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

Optimal. Leaf size=23 \[ e^{3-x-3 \log (5) \left (5+\log \left (x^2\right )\right )^2}-x \]

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Rubi [A]  time = 0.34, antiderivative size = 32, normalized size of antiderivative = 1.39, number of steps used = 5, number of rules used = 4, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {14, 2274, 15, 2288} \begin {gather*} e^{3-x} 5^{-3 \left (\log ^2\left (x^2\right )+25\right )} \left (x^2\right )^{-30 \log (5)}-x \end {gather*}

Antiderivative was successfully verified.

[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 2274

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.59, size = 32, normalized size = 1.39 \begin {gather*} 5^{-3 \left (25+\log ^2\left (x^2\right )\right )} e^{3-x-30 \log (5) \log \left (x^2\right )}-x \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.64, size = 32, normalized size = 1.39 \begin {gather*} -x + e^{\left (-3 \, \log \relax (5) \log \left (x^{2}\right )^{2} - 30 \, \log \relax (5) \log \left (x^{2}\right ) - x - 75 \, \log \relax (5) + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.40, size = 30, normalized size = 1.30 \begin {gather*} -x + \frac {1}{26469779601696885595885078146238811314105987548828125} \, e^{\left (-3 \, \log \relax (5) \log \left (x^{2}\right )^{2} - 30 \, \log \relax (5) \log \left (x^{2}\right ) - x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.10, size = 29, normalized size = 1.26

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

Verification of antiderivative is not currently implemented for this CAS.

[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(3-x)

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maxima [A]  time = 0.54, size = 26, normalized size = 1.13 \begin {gather*} -x + \frac {1}{26469779601696885595885078146238811314105987548828125} \, e^{\left (-12 \, \log \relax (5) \log \relax (x)^{2} - 60 \, \log \relax (5) \log \relax (x) - x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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
)

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mupad [B]  time = 4.31, size = 34, normalized size = 1.48 \begin {gather*} \frac {{\mathrm {e}}^{3-x}}{26469779601696885595885078146238811314105987548828125\,5^{3\,{\ln \left (x^2\right )}^2}\,{\left (x^2\right )}^{30\,\ln \relax (5)}}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.36, size = 29, normalized size = 1.26 \begin {gather*} - x + \frac {e^{- x - 3 \log {\relax (5 )} \log {\left (x^{2} \right )}^{2} - 30 \log {\relax (5 )} \log {\left (x^{2} \right )} + 3}}{26469779601696885595885078146238811314105987548828125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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