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

Optimal. Leaf size=16 \[ x+e^{x^2} \log (x)-2 (x+\log (x)) \]

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Rubi [A]  time = 0.05, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {14, 43, 2288} \begin {gather*} e^{x^2} \log (x)-x-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x - 2*Log[x] + E^x^2*Log[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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 16, normalized size = 1.00 \begin {gather*} -x-2 \log (x)+e^{x^2} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x - 2*Log[x] + E^x^2*Log[x]

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fricas [A]  time = 0.76, size = 13, normalized size = 0.81 \begin {gather*} {\left (e^{\left (x^{2}\right )} - 2\right )} \log \relax (x) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^(x^2) - 2)*log(x) - x

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giac [A]  time = 0.26, size = 15, normalized size = 0.94 \begin {gather*} e^{\left (x^{2}\right )} \log \relax (x) - x - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(x^2)*log(x) - x - 2*log(x)

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maple [A]  time = 0.06, size = 16, normalized size = 1.00

method	result	size
default	\({\mathrm e}^{x^{2}} \ln \relax (x )-x -2 \ln \relax (x )\)	\(16\)
norman	\({\mathrm e}^{x^{2}} \ln \relax (x )-x -2 \ln \relax (x )\)	\(16\)
risch	\({\mathrm e}^{x^{2}} \ln \relax (x )-x -2 \ln \relax (x )\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x^2)*ln(x)-x-2*ln(x)

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maxima [A]  time = 0.43, size = 15, normalized size = 0.94 \begin {gather*} e^{\left (x^{2}\right )} \log \relax (x) - x - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(x^2)*log(x) - x - 2*log(x)

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mupad [B]  time = 0.50, size = 15, normalized size = 0.94 \begin {gather*} {\mathrm {e}}^{x^2}\,\ln \relax (x)-2\,\ln \relax (x)-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x^2)*log(x) - 2*log(x) - x

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sympy [A]  time = 0.31, size = 14, normalized size = 0.88 \begin {gather*} - x + e^{x^{2}} \log {\relax (x )} - 2 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + exp(x**2)*log(x) - 2*log(x)

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