3.101.45 \(\int \frac {-50 e^{25-50 x} x^2-2 \log (x)+\log ^2(x)}{x^2} \, dx\)

Optimal. Leaf size=20 \[ 3+e^{25 (1-2 x)}-\frac {\log ^2(x)}{x} \]

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Rubi [A]  time = 0.05, antiderivative size = 27, normalized size of antiderivative = 1.35, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {14, 2194, 2304, 2366, 2303} \begin {gather*} e^{25-50 x}+\frac {(2-\log (x)) \log (x)}{x}-\frac {2 \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50*E^(25 - 50*x)*x^2 - 2*Log[x] + Log[x]^2)/x^2,x]

[Out]

E^(25 - 50*x) - (2*Log[x])/x + ((2 - Log[x])*Log[x])/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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2303

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

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-50 e^{25-50 x}+\frac {(-2+\log (x)) \log (x)}{x^2}\right ) \, dx\\ &=-\left (50 \int e^{25-50 x} \, dx\right )+\int \frac {(-2+\log (x)) \log (x)}{x^2} \, dx\\ &=e^{25-50 x}-\frac {\log (x)}{x}+\frac {(2-\log (x)) \log (x)}{x}-\int \frac {1-\log (x)}{x^2} \, dx\\ &=e^{25-50 x}-\frac {2 \log (x)}{x}+\frac {(2-\log (x)) \log (x)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 17, normalized size = 0.85 \begin {gather*} e^{25-50 x}-\frac {\log ^2(x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50*E^(25 - 50*x)*x^2 - 2*Log[x] + Log[x]^2)/x^2,x]

[Out]

E^(25 - 50*x) - Log[x]^2/x

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fricas [A]  time = 2.51, size = 19, normalized size = 0.95 \begin {gather*} \frac {x e^{\left (-50 \, x + 25\right )} - \log \relax (x)^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)^2-2*log(x)-50*x^2*exp(-50*x+25))/x^2,x, algorithm="fricas")

[Out]

(x*e^(-50*x + 25) - log(x)^2)/x

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giac [A]  time = 0.16, size = 19, normalized size = 0.95 \begin {gather*} \frac {x e^{\left (-50 \, x + 25\right )} - \log \relax (x)^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)^2-2*log(x)-50*x^2*exp(-50*x+25))/x^2,x, algorithm="giac")

[Out]

(x*e^(-50*x + 25) - log(x)^2)/x

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maple [A]  time = 0.02, size = 17, normalized size = 0.85

method	result	size
default	\(-\frac {\ln \relax (x )^{2}}{x}+{\mathrm e}^{-50 x +25}\)	\(17\)
risch	\(-\frac {\ln \relax (x )^{2}}{x}+{\mathrm e}^{-50 x +25}\)	\(17\)
norman	\(\frac {x \,{\mathrm e}^{-50 x +25}-\ln \relax (x )^{2}}{x}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(x)^2-2*ln(x)-50*x^2*exp(-50*x+25))/x^2,x,method=_RETURNVERBOSE)

[Out]

-ln(x)^2/x+exp(-50*x+25)

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maxima [A]  time = 0.37, size = 34, normalized size = 1.70 \begin {gather*} -\frac {\log \relax (x)^{2} + 2 \, \log \relax (x) + 2}{x} + \frac {2 \, \log \relax (x)}{x} + \frac {2}{x} + e^{\left (-50 \, x + 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)^2-2*log(x)-50*x^2*exp(-50*x+25))/x^2,x, algorithm="maxima")

[Out]

-(log(x)^2 + 2*log(x) + 2)/x + 2*log(x)/x + 2/x + e^(-50*x + 25)

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mupad [B]  time = 6.94, size = 17, normalized size = 0.85 \begin {gather*} {\mathrm {e}}^{-50\,x}\,{\mathrm {e}}^{25}-\frac {{\ln \relax (x)}^2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(x) - log(x)^2 + 50*x^2*exp(25 - 50*x))/x^2,x)

[Out]

exp(-50*x)*exp(25) - log(x)^2/x

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sympy [A]  time = 0.30, size = 12, normalized size = 0.60 \begin {gather*} e^{25 - 50 x} - \frac {\log {\relax (x )}^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(x)**2-2*ln(x)-50*x**2*exp(-50*x+25))/x**2,x)

[Out]

exp(25 - 50*x) - log(x)**2/x

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