3.45.50 \(\int \frac {12 e^3+x^2 \log (2)}{x^2 \log (2)} \, dx\)

Optimal. Leaf size=23 \[ 9-3 \left (-5-\frac {x}{3}+\frac {4 e^3}{x \log (2)}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 0.61, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 14} \begin {gather*} x-\frac {12 e^3}{x \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12*E^3 + x^2*Log[2])/(x^2*Log[2]),x]

[Out]

x - (12*E^3)/(x*Log[2])

Rule 12

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {12 e^3+x^2 \log (2)}{x^2} \, dx}{\log (2)}\\ &=\frac {\int \left (\frac {12 e^3}{x^2}+\log (2)\right ) \, dx}{\log (2)}\\ &=x-\frac {12 e^3}{x \log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.61 \begin {gather*} x-\frac {12 e^3}{x \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12*E^3 + x^2*Log[2])/(x^2*Log[2]),x]

[Out]

x - (12*E^3)/(x*Log[2])

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fricas [A]  time = 0.79, size = 19, normalized size = 0.83 \begin {gather*} \frac {x^{2} \log \relax (2) - 12 \, e^{3}}{x \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(2)+12*exp(3))/x^2/log(2),x, algorithm="fricas")

[Out]

(x^2*log(2) - 12*e^3)/(x*log(2))

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giac [A]  time = 0.15, size = 17, normalized size = 0.74 \begin {gather*} \frac {x \log \relax (2) - \frac {12 \, e^{3}}{x}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(2)+12*exp(3))/x^2/log(2),x, algorithm="giac")

[Out]

(x*log(2) - 12*e^3/x)/log(2)

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maple [A]  time = 0.04, size = 14, normalized size = 0.61

method	result	size
risch	\(x -\frac {12 \,{\mathrm e}^{3}}{x \ln \relax (2)}\)	\(14\)
norman	\(\frac {x^{2}-\frac {12 \,{\mathrm e}^{3}}{\ln \relax (2)}}{x}\)	\(17\)
default	\(\frac {x \ln \relax (2)-\frac {12 \,{\mathrm e}^{3}}{x}}{\ln \relax (2)}\)	\(18\)
gosper	\(\frac {x^{2} \ln \relax (2)-12 \,{\mathrm e}^{3}}{x \ln \relax (2)}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*ln(2)+12*exp(3))/x^2/ln(2),x,method=_RETURNVERBOSE)

[Out]

x-12*exp(3)/x/ln(2)

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maxima [A]  time = 0.37, size = 17, normalized size = 0.74 \begin {gather*} \frac {x \log \relax (2) - \frac {12 \, e^{3}}{x}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(2)+12*exp(3))/x^2/log(2),x, algorithm="maxima")

[Out]

(x*log(2) - 12*e^3/x)/log(2)

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mupad [B]  time = 3.21, size = 13, normalized size = 0.57 \begin {gather*} x-\frac {12\,{\mathrm {e}}^3}{x\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*exp(3) + x^2*log(2))/(x^2*log(2)),x)

[Out]

x - (12*exp(3))/(x*log(2))

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sympy [A]  time = 0.08, size = 14, normalized size = 0.61 \begin {gather*} \frac {x \log {\relax (2 )} - \frac {12 e^{3}}{x}}{\log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*ln(2)+12*exp(3))/x**2/ln(2),x)

[Out]

(x*log(2) - 12*exp(3)/x)/log(2)

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