\(\int \frac {(-4-e^3+2 x^2) \log (x)+(-16 x-32 x \log (x)) \log ^3(\frac {1}{16} x^4 \log ^2(x))}{2 x^2 \log (x)} \, dx\) [5064]

Optimal result

Integrand size = 51, antiderivative size = 32 \[ \int \frac {\left (-4-e^3+2 x^2\right ) \log (x)+(-16 x-32 x \log (x)) \log ^3\left (\frac {1}{16} x^4 \log ^2(x)\right )}{2 x^2 \log (x)} \, dx=-\frac {-4-e^3}{2 x}+x-\log ^4\left (\frac {1}{16} x^4 \log ^2(x)\right ) \]

[Out]

x-ln(1/16*x^4*ln(x)^2)^4-1/2*(-4-exp(3))/x

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {12, 6874, 14, 6818} \[ \int \frac {\left (-4-e^3+2 x^2\right ) \log (x)+(-16 x-32 x \log (x)) \log ^3\left (\frac {1}{16} x^4 \log ^2(x)\right )}{2 x^2 \log (x)} \, dx=-\log ^4\left (\frac {1}{16} x^4 \log ^2(x)\right )+x+\frac {4+e^3}{2 x} \]

[In]

Int[((-4 - E^3 + 2*x^2)*Log[x] + (-16*x - 32*x*Log[x])*Log[(x^4*Log[x]^2)/16]^3)/(2*x^2*Log[x]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {\left (-4-e^3+2 x^2\right ) \log (x)+(-16 x-32 x \log (x)) \log ^3\left (\frac {1}{16} x^4 \log ^2(x)\right )}{x^2 \log (x)} \, dx \\ & = \frac {1}{2} \int \left (\frac {-4-e^3+2 x^2}{x^2}-\frac {16 (1+2 \log (x)) \log ^3\left (\frac {1}{16} x^4 \log ^2(x)\right )}{x \log (x)}\right ) \, dx \\ & = \frac {1}{2} \int \frac {-4-e^3+2 x^2}{x^2} \, dx-8 \int \frac {(1+2 \log (x)) \log ^3\left (\frac {1}{16} x^4 \log ^2(x)\right )}{x \log (x)} \, dx \\ & = -\log ^4\left (\frac {1}{16} x^4 \log ^2(x)\right )+\frac {1}{2} \int \left (2+\frac {-4-e^3}{x^2}\right ) \, dx \\ & = \frac {4+e^3}{2 x}+x-\log ^4\left (\frac {1}{16} x^4 \log ^2(x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {\left (-4-e^3+2 x^2\right ) \log (x)+(-16 x-32 x \log (x)) \log ^3\left (\frac {1}{16} x^4 \log ^2(x)\right )}{2 x^2 \log (x)} \, dx=\frac {2}{x}+\frac {e^3}{2 x}+x-\log ^4\left (\frac {1}{16} x^4 \log ^2(x)\right ) \]

[In]

Integrate[((-4 - E^3 + 2*x^2)*Log[x] + (-16*x - 32*x*Log[x])*Log[(x^4*Log[x]^2)/16]^3)/(2*x^2*Log[x]),x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(597\) vs. \(2(27)=54\).

Time = 11.23 (sec) , antiderivative size = 598, normalized size of antiderivative = 18.69

method	result	size
default	\(\text {Expression too large to display}\)	\(598\)
parts	\(\text {Expression too large to display}\)	\(663\)
risch	\(\text {Expression too large to display}\)	\(106036\)

[In]

int(1/2*((-32*x*ln(x)-16*x)*ln(1/16*x^4*ln(x)^2)^3+(-exp(3)+2*x^2-4)*ln(x))/x^2/ln(x),x,method=_RETURNVERBOSE)

[Out]

-16*ln(ln(x))^4-16*ln(x)*(ln(x^4*ln(x)^2)-2*ln(ln(x))-4*ln(x))^3-192*(ln(x^4*ln(x)^2)-2*ln(ln(x))-4*ln(x))*(ln
(x)*ln(ln(x))^2-2*ln(x)*ln(ln(x))+2*ln(x))-512*ln(ln(x))*ln(x)^3-768*(ln(x^4*ln(x)^2)-2*ln(ln(x))-4*ln(x))*(1/
2*ln(x)^2*ln(ln(x))-1/4*ln(x)^2)-96*(ln(x^4*ln(x)^2)-2*ln(ln(x))-4*ln(x))^2*(ln(x)*ln(ln(x))-ln(x))-128*ln(x)*
ln(ln(x))^3-256*(ln(x^4*ln(x)^2)-2*ln(ln(x))-4*ln(x))*ln(x)^3-96*(ln(x^4*ln(x)^2)-2*ln(ln(x))-4*ln(x))^2*ln(x)
^2-32*(ln(x^4*ln(x)^2)-2*ln(ln(x))-4*ln(x))*ln(ln(x))^3-384*(ln(x^4*ln(x)^2)-2*ln(ln(x))-4*ln(x))*(ln(x)*ln(ln
(x))-ln(x))-24*(ln(x^4*ln(x)^2)-2*ln(ln(x))-4*ln(x))^2*ln(ln(x))^2-192*(ln(x^4*ln(x)^2)-2*ln(ln(x))-4*ln(x))*l
n(x)^2-96*ln(x)*(ln(x^4*ln(x)^2)-2*ln(ln(x))-4*ln(x))^2-8*(ln(x^4*ln(x)^2)-2*ln(ln(x))-4*ln(x))^3*ln(ln(x))+1/
2*(4+exp(3))/x+512*ln(2)^3*(ln(ln(x))+2*ln(x))+384*ln(2)^2*(-2*ln(x^4*ln(x)^2)*ln(x)+4*ln(x)^2-ln(x^4*ln(x)^2)
*ln(ln(x))+ln(ln(x))^2+4*ln(x)*ln(ln(x)))+x+768*ln(2)*ln(x)*ln(ln(x))^2-384*ln(x)^2*ln(ln(x))^2+1024*ln(2)*ln(
x)^3-768*ln(2)*ln(x)*ln(x^4*ln(x)^2)*ln(ln(x))-256*ln(x)^4-768*ln(2)*ln(x)^2*ln(x^4*ln(x)^2)+1536*ln(2)*ln(x)^
2*ln(ln(x))+192*ln(2)*ln(x^4*ln(x)^2)^2*ln(x)+96*ln(2)*ln(x^4*ln(x)^2)^2*ln(ln(x))-192*ln(2)*ln(ln(x))^2*ln(x^
4*ln(x)^2)+128*ln(2)*ln(ln(x))^3

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-4-e^3+2 x^2\right ) \log (x)+(-16 x-32 x \log (x)) \log ^3\left (\frac {1}{16} x^4 \log ^2(x)\right )}{2 x^2 \log (x)} \, dx=-\frac {2 \, x \log \left (\frac {1}{16} \, x^{4} \log \left (x\right )^{2}\right )^{4} - 2 \, x^{2} - e^{3} - 4}{2 \, x} \]

[In]

integrate(1/2*((-32*x*log(x)-16*x)*log(1/16*x^4*log(x)^2)^3+(-exp(3)+2*x^2-4)*log(x))/x^2/log(x),x, algorithm=
"fricas")

[Out]

-1/2*(2*x*log(1/16*x^4*log(x)^2)^4 - 2*x^2 - e^3 - 4)/x

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {\left (-4-e^3+2 x^2\right ) \log (x)+(-16 x-32 x \log (x)) \log ^3\left (\frac {1}{16} x^4 \log ^2(x)\right )}{2 x^2 \log (x)} \, dx=x - \log {\left (\frac {x^{4} \log {\left (x \right )}^{2}}{16} \right )}^{4} + \frac {4 + e^{3}}{2 x} \]

[In]

integrate(1/2*((-32*x*ln(x)-16*x)*ln(1/16*x**4*ln(x)**2)**3+(-exp(3)+2*x**2-4)*ln(x))/x**2/ln(x),x)

[Out]

x - log(x**4*log(x)**2/16)**4 + (4 + exp(3))/(2*x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (25) = 50\).

Time = 0.27 (sec) , antiderivative size = 188, normalized size of antiderivative = 5.88 \[ \int \frac {\left (-4-e^3+2 x^2\right ) \log (x)+(-16 x-32 x \log (x)) \log ^3\left (\frac {1}{16} x^4 \log ^2(x)\right )}{2 x^2 \log (x)} \, dx=-16 \, \log \left (\frac {1}{16} \, x^{4} \log \left (x\right )^{2}\right )^{3} \log \left (x\right ) + 256 \, \log \left (x\right )^{4} + 128 \, \log \left (2\right ) \log \left (\log \left (x\right )\right )^{3} - 16 \, \log \left (\log \left (x\right )\right )^{4} + 96 \, {\left (\log \left (x\right )^{2} + \log \left (x\right )\right )} \log \left (\frac {1}{16} \, x^{4} \log \left (x\right )^{2}\right )^{2} + 192 \, {\left (4 \, \log \left (2\right ) + 1\right )} \log \left (x\right )^{2} + 768 \, \log \left (x\right )^{3} - 384 \, {\left (\log \left (2\right )^{2} + \log \left (x\right )\right )} \log \left (\log \left (x\right )\right )^{2} - 64 \, {\left (4 \, \log \left (x\right )^{3} + 9 \, \log \left (x\right )^{2} + 6 \, \log \left (x\right )\right )} \log \left (\frac {1}{16} \, x^{4} \log \left (x\right )^{2}\right ) - 768 \, {\left (2 \, \log \left (2\right )^{2} + 2 \, \log \left (2\right ) + 1\right )} \log \left (x\right ) + 1344 \, \log \left (x\right )^{2} + 128 \, {\left (4 \, \log \left (2\right )^{3} + 6 \, {\left (2 \, \log \left (2\right ) + 1\right )} \log \left (x\right ) - 3 \, \log \left (x\right )^{2}\right )} \log \left (\log \left (x\right )\right ) + x + \frac {e^{3}}{2 \, x} + \frac {2}{x} + 768 \, \log \left (x\right ) \]

[In]

integrate(1/2*((-32*x*log(x)-16*x)*log(1/16*x^4*log(x)^2)^3+(-exp(3)+2*x^2-4)*log(x))/x^2/log(x),x, algorithm=
"maxima")

[Out]

-16*log(1/16*x^4*log(x)^2)^3*log(x) + 256*log(x)^4 + 128*log(2)*log(log(x))^3 - 16*log(log(x))^4 + 96*(log(x)^
2 + log(x))*log(1/16*x^4*log(x)^2)^2 + 192*(4*log(2) + 1)*log(x)^2 + 768*log(x)^3 - 384*(log(2)^2 + log(x))*lo
g(log(x))^2 - 64*(4*log(x)^3 + 9*log(x)^2 + 6*log(x))*log(1/16*x^4*log(x)^2) - 768*(2*log(2)^2 + 2*log(2) + 1)
*log(x) + 1344*log(x)^2 + 128*(4*log(2)^3 + 6*(2*log(2) + 1)*log(x) - 3*log(x)^2)*log(log(x)) + x + 1/2*e^3/x
+ 2/x + 768*log(x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (25) = 50\).

Time = 0.41 (sec) , antiderivative size = 178, normalized size of antiderivative = 5.56 \[ \int \frac {\left (-4-e^3+2 x^2\right ) \log (x)+(-16 x-32 x \log (x)) \log ^3\left (\frac {1}{16} x^4 \log ^2(x)\right )}{2 x^2 \log (x)} \, dx=-\frac {1536 \, x \log \left (2\right )^{2} \log \left (x\right ) \log \left (\log \left (x\right )^{2}\right ) - 1536 \, x \log \left (2\right ) \log \left (x\right )^{2} \log \left (\log \left (x\right )^{2}\right ) + 512 \, x \log \left (x\right )^{3} \log \left (\log \left (x\right )^{2}\right ) + 192 \, x \log \left (2\right )^{2} \log \left (\log \left (x\right )^{2}\right )^{2} - 384 \, x \log \left (2\right ) \log \left (x\right ) \log \left (\log \left (x\right )^{2}\right )^{2} + 192 \, x \log \left (x\right )^{2} \log \left (\log \left (x\right )^{2}\right )^{2} - 32 \, x \log \left (2\right ) \log \left (\log \left (x\right )^{2}\right )^{3} + 32 \, x \log \left (x\right ) \log \left (\log \left (x\right )^{2}\right )^{3} + 2 \, x \log \left (\log \left (x\right )^{2}\right )^{4} - 2048 \, x \log \left (2\right )^{3} \log \left (x\right ) + 3072 \, x \log \left (2\right )^{2} \log \left (x\right )^{2} - 2048 \, x \log \left (2\right ) \log \left (x\right )^{3} + 512 \, x \log \left (x\right )^{4} - 1024 \, x \log \left (2\right )^{3} \log \left (\log \left (x\right )\right ) - 2 \, x^{2} - e^{3} - 4}{2 \, x} \]

[In]

integrate(1/2*((-32*x*log(x)-16*x)*log(1/16*x^4*log(x)^2)^3+(-exp(3)+2*x^2-4)*log(x))/x^2/log(x),x, algorithm=
"giac")

[Out]

-1/2*(1536*x*log(2)^2*log(x)*log(log(x)^2) - 1536*x*log(2)*log(x)^2*log(log(x)^2) + 512*x*log(x)^3*log(log(x)^
2) + 192*x*log(2)^2*log(log(x)^2)^2 - 384*x*log(2)*log(x)*log(log(x)^2)^2 + 192*x*log(x)^2*log(log(x)^2)^2 - 3
2*x*log(2)*log(log(x)^2)^3 + 32*x*log(x)*log(log(x)^2)^3 + 2*x*log(log(x)^2)^4 - 2048*x*log(2)^3*log(x) + 3072
*x*log(2)^2*log(x)^2 - 2048*x*log(2)*log(x)^3 + 512*x*log(x)^4 - 1024*x*log(2)^3*log(log(x)) - 2*x^2 - e^3 - 4
)/x

Mupad [B] (verification not implemented)

Time = 12.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {\left (-4-e^3+2 x^2\right ) \log (x)+(-16 x-32 x \log (x)) \log ^3\left (\frac {1}{16} x^4 \log ^2(x)\right )}{2 x^2 \log (x)} \, dx=x-{\ln \left (\frac {x^4\,{\ln \left (x\right )}^2}{16}\right )}^4+\frac {\frac {{\mathrm {e}}^3}{2}+2}{x} \]

[In]

int(-((log(x)*(exp(3) - 2*x^2 + 4))/2 + (log((x^4*log(x)^2)/16)^3*(16*x + 32*x*log(x)))/2)/(x^2*log(x)),x)

[Out]

x - log((x^4*log(x)^2)/16)^4 + (exp(3)/2 + 2)/x