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

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

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Rubi [A]  time = 0.17, antiderivative size = 45, normalized size of antiderivative = 1.88, number of steps used = 6, number of rules used = 3, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {14, 2295, 2288} \begin {gather*} 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) \end {gather*}

Antiderivative was successfully verified.

[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 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 2295

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 25, normalized size = 1.04 \begin {gather*} \left (e^{x^2}+5 x+e^{\frac {1}{x^2}-x^2} x\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.83, size = 24, normalized size = 1.00 \begin {gather*} {\left (x e^{\left (-\frac {x^{4} - 1}{x^{2}}\right )} + 5 \, x + e^{\left (x^{2}\right )}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.20, size = 28, normalized size = 1.17 \begin {gather*} x e^{\left (-\frac {x^{4} - 1}{x^{2}}\right )} \log \relax (x) + 5 \, x \log \relax (x) + e^{\left (x^{2}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.08, size = 30, normalized size = 1.25

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

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{4} \, \sqrt {\pi } {\left (\operatorname {erfc}\left (x + \frac {i}{x}\right ) e^{\left (4 i\right )} - \operatorname {erfc}\left (-x + \frac {i}{x}\right )\right )} e^{\left (-2 i\right )} + 5 \, x \log \relax (x) + e^{\left (x^{2}\right )} \log \relax (x) + \frac {1}{2} \, {\rm Ei}\left (x^{2}\right ) - \int \frac {{\left (2 \, x^{4} - x^{2} + 2\right )} e^{\left (-x^{2} + \frac {1}{x^{2}}\right )} \log \relax (x)}{x^{2}}\,{d x} - \int \frac {e^{\left (x^{2}\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

-1/4*sqrt(pi)*(erfc(x + I/x)*e^(4*I) - erfc(-x + I/x))*e^(-2*I) + 5*x*log(x) + e^(x^2)*log(x) + 1/2*Ei(x^2) -
integrate((2*x^4 - x^2 + 2)*e^(-x^2 + 1/x^2)*log(x)/x^2, x) - integrate(e^(x^2)/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {x^2\,{\mathrm {e}}^{-\frac {x^4-1}{x^2}}+x\,{\mathrm {e}}^{x^2}+\ln \relax (x)\,\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.58, size = 29, normalized size = 1.21 \begin {gather*} x e^{\frac {1 - x^{4}}{x^{2}}} \log {\relax (x )} + 5 x \log {\relax (x )} + e^{x^{2}} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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