3.70.35 \(\int \frac {21474836480 e-5 x}{4294967296 x^2} \, dx\)

Optimal. Leaf size=21 \[ \frac {5 \left (-e+x+\frac {x (7-\log (x))}{4294967296}\right )}{x} \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.62, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 43} \begin {gather*} -\frac {5 e}{x}-\frac {5 \log (x)}{4294967296} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(21474836480*E - 5*x)/(4294967296*x^2),x]

[Out]

(-5*E)/x - (5*Log[x])/4294967296

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {21474836480 e-5 x}{x^2} \, dx}{4294967296}\\ &=\frac {\int \left (\frac {21474836480 e}{x^2}-\frac {5}{x}\right ) \, dx}{4294967296}\\ &=-\frac {5 e}{x}-\frac {5 \log (x)}{4294967296}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.62 \begin {gather*} -\frac {5 e}{x}-\frac {5 \log (x)}{4294967296} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(21474836480*E - 5*x)/(4294967296*x^2),x]

[Out]

(-5*E)/x - (5*Log[x])/4294967296

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fricas [A]  time = 0.54, size = 14, normalized size = 0.67 \begin {gather*} -\frac {5 \, {\left (x \log \relax (x) + 4294967296 \, e\right )}}{4294967296 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4294967296*(21474836480*exp(1)-5*x)/x^2,x, algorithm="fricas")

[Out]

-5/4294967296*(x*log(x) + 4294967296*e)/x

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giac [A]  time = 0.94, size = 13, normalized size = 0.62 \begin {gather*} -\frac {5 \, e}{x} - \frac {5}{4294967296} \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4294967296*(21474836480*exp(1)-5*x)/x^2,x, algorithm="giac")

[Out]

-5*e/x - 5/4294967296*log(abs(x))

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maple [A]  time = 0.03, size = 13, normalized size = 0.62

method	result	size
default	\(-\frac {5 \ln \relax (x )}{4294967296}-\frac {5 \,{\mathrm e}}{x}\)	\(13\)
norman	\(-\frac {5 \ln \relax (x )}{4294967296}-\frac {5 \,{\mathrm e}}{x}\)	\(13\)
risch	\(-\frac {5 \ln \relax (x )}{4294967296}-\frac {5 \,{\mathrm e}}{x}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4294967296*(21474836480*exp(1)-5*x)/x^2,x,method=_RETURNVERBOSE)

[Out]

-5/4294967296*ln(x)-5*exp(1)/x

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maxima [A]  time = 0.39, size = 12, normalized size = 0.57 \begin {gather*} -\frac {5 \, e}{x} - \frac {5}{4294967296} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4294967296*(21474836480*exp(1)-5*x)/x^2,x, algorithm="maxima")

[Out]

-5*e/x - 5/4294967296*log(x)

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mupad [B]  time = 4.12, size = 12, normalized size = 0.57 \begin {gather*} -\frac {5\,\ln \relax (x)}{4294967296}-\frac {5\,\mathrm {e}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((5*x)/4294967296 - 5*exp(1))/x^2,x)

[Out]

- (5*log(x))/4294967296 - (5*exp(1))/x

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sympy [A]  time = 0.08, size = 14, normalized size = 0.67 \begin {gather*} - \frac {5 \log {\relax (x )}}{4294967296} - \frac {5 e}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4294967296*(21474836480*exp(1)-5*x)/x**2,x)

[Out]

-5*log(x)/4294967296 - 5*E/x

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