3.60.87 \(\int \frac {e^{\frac {2 (-e^5 x+x^2+(1-x) \log (4))}{\log (4)}} (-50 e^5 x+100 x^2+(25-50 x) \log (4))}{9 \log (4)} \, dx\)

Optimal. Leaf size=26 \[ \frac {25}{9} e^{2-2 x+\frac {2 x \left (-e^5+x\right )}{\log (4)}} x \]

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Rubi [B]  time = 0.26, antiderivative size = 53, normalized size of antiderivative = 2.04, number of steps used = 5, number of rules used = 4, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 6741, 2274, 2288} \begin {gather*} -\frac {25 e^{\frac {2 \left (x^2-x \left (e^5+\log (4)\right )\right )}{\log (4)}+2} \left (2 x^2-x \left (e^5+\log (4)\right )\right )}{9 \left (-2 x+e^5+\log (4)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((2*(-(E^5*x) + x^2 + (1 - x)*Log[4]))/Log[4])*(-50*E^5*x + 100*x^2 + (25 - 50*x)*Log[4]))/(9*Log[4]),x
]

[Out]

(-25*E^(2 + (2*(x^2 - x*(E^5 + Log[4])))/Log[4])*(2*x^2 - x*(E^5 + Log[4])))/(9*(E^5 - 2*x + Log[4]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int e^{\frac {2 \left (-e^5 x+x^2+(1-x) \log (4)\right )}{\log (4)}} \left (-50 e^5 x+100 x^2+(25-50 x) \log (4)\right ) \, dx}{9 \log (4)}\\ &=\frac {\int e^{\frac {2 \left (x^2+\log (4)-x \left (e^5+\log (4)\right )\right )}{\log (4)}} \left (100 x^2+25 \log (4)-50 x \left (e^5+\log (4)\right )\right ) \, dx}{9 \log (4)}\\ &=\frac {\int 4^{\frac {2}{\log (4)}} e^{\frac {2 \left (x^2-x \left (e^5+\log (4)\right )\right )}{\log (4)}} \left (100 x^2+25 \log (4)-50 x \left (e^5+\log (4)\right )\right ) \, dx}{9 \log (4)}\\ &=\frac {e^2 \int e^{\frac {2 \left (x^2-x \left (e^5+\log (4)\right )\right )}{\log (4)}} \left (100 x^2+25 \log (4)-50 x \left (e^5+\log (4)\right )\right ) \, dx}{9 \log (4)}\\ &=-\frac {25 e^{2+\frac {2 \left (x^2-x \left (e^5+\log (4)\right )\right )}{\log (4)}} \left (2 x^2-x \left (e^5+\log (4)\right )\right )}{9 \left (e^5-2 x+\log (4)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 52, normalized size = 2.00 \begin {gather*} \frac {25 e^{\frac {2 \left (-e^5 x+x^2+\log (4)-x \log (4)\right )}{\log (4)}} x \left (2 e^5-4 x+\log (16)\right )}{18 \left (e^5-2 x+\log (4)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*(-(E^5*x) + x^2 + (1 - x)*Log[4]))/Log[4])*(-50*E^5*x + 100*x^2 + (25 - 50*x)*Log[4]))/(9*Log
[4]),x]

[Out]

(25*E^((2*(-(E^5*x) + x^2 + Log[4] - x*Log[4]))/Log[4])*x*(2*E^5 - 4*x + Log[16]))/(18*(E^5 - 2*x + Log[4]))

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fricas [A]  time = 0.91, size = 25, normalized size = 0.96 \begin {gather*} \frac {25}{9} \, x e^{\left (\frac {x^{2} - x e^{5} - 2 \, {\left (x - 1\right )} \log \relax (2)}{\log \relax (2)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(2*(-50*x+25)*log(2)-50*x*exp(5)+100*x^2)*exp(1/2*(2*(-x+1)*log(2)-x*exp(5)+x^2)/log(2))^2/log(
2),x, algorithm="fricas")

[Out]

25/9*x*e^((x^2 - x*e^5 - 2*(x - 1)*log(2))/log(2))

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giac [C]  time = 0.20, size = 213, normalized size = 8.19 \begin {gather*} -\frac {25 \, {\left (-i \, \sqrt {\pi } {\left (2 \, e^{5} \log \relax (2) + e^{10}\right )} \operatorname {erf}\left (-\frac {i \, {\left (2 \, x - e^{5} - 2 \, \log \relax (2)\right )}}{2 \, \sqrt {\log \relax (2)}}\right ) e^{\left (-\frac {4 \, e^{5} \log \relax (2) + 4 \, \log \relax (2)^{2} + e^{10} - 8 \, \log \relax (2)}{4 \, \log \relax (2)}\right )} \sqrt {\log \relax (2)} + i \, \sqrt {\pi } {\left (e^{5} + 2 \, \log \relax (2)\right )} \operatorname {erf}\left (-\frac {i \, {\left (2 \, x - e^{5} - 2 \, \log \relax (2)\right )}}{2 \, \sqrt {\log \relax (2)}}\right ) e^{\left (-\frac {4 \, e^{5} \log \relax (2) + 4 \, \log \relax (2)^{2} + e^{10} - 28 \, \log \relax (2)}{4 \, \log \relax (2)}\right )} \sqrt {\log \relax (2)} - 2 \, {\left ({\left (2 \, x - e^{5} - 2 \, \log \relax (2)\right )} \log \relax (2) + 2 \, e^{5} \log \relax (2) + 2 \, \log \relax (2)^{2}\right )} e^{\left (\frac {x^{2} - x e^{5} - 2 \, x \log \relax (2) + 2 \, \log \relax (2)}{\log \relax (2)}\right )} + 2 \, e^{\left (\frac {x^{2} - x e^{5} - 2 \, x \log \relax (2) + 7 \, \log \relax (2)}{\log \relax (2)}\right )} \log \relax (2)\right )}}{36 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(2*(-50*x+25)*log(2)-50*x*exp(5)+100*x^2)*exp(1/2*(2*(-x+1)*log(2)-x*exp(5)+x^2)/log(2))^2/log(
2),x, algorithm="giac")

[Out]

-25/36*(-I*sqrt(pi)*(2*e^5*log(2) + e^10)*erf(-1/2*I*(2*x - e^5 - 2*log(2))/sqrt(log(2)))*e^(-1/4*(4*e^5*log(2
) + 4*log(2)^2 + e^10 - 8*log(2))/log(2))*sqrt(log(2)) + I*sqrt(pi)*(e^5 + 2*log(2))*erf(-1/2*I*(2*x - e^5 - 2
*log(2))/sqrt(log(2)))*e^(-1/4*(4*e^5*log(2) + 4*log(2)^2 + e^10 - 28*log(2))/log(2))*sqrt(log(2)) - 2*((2*x -
e^5 - 2*log(2))*log(2) + 2*e^5*log(2) + 2*log(2)^2)*e^((x^2 - x*e^5 - 2*x*log(2) + 2*log(2))/log(2)) + 2*e^((
x^2 - x*e^5 - 2*x*log(2) + 7*log(2))/log(2))*log(2))/log(2)

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maple [A]  time = 0.09, size = 30, normalized size = 1.15

method	result	size
risch	\(\frac {25 x \,{\mathrm e}^{-\frac {x \,{\mathrm e}^{5}+2 x \ln \relax (2)-x^{2}-2 \ln \relax (2)}{\ln \relax (2)}}}{9}\)	\(30\)
norman	\(\frac {25 x \,{\mathrm e}^{\frac {2 \left (1-x \right ) \ln \relax (2)-x \,{\mathrm e}^{5}+x^{2}}{\ln \relax (2)}}}{9}\)	\(31\)
gosper	\(\frac {25 \,{\mathrm e}^{-\frac {x \ln \relax (4)-2 \ln \relax (2)+x \,{\mathrm e}^{5}-x^{2}}{\ln \relax (2)}} x}{9}\)	\(32\)
default	\(\frac {25 \ln \relax (2) x \,{\mathrm e}^{2+\frac {x^{2}}{\ln \relax (2)}+\left (-2-\frac {{\mathrm e}^{5}}{\ln \relax (2)}\right ) x}-\frac {25 i \ln \relax (2)^{\frac {3}{2}} \sqrt {\pi }\, {\mathrm e}^{2-\frac {\left (-2-\frac {{\mathrm e}^{5}}{\ln \relax (2)}\right )^{2} \ln \relax (2)}{4}} \erf \left (\frac {i x}{\sqrt {\ln \relax (2)}}+\frac {i \left (-2-\frac {{\mathrm e}^{5}}{\ln \relax (2)}\right ) \sqrt {\ln \relax (2)}}{2}\right ) {\mathrm e}^{5}}{2}+\frac {25 \ln \relax (2) {\mathrm e}^{5} {\mathrm e}^{2+\frac {x^{2}}{\ln \relax (2)}+\left (-2-\frac {{\mathrm e}^{5}}{\ln \relax (2)}\right ) x}}{2}-\frac {25 i {\mathrm e}^{10} \sqrt {\ln \relax (2)}\, \sqrt {\pi }\, {\mathrm e}^{2-\frac {\left (-2-\frac {{\mathrm e}^{5}}{\ln \relax (2)}\right )^{2} \ln \relax (2)}{4}} \erf \left (\frac {i x}{\sqrt {\ln \relax (2)}}+\frac {i \left (-2-\frac {{\mathrm e}^{5}}{\ln \relax (2)}\right ) \sqrt {\ln \relax (2)}}{2}\right )}{4}-\frac {25 \ln \relax (2) {\mathrm e}^{7+\frac {x^{2}}{\ln \relax (2)}+\left (-2-\frac {{\mathrm e}^{5}}{\ln \relax (2)}\right ) x}}{2}+\frac {25 i \ln \relax (2)^{\frac {3}{2}} \sqrt {\pi }\, {\mathrm e}^{7-\frac {\left (-2-\frac {{\mathrm e}^{5}}{\ln \relax (2)}\right )^{2} \ln \relax (2)}{4}} \erf \left (\frac {i x}{\sqrt {\ln \relax (2)}}+\frac {i \left (-2-\frac {{\mathrm e}^{5}}{\ln \relax (2)}\right ) \sqrt {\ln \relax (2)}}{2}\right )}{2}+\frac {25 i \sqrt {\ln \relax (2)}\, \sqrt {\pi }\, {\mathrm e}^{7-\frac {\left (-2-\frac {{\mathrm e}^{5}}{\ln \relax (2)}\right )^{2} \ln \relax (2)}{4}} \erf \left (\frac {i x}{\sqrt {\ln \relax (2)}}+\frac {i \left (-2-\frac {{\mathrm e}^{5}}{\ln \relax (2)}\right ) \sqrt {\ln \relax (2)}}{2}\right ) {\mathrm e}^{5}}{4}}{9 \ln \relax (2)}\)	\(324\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/18*(2*(-50*x+25)*ln(2)-50*x*exp(5)+100*x^2)*exp(1/2*(2*(1-x)*ln(2)-x*exp(5)+x^2)/ln(2))^2/ln(2),x,method
=_RETURNVERBOSE)

[Out]

25/9*x*exp(-(x*exp(5)+2*x*ln(2)-x^2-2*ln(2))/ln(2))

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maxima [C]  time = 0.54, size = 639, normalized size = 24.58 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(2*(-50*x+25)*log(2)-50*x*exp(5)+100*x^2)*exp(1/2*(2*(-x+1)*log(2)-x*exp(5)+x^2)/log(2))^2/log(
2),x, algorithm="maxima")

[Out]

-25/36*(2*(sqrt(pi)*(2*x/log(2) - (e^5 + 2*log(2))/log(2))*(erf(1/2*sqrt(-(2*x/log(2) - (e^5 + 2*log(2))/log(2
))^2*log(2))) - 1)*(e^5 + 2*log(2))*sqrt(log(2))/sqrt(-(2*x/log(2) - (e^5 + 2*log(2))/log(2))^2*log(2)) + 2*2^
(1/4*(2*x/log(2) - (e^5 + 2*log(2))/log(2))^2)*sqrt(log(2)))*e^(-1/4*(e^5 + 2*log(2))^2/log(2) + 2)*log(2)^(3/
2) - 2*sqrt(pi)*erf(x*sqrt(-1/log(2)) + 1/2*(e^5 + 2*log(2))/(sqrt(-1/log(2))*log(2)))*e^(-1/4*(e^5 + 2*log(2)
)^2/log(2) + 2)*log(2)/sqrt(-1/log(2)) + (sqrt(pi)*(2*x/log(2) - (e^5 + 2*log(2))/log(2))*(erf(1/2*sqrt(-(2*x/
log(2) - (e^5 + 2*log(2))/log(2))^2*log(2))) - 1)*(e^5 + 2*log(2))*sqrt(log(2))/sqrt(-(2*x/log(2) - (e^5 + 2*l
og(2))/log(2))^2*log(2)) + 2*2^(1/4*(2*x/log(2) - (e^5 + 2*log(2))/log(2))^2)*sqrt(log(2)))*e^(-1/4*(e^5 + 2*l
og(2))^2/log(2) + 7)*sqrt(log(2)) + (4*(2*x/log(2) - (e^5 + 2*log(2))/log(2))^3*gamma(3/2, -1/4*(2*x/log(2) -
(e^5 + 2*log(2))/log(2))^2*log(2))*log(2)^(5/2)/(-(2*x/log(2) - (e^5 + 2*log(2))/log(2))^2*log(2))^(3/2) - sqr
t(pi)*(2*x/log(2) - (e^5 + 2*log(2))/log(2))*(erf(1/2*sqrt(-(2*x/log(2) - (e^5 + 2*log(2))/log(2))^2*log(2)))
- 1)*(e^5 + 2*log(2))^2*sqrt(log(2))/sqrt(-(2*x/log(2) - (e^5 + 2*log(2))/log(2))^2*log(2)) - 4*2^(1/4*(2*x/lo
g(2) - (e^5 + 2*log(2))/log(2))^2)*(e^5 + 2*log(2))*sqrt(log(2)))*e^(-1/4*(e^5 + 2*log(2))^2/log(2) + 2)*sqrt(
log(2)))/log(2)

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mupad [B]  time = 0.34, size = 28, normalized size = 1.08 \begin {gather*} \frac {25\,x\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^2\,{\mathrm {e}}^{\frac {x^2}{\ln \relax (2)}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^5}{\ln \relax (2)}}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(2*((x*exp(5))/2 + log(2)*(x - 1) - x^2/2))/log(2))*(2*log(2)*(50*x - 25) + 50*x*exp(5) - 100*x^2))
/(18*log(2)),x)

[Out]

(25*x*exp(-2*x)*exp(2)*exp(x^2/log(2))*exp(-(x*exp(5))/log(2)))/9

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sympy [A]  time = 0.16, size = 26, normalized size = 1.00 \begin {gather*} \frac {25 x e^{\frac {2 \left (\frac {x^{2}}{2} - \frac {x e^{5}}{2} + \frac {\left (2 - 2 x\right ) \log {\relax (2 )}}{2}\right )}{\log {\relax (2 )}}}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*(2*(-50*x+25)*ln(2)-50*x*exp(5)+100*x**2)*exp(1/2*(2*(-x+1)*ln(2)-x*exp(5)+x**2)/ln(2))**2/ln(2
),x)

[Out]

25*x*exp(2*(x**2/2 - x*exp(5)/2 + (2 - 2*x)*log(2)/2)/log(2))/9

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