3.89.80 \(\int (e^4 x \log (4)+2 e^4 x \log (4) \log (x)) \, dx\)

Optimal. Leaf size=15 \[ \frac {9}{5}+e^4 x^2 \log (4) \log (x) \]

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Rubi [A]  time = 0.01, antiderivative size = 11, normalized size of antiderivative = 0.73, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2304} \begin {gather*} e^4 x^2 \log (4) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^4*x*Log[4] + 2*E^4*x*Log[4]*Log[x],x]

[Out]

E^4*x^2*Log[4]*Log[x]

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

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Mathematica [A]  time = 0.00, size = 11, normalized size = 0.73 \begin {gather*} e^4 x^2 \log (4) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^4*x*Log[4] + 2*E^4*x*Log[4]*Log[x],x]

[Out]

E^4*x^2*Log[4]*Log[x]

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fricas [A]  time = 0.55, size = 11, normalized size = 0.73 \begin {gather*} 2 \, x^{2} e^{4} \log \relax (2) \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.16, size = 27, normalized size = 1.80 \begin {gather*} x^{2} e^{4} \log \relax (2) + {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} e^{4} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 12, normalized size = 0.80

method	result	size
risch	\(2 \ln \relax (x ) \ln \relax (2) {\mathrm e}^{4} x^{2}\)	\(12\)
default	\(2 \ln \relax (x ) \ln \relax (2) {\mathrm e}^{4} x^{2}\)	\(14\)
norman	\(2 \ln \relax (x ) \ln \relax (2) {\mathrm e}^{4} x^{2}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*ln(x)*ln(2)*exp(4)*x^2

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maxima [B]  time = 0.37, size = 27, normalized size = 1.80 \begin {gather*} x^{2} e^{4} \log \relax (2) + {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} e^{4} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.12, size = 11, normalized size = 0.73 \begin {gather*} 2\,x^2\,{\mathrm {e}}^4\,\ln \relax (2)\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x*exp(4)*log(2) + 4*x*exp(4)*log(2)*log(x),x)

[Out]

2*x^2*exp(4)*log(2)*log(x)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.93 \begin {gather*} 2 x^{2} e^{4} \log {\relax (2 )} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4*x*exp(1)**4*ln(2)*ln(x)+2*x*exp(1)**4*ln(2),x)

[Out]

2*x**2*exp(4)*log(2)*log(x)

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