3.12.32 \(\int \frac {-2 x-10 x^2+2 e^{\log ^2(x)} \log (x)}{x} \, dx\) [1132]

Optimal. Leaf size=16 \[ 3+e^{\log ^2(x)}-2 x-5 x^2 \]

[Out]

3-2*x+exp(ln(x)^2)-5*x^2

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Rubi [A]
time = 0.04, antiderivative size = 18, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {14, 2308, 2235, 2240} \begin {gather*} e^{\log ^2(x)}-\frac {1}{5} (5 x+1)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x - 10*x^2 + 2*E^Log[x]^2*Log[x])/x,x]

[Out]

E^Log[x]^2 - (1 + 5*x)^2/5

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 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2308

Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]^2*(b_.))*(f_.))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol]
:> Dist[(g + h*x)^(m + 1)/(h*n*(c*(d + e*x)^n)^((m + 1)/n)), Subst[Int[E^(a*f*Log[F] + ((m + 1)*x)/n + b*f*Lo
g[F]*x^2), x], x, Log[c*(d + e*x)^n]], x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m, n}, x] && EqQ[e*g - d*h, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-2 (1+5 x)+\frac {2 e^{\log ^2(x)} \log (x)}{x}\right ) \, dx\\ &=-\frac {1}{5} (1+5 x)^2+2 \int \frac {e^{\log ^2(x)} \log (x)}{x} \, dx\\ &=-\frac {1}{5} (1+5 x)^2+2 \text {Subst}\left (\int e^{x^2} x \, dx,x,\log (x)\right )\\ &=e^{\log ^2(x)}-\frac {1}{5} (1+5 x)^2\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 15, normalized size = 0.94 \begin {gather*} e^{\log ^2(x)}-2 x-5 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x - 10*x^2 + 2*E^Log[x]^2*Log[x])/x,x]

[Out]

E^Log[x]^2 - 2*x - 5*x^2

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Maple [A]
time = 0.03, size = 15, normalized size = 0.94

method	result	size
default	\(-2 x +{\mathrm e}^{\ln \left (x \right )^{2}}-5 x^{2}\)	\(15\)
norman	\(-2 x +{\mathrm e}^{\ln \left (x \right )^{2}}-5 x^{2}\)	\(15\)
risch	\(-2 x +{\mathrm e}^{\ln \left (x \right )^{2}}-5 x^{2}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(x)*exp(ln(x)^2)-10*x^2-2*x)/x,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 14, normalized size = 0.88 \begin {gather*} -5 \, x^{2} - 2 \, x + e^{\left (\log \left (x\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)*exp(log(x)^2)-10*x^2-2*x)/x,x, algorithm="maxima")

[Out]

-5*x^2 - 2*x + e^(log(x)^2)

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Fricas [A]
time = 0.35, size = 14, normalized size = 0.88 \begin {gather*} -5 \, x^{2} - 2 \, x + e^{\left (\log \left (x\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)*exp(log(x)^2)-10*x^2-2*x)/x,x, algorithm="fricas")

[Out]

-5*x^2 - 2*x + e^(log(x)^2)

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Sympy [A]
time = 0.06, size = 14, normalized size = 0.88 \begin {gather*} - 5 x^{2} - 2 x + e^{\log {\left (x \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(x)*exp(ln(x)**2)-10*x**2-2*x)/x,x)

[Out]

-5*x**2 - 2*x + exp(log(x)**2)

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Giac [A]
time = 0.40, size = 14, normalized size = 0.88 \begin {gather*} -5 \, x^{2} - 2 \, x + e^{\left (\log \left (x\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)*exp(log(x)^2)-10*x^2-2*x)/x,x, algorithm="giac")

[Out]

-5*x^2 - 2*x + e^(log(x)^2)

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Mupad [B]
time = 0.75, size = 14, normalized size = 0.88 \begin {gather*} {\mathrm {e}}^{{\ln \left (x\right )}^2}-2\,x-5\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + 10*x^2 - 2*exp(log(x)^2)*log(x))/x,x)

[Out]

exp(log(x)^2) - 2*x - 5*x^2

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