3.45.42 \(\int e^{\frac {127-250 x+125 x^2+5 \log (2)+(25-50 x+25 x^2+\log (2)) \log (3) \log ^2(4)}{5+\log (3) \log ^2(4)}} (-50+50 x) \, dx\)

Optimal. Leaf size=25 \[ 2 e^{25 (-1+x)^2+\frac {2}{5+\log (3) \log ^2(4)}} \]

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Rubi [A]  time = 0.33, antiderivative size = 38, normalized size of antiderivative = 1.52, number of steps used = 2, number of rules used = 2, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {2244, 2236} \begin {gather*} \exp \left (25 x^2-50 x+\frac {127+(25+\log (2)) \log (3) \log ^2(4)+\log (32)}{5+\log (3) \log ^2(4)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^((127 - 250*x + 125*x^2 + 5*Log[2] + (25 - 50*x + 25*x^2 + Log[2])*Log[3]*Log[4]^2)/(5 + Log[3]*Log[4]^2
))*(-50 + 50*x),x]

[Out]

E^(-50*x + 25*x^2 + (127 + (25 + Log[2])*Log[3]*Log[4]^2 + Log[32])/(5 + Log[3]*Log[4]^2))

Rule 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2244

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \exp \left (-50 x+25 x^2+\frac {127+(25+\log (2)) \log (3) \log ^2(4)+\log (32)}{5+\log (3) \log ^2(4)}\right ) (-50+50 x) \, dx\\ &=\exp \left (-50 x+25 x^2+\frac {127+(25+\log (2)) \log (3) \log ^2(4)+\log (32)}{5+\log (3) \log ^2(4)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.40, size = 38, normalized size = 1.52 \begin {gather*} e^{-50 x+25 x^2+\frac {127+(25+\log (2)) \log (3) \log ^2(4)+\log (32)}{5+\log (3) \log ^2(4)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^((127 - 250*x + 125*x^2 + 5*Log[2] + (25 - 50*x + 25*x^2 + Log[2])*Log[3]*Log[4]^2)/(5 + Log[3]*Lo
g[4]^2))*(-50 + 50*x),x]

[Out]

E^(-50*x + 25*x^2 + (127 + (25 + Log[2])*Log[3]*Log[4]^2 + Log[32])/(5 + Log[3]*Log[4]^2))

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fricas [B]  time = 0.50, size = 51, normalized size = 2.04 \begin {gather*} e^{\left (\frac {125 \, x^{2} + 4 \, {\left (25 \, {\left (x^{2} - 2 \, x + 1\right )} \log \relax (2)^{2} + \log \relax (2)^{3}\right )} \log \relax (3) - 250 \, x + 5 \, \log \relax (2) + 127}{4 \, \log \relax (3) \log \relax (2)^{2} + 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*x-50)*exp((4*(log(2)+25*x^2-50*x+25)*log(3)*log(2)^2+5*log(2)+125*x^2-250*x+127)/(4*log(2)^2*log
(3)+5)),x, algorithm="fricas")

[Out]

e^((125*x^2 + 4*(25*(x^2 - 2*x + 1)*log(2)^2 + log(2)^3)*log(3) - 250*x + 5*log(2) + 127)/(4*log(3)*log(2)^2 +
5))

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giac [B]  time = 0.17, size = 148, normalized size = 5.92 \begin {gather*} e^{\left (\frac {100 \, x^{2} \log \relax (3) \log \relax (2)^{2}}{4 \, \log \relax (3) \log \relax (2)^{2} + 5} - \frac {200 \, x \log \relax (3) \log \relax (2)^{2}}{4 \, \log \relax (3) \log \relax (2)^{2} + 5} + \frac {4 \, \log \relax (3) \log \relax (2)^{3}}{4 \, \log \relax (3) \log \relax (2)^{2} + 5} + \frac {100 \, \log \relax (3) \log \relax (2)^{2}}{4 \, \log \relax (3) \log \relax (2)^{2} + 5} + \frac {125 \, x^{2}}{4 \, \log \relax (3) \log \relax (2)^{2} + 5} - \frac {250 \, x}{4 \, \log \relax (3) \log \relax (2)^{2} + 5} + \frac {5 \, \log \relax (2)}{4 \, \log \relax (3) \log \relax (2)^{2} + 5} + \frac {127}{4 \, \log \relax (3) \log \relax (2)^{2} + 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*x-50)*exp((4*(log(2)+25*x^2-50*x+25)*log(3)*log(2)^2+5*log(2)+125*x^2-250*x+127)/(4*log(2)^2*log
(3)+5)),x, algorithm="giac")

[Out]

e^(100*x^2*log(3)*log(2)^2/(4*log(3)*log(2)^2 + 5) - 200*x*log(3)*log(2)^2/(4*log(3)*log(2)^2 + 5) + 4*log(3)*
log(2)^3/(4*log(3)*log(2)^2 + 5) + 100*log(3)*log(2)^2/(4*log(3)*log(2)^2 + 5) + 125*x^2/(4*log(3)*log(2)^2 +
5) - 250*x/(4*log(3)*log(2)^2 + 5) + 5*log(2)/(4*log(3)*log(2)^2 + 5) + 127/(4*log(3)*log(2)^2 + 5))

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maple [A]  time = 0.13, size = 49, normalized size = 1.96

method	result	size
norman	\({\mathrm e}^{\frac {4 \left (\ln \relax (2)+25 x^{2}-50 x +25\right ) \ln \relax (3) \ln \relax (2)^{2}+5 \ln \relax (2)+125 x^{2}-250 x +127}{4 \ln \relax (2)^{2} \ln \relax (3)+5}}\)	\(49\)
gosper	\({\mathrm e}^{\frac {100 x^{2} \ln \relax (3) \ln \relax (2)^{2}+4 \ln \relax (3) \ln \relax (2)^{3}-200 \ln \relax (2)^{2} \ln \relax (3) x +100 \ln \relax (2)^{2} \ln \relax (3)+125 x^{2}+5 \ln \relax (2)-250 x +127}{4 \ln \relax (2)^{2} \ln \relax (3)+5}}\)	\(65\)
risch	\({\mathrm e}^{\frac {100 x^{2} \ln \relax (3) \ln \relax (2)^{2}+4 \ln \relax (3) \ln \relax (2)^{3}-200 \ln \relax (2)^{2} \ln \relax (3) x +100 \ln \relax (2)^{2} \ln \relax (3)+125 x^{2}+5 \ln \relax (2)-250 x +127}{4 \ln \relax (2)^{2} \ln \relax (3)+5}}\)	\(65\)
default	\(\frac {25 \left (4 \ln \relax (2)^{2} \ln \relax (3)+5\right ) {\mathrm e}^{\frac {\left (100 \ln \relax (2)^{2} \ln \relax (3)+125\right ) x^{2}}{4 \ln \relax (2)^{2} \ln \relax (3)+5}+\frac {\left (-200 \ln \relax (2)^{2} \ln \relax (3)-250\right ) x}{4 \ln \relax (2)^{2} \ln \relax (3)+5}+\frac {4 \left (\ln \relax (2)+25\right ) \ln \relax (3) \ln \relax (2)^{2}+5 \ln \relax (2)+127}{4 \ln \relax (2)^{2} \ln \relax (3)+5}}}{100 \ln \relax (2)^{2} \ln \relax (3)+125}+\frac {25 i \left (-200 \ln \relax (2)^{2} \ln \relax (3)-250\right ) \sqrt {\pi }\, {\mathrm e}^{\frac {4 \left (\ln \relax (2)+25\right ) \ln \relax (3) \ln \relax (2)^{2}+5 \ln \relax (2)+127}{4 \ln \relax (2)^{2} \ln \relax (3)+5}-\frac {\left (-200 \ln \relax (2)^{2} \ln \relax (3)-250\right )^{2}}{4 \left (4 \ln \relax (2)^{2} \ln \relax (3)+5\right ) \left (100 \ln \relax (2)^{2} \ln \relax (3)+125\right )}} \erf \left (i \sqrt {\frac {100 \ln \relax (2)^{2} \ln \relax (3)+125}{4 \ln \relax (2)^{2} \ln \relax (3)+5}}\, x +\frac {i \left (-200 \ln \relax (2)^{2} \ln \relax (3)-250\right )}{2 \left (4 \ln \relax (2)^{2} \ln \relax (3)+5\right ) \sqrt {\frac {100 \ln \relax (2)^{2} \ln \relax (3)+125}{4 \ln \relax (2)^{2} \ln \relax (3)+5}}}\right )}{2 \left (100 \ln \relax (2)^{2} \ln \relax (3)+125\right ) \sqrt {\frac {100 \ln \relax (2)^{2} \ln \relax (3)+125}{4 \ln \relax (2)^{2} \ln \relax (3)+5}}}+\frac {25 i \sqrt {\pi }\, {\mathrm e}^{\frac {4 \left (\ln \relax (2)+25\right ) \ln \relax (3) \ln \relax (2)^{2}+5 \ln \relax (2)+127}{4 \ln \relax (2)^{2} \ln \relax (3)+5}-\frac {\left (-200 \ln \relax (2)^{2} \ln \relax (3)-250\right )^{2}}{4 \left (4 \ln \relax (2)^{2} \ln \relax (3)+5\right ) \left (100 \ln \relax (2)^{2} \ln \relax (3)+125\right )}} \erf \left (i \sqrt {\frac {100 \ln \relax (2)^{2} \ln \relax (3)+125}{4 \ln \relax (2)^{2} \ln \relax (3)+5}}\, x +\frac {i \left (-200 \ln \relax (2)^{2} \ln \relax (3)-250\right )}{2 \left (4 \ln \relax (2)^{2} \ln \relax (3)+5\right ) \sqrt {\frac {100 \ln \relax (2)^{2} \ln \relax (3)+125}{4 \ln \relax (2)^{2} \ln \relax (3)+5}}}\right )}{\sqrt {\frac {100 \ln \relax (2)^{2} \ln \relax (3)+125}{4 \ln \relax (2)^{2} \ln \relax (3)+5}}}\)	\(497\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*x-50)*exp((4*(ln(2)+25*x^2-50*x+25)*ln(3)*ln(2)^2+5*ln(2)+125*x^2-250*x+127)/(4*ln(2)^2*ln(3)+5)),x,me
thod=_RETURNVERBOSE)

[Out]

exp((4*(ln(2)+25*x^2-50*x+25)*ln(3)*ln(2)^2+5*ln(2)+125*x^2-250*x+127)/(4*ln(2)^2*ln(3)+5))

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maxima [C]  time = 0.56, size = 200, normalized size = 8.00 \begin {gather*} 10 i \, \sqrt {\pi } 3^{\frac {4 \, \log \relax (2)^{3}}{4 \, \log \relax (3) \log \relax (2)^{2} + 5} + \frac {100 \, \log \relax (2)^{2}}{4 \, \log \relax (3) \log \relax (2)^{2} + 5}} 2^{\frac {5}{4 \, \log \relax (3) \log \relax (2)^{2} + 5} - 1} \operatorname {erf}\left (5 i \, x - 5 i\right ) e^{\left (\frac {127}{4 \, \log \relax (3) \log \relax (2)^{2} + 5} - 25\right )} + {\left (\frac {5 \, \sqrt {\pi } {\left (x - 1\right )} {\left (\operatorname {erf}\left (5 \, \sqrt {-{\left (x - 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (x - 1\right )}^{2}}} + e^{\left (25 \, {\left (x - 1\right )}^{2}\right )}\right )} e^{\left (\frac {4 \, \log \relax (3) \log \relax (2)^{3}}{4 \, \log \relax (3) \log \relax (2)^{2} + 5} + \frac {100 \, \log \relax (3) \log \relax (2)^{2}}{4 \, \log \relax (3) \log \relax (2)^{2} + 5} + \frac {5 \, \log \relax (2)}{4 \, \log \relax (3) \log \relax (2)^{2} + 5} + \frac {127}{4 \, \log \relax (3) \log \relax (2)^{2} + 5} - 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*x-50)*exp((4*(log(2)+25*x^2-50*x+25)*log(3)*log(2)^2+5*log(2)+125*x^2-250*x+127)/(4*log(2)^2*log
(3)+5)),x, algorithm="maxima")

[Out]

10*I*sqrt(pi)*3^(4*log(2)^3/(4*log(3)*log(2)^2 + 5) + 100*log(2)^2/(4*log(3)*log(2)^2 + 5))*2^(5/(4*log(3)*log
(2)^2 + 5) - 1)*erf(5*I*x - 5*I)*e^(127/(4*log(3)*log(2)^2 + 5) - 25) + (5*sqrt(pi)*(x - 1)*(erf(5*sqrt(-(x -
1)^2)) - 1)/sqrt(-(x - 1)^2) + e^(25*(x - 1)^2))*e^(4*log(3)*log(2)^3/(4*log(3)*log(2)^2 + 5) + 100*log(3)*log
(2)^2/(4*log(3)*log(2)^2 + 5) + 5*log(2)/(4*log(3)*log(2)^2 + 5) + 127/(4*log(3)*log(2)^2 + 5) - 25)

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mupad [B]  time = 0.33, size = 106, normalized size = 4.24 \begin {gather*} {32}^{\frac {1}{4\,{\ln \relax (2)}^2\,\ln \relax (3)+5}}\,{81}^{\frac {25\,{\ln \relax (2)}^2\,x^2-50\,{\ln \relax (2)}^2\,x+25\,{\ln \relax (2)}^2+{\ln \relax (2)}^3}{4\,{\ln \relax (2)}^2\,\ln \relax (3)+5}}\,{\mathrm {e}}^{\frac {127}{4\,{\ln \relax (2)}^2\,\ln \relax (3)+5}}\,{\mathrm {e}}^{-\frac {250\,x}{4\,{\ln \relax (2)}^2\,\ln \relax (3)+5}}\,{\mathrm {e}}^{\frac {125\,x^2}{4\,{\ln \relax (2)}^2\,\ln \relax (3)+5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp((5*log(2) - 250*x + 125*x^2 + 4*log(2)^2*log(3)*(log(2) - 50*x + 25*x^2 + 25) + 127)/(4*log(2)^2*log(3
) + 5))*(50*x - 50),x)

[Out]

32^(1/(4*log(2)^2*log(3) + 5))*81^((25*x^2*log(2)^2 - 50*x*log(2)^2 + 25*log(2)^2 + log(2)^3)/(4*log(2)^2*log(
3) + 5))*exp(127/(4*log(2)^2*log(3) + 5))*exp(-(250*x)/(4*log(2)^2*log(3) + 5))*exp((125*x^2)/(4*log(2)^2*log(
3) + 5))

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sympy [B]  time = 0.15, size = 51, normalized size = 2.04 \begin {gather*} e^{\frac {125 x^{2} - 250 x + \left (100 x^{2} - 200 x + 4 \log {\relax (2 )} + 100\right ) \log {\relax (2 )}^{2} \log {\relax (3 )} + 5 \log {\relax (2 )} + 127}{4 \log {\relax (2 )}^{2} \log {\relax (3 )} + 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((50*x-50)*exp((4*(ln(2)+25*x**2-50*x+25)*ln(3)*ln(2)**2+5*ln(2)+125*x**2-250*x+127)/(4*ln(2)**2*ln(3
)+5)),x)

[Out]

exp((125*x**2 - 250*x + (100*x**2 - 200*x + 4*log(2) + 100)*log(2)**2*log(3) + 5*log(2) + 127)/(4*log(2)**2*lo
g(3) + 5))

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