3.43.90 \(\int \frac {5+5 x \log (4)}{x \log (4)} \, dx\)

Optimal. Leaf size=19 \[ 3+5 x-\log (3)+\log (5)+\frac {5 \log (x)}{\log (4)} \]

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Rubi [A]  time = 0.01, antiderivative size = 12, normalized size of antiderivative = 0.63, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 43} \begin {gather*} 5 x+\frac {5 \log (x)}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + 5*x*Log[4])/(x*Log[4]),x]

[Out]

5*x + (5*Log[x])/Log[4]

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 {5+5 x \log (4)}{x} \, dx}{\log (4)}\\ &=\frac {\int \left (\frac {5}{x}+5 \log (4)\right ) \, dx}{\log (4)}\\ &=5 x+\frac {5 \log (x)}{\log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 0.63 \begin {gather*} 5 x+\frac {5 \log (x)}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + 5*x*Log[4])/(x*Log[4]),x]

[Out]

5*x + (5*Log[x])/Log[4]

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fricas [A]  time = 0.56, size = 14, normalized size = 0.74 \begin {gather*} \frac {5 \, {\left (2 \, x \log \relax (2) + \log \relax (x)\right )}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(10*x*log(2)+5)/x/log(2),x, algorithm="fricas")

[Out]

5/2*(2*x*log(2) + log(x))/log(2)

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giac [A]  time = 0.12, size = 15, normalized size = 0.79 \begin {gather*} \frac {5 \, {\left (2 \, x \log \relax (2) + \log \left ({\left | x \right |}\right )\right )}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(10*x*log(2)+5)/x/log(2),x, algorithm="giac")

[Out]

5/2*(2*x*log(2) + log(abs(x)))/log(2)

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maple [A]  time = 0.02, size = 13, normalized size = 0.68

method	result	size
norman	\(5 x +\frac {5 \ln \relax (x )}{2 \ln \relax (2)}\)	\(13\)
risch	\(5 x +\frac {5 \ln \relax (x )}{2 \ln \relax (2)}\)	\(13\)
default	\(\frac {5 x \ln \relax (2)+\frac {5 \ln \relax (x )}{2}}{\ln \relax (2)}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(10*x*ln(2)+5)/x/ln(2),x,method=_RETURNVERBOSE)

[Out]

5*x+5/2*ln(x)/ln(2)

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maxima [A]  time = 0.36, size = 14, normalized size = 0.74 \begin {gather*} \frac {5 \, {\left (2 \, x \log \relax (2) + \log \relax (x)\right )}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(10*x*log(2)+5)/x/log(2),x, algorithm="maxima")

[Out]

5/2*(2*x*log(2) + log(x))/log(2)

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mupad [B]  time = 0.04, size = 12, normalized size = 0.63 \begin {gather*} 5\,x+\frac {5\,\ln \relax (x)}{2\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x*log(2) + 5/2)/(x*log(2)),x)

[Out]

5*x + (5*log(x))/(2*log(2))

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sympy [A]  time = 0.07, size = 15, normalized size = 0.79 \begin {gather*} \frac {10 x \log {\relax (2 )} + 5 \log {\relax (x )}}{2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(10*x*ln(2)+5)/x/ln(2),x)

[Out]

(10*x*log(2) + 5*log(x))/(2*log(2))

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