3.62.6 \(\int \frac {4+16 e^2 x^2+32 e^2 x^2 \log (x)}{e^2 x} \, dx\)

Optimal. Leaf size=25 \[ 4 \left (e^{\frac {e^4}{8}}+\frac {\log (x)}{e^2}+4 x^2 \log (x)\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 0.60, number of steps used = 6, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 14, 2304} \begin {gather*} 16 x^2 \log (x)+\frac {4 \log (x)}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + 16*E^2*x^2 + 32*E^2*x^2*Log[x])/(E^2*x),x]

[Out]

(4*Log[x])/E^2 + 16*x^2*Log[x]

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]]

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
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {4+16 e^2 x^2+32 e^2 x^2 \log (x)}{x} \, dx}{e^2}\\ &=\frac {\int \left (\frac {4 \left (1+4 e^2 x^2\right )}{x}+32 e^2 x \log (x)\right ) \, dx}{e^2}\\ &=32 \int x \log (x) \, dx+\frac {4 \int \frac {1+4 e^2 x^2}{x} \, dx}{e^2}\\ &=-8 x^2+16 x^2 \log (x)+\frac {4 \int \left (\frac {1}{x}+4 e^2 x\right ) \, dx}{e^2}\\ &=\frac {4 \log (x)}{e^2}+16 x^2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.60 \begin {gather*} \frac {4 \log (x)}{e^2}+16 x^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + 16*E^2*x^2 + 32*E^2*x^2*Log[x])/(E^2*x),x]

[Out]

(4*Log[x])/E^2 + 16*x^2*Log[x]

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fricas [A]  time = 0.73, size = 15, normalized size = 0.60 \begin {gather*} 4 \, {\left (4 \, x^{2} e^{2} + 1\right )} e^{\left (-2\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^2*exp(2)*log(x)+16*x^2*exp(2)+4)/exp(2)/x,x, algorithm="fricas")

[Out]

4*(4*x^2*e^2 + 1)*e^(-2)*log(x)

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giac [A]  time = 0.13, size = 16, normalized size = 0.64 \begin {gather*} 4 \, {\left (4 \, x^{2} e^{2} \log \relax (x) + \log \relax (x)\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^2*exp(2)*log(x)+16*x^2*exp(2)+4)/exp(2)/x,x, algorithm="giac")

[Out]

4*(4*x^2*e^2*log(x) + log(x))*e^(-2)

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maple [A]  time = 0.06, size = 15, normalized size = 0.60

method	result	size
risch	\(4 \ln \relax (x ) {\mathrm e}^{-2}+16 x^{2} \ln \relax (x )\)	\(15\)
norman	\(4 \ln \relax (x ) {\mathrm e}^{-2}+16 x^{2} \ln \relax (x )\)	\(17\)
default	\(4 \,{\mathrm e}^{-2} \left (8 \,{\mathrm e}^{2} \left (\frac {x^{2} \ln \relax (x )}{2}-\frac {x^{2}}{4}\right )+2 x^{2} {\mathrm e}^{2}+\ln \relax (x )\right )\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x^2*exp(2)*ln(x)+16*x^2*exp(2)+4)/exp(2)/x,x,method=_RETURNVERBOSE)

[Out]

4*ln(x)*exp(-2)+16*x^2*ln(x)

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maxima [A]  time = 0.45, size = 31, normalized size = 1.24 \begin {gather*} 4 \, {\left (2 \, x^{2} e^{2} + 2 \, {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} e^{2} + \log \relax (x)\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^2*exp(2)*log(x)+16*x^2*exp(2)+4)/exp(2)/x,x, algorithm="maxima")

[Out]

4*(2*x^2*e^2 + 2*(2*x^2*log(x) - x^2)*e^2 + log(x))*e^(-2)

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mupad [B]  time = 4.41, size = 12, normalized size = 0.48 \begin {gather*} 4\,\ln \relax (x)\,\left (4\,x^2+{\mathrm {e}}^{-2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2)*(16*x^2*exp(2) + 32*x^2*exp(2)*log(x) + 4))/x,x)

[Out]

4*log(x)*(exp(-2) + 4*x^2)

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sympy [A]  time = 0.09, size = 15, normalized size = 0.60 \begin {gather*} 16 x^{2} \log {\relax (x )} + \frac {4 \log {\relax (x )}}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x**2*exp(2)*ln(x)+16*x**2*exp(2)+4)/exp(2)/x,x)

[Out]

16*x**2*log(x) + 4*exp(-2)*log(x)

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