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

Optimal. Leaf size=32 \[ -\frac {-4-e^3}{2 x}+x-\log ^4\left (\frac {1}{16} x^4 \log ^2(x)\right ) \]

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Rubi [A]  time = 0.26, antiderivative size = 30, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 4, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {12, 6742, 14, 6686} \begin {gather*} -\log ^4\left (\frac {1}{16} x^4 \log ^2(x)\right )+x+\frac {4+e^3}{2 x} \end {gather*}

Antiderivative was successfully verified.

[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 6686

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 33, normalized size = 1.03 \begin {gather*} \frac {2}{x}+\frac {e^3}{2 x}+x-\log ^4\left (\frac {1}{16} x^4 \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 1.00, size = 31, normalized size = 0.97 \begin {gather*} -\frac {2 \, x \log \left (\frac {1}{16} \, x^{4} \log \relax (x)^{2}\right )^{4} - 2 \, x^{2} - e^{3} - 4}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 1.84, size = 178, normalized size = 5.56 \begin {gather*} -\frac {1536 \, x \log \relax (2)^{2} \log \relax (x) \log \left (\log \relax (x)^{2}\right ) - 1536 \, x \log \relax (2) \log \relax (x)^{2} \log \left (\log \relax (x)^{2}\right ) + 512 \, x \log \relax (x)^{3} \log \left (\log \relax (x)^{2}\right ) + 192 \, x \log \relax (2)^{2} \log \left (\log \relax (x)^{2}\right )^{2} - 384 \, x \log \relax (2) \log \relax (x) \log \left (\log \relax (x)^{2}\right )^{2} + 192 \, x \log \relax (x)^{2} \log \left (\log \relax (x)^{2}\right )^{2} - 32 \, x \log \relax (2) \log \left (\log \relax (x)^{2}\right )^{3} + 32 \, x \log \relax (x) \log \left (\log \relax (x)^{2}\right )^{3} + 2 \, x \log \left (\log \relax (x)^{2}\right )^{4} - 2048 \, x \log \relax (2)^{3} \log \relax (x) + 3072 \, x \log \relax (2)^{2} \log \relax (x)^{2} - 2048 \, x \log \relax (2) \log \relax (x)^{3} + 512 \, x \log \relax (x)^{4} - 1024 \, x \log \relax (2)^{3} \log \left (\log \relax (x)\right ) - 2 \, x^{2} - e^{3} - 4}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [C]  time = 19.13, size = 106036, normalized size = 3313.62

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

Verification of antiderivative is not currently implemented for this CAS.

[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]

result too large to display

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maxima [B]  time = 0.43, size = 188, normalized size = 5.88 \begin {gather*} -16 \, \log \left (\frac {1}{16} \, x^{4} \log \relax (x)^{2}\right )^{3} \log \relax (x) + 256 \, \log \relax (x)^{4} + 128 \, \log \relax (2) \log \left (\log \relax (x)\right )^{3} - 16 \, \log \left (\log \relax (x)\right )^{4} + 96 \, {\left (\log \relax (x)^{2} + \log \relax (x)\right )} \log \left (\frac {1}{16} \, x^{4} \log \relax (x)^{2}\right )^{2} + 192 \, {\left (4 \, \log \relax (2) + 1\right )} \log \relax (x)^{2} + 768 \, \log \relax (x)^{3} - 384 \, {\left (\log \relax (2)^{2} + \log \relax (x)\right )} \log \left (\log \relax (x)\right )^{2} - 64 \, {\left (4 \, \log \relax (x)^{3} + 9 \, \log \relax (x)^{2} + 6 \, \log \relax (x)\right )} \log \left (\frac {1}{16} \, x^{4} \log \relax (x)^{2}\right ) - 768 \, {\left (2 \, \log \relax (2)^{2} + 2 \, \log \relax (2) + 1\right )} \log \relax (x) + 1344 \, \log \relax (x)^{2} + 128 \, {\left (4 \, \log \relax (2)^{3} + 6 \, {\left (2 \, \log \relax (2) + 1\right )} \log \relax (x) - 3 \, \log \relax (x)^{2}\right )} \log \left (\log \relax (x)\right ) + x + \frac {e^{3}}{2 \, x} + \frac {2}{x} + 768 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.81, size = 26, normalized size = 0.81 \begin {gather*} x-{\ln \left (\frac {x^4\,{\ln \relax (x)}^2}{16}\right )}^4+\frac {\frac {{\mathrm {e}}^3}{2}+2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.32, size = 22, normalized size = 0.69 \begin {gather*} x - \log {\left (\frac {x^{4} \log {\relax (x )}^{2}}{16} \right )}^{4} + \frac {4 + e^{3}}{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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