3.16.14 \(\int \frac {-\log (x)+e^{10+2 x-x^2 \log (\log ^2(x))} (-2 x+2 \log (x)-2 x \log (x) \log (\log ^2(x)))}{\log (x)} \, dx\)

Optimal. Leaf size=22 \[ 8+e^{10+2 x-x^2 \log \left (\log ^2(x)\right )}-x \]

________________________________________________________________________________________

Rubi [A]  time = 0.33, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6742, 2288} \begin {gather*} e^{2 x+10} \log ^2(x)^{-x^2}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-Log[x] + E^(10 + 2*x - x^2*Log[Log[x]^2])*(-2*x + 2*Log[x] - 2*x*Log[x]*Log[Log[x]^2]))/Log[x],x]

[Out]

-x + E^(10 + 2*x)/(Log[x]^2)^x^2

Rule 2288

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]

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 \left (-1-\frac {2 e^{10+2 x} \log ^2(x)^{-x^2} \left (x-\log (x)+x \log (x) \log \left (\log ^2(x)\right )\right )}{\log (x)}\right ) \, dx\\ &=-x-2 \int \frac {e^{10+2 x} \log ^2(x)^{-x^2} \left (x-\log (x)+x \log (x) \log \left (\log ^2(x)\right )\right )}{\log (x)} \, dx\\ &=-x+e^{10+2 x} \log ^2(x)^{-x^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 22, normalized size = 1.00 \begin {gather*} -x+e^{10+2 x} \log ^2(x)^{-x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-Log[x] + E^(10 + 2*x - x^2*Log[Log[x]^2])*(-2*x + 2*Log[x] - 2*x*Log[x]*Log[Log[x]^2]))/Log[x],x]

[Out]

-x + E^(10 + 2*x)/(Log[x]^2)^x^2

________________________________________________________________________________________

fricas [A]  time = 0.63, size = 20, normalized size = 0.91 \begin {gather*} -x + e^{\left (-x^{2} \log \left (\log \relax (x)^{2}\right ) + 2 \, x + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(x)*log(log(x)^2)+2*log(x)-2*x)*exp(-x^2*log(log(x)^2)+2*x+10)-log(x))/log(x),x, algorithm
="fricas")

[Out]

-x + e^(-x^2*log(log(x)^2) + 2*x + 10)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(x)*log(log(x)^2)+2*log(x)-2*x)*exp(-x^2*log(log(x)^2)+2*x+10)-log(x))/log(x),x, algorithm
="giac")

[Out]

undef

________________________________________________________________________________________

maple [C]  time = 0.44, size = 86, normalized size = 3.91

method	result	size
risch	\(-x +\ln \relax (x )^{-2 x^{2}} {\mathrm e}^{2 x +10} {\mathrm e}^{\frac {i x^{2} \pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}}{2}} {\mathrm e}^{-i x^{2} \pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )\right )} {\mathrm e}^{\frac {i x^{2} \pi \,\mathrm {csgn}\left (i \ln \relax (x )^{2}\right ) \mathrm {csgn}\left (i \ln \relax (x )\right )^{2}}{2}}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*ln(x)*ln(ln(x)^2)+2*ln(x)-2*x)*exp(-x^2*ln(ln(x)^2)+2*x+10)-ln(x))/ln(x),x,method=_RETURNVERBOSE)

[Out]

-x+ln(x)^(-2*x^2)*exp(2*x+10)*exp(1/2*I*x^2*Pi*csgn(I*ln(x)^2)^3)*exp(-I*x^2*Pi*csgn(I*ln(x)^2)^2*csgn(I*ln(x)
))*exp(1/2*I*x^2*Pi*csgn(I*ln(x)^2)*csgn(I*ln(x))^2)

________________________________________________________________________________________

maxima [A]  time = 1.00, size = 18, normalized size = 0.82 \begin {gather*} -x + e^{\left (-2 \, x^{2} \log \left (\log \relax (x)\right ) + 2 \, x + 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(x)*log(log(x)^2)+2*log(x)-2*x)*exp(-x^2*log(log(x)^2)+2*x+10)-log(x))/log(x),x, algorithm
="maxima")

[Out]

-x + e^(-2*x^2*log(log(x)) + 2*x + 10)

________________________________________________________________________________________

mupad [B]  time = 1.03, size = 21, normalized size = 0.95 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{10}}{{\left ({\ln \relax (x)}^2\right )}^{x^2}}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x) + exp(2*x - x^2*log(log(x)^2) + 10)*(2*x - 2*log(x) + 2*x*log(log(x)^2)*log(x)))/log(x),x)

[Out]

(exp(2*x)*exp(10))/(log(x)^2)^(x^2) - x

________________________________________________________________________________________

sympy [A]  time = 0.47, size = 17, normalized size = 0.77 \begin {gather*} - x + e^{- x^{2} \log {\left (\log {\relax (x )}^{2} \right )} + 2 x + 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*ln(x)*ln(ln(x)**2)+2*ln(x)-2*x)*exp(-x**2*ln(ln(x)**2)+2*x+10)-ln(x))/ln(x),x)

[Out]

-x + exp(-x**2*log(log(x)**2) + 2*x + 10)

________________________________________________________________________________________