3.63.52 \(\int \frac {4 x^2-4 e^{16} \log (4)}{x^2 \log (4)} \, dx\) [6252]

Optimal. Leaf size=19 \[ -1+\frac {4 \left (e^{16}+x\right )}{x}+\frac {4 x}{\log (4)} \]

[Out]

4*(x+exp(16))/x-1+2*x/ln(2)

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Rubi [A]
time = 0.01, antiderivative size = 16, normalized size of antiderivative = 0.84, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 14} \begin {gather*} \frac {4 e^{16}}{x}+\frac {4 x}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x^2 - 4*E^16*Log[4])/(x^2*Log[4]),x]

[Out]

(4*E^16)/x + (4*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 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 {4 x^2-4 e^{16} \log (4)}{x^2} \, dx}{\log (4)}\\ &=\frac {\int \left (4-\frac {4 e^{16} \log (4)}{x^2}\right ) \, dx}{\log (4)}\\ &=\frac {4 e^{16}}{x}+\frac {4 x}{\log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 16, normalized size = 0.84 \begin {gather*} \frac {4 e^{16}}{x}+\frac {4 x}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x^2 - 4*E^16*Log[4])/(x^2*Log[4]),x]

[Out]

(4*E^16)/x + (4*x)/Log[4]

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Maple [A]
time = 0.62, size = 18, normalized size = 0.95

method	result	size
risch	\(\frac {2 x}{\ln \left (2\right )}+\frac {4 \,{\mathrm e}^{16}}{x}\)	\(16\)
default	\(\frac {2 x +\frac {4 \,{\mathrm e}^{16} \ln \left (2\right )}{x}}{\ln \left (2\right )}\)	\(18\)
norman	\(\frac {\frac {2 x^{2}}{\ln \left (2\right )}+4 \,{\mathrm e}^{16}}{x}\)	\(19\)
gosper	\(\frac {4 \,{\mathrm e}^{16} \ln \left (2\right )+2 x^{2}}{x \ln \left (2\right )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-8*exp(16)*ln(2)+4*x^2)/x^2/ln(2),x,method=_RETURNVERBOSE)

[Out]

2/ln(2)*(x+2*exp(16)*ln(2)/x)

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Maxima [A]
time = 0.30, size = 17, normalized size = 0.89 \begin {gather*} \frac {2 \, {\left (x + \frac {2 \, e^{16} \log \left (2\right )}{x}\right )}}{\log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-8*exp(16)*log(2)+4*x^2)/x^2/log(2),x, algorithm="maxima")

[Out]

2*(x + 2*e^16*log(2)/x)/log(2)

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Fricas [A]
time = 0.35, size = 19, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (x^{2} + 2 \, e^{16} \log \left (2\right )\right )}}{x \log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-8*exp(16)*log(2)+4*x^2)/x^2/log(2),x, algorithm="fricas")

[Out]

2*(x^2 + 2*e^16*log(2))/(x*log(2))

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Sympy [A]
time = 0.03, size = 15, normalized size = 0.79 \begin {gather*} \frac {2 x + \frac {4 e^{16} \log {\left (2 \right )}}{x}}{\log {\left (2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-8*exp(16)*ln(2)+4*x**2)/x**2/ln(2),x)

[Out]

(2*x + 4*exp(16)*log(2)/x)/log(2)

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Giac [A]
time = 0.38, size = 17, normalized size = 0.89 \begin {gather*} \frac {2 \, {\left (x + \frac {2 \, e^{16} \log \left (2\right )}{x}\right )}}{\log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-8*exp(16)*log(2)+4*x^2)/x^2/log(2),x, algorithm="giac")

[Out]

2*(x + 2*e^16*log(2)/x)/log(2)

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Mupad [B]
time = 0.04, size = 15, normalized size = 0.79 \begin {gather*} \frac {4\,{\mathrm {e}}^{16}}{x}+\frac {2\,x}{\ln \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(16)*log(2) - 2*x^2)/(x^2*log(2)),x)

[Out]

(4*exp(16))/x + (2*x)/log(2)

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