\(\int \frac {-x+\frac {e^{3-x-30 \log (5) \log (x^2)-3 \log (5) \log ^2(x^2)} (-x-60 \log (5)-12 \log (5) \log (x^2))}{26469779601696885595885078146238811314105987548828125}}{x} \, dx\) [6787]

Optimal result

Integrand size = 53, antiderivative size = 23 \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=e^{3-x-3 \log (5) \left (5+\log \left (x^2\right )\right )^2}-x \]

[Out]

exp(3-3*(5+ln(x^2))^2*ln(5)-x)-x

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {14, 2306, 15, 2326} \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=e^{3-x} 5^{-3 \left (\log ^2\left (x^2\right )+25\right )} \left (x^2\right )^{-30 \log (5)}-x \]

[In]

Int[(-x + (E^(3 - x - 30*Log[5]*Log[x^2] - 3*Log[5]*Log[x^2]^2)*(-x - 60*Log[5] - 12*Log[5]*Log[x^2]))/2646977
9601696885595885078146238811314105987548828125)/x,x]

[Out]

-x + E^(3 - x)/(5^(3*(25 + Log[x^2]^2))*(x^2)^(30*Log[5]))

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 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 2306

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

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 (-1+\frac {5^{-3 \left (25+\log ^2\left (x^2\right )\right )} e^{3-x-30 \log (5) \log \left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{x}\right ) \, dx \\ & = -x+\int \frac {5^{-3 \left (25+\log ^2\left (x^2\right )\right )} e^{3-x-30 \log (5) \log \left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{x} \, dx \\ & = -x+\int \frac {5^{-3 \left (25+\log ^2\left (x^2\right )\right )} e^{3-x} \left (x^2\right )^{-30 \log (5)} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{x} \, dx \\ & = -x+\left (x^{60 \log (5)} \left (x^2\right )^{-30 \log (5)}\right ) \int 5^{-3 \left (25+\log ^2\left (x^2\right )\right )} e^{3-x} x^{-1-60 \log (5)} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right ) \, dx \\ & = -x+5^{-3 \left (25+\log ^2\left (x^2\right )\right )} e^{3-x} \left (x^2\right )^{-30 \log (5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=5^{-3 \left (25+\log ^2\left (x^2\right )\right )} e^{3-x-30 \log (5) \log \left (x^2\right )}-x \]

[In]

Integrate[(-x + (E^(3 - x - 30*Log[5]*Log[x^2] - 3*Log[5]*Log[x^2]^2)*(-x - 60*Log[5] - 12*Log[5]*Log[x^2]))/2
6469779601696885595885078146238811314105987548828125)/x,x]

[Out]

E^(3 - x - 30*Log[5]*Log[x^2])/5^(3*(25 + Log[x^2]^2)) - x

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26

method	result	size
risch	\(-x +\frac {\left (\frac {1}{125}\right )^{\ln \left (x^{2}\right )^{2}} \left (x^{2}\right )^{-30 \ln \left (5\right )} {\mathrm e}^{-x +3}}{26469779601696885595885078146238811314105987548828125}\)	\(29\)
default	\(-x +{\mathrm e}^{-3 \ln \left (5\right ) \ln \left (x^{2}\right )^{2}-30 \ln \left (5\right ) \ln \left (x^{2}\right )-75 \ln \left (5\right )+3-x}\)	\(33\)
norman	\(-x +{\mathrm e}^{-3 \ln \left (5\right ) \ln \left (x^{2}\right )^{2}-30 \ln \left (5\right ) \ln \left (x^{2}\right )-75 \ln \left (5\right )+3-x}\)	\(33\)
parallelrisch	\(-x +{\mathrm e}^{-3 \ln \left (5\right ) \ln \left (x^{2}\right )^{2}-30 \ln \left (5\right ) \ln \left (x^{2}\right )-75 \ln \left (5\right )+3-x}\)	\(33\)
parts	\(-x +{\mathrm e}^{-3 \ln \left (5\right ) \ln \left (x^{2}\right )^{2}-30 \ln \left (5\right ) \ln \left (x^{2}\right )-75 \ln \left (5\right )+3-x}\)	\(33\)

[In]

int(((-12*ln(5)*ln(x^2)-60*ln(5)-x)*exp(-3*ln(5)*ln(x^2)^2-30*ln(5)*ln(x^2)-75*ln(5)+3-x)-x)/x,x,method=_RETUR
NVERBOSE)

[Out]

-x+1/26469779601696885595885078146238811314105987548828125*(1/125)^(ln(x^2)^2)*(x^2)^(-30*ln(5))*exp(-x+3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=-x + e^{\left (-3 \, \log \left (5\right ) \log \left (x^{2}\right )^{2} - 30 \, \log \left (5\right ) \log \left (x^{2}\right ) - x - 75 \, \log \left (5\right ) + 3\right )} \]

[In]

integrate(((-12*log(5)*log(x^2)-60*log(5)-x)*exp(-3*log(5)*log(x^2)^2-30*log(5)*log(x^2)-75*log(5)+3-x)-x)/x,x
, algorithm="fricas")

[Out]

-x + e^(-3*log(5)*log(x^2)^2 - 30*log(5)*log(x^2) - x - 75*log(5) + 3)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=- x + \frac {e^{- x - 3 \log {\left (5 \right )} \log {\left (x^{2} \right )}^{2} - 30 \log {\left (5 \right )} \log {\left (x^{2} \right )} + 3}}{26469779601696885595885078146238811314105987548828125} \]

[In]

integrate(((-12*ln(5)*ln(x**2)-60*ln(5)-x)*exp(-3*ln(5)*ln(x**2)**2-30*ln(5)*ln(x**2)-75*ln(5)+3-x)-x)/x,x)

[Out]

-x + exp(-x - 3*log(5)*log(x**2)**2 - 30*log(5)*log(x**2) + 3)/26469779601696885595885078146238811314105987548
828125

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=-x + \frac {1}{26469779601696885595885078146238811314105987548828125} \, e^{\left (-12 \, \log \left (5\right ) \log \left (x\right )^{2} - 60 \, \log \left (5\right ) \log \left (x\right ) - x + 3\right )} \]

[In]

integrate(((-12*log(5)*log(x^2)-60*log(5)-x)*exp(-3*log(5)*log(x^2)^2-30*log(5)*log(x^2)-75*log(5)+3-x)-x)/x,x
, algorithm="maxima")

[Out]

-x + 1/26469779601696885595885078146238811314105987548828125*e^(-12*log(5)*log(x)^2 - 60*log(5)*log(x) - x + 3
)

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=-x + \frac {1}{26469779601696885595885078146238811314105987548828125} \, e^{\left (-3 \, \log \left (5\right ) \log \left (x^{2}\right )^{2} - 30 \, \log \left (5\right ) \log \left (x^{2}\right ) - x + 3\right )} \]

[In]

integrate(((-12*log(5)*log(x^2)-60*log(5)-x)*exp(-3*log(5)*log(x^2)^2-30*log(5)*log(x^2)-75*log(5)+3-x)-x)/x,x
, algorithm="giac")

[Out]

-x + 1/26469779601696885595885078146238811314105987548828125*e^(-3*log(5)*log(x^2)^2 - 30*log(5)*log(x^2) - x
+ 3)

Mupad [B] (verification not implemented)

Time = 11.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {-x+\frac {e^{3-x-30 \log (5) \log \left (x^2\right )-3 \log (5) \log ^2\left (x^2\right )} \left (-x-60 \log (5)-12 \log (5) \log \left (x^2\right )\right )}{26469779601696885595885078146238811314105987548828125}}{x} \, dx=\frac {{\mathrm {e}}^{3-x}}{26469779601696885595885078146238811314105987548828125\,5^{3\,{\ln \left (x^2\right )}^2}\,{\left (x^2\right )}^{30\,\ln \left (5\right )}}-x \]

[In]

int(-(x + exp(3 - 75*log(5) - 30*log(x^2)*log(5) - 3*log(x^2)^2*log(5) - x)*(x + 60*log(5) + 12*log(x^2)*log(5
)))/x,x)

[Out]

exp(3 - x)/(26469779601696885595885078146238811314105987548828125*5^(3*log(x^2)^2)*(x^2)^(30*log(5))) - x