3.31.81 \(\int (e^{\frac {4 \log ^2(4)}{5}} x+2 e^{\frac {4 \log ^2(4)}{5}} x \log (\frac {x}{e^4})) \, dx\) [3081]

Optimal. Leaf size=20 \[ e^{\frac {4 \log ^2(4)}{5}} x^2 \log \left (\frac {x}{e^4}\right ) \]

[Out]

x^2*ln(x/exp(4))*exp(16/5*ln(2)^2)

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Rubi [A]
time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {2341} \begin {gather*} x^2 e^{\frac {4 \log ^2(4)}{5}} \log \left (\frac {x}{e^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^((4*Log[4]^2)/5)*x + 2*E^((4*Log[4]^2)/5)*x*Log[x/E^4],x]

[Out]

E^((4*Log[4]^2)/5)*x^2*Log[x/E^4]

Rule 2341

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 {1}{2} e^{\frac {4 \log ^2(4)}{5}} x^2+\left (2 e^{\frac {4 \log ^2(4)}{5}}\right ) \int x \log \left (\frac {x}{e^4}\right ) \, dx\\ &=e^{\frac {4 \log ^2(4)}{5}} x^2 \log \left (\frac {x}{e^4}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 23, normalized size = 1.15 \begin {gather*} e^{\frac {4 \log ^2(4)}{5}} \left (-4 x^2+x^2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^((4*Log[4]^2)/5)*x + 2*E^((4*Log[4]^2)/5)*x*Log[x/E^4],x]

[Out]

E^((4*Log[4]^2)/5)*(-4*x^2 + x^2*Log[x])

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Maple [A]
time = 0.37, size = 19, normalized size = 0.95

method	result	size
risch	\({\mathrm e}^{\frac {16 \ln \left (2\right )^{2}}{5}} x^{2} \ln \left (x \,{\mathrm e}^{-4}\right )\)	\(17\)
derivativedivides	\({\mathrm e}^{\frac {16 \ln \left (2\right )^{2}}{5}} x^{2} \ln \left (x \,{\mathrm e}^{-4}\right )\)	\(19\)
default	\({\mathrm e}^{\frac {16 \ln \left (2\right )^{2}}{5}} x^{2} \ln \left (x \,{\mathrm e}^{-4}\right )\)	\(19\)
norman	\({\mathrm e}^{\frac {16 \ln \left (2\right )^{2}}{5}} x^{2} \ln \left (x \,{\mathrm e}^{-4}\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x*exp(16/5*ln(2)^2)*ln(x/exp(4))+x*exp(16/5*ln(2)^2),x,method=_RETURNVERBOSE)

[Out]

x^2*ln(x/exp(4))*exp(16/5*ln(2)^2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (16) = 32\).
time = 0.26, size = 38, normalized size = 1.90 \begin {gather*} \frac {1}{2} \, x^{2} e^{\left (\frac {16}{5} \, \log \left (2\right )^{2}\right )} + \frac {1}{2} \, {\left (2 \, x^{2} \log \left (x e^{\left (-4\right )}\right ) - x^{2}\right )} e^{\left (\frac {16}{5} \, \log \left (2\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(16/5*log(2)^2)*log(x/exp(4))+x*exp(16/5*log(2)^2),x, algorithm="maxima")

[Out]

1/2*x^2*e^(16/5*log(2)^2) + 1/2*(2*x^2*log(x*e^(-4)) - x^2)*e^(16/5*log(2)^2)

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Fricas [A]
time = 0.39, size = 16, normalized size = 0.80 \begin {gather*} x^{2} e^{\left (\frac {16}{5} \, \log \left (2\right )^{2}\right )} \log \left (x e^{\left (-4\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(16/5*log(2)^2)*log(x/exp(4))+x*exp(16/5*log(2)^2),x, algorithm="fricas")

[Out]

x^2*e^(16/5*log(2)^2)*log(x*e^(-4))

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Sympy [A]
time = 0.04, size = 19, normalized size = 0.95 \begin {gather*} x^{2} e^{\frac {16 \log {\left (2 \right )}^{2}}{5}} \log {\left (\frac {x}{e^{4}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(16/5*ln(2)**2)*ln(x/exp(4))+x*exp(16/5*ln(2)**2),x)

[Out]

x**2*exp(16*log(2)**2/5)*log(x*exp(-4))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (16) = 32\).
time = 0.36, size = 38, normalized size = 1.90 \begin {gather*} \frac {1}{2} \, x^{2} e^{\left (\frac {16}{5} \, \log \left (2\right )^{2}\right )} + \frac {1}{2} \, {\left (2 \, x^{2} \log \left (x e^{\left (-4\right )}\right ) - x^{2}\right )} e^{\left (\frac {16}{5} \, \log \left (2\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(16/5*log(2)^2)*log(x/exp(4))+x*exp(16/5*log(2)^2),x, algorithm="giac")

[Out]

1/2*x^2*e^(16/5*log(2)^2) + 1/2*(2*x^2*log(x*e^(-4)) - x^2)*e^(16/5*log(2)^2)

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Mupad [B]
time = 1.81, size = 15, normalized size = 0.75 \begin {gather*} x^2\,{\mathrm {e}}^{\frac {16\,{\ln \left (2\right )}^2}{5}}\,\left (\ln \left (x\right )-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*exp((16*log(2)^2)/5) + 2*x*exp((16*log(2)^2)/5)*log(x*exp(-4)),x)

[Out]

x^2*exp((16*log(2)^2)/5)*(log(x) - 4)

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