3.13.85 \(\int \frac {2 x+2 x^2+\frac {e^{-8+2 x^2} (-2+4 x^2)}{x^2}+x \log (e^{2 x} x^2)}{x} \, dx\)

Optimal. Leaf size=26 \[ \frac {e^{-8+2 x^2}}{x^2}+x \log \left (e^{2 x} x^2\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 39, normalized size of antiderivative = 1.50, number of steps used = 5, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {14, 2288, 2548, 9} \begin {gather*} x^2+\frac {e^{2 x^2-8}}{x^2}+x \log \left (e^{2 x} x^2\right )+2 x-(x+1)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-8 + 2*x^2)/x^2 + 2*x + x^2 - (1 + x)^2 + x*Log[E^(2*x)*x^2]

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[(a*(b + c*x)^2)/(2*c), x] /; FreeQ[{a, b, c}, 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 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2+2 x+\frac {2 e^{-8+2 x^2} \left (-1+2 x^2\right )}{x^3}+\log \left (e^{2 x} x^2\right )\right ) \, dx\\ &=2 x+x^2+2 \int \frac {e^{-8+2 x^2} \left (-1+2 x^2\right )}{x^3} \, dx+\int \log \left (e^{2 x} x^2\right ) \, dx\\ &=\frac {e^{-8+2 x^2}}{x^2}+2 x+x^2+x \log \left (e^{2 x} x^2\right )-\int 2 (1+x) \, dx\\ &=\frac {e^{-8+2 x^2}}{x^2}+2 x+x^2-(1+x)^2+x \log \left (e^{2 x} x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 27, normalized size = 1.04 \begin {gather*} 1+\frac {e^{-8+2 x^2}}{x^2}+x \log \left (e^{2 x} x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1 + E^(-8 + 2*x^2)/x^2 + x*Log[E^(2*x)*x^2]

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fricas [A]  time = 0.89, size = 23, normalized size = 0.88 \begin {gather*} 2 \, x^{2} + 2 \, x \log \relax (x) + e^{\left (2 \, x^{2} - 2 \, \log \relax (x) - 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x^2 + 2*x*log(x) + e^(2*x^2 - 2*log(x) - 8)

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giac [A]  time = 0.26, size = 30, normalized size = 1.15 \begin {gather*} 2 \, x^{2} \mathrm {sgn}\relax (x)^{2} + 2 \, x \log \left (x \mathrm {sgn}\relax (x)\right ) + \frac {e^{\left (2 \, x^{2} - 8\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x^2*sgn(x)^2 + 2*x*log(x*sgn(x)) + e^(2*x^2 - 8)/x^2

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maple [C]  time = 0.10, size = 49, normalized size = 1.88

method	result	size
default	\(-2 \,{\mathrm e}^{-8} \left (-\frac {{\mathrm e}^{2 x^{2}}}{2 x^{2}}-\expIntegralEi \left (1, -2 x^{2}\right )\right )-2 \,{\mathrm e}^{-8} \expIntegralEi \left (1, -2 x^{2}\right )+x \ln \left ({\mathrm e}^{2 x} x^{2}\right )\)	\(49\)
risch	\(2 x \ln \left ({\mathrm e}^{x}\right )+2 x \ln \relax (x )-\frac {i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}+i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-\frac {i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}-\frac {i x \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )}{2}+\frac {i x \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )^{2}}{2}-\frac {i x \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )}{2}+i x \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}-\frac {i x \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}}{2}+\frac {i x \pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 x}\right ) \mathrm {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )^{2}}{2}-\frac {i x \pi \mathrm {csgn}\left (i x^{2} {\mathrm e}^{2 x}\right )^{3}}{2}+\frac {{\mathrm e}^{2 \left (x -2\right ) \left (2+x \right )}}{x^{2}}\)	\(235\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*exp(-8)*(-1/2*exp(x^2)^2/x^2-Ei(1,-2*x^2))-2*exp(-8)*Ei(1,-2*x^2)+x*ln(exp(x)^2*x^2)

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maxima [C]  time = 0.41, size = 33, normalized size = 1.27 \begin {gather*} 2 \, {\rm Ei}\left (2 \, x^{2}\right ) e^{\left (-8\right )} - 2 \, e^{\left (-8\right )} \Gamma \left (-1, -2 \, x^{2}\right ) + x \log \left (x^{2} e^{\left (2 \, x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*Ei(2*x^2)*e^(-8) - 2*e^(-8)*gamma(-1, -2*x^2) + x*log(x^2*e^(2*x))

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mupad [B]  time = 1.08, size = 24, normalized size = 0.92 \begin {gather*} x\,\ln \left (x^2\right )+2\,x^2+\frac {{\mathrm {e}}^{-8}\,{\mathrm {e}}^{2\,x^2}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(x^2) + 2*x^2 + (exp(-8)*exp(2*x^2))/x^2

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sympy [A]  time = 0.27, size = 22, normalized size = 0.85 \begin {gather*} x \log {\left (x^{2} e^{2 x} \right )} + \frac {e^{2 x^{2} - 8}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(x**2*exp(2*x)) + exp(2*x**2 - 8)/x**2

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