3.4.45 \(\int e^{e^{5+e^x+x}} (e^{5+e^x+x} (e^{\frac {100-4 x^2 \log ^2(5)}{\log ^2(5)}} x+e^{x+\frac {100-4 x^2 \log ^2(5)}{\log ^2(5)}} x)+e^{\frac {100-4 x^2 \log ^2(5)}{\log ^2(5)}} (1-8 x^2)) \, dx\)

Optimal. Leaf size=27 \[ e^{e^{5+e^x+x}+4 \left (-x^2+\frac {25}{\log ^2(5)}\right )} x \]

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Rubi [B]  time = 0.08, antiderivative size = 63, normalized size of antiderivative = 2.33, number of steps used = 1, number of rules used = 1, integrand size = 90, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {2288} \begin {gather*} \frac {e^{e^{x+e^x+5}} \left (x e^{\frac {4 \left (25-x^2 \log ^2(5)\right )}{\log ^2(5)}}+x e^{\frac {4 \left (25-x^2 \log ^2(5)\right )}{\log ^2(5)}+x}\right )}{e^x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^E^(5 + E^x + x)*(E^(5 + E^x + x)*(E^((100 - 4*x^2*Log[5]^2)/Log[5]^2)*x + E^(x + (100 - 4*x^2*Log[5]^2)/
Log[5]^2)*x) + E^((100 - 4*x^2*Log[5]^2)/Log[5]^2)*(1 - 8*x^2)),x]

[Out]

(E^E^(5 + E^x + x)*(E^((4*(25 - x^2*Log[5]^2))/Log[5]^2)*x + E^(x + (4*(25 - x^2*Log[5]^2))/Log[5]^2)*x))/(1 +
E^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 {e^{e^{5+e^x+x}} \left (e^{\frac {4 \left (25-x^2 \log ^2(5)\right )}{\log ^2(5)}} x+e^{x+\frac {4 \left (25-x^2 \log ^2(5)\right )}{\log ^2(5)}} x\right )}{1+e^x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 24, normalized size = 0.89 \begin {gather*} e^{e^{5+e^x+x}-4 x^2+\frac {100}{\log ^2(5)}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^E^(5 + E^x + x)*(E^(5 + E^x + x)*(E^((100 - 4*x^2*Log[5]^2)/Log[5]^2)*x + E^(x + (100 - 4*x^2*Log[
5]^2)/Log[5]^2)*x) + E^((100 - 4*x^2*Log[5]^2)/Log[5]^2)*(1 - 8*x^2)),x]

[Out]

E^(E^(5 + E^x + x) - 4*x^2 + 100/Log[5]^2)*x

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fricas [A]  time = 0.59, size = 35, normalized size = 1.30 \begin {gather*} x e^{\left (-x - \frac {{\left (4 \, x^{2} - x\right )} \log \relax (5)^{2} - 100}{\log \relax (5)^{2}} + e^{\left (x + e^{x} + 5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp((-4*x^2*log(5)^2+100)/log(5)^2)*exp(x)+x*exp((-4*x^2*log(5)^2+100)/log(5)^2))*exp(exp(x)+5+x
)+(-8*x^2+1)*exp((-4*x^2*log(5)^2+100)/log(5)^2))*exp(exp(exp(x)+5+x)),x, algorithm="fricas")

[Out]

x*e^(-x - ((4*x^2 - x)*log(5)^2 - 100)/log(5)^2 + e^(x + e^x + 5))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp((-4*x^2*log(5)^2+100)/log(5)^2)*exp(x)+x*exp((-4*x^2*log(5)^2+100)/log(5)^2))*exp(exp(x)+5+x
)+(-8*x^2+1)*exp((-4*x^2*log(5)^2+100)/log(5)^2))*exp(exp(exp(x)+5+x)),x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.08, size = 33, normalized size = 1.22

method	result	size
risch	\(x \,{\mathrm e}^{-\frac {4 x^{2} \ln \relax (5)^{2}-{\mathrm e}^{{\mathrm e}^{x}+5+x} \ln \relax (5)^{2}-100}{\ln \relax (5)^{2}}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp((-4*x^2*ln(5)^2+100)/ln(5)^2)*exp(x)+x*exp((-4*x^2*ln(5)^2+100)/ln(5)^2))*exp(exp(x)+5+x)+(-8*x^2+
1)*exp((-4*x^2*ln(5)^2+100)/ln(5)^2))*exp(exp(exp(x)+5+x)),x,method=_RETURNVERBOSE)

[Out]

x*exp(-(4*x^2*ln(5)^2-exp(exp(x)+5+x)*ln(5)^2-100)/ln(5)^2)

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maxima [A]  time = 0.82, size = 21, normalized size = 0.78 \begin {gather*} x e^{\left (-4 \, x^{2} + \frac {100}{\log \relax (5)^{2}} + e^{\left (x + e^{x} + 5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp((-4*x^2*log(5)^2+100)/log(5)^2)*exp(x)+x*exp((-4*x^2*log(5)^2+100)/log(5)^2))*exp(exp(x)+5+x
)+(-8*x^2+1)*exp((-4*x^2*log(5)^2+100)/log(5)^2))*exp(exp(exp(x)+5+x)),x, algorithm="maxima")

[Out]

x*e^(-4*x^2 + 100/log(5)^2 + e^(x + e^x + 5))

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mupad [B]  time = 0.50, size = 24, normalized size = 0.89 \begin {gather*} x\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^5\,{\mathrm {e}}^x}\,{\mathrm {e}}^{\frac {100}{{\ln \relax (5)}^2}}\,{\mathrm {e}}^{-4\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(exp(x + exp(x) + 5))*(exp(x + exp(x) + 5)*(x*exp(-(4*x^2*log(5)^2 - 100)/log(5)^2) + x*exp(x)*exp(-(4*
x^2*log(5)^2 - 100)/log(5)^2)) - exp(-(4*x^2*log(5)^2 - 100)/log(5)^2)*(8*x^2 - 1)),x)

[Out]

x*exp(exp(exp(x))*exp(5)*exp(x))*exp(100/log(5)^2)*exp(-4*x^2)

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sympy [A]  time = 22.08, size = 29, normalized size = 1.07 \begin {gather*} x e^{\frac {- 4 x^{2} \log {\relax (5 )}^{2} + 100}{\log {\relax (5 )}^{2}}} e^{e^{x + e^{x} + 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp((-4*x**2*ln(5)**2+100)/ln(5)**2)*exp(x)+x*exp((-4*x**2*ln(5)**2+100)/ln(5)**2))*exp(exp(x)+5
+x)+(-8*x**2+1)*exp((-4*x**2*ln(5)**2+100)/ln(5)**2))*exp(exp(exp(x)+5+x)),x)

[Out]

x*exp((-4*x**2*log(5)**2 + 100)/log(5)**2)*exp(exp(x + exp(x) + 5))

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