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3.67.98 \(\int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} (-5 e^{3+x} x^2+50 e^{x^2} x^3)+(-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3) \log (\frac {1+e^{e^{3+x}} x}{x})+(2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3) \log ^2(\frac {1+e^{e^{3+x}} x}{x})}{25 x+25 e^{e^{3+x}} x^2+(-10 x-10 e^{e^{3+x}} x^2) \log (\frac {1+e^{e^{3+x}} x}{x})+(x+e^{e^{3+x}} x^2) \log ^2(\frac {1+e^{e^{3+x}} x}{x})} \, dx\)

Optimal. Leaf size=24 \[ e^{x^2}+\frac {5}{-5+\log \left (e^{e^{3+x}}+\frac {1}{x}\right )} \]

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Rubi [A]  time = 3.00, antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, integrand size = 215, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6688, 6742, 2209, 6686} \begin {gather*} e^{x^2}-\frac {5}{5-\log \left (e^{e^{x+3}}+\frac {1}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + 50*E^x^2*x^2 + E^E^(3 + x)*(-5*E^(3 + x)*x^2 + 50*E^x^2*x^3) + (-20*E^x^2*x^2 - 20*E^(E^(3 + x) + x^2
)*x^3)*Log[(1 + E^E^(3 + x)*x)/x] + (2*E^x^2*x^2 + 2*E^(E^(3 + x) + x^2)*x^3)*Log[(1 + E^E^(3 + x)*x)/x]^2)/(2
5*x + 25*E^E^(3 + x)*x^2 + (-10*x - 10*E^E^(3 + x)*x^2)*Log[(1 + E^E^(3 + x)*x)/x] + (x + E^E^(3 + x)*x^2)*Log
[(1 + E^E^(3 + x)*x)/x]^2),x]

[Out]

E^x^2 - 5/(5 - Log[E^E^(3 + x) + x^(-1)])

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5+50 e^{x^2} x^2-5 e^{3+e^{3+x}+x} x^2+50 e^{e^{3+x}+x^2} x^3-20 e^{x^2} x^2 \left (1+e^{e^{3+x}} x\right ) \log \left (e^{e^{3+x}}+\frac {1}{x}\right )+2 e^{x^2} x^2 \left (1+e^{e^{3+x}} x\right ) \log ^2\left (e^{e^{3+x}}+\frac {1}{x}\right )}{x \left (1+e^{e^{3+x}} x\right ) \left (5-\log \left (e^{e^{3+x}}+\frac {1}{x}\right )\right )^2} \, dx\\ &=\int \left (2 e^{x^2} x-\frac {5 \left (-1+e^{3+e^{3+x}+x} x^2\right )}{x \left (1+e^{e^{3+x}} x\right ) \left (-5+\log \left (e^{e^{3+x}}+\frac {1}{x}\right )\right )^2}\right ) \, dx\\ &=2 \int e^{x^2} x \, dx-5 \int \frac {-1+e^{3+e^{3+x}+x} x^2}{x \left (1+e^{e^{3+x}} x\right ) \left (-5+\log \left (e^{e^{3+x}}+\frac {1}{x}\right )\right )^2} \, dx\\ &=e^{x^2}-\frac {5}{5-\log \left (e^{e^{3+x}}+\frac {1}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 24, normalized size = 1.00 \begin {gather*} e^{x^2}+\frac {5}{-5+\log \left (e^{e^{3+x}}+\frac {1}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + 50*E^x^2*x^2 + E^E^(3 + x)*(-5*E^(3 + x)*x^2 + 50*E^x^2*x^3) + (-20*E^x^2*x^2 - 20*E^(E^(3 + x)
 + x^2)*x^3)*Log[(1 + E^E^(3 + x)*x)/x] + (2*E^x^2*x^2 + 2*E^(E^(3 + x) + x^2)*x^3)*Log[(1 + E^E^(3 + x)*x)/x]
^2)/(25*x + 25*E^E^(3 + x)*x^2 + (-10*x - 10*E^E^(3 + x)*x^2)*Log[(1 + E^E^(3 + x)*x)/x] + (x + E^E^(3 + x)*x^
2)*Log[(1 + E^E^(3 + x)*x)/x]^2),x]

[Out]

E^x^2 + 5/(-5 + Log[E^E^(3 + x) + x^(-1)])

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fricas [B]  time = 0.56, size = 72, normalized size = 3.00 \begin {gather*} \frac {e^{\left (x^{2}\right )} \log \left (\frac {{\left (x e^{\left (x^{2} + e^{\left (x + 3\right )}\right )} + e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}}{x}\right ) - 5 \, e^{\left (x^{2}\right )} + 5}{\log \left (\frac {{\left (x e^{\left (x^{2} + e^{\left (x + 3\right )}\right )} + e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}}{x}\right ) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(x^2)*exp(exp(3+x))+2*x^2*exp(x^2))*log((x*exp(exp(3+x))+1)/x)^2+(-20*x^3*exp(x^2)*exp(ex
p(3+x))-20*x^2*exp(x^2))*log((x*exp(exp(3+x))+1)/x)+(50*x^3*exp(x^2)-5*x^2*exp(3+x))*exp(exp(3+x))+50*x^2*exp(
x^2)+5)/((x^2*exp(exp(3+x))+x)*log((x*exp(exp(3+x))+1)/x)^2+(-10*x^2*exp(exp(3+x))-10*x)*log((x*exp(exp(3+x))+
1)/x)+25*x^2*exp(exp(3+x))+25*x),x, algorithm="fricas")

[Out]

(e^(x^2)*log((x*e^(x^2 + e^(x + 3)) + e^(x^2))*e^(-x^2)/x) - 5*e^(x^2) + 5)/(log((x*e^(x^2 + e^(x + 3)) + e^(x
^2))*e^(-x^2)/x) - 5)

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giac [B]  time = 0.50, size = 932, normalized size = 38.83 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(x^2)*exp(exp(3+x))+2*x^2*exp(x^2))*log((x*exp(exp(3+x))+1)/x)^2+(-20*x^3*exp(x^2)*exp(ex
p(3+x))-20*x^2*exp(x^2))*log((x*exp(exp(3+x))+1)/x)+(50*x^3*exp(x^2)-5*x^2*exp(3+x))*exp(exp(3+x))+50*x^2*exp(
x^2)+5)/((x^2*exp(exp(3+x))+x)*log((x*exp(exp(3+x))+1)/x)^2+(-10*x^2*exp(exp(3+x))-10*x)*log((x*exp(exp(3+x))+
1)/x)+25*x^2*exp(exp(3+x))+25*x),x, algorithm="giac")

[Out]

(x^4*e^(x^2 + 2*x + 2*e^(x + 3) + 6)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x)*log((x*e^(e^(x +
3)) + 1)/x) - 5*x^4*e^(x^2 + 2*x + 2*e^(x + 3) + 6)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x) -
5*x^4*e^(x^2 + 2*x + 2*e^(x + 3) + 6)*log((x*e^(e^(x + 3)) + 1)/x) + 5*x^4*e^(2*x + 2*e^(x + 3) + 6)*log((x*e^
(e^(x + 3)) + 1)/x) + 25*x^4*e^(x^2 + 2*x + 2*e^(x + 3) + 6) - 25*x^4*e^(2*x + 2*e^(x + 3) + 6) - 2*x^2*e^(x^2
 + x + e^(x + 3) + 3)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x)*log((x*e^(e^(x + 3)) + 1)/x) + 1
0*x^2*e^(x^2 + x + e^(x + 3) + 3)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x) - 5*x^2*e^(x + e^(x
+ 3) + 3)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x) + 10*x^2*e^(x^2 + x + e^(x + 3) + 3)*log((x*
e^(e^(x + 3)) + 1)/x) - 5*x^2*e^(x + e^(x + 3) + 3)*log((x*e^(e^(x + 3)) + 1)/x) - 50*x^2*e^(x^2 + x + e^(x +
3) + 3) + 50*x^2*e^(x + e^(x + 3) + 3) + e^(x^2)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x)*log((
x*e^(e^(x + 3)) + 1)/x) - 5*e^(x^2)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x) - 5*e^(x^2)*log((x
*e^(e^(x + 3)) + 1)/x) + 25*e^(x^2) + 5*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x) - 25)/(x^4*e^(
2*x + 2*e^(x + 3) + 6)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x)*log((x*e^(e^(x + 3)) + 1)/x) -
5*x^4*e^(2*x + 2*e^(x + 3) + 6)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x) - 5*x^4*e^(2*x + 2*e^(
x + 3) + 6)*log((x*e^(e^(x + 3)) + 1)/x) + 25*x^4*e^(2*x + 2*e^(x + 3) + 6) - 2*x^2*e^(x + e^(x + 3) + 3)*log(
(x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x)*log((x*e^(e^(x + 3)) + 1)/x) + 10*x^2*e^(x + e^(x + 3) + 3
)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x) + 10*x^2*e^(x + e^(x + 3) + 3)*log((x*e^(e^(x + 3))
+ 1)/x) - 50*x^2*e^(x + e^(x + 3) + 3) + log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x)*log((x*e^(e^(
x + 3)) + 1)/x) - 5*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x) - 5*log((x*e^(e^(x + 3)) + 1)/x) +
 25)

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maple [C]  time = 0.16, size = 152, normalized size = 6.33

method result size
risch \({\mathrm e}^{x^{2}}+\frac {10 i}{\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )}{x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )}{x}\right )^{3}-2 i \ln \relax (x )+2 i \ln \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )-10 i}\) \(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3*exp(x^2)*exp(exp(3+x))+2*x^2*exp(x^2))*ln((x*exp(exp(3+x))+1)/x)^2+(-20*x^3*exp(x^2)*exp(exp(3+x))
-20*x^2*exp(x^2))*ln((x*exp(exp(3+x))+1)/x)+(50*x^3*exp(x^2)-5*x^2*exp(3+x))*exp(exp(3+x))+50*x^2*exp(x^2)+5)/
((x^2*exp(exp(3+x))+x)*ln((x*exp(exp(3+x))+1)/x)^2+(-10*x^2*exp(exp(3+x))-10*x)*ln((x*exp(exp(3+x))+1)/x)+25*x
^2*exp(exp(3+x))+25*x),x,method=_RETURNVERBOSE)

[Out]

exp(x^2)+10*I/(Pi*csgn(I/x)*csgn(I*(x*exp(exp(3+x))+1))*csgn(I/x*(x*exp(exp(3+x))+1))-Pi*csgn(I/x)*csgn(I/x*(x
*exp(exp(3+x))+1))^2-Pi*csgn(I*(x*exp(exp(3+x))+1))*csgn(I/x*(x*exp(exp(3+x))+1))^2+Pi*csgn(I/x*(x*exp(exp(3+x
))+1))^3-2*I*ln(x)+2*I*ln(x*exp(exp(3+x))+1)-10*I)

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maxima [B]  time = 0.49, size = 47, normalized size = 1.96 \begin {gather*} -\frac {{\left (\log \relax (x) + 5\right )} e^{\left (x^{2}\right )} - e^{\left (x^{2}\right )} \log \left (x e^{\left (e^{\left (x + 3\right )}\right )} + 1\right ) - 5}{\log \left (x e^{\left (e^{\left (x + 3\right )}\right )} + 1\right ) - \log \relax (x) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(x^2)*exp(exp(3+x))+2*x^2*exp(x^2))*log((x*exp(exp(3+x))+1)/x)^2+(-20*x^3*exp(x^2)*exp(ex
p(3+x))-20*x^2*exp(x^2))*log((x*exp(exp(3+x))+1)/x)+(50*x^3*exp(x^2)-5*x^2*exp(3+x))*exp(exp(3+x))+50*x^2*exp(
x^2)+5)/((x^2*exp(exp(3+x))+x)*log((x*exp(exp(3+x))+1)/x)^2+(-10*x^2*exp(exp(3+x))-10*x)*log((x*exp(exp(3+x))+
1)/x)+25*x^2*exp(exp(3+x))+25*x),x, algorithm="maxima")

[Out]

-((log(x) + 5)*e^(x^2) - e^(x^2)*log(x*e^(e^(x + 3)) + 1) - 5)/(log(x*e^(e^(x + 3)) + 1) - log(x) - 5)

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mupad [B]  time = 4.43, size = 26, normalized size = 1.08 \begin {gather*} {\mathrm {e}}^{x^2}+\frac {5}{\ln \left (\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^3\,{\mathrm {e}}^x}+1}{x}\right )-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((x*exp(exp(x + 3)) + 1)/x)^2*(2*x^2*exp(x^2) + 2*x^3*exp(x^2)*exp(exp(x + 3))) - exp(exp(x + 3))*(5*x
^2*exp(x + 3) - 50*x^3*exp(x^2)) + 50*x^2*exp(x^2) - log((x*exp(exp(x + 3)) + 1)/x)*(20*x^2*exp(x^2) + 20*x^3*
exp(x^2)*exp(exp(x + 3))) + 5)/(25*x + 25*x^2*exp(exp(x + 3)) - log((x*exp(exp(x + 3)) + 1)/x)*(10*x + 10*x^2*
exp(exp(x + 3))) + log((x*exp(exp(x + 3)) + 1)/x)^2*(x + x^2*exp(exp(x + 3)))),x)

[Out]

exp(x^2) + 5/(log((x*exp(exp(3)*exp(x)) + 1)/x) - 5)

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sympy [A]  time = 1.01, size = 20, normalized size = 0.83 \begin {gather*} e^{x^{2}} + \frac {5}{\log {\left (\frac {x e^{e^{x + 3}} + 1}{x} \right )} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3*exp(x**2)*exp(exp(3+x))+2*x**2*exp(x**2))*ln((x*exp(exp(3+x))+1)/x)**2+(-20*x**3*exp(x**2)*
exp(exp(3+x))-20*x**2*exp(x**2))*ln((x*exp(exp(3+x))+1)/x)+(50*x**3*exp(x**2)-5*x**2*exp(3+x))*exp(exp(3+x))+5
0*x**2*exp(x**2)+5)/((x**2*exp(exp(3+x))+x)*ln((x*exp(exp(3+x))+1)/x)**2+(-10*x**2*exp(exp(3+x))-10*x)*ln((x*e
xp(exp(3+x))+1)/x)+25*x**2*exp(exp(3+x))+25*x),x)

[Out]

exp(x**2) + 5/(log((x*exp(exp(x + 3)) + 1)/x) - 5)

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