\(\int \frac {e^{x^2} x+5 x^2+e^{\frac {1-x^4}{x^2}} x^2+(5 x^2+2 e^{x^2} x^3+e^{\frac {1-x^4}{x^2}} (-2+x^2-2 x^4)) \log (x)}{x^2} \, dx\) [8419]

Optimal result

Integrand size = 77, antiderivative size = 24 \[ \int \frac {e^{x^2} x+5 x^2+e^{\frac {1-x^4}{x^2}} x^2+\left (5 x^2+2 e^{x^2} x^3+e^{\frac {1-x^4}{x^2}} \left (-2+x^2-2 x^4\right )\right ) \log (x)}{x^2} \, dx=\left (e^{x^2}+\left (5+e^{\frac {1}{x^2}-x^2}\right ) x\right ) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {14, 2332, 2326} \[ \int \frac {e^{x^2} x+5 x^2+e^{\frac {1-x^4}{x^2}} x^2+\left (5 x^2+2 e^{x^2} x^3+e^{\frac {1-x^4}{x^2}} \left (-2+x^2-2 x^4\right )\right ) \log (x)}{x^2} \, dx=e^{x^2} \log (x)+\frac {e^{\frac {1}{x^2}-x^2} \left (x^4 \log (x)+\log (x)\right )}{x^2 \left (\frac {1}{x^3}+x\right )}+5 x \log (x) \]

[In]

Int[(E^x^2*x + 5*x^2 + E^((1 - x^4)/x^2)*x^2 + (5*x^2 + 2*E^x^2*x^3 + E^((1 - x^4)/x^2)*(-2 + x^2 - 2*x^4))*Lo
g[x])/x^2,x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (5 (1+\log (x))+\frac {e^{x^2} \left (1+2 x^2 \log (x)\right )}{x}-\frac {e^{\frac {1}{x^2}-x^2} \left (-x^2+2 \log (x)-x^2 \log (x)+2 x^4 \log (x)\right )}{x^2}\right ) \, dx \\ & = 5 \int (1+\log (x)) \, dx+\int \frac {e^{x^2} \left (1+2 x^2 \log (x)\right )}{x} \, dx-\int \frac {e^{\frac {1}{x^2}-x^2} \left (-x^2+2 \log (x)-x^2 \log (x)+2 x^4 \log (x)\right )}{x^2} \, dx \\ & = 5 x+e^{x^2} \log (x)+\frac {e^{\frac {1}{x^2}-x^2} \left (\log (x)+x^4 \log (x)\right )}{x^2 \left (\frac {1}{x^3}+x\right )}+5 \int \log (x) \, dx \\ & = e^{x^2} \log (x)+5 x \log (x)+\frac {e^{\frac {1}{x^2}-x^2} \left (\log (x)+x^4 \log (x)\right )}{x^2 \left (\frac {1}{x^3}+x\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{x^2} x+5 x^2+e^{\frac {1-x^4}{x^2}} x^2+\left (5 x^2+2 e^{x^2} x^3+e^{\frac {1-x^4}{x^2}} \left (-2+x^2-2 x^4\right )\right ) \log (x)}{x^2} \, dx=\left (e^{x^2}+5 x+e^{\frac {1}{x^2}-x^2} x\right ) \log (x) \]

[In]

Integrate[(E^x^2*x + 5*x^2 + E^((1 - x^4)/x^2)*x^2 + (5*x^2 + 2*E^x^2*x^3 + E^((1 - x^4)/x^2)*(-2 + x^2 - 2*x^
4))*Log[x])/x^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21

method	result	size
parallelrisch	\(\ln \left (x \right ) {\mathrm e}^{-\frac {x^{4}-1}{x^{2}}} x +5 x \ln \left (x \right )+{\mathrm e}^{x^{2}} \ln \left (x \right )\)	\(29\)
default	\(x \,{\mathrm e}^{\frac {-x^{4}+1}{x^{2}}} \ln \left (x \right )+{\mathrm e}^{x^{2}} \ln \left (x \right )+5 x \ln \left (x \right )\)	\(30\)
parts	\(x \,{\mathrm e}^{\frac {-x^{4}+1}{x^{2}}} \ln \left (x \right )+{\mathrm e}^{x^{2}} \ln \left (x \right )+5 x \ln \left (x \right )\)	\(30\)
risch	\(\left (x \,{\mathrm e}^{-\frac {\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}{x^{2}}}+5 x +{\mathrm e}^{x^{2}}\right ) \ln \left (x \right )\)	\(31\)

[In]

int(((2*x^3*exp(x^2)+(-2*x^4+x^2-2)*exp((-x^4+1)/x^2)+5*x^2)*ln(x)+exp(x^2)*x+x^2*exp((-x^4+1)/x^2)+5*x^2)/x^2
,x,method=_RETURNVERBOSE)

[Out]

ln(x)*exp(-(x^4-1)/x^2)*x+5*x*ln(x)+exp(x^2)*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{x^2} x+5 x^2+e^{\frac {1-x^4}{x^2}} x^2+\left (5 x^2+2 e^{x^2} x^3+e^{\frac {1-x^4}{x^2}} \left (-2+x^2-2 x^4\right )\right ) \log (x)}{x^2} \, dx={\left (x e^{\left (-\frac {x^{4} - 1}{x^{2}}\right )} + 5 \, x + e^{\left (x^{2}\right )}\right )} \log \left (x\right ) \]

[In]

integrate(((2*x^3*exp(x^2)+(-2*x^4+x^2-2)*exp((-x^4+1)/x^2)+5*x^2)*log(x)+exp(x^2)*x+x^2*exp((-x^4+1)/x^2)+5*x
^2)/x^2,x, algorithm="fricas")

[Out]

(x*e^(-(x^4 - 1)/x^2) + 5*x + e^(x^2))*log(x)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^{x^2} x+5 x^2+e^{\frac {1-x^4}{x^2}} x^2+\left (5 x^2+2 e^{x^2} x^3+e^{\frac {1-x^4}{x^2}} \left (-2+x^2-2 x^4\right )\right ) \log (x)}{x^2} \, dx=x e^{\frac {1 - x^{4}}{x^{2}}} \log {\left (x \right )} + 5 x \log {\left (x \right )} + e^{x^{2}} \log {\left (x \right )} \]

[In]

integrate(((2*x**3*exp(x**2)+(-2*x**4+x**2-2)*exp((-x**4+1)/x**2)+5*x**2)*ln(x)+exp(x**2)*x+x**2*exp((-x**4+1)
/x**2)+5*x**2)/x**2,x)

[Out]

x*exp((1 - x**4)/x**2)*log(x) + 5*x*log(x) + exp(x**2)*log(x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{x^2} x+5 x^2+e^{\frac {1-x^4}{x^2}} x^2+\left (5 x^2+2 e^{x^2} x^3+e^{\frac {1-x^4}{x^2}} \left (-2+x^2-2 x^4\right )\right ) \log (x)}{x^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(((2*x^3*exp(x^2)+(-2*x^4+x^2-2)*exp((-x^4+1)/x^2)+5*x^2)*log(x)+exp(x^2)*x+x^2*exp((-x^4+1)/x^2)+5*x
^2)/x^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> Encountered operator mismatch in maxima-to-sr translation

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{x^2} x+5 x^2+e^{\frac {1-x^4}{x^2}} x^2+\left (5 x^2+2 e^{x^2} x^3+e^{\frac {1-x^4}{x^2}} \left (-2+x^2-2 x^4\right )\right ) \log (x)}{x^2} \, dx=x e^{\left (-\frac {x^{4} - 1}{x^{2}}\right )} \log \left (x\right ) + 5 \, x \log \left (x\right ) + e^{\left (x^{2}\right )} \log \left (x\right ) \]

[In]

integrate(((2*x^3*exp(x^2)+(-2*x^4+x^2-2)*exp((-x^4+1)/x^2)+5*x^2)*log(x)+exp(x^2)*x+x^2*exp((-x^4+1)/x^2)+5*x
^2)/x^2,x, algorithm="giac")

[Out]

x*e^(-(x^4 - 1)/x^2)*log(x) + 5*x*log(x) + e^(x^2)*log(x)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{x^2} x+5 x^2+e^{\frac {1-x^4}{x^2}} x^2+\left (5 x^2+2 e^{x^2} x^3+e^{\frac {1-x^4}{x^2}} \left (-2+x^2-2 x^4\right )\right ) \log (x)}{x^2} \, dx=\int \frac {x^2\,{\mathrm {e}}^{-\frac {x^4-1}{x^2}}+x\,{\mathrm {e}}^{x^2}+\ln \left (x\right )\,\left (2\,x^3\,{\mathrm {e}}^{x^2}+5\,x^2-{\mathrm {e}}^{-\frac {x^4-1}{x^2}}\,\left (2\,x^4-x^2+2\right )\right )+5\,x^2}{x^2} \,d x \]

[In]

int((x^2*exp(-(x^4 - 1)/x^2) + x*exp(x^2) + log(x)*(2*x^3*exp(x^2) + 5*x^2 - exp(-(x^4 - 1)/x^2)*(2*x^4 - x^2
+ 2)) + 5*x^2)/x^2,x)

[Out]

int((x^2*exp(-(x^4 - 1)/x^2) + x*exp(x^2) + log(x)*(2*x^3*exp(x^2) + 5*x^2 - exp(-(x^4 - 1)/x^2)*(2*x^4 - x^2
+ 2)) + 5*x^2)/x^2, x)