3.15.33 \(\int e^{-1-20 x \log (2)-25 x^2 \log ^2(2)} (1-20 x \log (2)-50 x^2 \log ^2(2)) \, dx\)

Optimal. Leaf size=17 \[ e^{3-(2+5 x \log (2))^2} x \]

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Rubi [B]  time = 0.06, antiderivative size = 47, normalized size of antiderivative = 2.76, number of steps used = 1, number of rules used = 1, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {2288} \begin {gather*} \frac {2^{-20 x} e^{-25 x^2 \log ^2(2)-1} \left (5 x^2 \log ^2(2)+2 x \log (2)\right )}{\log (2) (5 x \log (2)+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(-1 - 20*x*Log[2] - 25*x^2*Log[2]^2)*(1 - 20*x*Log[2] - 50*x^2*Log[2]^2),x]

[Out]

(E^(-1 - 25*x^2*Log[2]^2)*(2*x*Log[2] + 5*x^2*Log[2]^2))/(2^(20*x)*Log[2]*(2 + 5*x*Log[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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {2^{-20 x} e^{-1-25 x^2 \log ^2(2)} \left (2 x \log (2)+5 x^2 \log ^2(2)\right )}{\log (2) (2+5 x \log (2))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 20, normalized size = 1.18 \begin {gather*} 2^{-20 x} e^{-1-25 x^2 \log ^2(2)} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(-1 - 20*x*Log[2] - 25*x^2*Log[2]^2)*(1 - 20*x*Log[2] - 50*x^2*Log[2]^2),x]

[Out]

(E^(-1 - 25*x^2*Log[2]^2)*x)/2^(20*x)

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fricas [A]  time = 1.02, size = 19, normalized size = 1.12 \begin {gather*} e^{\left (-25 \, x^{2} \log \relax (2)^{2} - 20 \, x \log \relax (2) + \log \relax (x) - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*x^2*log(2)^2-20*x*log(2)+1)*exp(log(x)-25*x^2*log(2)^2-20*x*log(2)-1)/x,x, algorithm="fricas")

[Out]

e^(-25*x^2*log(2)^2 - 20*x*log(2) + log(x) - 1)

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giac [A]  time = 0.37, size = 19, normalized size = 1.12 \begin {gather*} e^{\left (-25 \, x^{2} \log \relax (2)^{2} - 20 \, x \log \relax (2) + \log \relax (x) - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*x^2*log(2)^2-20*x*log(2)+1)*exp(log(x)-25*x^2*log(2)^2-20*x*log(2)-1)/x,x, algorithm="giac")

[Out]

e^(-25*x^2*log(2)^2 - 20*x*log(2) + log(x) - 1)

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maple [A]  time = 0.07, size = 18, normalized size = 1.06

method	result	size
risch	\(x \left (\frac {1}{1048576}\right )^{x} {\mathrm e}^{-1-25 x^{2} \ln \relax (2)^{2}}\)	\(18\)
gosper	\({\mathrm e}^{\ln \relax (x )-25 x^{2} \ln \relax (2)^{2}-20 x \ln \relax (2)-1}\)	\(20\)
norman	\({\mathrm e}^{\ln \relax (x )-25 x^{2} \ln \relax (2)^{2}-20 x \ln \relax (2)-1}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-50*x^2*ln(2)^2-20*x*ln(2)+1)*exp(ln(x)-25*x^2*ln(2)^2-20*x*ln(2)-1)/x,x,method=_RETURNVERBOSE)

[Out]

x*(1/1048576)^x*exp(-1-25*x^2*ln(2)^2)

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maxima [C]  time = 0.73, size = 323, normalized size = 19.00 \begin {gather*} -\frac {{\left (\frac {{\left (5 \, x \log \relax (2)^{2} + 2 \, \log \relax (2)\right )}^{3} \Gamma \left (\frac {3}{2}, \frac {{\left (5 \, x \log \relax (2)^{2} + 2 \, \log \relax (2)\right )}^{2}}{\log \relax (2)^{2}}\right ) \log \relax (2)^{3}}{{\left ({\left (5 \, x \log \relax (2)^{2} + 2 \, \log \relax (2)\right )}^{2}\right )}^{\frac {3}{2}} \left (-\log \relax (2)^{2}\right )^{\frac {5}{2}}} - \frac {4 \, \sqrt {\pi } {\left (5 \, x \log \relax (2)^{2} + 2 \, \log \relax (2)\right )} {\left (\operatorname {erf}\left (\frac {\sqrt {{\left (5 \, x \log \relax (2)^{2} + 2 \, \log \relax (2)\right )}^{2}}}{\log \relax (2)}\right ) - 1\right )} \log \relax (2)^{3}}{\sqrt {{\left (5 \, x \log \relax (2)^{2} + 2 \, \log \relax (2)\right )}^{2}} \left (-\log \relax (2)^{2}\right )^{\frac {5}{2}}} - \frac {4 \, e^{\left (-\frac {{\left (5 \, x \log \relax (2)^{2} + 2 \, \log \relax (2)\right )}^{2}}{\log \relax (2)^{2}}\right )} \log \relax (2)^{3}}{\left (-\log \relax (2)^{2}\right )^{\frac {5}{2}}}\right )} e^{3} \log \relax (2)^{2}}{5 \, \sqrt {-\log \relax (2)^{2}}} + \frac {2 \, {\left (\frac {2 \, \sqrt {\pi } {\left (5 \, x \log \relax (2)^{2} + 2 \, \log \relax (2)\right )} {\left (\operatorname {erf}\left (\frac {\sqrt {{\left (5 \, x \log \relax (2)^{2} + 2 \, \log \relax (2)\right )}^{2}}}{\log \relax (2)}\right ) - 1\right )} \log \relax (2)^{2}}{\sqrt {{\left (5 \, x \log \relax (2)^{2} + 2 \, \log \relax (2)\right )}^{2}} \left (-\log \relax (2)^{2}\right )^{\frac {3}{2}}} + \frac {e^{\left (-\frac {{\left (5 \, x \log \relax (2)^{2} + 2 \, \log \relax (2)\right )}^{2}}{\log \relax (2)^{2}}\right )} \log \relax (2)^{2}}{\left (-\log \relax (2)^{2}\right )^{\frac {3}{2}}}\right )} e^{3} \log \relax (2)}{5 \, \sqrt {-\log \relax (2)^{2}}} + \frac {\sqrt {\pi } \operatorname {erf}\left (5 \, x \log \relax (2) + 2\right ) e^{3}}{10 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*x^2*log(2)^2-20*x*log(2)+1)*exp(log(x)-25*x^2*log(2)^2-20*x*log(2)-1)/x,x, algorithm="maxima")

[Out]

-1/5*((5*x*log(2)^2 + 2*log(2))^3*gamma(3/2, (5*x*log(2)^2 + 2*log(2))^2/log(2)^2)*log(2)^3/(((5*x*log(2)^2 +
2*log(2))^2)^(3/2)*(-log(2)^2)^(5/2)) - 4*sqrt(pi)*(5*x*log(2)^2 + 2*log(2))*(erf(sqrt((5*x*log(2)^2 + 2*log(2
))^2)/log(2)) - 1)*log(2)^3/(sqrt((5*x*log(2)^2 + 2*log(2))^2)*(-log(2)^2)^(5/2)) - 4*e^(-(5*x*log(2)^2 + 2*lo
g(2))^2/log(2)^2)*log(2)^3/(-log(2)^2)^(5/2))*e^3*log(2)^2/sqrt(-log(2)^2) + 2/5*(2*sqrt(pi)*(5*x*log(2)^2 + 2
*log(2))*(erf(sqrt((5*x*log(2)^2 + 2*log(2))^2)/log(2)) - 1)*log(2)^2/(sqrt((5*x*log(2)^2 + 2*log(2))^2)*(-log
(2)^2)^(3/2)) + e^(-(5*x*log(2)^2 + 2*log(2))^2/log(2)^2)*log(2)^2/(-log(2)^2)^(3/2))*e^3*log(2)/sqrt(-log(2)^
2) + 1/10*sqrt(pi)*erf(5*x*log(2) + 2)*e^3/log(2)

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mupad [B]  time = 1.01, size = 21, normalized size = 1.24 \begin {gather*} \frac {x\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{-25\,x^2\,{\ln \relax (2)}^2}}{2^{20\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(x) - 25*x^2*log(2)^2 - 20*x*log(2) - 1)*(50*x^2*log(2)^2 + 20*x*log(2) - 1))/x,x)

[Out]

(x*exp(-1)*exp(-25*x^2*log(2)^2))/2^(20*x)

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sympy [A]  time = 0.13, size = 22, normalized size = 1.29 \begin {gather*} x e^{- 25 x^{2} \log {\relax (2 )}^{2} - 20 x \log {\relax (2 )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-50*x**2*ln(2)**2-20*x*ln(2)+1)*exp(ln(x)-25*x**2*ln(2)**2-20*x*ln(2)-1)/x,x)

[Out]

x*exp(-25*x**2*log(2)**2 - 20*x*log(2) - 1)

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