x \log \left (\frac {16}{x^2}\right )\right )}{x \log \left (\frac {16}{x^2}\right )}-2 \int \left (1+\frac {18}{x^3}-\frac {6}{x^2}\right ) \, dx+8 \text {Subst}\left (\int e^{2 x^2} x \, dx,x,\log \left (\frac {16}{x^2}\right )\right ) \\ & = 2 e^{2 \log ^2\left (\frac {16}{x^2}\right )}+\frac {18}{x^2}-\frac {12}{x}-2 x-\frac {4 e^{\log ^2\left (\frac {16}{x^2}\right )} \left (3 \log \left (\frac {16}{x^2}\right )-x \log \left (\frac {16}{x^2}\right )\right )}{x \log \left (\frac {16}{x^2}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {-36+12 x-2 x^3-16 e^{2 \log ^2\left (\frac {16}{x^2}\right )} x^2 \log \left (\frac {16}{x^2}\right )+e^{\log ^2\left (\frac {16}{x^2}\right )} \left (12 x+\left (48 x-16 x^2\right ) \log \left (\frac {16}{x^2}\right )\right )}{x^3} \, dx=-2 \left (-e^{2 \log ^2\left (\frac {16}{x^2}\right )}-\frac {9}{x^2}+\frac {6}{x}-\frac {2 e^{\log ^2\left (\frac {16}{x^2}\right )} (-3+x)}{x}+x\right ) \]

[In]

Integrate[(-36 + 12*x - 2*x^3 - 16*E^(2*Log[16/x^2]^2)*x^2*Log[16/x^2] + E^Log[16/x^2]^2*(12*x + (48*x - 16*x^
2)*Log[16/x^2]))/x^3,x]

[Out]

-2*(-E^(2*Log[16/x^2]^2) - 9/x^2 + 6/x - (2*E^Log[16/x^2]^2*(-3 + x))/x + x)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69

method	result	size
risch	\(-2 x +\frac {-12 x +18}{x^{2}}+2 \,{\mathrm e}^{2 \ln \left (\frac {16}{x^{2}}\right )^{2}}+\frac {4 \left (-3+x \right ) {\mathrm e}^{\ln \left (\frac {16}{x^{2}}\right )^{2}}}{x}\)	\(44\)
default	\(\frac {4 \,{\mathrm e}^{\ln \left (\frac {16}{x^{2}}\right )^{2}} x -12 \,{\mathrm e}^{\ln \left (\frac {16}{x^{2}}\right )^{2}}}{x}-2 x -\frac {12}{x}+\frac {18}{x^{2}}+2 \,{\mathrm e}^{2 \ln \left (\frac {16}{x^{2}}\right )^{2}}\)	\(56\)
parts	\(\frac {4 \,{\mathrm e}^{\ln \left (\frac {16}{x^{2}}\right )^{2}} x -12 \,{\mathrm e}^{\ln \left (\frac {16}{x^{2}}\right )^{2}}}{x}-2 x -\frac {12}{x}+\frac {18}{x^{2}}+2 \,{\mathrm e}^{2 \ln \left (\frac {16}{x^{2}}\right )^{2}}\)	\(56\)
parallelrisch	\(\frac {36+4 \,{\mathrm e}^{2 \ln \left (\frac {16}{x^{2}}\right )^{2}} x^{2}-4 x^{3}+8 \,{\mathrm e}^{\ln \left (\frac {16}{x^{2}}\right )^{2}} x^{2}-24 \,{\mathrm e}^{\ln \left (\frac {16}{x^{2}}\right )^{2}} x -24 x}{2 x^{2}}\)	\(58\)

[In]

int((-16*x^2*ln(16/x^2)*exp(ln(16/x^2)^2)^2+((-16*x^2+48*x)*ln(16/x^2)+12*x)*exp(ln(16/x^2)^2)-2*x^3+12*x-36)/
x^3,x,method=_RETURNVERBOSE)

[Out]

-2*x+(-12*x+18)/x^2+2*exp(2*ln(16/x^2)^2)+4*(-3+x)/x*exp(ln(16/x^2)^2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81 \[ \int \frac {-36+12 x-2 x^3-16 e^{2 \log ^2\left (\frac {16}{x^2}\right )} x^2 \log \left (\frac {16}{x^2}\right )+e^{\log ^2\left (\frac {16}{x^2}\right )} \left (12 x+\left (48 x-16 x^2\right ) \log \left (\frac {16}{x^2}\right )\right )}{x^3} \, dx=-\frac {2 \, {\left (x^{3} - x^{2} e^{\left (2 \, \log \left (\frac {16}{x^{2}}\right )^{2}\right )} - 2 \, {\left (x^{2} - 3 \, x\right )} e^{\left (\log \left (\frac {16}{x^{2}}\right )^{2}\right )} + 6 \, x - 9\right )}}{x^{2}} \]

[In]

integrate((-16*x^2*log(16/x^2)*exp(log(16/x^2)^2)^2+((-16*x^2+48*x)*log(16/x^2)+12*x)*exp(log(16/x^2)^2)-2*x^3
+12*x-36)/x^3,x, algorithm="fricas")

[Out]

-2*(x^3 - x^2*e^(2*log(16/x^2)^2) - 2*(x^2 - 3*x)*e^(log(16/x^2)^2) + 6*x - 9)/x^2

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {-36+12 x-2 x^3-16 e^{2 \log ^2\left (\frac {16}{x^2}\right )} x^2 \log \left (\frac {16}{x^2}\right )+e^{\log ^2\left (\frac {16}{x^2}\right )} \left (12 x+\left (48 x-16 x^2\right ) \log \left (\frac {16}{x^2}\right )\right )}{x^3} \, dx=- 2 x + \frac {2 x e^{2 \log {\left (\frac {16}{x^{2}} \right )}^{2}} + \left (4 x - 12\right ) e^{\log {\left (\frac {16}{x^{2}} \right )}^{2}}}{x} - \frac {12 x - 18}{x^{2}} \]

[In]

integrate((-16*x**2*ln(16/x**2)*exp(ln(16/x**2)**2)**2+((-16*x**2+48*x)*ln(16/x**2)+12*x)*exp(ln(16/x**2)**2)-
2*x**3+12*x-36)/x**3,x)

[Out]

-2*x + (2*x*exp(2*log(16/x**2)**2) + (4*x - 12)*exp(log(16/x**2)**2))/x - (12*x - 18)/x**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (27) = 54\).

Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.96 \[ \int \frac {-36+12 x-2 x^3-16 e^{2 \log ^2\left (\frac {16}{x^2}\right )} x^2 \log \left (\frac {16}{x^2}\right )+e^{\log ^2\left (\frac {16}{x^2}\right )} \left (12 x+\left (48 x-16 x^2\right ) \log \left (\frac {16}{x^2}\right )\right )}{x^3} \, dx=-2 \, x + \frac {2 \, {\left (x e^{\left (32 \, \log \left (2\right )^{2} - 32 \, \log \left (2\right ) \log \left (x\right ) + 8 \, \log \left (x\right )^{2}\right )} + 2 \, {\left (x e^{\left (16 \, \log \left (2\right )^{2}\right )} - 3 \, e^{\left (16 \, \log \left (2\right )^{2}\right )}\right )} e^{\left (-16 \, \log \left (2\right ) \log \left (x\right ) + 4 \, \log \left (x\right )^{2}\right )}\right )}}{x} - \frac {12}{x} + \frac {18}{x^{2}} \]

[In]

integrate((-16*x^2*log(16/x^2)*exp(log(16/x^2)^2)^2+((-16*x^2+48*x)*log(16/x^2)+12*x)*exp(log(16/x^2)^2)-2*x^3
+12*x-36)/x^3,x, algorithm="maxima")

[Out]

-2*x + 2*(x*e^(32*log(2)^2 - 32*log(2)*log(x) + 8*log(x)^2) + 2*(x*e^(16*log(2)^2) - 3*e^(16*log(2)^2))*e^(-16
*log(2)*log(x) + 4*log(x)^2))/x - 12/x + 18/x^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (27) = 54\).

Time = 0.39 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.12 \[ \int \frac {-36+12 x-2 x^3-16 e^{2 \log ^2\left (\frac {16}{x^2}\right )} x^2 \log \left (\frac {16}{x^2}\right )+e^{\log ^2\left (\frac {16}{x^2}\right )} \left (12 x+\left (48 x-16 x^2\right ) \log \left (\frac {16}{x^2}\right )\right )}{x^3} \, dx=-\frac {2 \, {\left (x^{3} - x^{2} e^{\left (2 \, \log \left (\frac {16}{x^{2}}\right )^{2}\right )} - 2 \, x^{2} e^{\left (\log \left (\frac {16}{x^{2}}\right )^{2}\right )} + 6 \, x e^{\left (\log \left (\frac {16}{x^{2}}\right )^{2}\right )} + 6 \, x - 9\right )}}{x^{2}} \]

[In]

integrate((-16*x^2*log(16/x^2)*exp(log(16/x^2)^2)^2+((-16*x^2+48*x)*log(16/x^2)+12*x)*exp(log(16/x^2)^2)-2*x^3
+12*x-36)/x^3,x, algorithm="giac")

[Out]

-2*(x^3 - x^2*e^(2*log(16/x^2)^2) - 2*x^2*e^(log(16/x^2)^2) + 6*x*e^(log(16/x^2)^2) + 6*x - 9)/x^2

Mupad [B] (verification not implemented)

Time = 8.51 (sec) , antiderivative size = 93, normalized size of antiderivative = 3.58 \[ \int \frac {-36+12 x-2 x^3-16 e^{2 \log ^2\left (\frac {16}{x^2}\right )} x^2 \log \left (\frac {16}{x^2}\right )+e^{\log ^2\left (\frac {16}{x^2}\right )} \left (12 x+\left (48 x-16 x^2\right ) \log \left (\frac {16}{x^2}\right )\right )}{x^3} \, dx=2\,{\mathrm {e}}^{2\,{\ln \left (\frac {1}{x^2}\right )}^2+32\,{\ln \left (2\right )}^2}\,{\left (\frac {1}{x^2}\right )}^{16\,\ln \left (2\right )}-2\,x+4\,{\mathrm {e}}^{{\ln \left (\frac {1}{x^2}\right )}^2+16\,{\ln \left (2\right )}^2}\,{\left (\frac {1}{x^2}\right )}^{8\,\ln \left (2\right )}+\frac {18\,x-x^2\,\left (12\,{\mathrm {e}}^{{\ln \left (\frac {1}{x^2}\right )}^2+16\,{\ln \left (2\right )}^2}\,{\left (\frac {1}{x^2}\right )}^{8\,\ln \left (2\right )}+12\right )}{x^3} \]

[In]

int(-(2*x^3 - exp(log(16/x^2)^2)*(12*x + log(16/x^2)*(48*x - 16*x^2)) - 12*x + 16*x^2*exp(2*log(16/x^2)^2)*log
(16/x^2) + 36)/x^3,x)

[Out]

2*exp(2*log(1/x^2)^2 + 32*log(2)^2)*(1/x^2)^(16*log(2)) - 2*x + 4*exp(log(1/x^2)^2 + 16*log(2)^2)*(1/x^2)^(8*l
og(2)) + (18*x - x^2*(12*exp(log(1/x^2)^2 + 16*log(2)^2)*(1/x^2)^(8*log(2)) + 12))/x^3