\(\int \frac {e^{-x^2} (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+(-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3) \log (19+e^{4 x}+x-2 e^{2 x} x+x^2))}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx\) [507]

Optimal result

Integrand size = 105, antiderivative size = 23 \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx=e^{-x^2} \log \left (19+x+\left (-e^{2 x}+x\right )^2\right ) \]

[Out]

ln(19+x+(x-exp(2*x))^2)/exp(x^2)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(23)=46\).

Time = 0.16 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.43, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {2326} \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx=\frac {e^{-x^2} \left (x^3-2 e^{2 x} x^2+x^2+e^{4 x} x+19 x\right ) \log \left (x^2-2 e^{2 x} x+x+e^{4 x}+19\right )}{x \left (x^2-2 e^{2 x} x+x+e^{4 x}+19\right )} \]

[In]

Int[(1 + 4*E^(4*x) + E^(2*x)*(-2 - 4*x) + 2*x + (-38*x - 2*E^(4*x)*x - 2*x^2 + 4*E^(2*x)*x^2 - 2*x^3)*Log[19 +
E^(4*x) + x - 2*E^(2*x)*x + x^2])/(E^x^2*(19 + E^(4*x) + x - 2*E^(2*x)*x + x^2)),x]

[Out]

((19*x + E^(4*x)*x + x^2 - 2*E^(2*x)*x^2 + x^3)*Log[19 + E^(4*x) + x - 2*E^(2*x)*x + x^2])/(E^x^2*x*(19 + E^(4
*x) + x - 2*E^(2*x)*x + x^2))

Rule 2326

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{align*} \text {integral}& = \frac {e^{-x^2} \left (19 x+e^{4 x} x+x^2-2 e^{2 x} x^2+x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )}{x \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx=e^{-x^2} \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right ) \]

[In]

Integrate[(1 + 4*E^(4*x) + E^(2*x)*(-2 - 4*x) + 2*x + (-38*x - 2*E^(4*x)*x - 2*x^2 + 4*E^(2*x)*x^2 - 2*x^3)*Lo
g[19 + E^(4*x) + x - 2*E^(2*x)*x + x^2])/(E^x^2*(19 + E^(4*x) + x - 2*E^(2*x)*x + x^2)),x]

[Out]

Log[19 + E^(4*x) + x - 2*E^(2*x)*x + x^2]/E^x^2

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13

method	result	size
risch	\({\mathrm e}^{-x^{2}} \ln \left ({\mathrm e}^{4 x}-2 x \,{\mathrm e}^{2 x}+x^{2}+x +19\right )\)	\(26\)
parallelrisch	\({\mathrm e}^{-x^{2}} \ln \left ({\mathrm e}^{4 x}-2 x \,{\mathrm e}^{2 x}+x^{2}+x +19\right )\)	\(28\)

[In]

int(((-2*x*exp(2*x)^2+4*exp(2*x)*x^2-2*x^3-2*x^2-38*x)*ln(exp(2*x)^2-2*x*exp(2*x)+x^2+x+19)+4*exp(2*x)^2+(-4*x
-2)*exp(2*x)+2*x+1)/(exp(2*x)^2-2*x*exp(2*x)+x^2+x+19)/exp(x^2),x,method=_RETURNVERBOSE)

[Out]

exp(-x^2)*ln(exp(4*x)-2*x*exp(2*x)+x^2+x+19)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx=e^{\left (-x^{2}\right )} \log \left (x^{2} - 2 \, x e^{\left (2 \, x\right )} + x + e^{\left (4 \, x\right )} + 19\right ) \]

[In]

integrate(((-2*x*exp(2*x)^2+4*exp(2*x)*x^2-2*x^3-2*x^2-38*x)*log(exp(2*x)^2-2*x*exp(2*x)+x^2+x+19)+4*exp(2*x)^
2+(-4*x-2)*exp(2*x)+2*x+1)/(exp(2*x)^2-2*x*exp(2*x)+x^2+x+19)/exp(x^2),x, algorithm="fricas")

[Out]

e^(-x^2)*log(x^2 - 2*x*e^(2*x) + x + e^(4*x) + 19)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx=\text {Timed out} \]

[In]

integrate(((-2*x*exp(2*x)**2+4*exp(2*x)*x**2-2*x**3-2*x**2-38*x)*ln(exp(2*x)**2-2*x*exp(2*x)+x**2+x+19)+4*exp(
2*x)**2+(-4*x-2)*exp(2*x)+2*x+1)/(exp(2*x)**2-2*x*exp(2*x)+x**2+x+19)/exp(x**2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx=e^{\left (-x^{2}\right )} \log \left (x^{2} - 2 \, x e^{\left (2 \, x\right )} + x + e^{\left (4 \, x\right )} + 19\right ) \]

[In]

integrate(((-2*x*exp(2*x)^2+4*exp(2*x)*x^2-2*x^3-2*x^2-38*x)*log(exp(2*x)^2-2*x*exp(2*x)+x^2+x+19)+4*exp(2*x)^
2+(-4*x-2)*exp(2*x)+2*x+1)/(exp(2*x)^2-2*x*exp(2*x)+x^2+x+19)/exp(x^2),x, algorithm="maxima")

[Out]

e^(-x^2)*log(x^2 - 2*x*e^(2*x) + x + e^(4*x) + 19)

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx=e^{\left (-x^{2}\right )} \log \left (x^{2} - 2 \, x e^{\left (2 \, x\right )} + x + e^{\left (4 \, x\right )} + 19\right ) \]

[In]

integrate(((-2*x*exp(2*x)^2+4*exp(2*x)*x^2-2*x^3-2*x^2-38*x)*log(exp(2*x)^2-2*x*exp(2*x)+x^2+x+19)+4*exp(2*x)^
2+(-4*x-2)*exp(2*x)+2*x+1)/(exp(2*x)^2-2*x*exp(2*x)+x^2+x+19)/exp(x^2),x, algorithm="giac")

[Out]

e^(-x^2)*log(x^2 - 2*x*e^(2*x) + x + e^(4*x) + 19)

Mupad [B] (verification not implemented)

Time = 7.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-x^2} \left (1+4 e^{4 x}+e^{2 x} (-2-4 x)+2 x+\left (-38 x-2 e^{4 x} x-2 x^2+4 e^{2 x} x^2-2 x^3\right ) \log \left (19+e^{4 x}+x-2 e^{2 x} x+x^2\right )\right )}{19+e^{4 x}+x-2 e^{2 x} x+x^2} \, dx={\mathrm {e}}^{-x^2}\,\ln \left (x+{\mathrm {e}}^{4\,x}-2\,x\,{\mathrm {e}}^{2\,x}+x^2+19\right ) \]

[In]

int((exp(-x^2)*(2*x + 4*exp(4*x) - exp(2*x)*(4*x + 2) - log(x + exp(4*x) - 2*x*exp(2*x) + x^2 + 19)*(38*x + 2*
x*exp(4*x) - 4*x^2*exp(2*x) + 2*x^2 + 2*x^3) + 1))/(x + exp(4*x) - 2*x*exp(2*x) + x^2 + 19),x)

[Out]

exp(-x^2)*log(x + exp(4*x) - 2*x*exp(2*x) + x^2 + 19)