3.99.36 \(\int e^{60 x^2 \log (\frac {3}{x})+4 x^2 \log (\frac {3}{x}) \log (x)} (1-60 x^2+124 x^2 \log (\frac {3}{x})+(-4 x^2+8 x^2 \log (\frac {3}{x})) \log (x)) \, dx\)

Optimal. Leaf size=27 \[ 3^{4 x^2 (15+\log (x))} \left (\frac {1}{x}\right )^{-1+4 x^2 (15+\log (x))} \]

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Rubi [B]  time = 0.44, antiderivative size = 97, normalized size of antiderivative = 3.59, number of steps used = 1, number of rules used = 1, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {2288} \begin {gather*} \frac {3^{60 x^2} \left (\frac {1}{x}\right )^{60 x^2} e^{4 x^2 \log \left (\frac {3}{x}\right ) \log (x)} \left (15 x^2-31 x^2 \log \left (\frac {3}{x}\right )+\left (x^2-2 x^2 \log \left (\frac {3}{x}\right )\right ) \log (x)\right )}{15 x-31 x \log \left (\frac {3}{x}\right )-2 x \log \left (\frac {3}{x}\right ) \log (x)+x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(60*x^2*Log[3/x] + 4*x^2*Log[3/x]*Log[x])*(1 - 60*x^2 + 124*x^2*Log[3/x] + (-4*x^2 + 8*x^2*Log[3/x])*Log
[x]),x]

[Out]

(3^(60*x^2)*E^(4*x^2*Log[3/x]*Log[x])*(x^(-1))^(60*x^2)*(15*x^2 - 31*x^2*Log[3/x] + (x^2 - 2*x^2*Log[3/x])*Log
[x]))/(15*x - 31*x*Log[3/x] + x*Log[x] - 2*x*Log[3/x]*Log[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} &=\frac {3^{60 x^2} e^{4 x^2 \log \left (\frac {3}{x}\right ) \log (x)} \left (\frac {1}{x}\right )^{60 x^2} \left (15 x^2-31 x^2 \log \left (\frac {3}{x}\right )+\left (x^2-2 x^2 \log \left (\frac {3}{x}\right )\right ) \log (x)\right )}{15 x-31 x \log \left (\frac {3}{x}\right )+x \log (x)-2 x \log \left (\frac {3}{x}\right ) \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 1.95, size = 0, normalized size = 0.00 \begin {gather*} \int e^{60 x^2 \log \left (\frac {3}{x}\right )+4 x^2 \log \left (\frac {3}{x}\right ) \log (x)} \left (1-60 x^2+124 x^2 \log \left (\frac {3}{x}\right )+\left (-4 x^2+8 x^2 \log \left (\frac {3}{x}\right )\right ) \log (x)\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[E^(60*x^2*Log[3/x] + 4*x^2*Log[3/x]*Log[x])*(1 - 60*x^2 + 124*x^2*Log[3/x] + (-4*x^2 + 8*x^2*Log[3/x
])*Log[x]),x]

[Out]

Integrate[E^(60*x^2*Log[3/x] + 4*x^2*Log[3/x]*Log[x])*(1 - 60*x^2 + 124*x^2*Log[3/x] + (-4*x^2 + 8*x^2*Log[3/x
])*Log[x]), x]

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fricas [B]  time = 0.91, size = 37, normalized size = 1.37 \begin {gather*} x e^{\left (-4 \, x^{2} \log \left (\frac {3}{x}\right )^{2} + 4 \, {\left (x^{2} \log \relax (3) + 15 \, x^{2}\right )} \log \left (\frac {3}{x}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2*log(3/x)-4*x^2)*log(x)+124*x^2*log(3/x)-60*x^2+1)*exp(4*x^2*log(3/x)*log(x)+60*x^2*log(3/x))
,x, algorithm="fricas")

[Out]

x*e^(-4*x^2*log(3/x)^2 + 4*(x^2*log(3) + 15*x^2)*log(3/x))

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giac [B]  time = 0.29, size = 36, normalized size = 1.33 \begin {gather*} x e^{\left (4 \, x^{2} \log \relax (3) \log \relax (x) - 4 \, x^{2} \log \relax (x)^{2} + 60 \, x^{2} \log \relax (3) - 60 \, x^{2} \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2*log(3/x)-4*x^2)*log(x)+124*x^2*log(3/x)-60*x^2+1)*exp(4*x^2*log(3/x)*log(x)+60*x^2*log(3/x))
,x, algorithm="giac")

[Out]

x*e^(4*x^2*log(3)*log(x) - 4*x^2*log(x)^2 + 60*x^2*log(3) - 60*x^2*log(x))

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maple [A]  time = 0.08, size = 29, normalized size = 1.07

method	result	size
default	\({\mathrm e}^{4 x^{2} \ln \left (\frac {3}{x}\right ) \ln \relax (x )+60 x^{2} \ln \left (\frac {3}{x}\right )} x\)	\(29\)
risch	\(x \,x^{-4 x^{2} \left (\ln \relax (x )-\ln \relax (3)\right )} {\mathrm e}^{60 x^{2} \left (\ln \relax (3)-\ln \relax (x )\right )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^2*ln(3/x)-4*x^2)*ln(x)+124*x^2*ln(3/x)-60*x^2+1)*exp(4*x^2*ln(3/x)*ln(x)+60*x^2*ln(3/x)),x,method=_R
ETURNVERBOSE)

[Out]

exp(4*x^2*ln(3/x)*ln(x)+60*x^2*ln(3/x))*x

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maxima [B]  time = 0.53, size = 36, normalized size = 1.33 \begin {gather*} x e^{\left (4 \, x^{2} \log \relax (3) \log \relax (x) - 4 \, x^{2} \log \relax (x)^{2} + 60 \, x^{2} \log \relax (3) - 60 \, x^{2} \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2*log(3/x)-4*x^2)*log(x)+124*x^2*log(3/x)-60*x^2+1)*exp(4*x^2*log(3/x)*log(x)+60*x^2*log(3/x))
,x, algorithm="maxima")

[Out]

x*e^(4*x^2*log(3)*log(x) - 4*x^2*log(x)^2 + 60*x^2*log(3) - 60*x^2*log(x))

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mupad [B]  time = 6.02, size = 38, normalized size = 1.41 \begin {gather*} 3^{60\,x^2}\,x\,x^{4\,x^2\,\ln \left (\frac {1}{x}\right )}\,x^{4\,x^2\,\ln \relax (3)}\,{\left (\frac {1}{x}\right )}^{60\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(60*x^2*log(3/x) + 4*x^2*log(3/x)*log(x))*(log(x)*(4*x^2 - 8*x^2*log(3/x)) + 60*x^2 - 124*x^2*log(3/x)
- 1),x)

[Out]

3^(60*x^2)*x*x^(4*x^2*log(1/x))*x^(4*x^2*log(3))*(1/x)^(60*x^2)

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sympy [A]  time = 0.43, size = 29, normalized size = 1.07 \begin {gather*} x e^{4 x^{2} \left (- \log {\relax (x )} + \log {\relax (3 )}\right ) \log {\relax (x )} + 60 x^{2} \left (- \log {\relax (x )} + \log {\relax (3 )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**2*ln(3/x)-4*x**2)*ln(x)+124*x**2*ln(3/x)-60*x**2+1)*exp(4*x**2*ln(3/x)*ln(x)+60*x**2*ln(3/x))
,x)

[Out]

x*exp(4*x**2*(-log(x) + log(3))*log(x) + 60*x**2*(-log(x) + log(3)))

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