3.91.98 \(\int \frac {1}{18} (e^4 x+e^4 x \log (\frac {x^2}{5})) \, dx\)

Optimal. Leaf size=18 \[ \frac {1}{36} e^4 x^2 \log \left (\frac {x^2}{5}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {12, 2304} \begin {gather*} \frac {1}{36} e^4 x^2 \log \left (\frac {x^2}{5}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^4*x + E^4*x*Log[x^2/5])/18,x]

[Out]

(E^4*x^2*Log[x^2/5])/36

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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}{18} \int \left (e^4 x+e^4 x \log \left (\frac {x^2}{5}\right )\right ) \, dx\\ &=\frac {e^4 x^2}{36}+\frac {1}{18} e^4 \int x \log \left (\frac {x^2}{5}\right ) \, dx\\ &=\frac {1}{36} e^4 x^2 \log \left (\frac {x^2}{5}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 18, normalized size = 1.00 \begin {gather*} \frac {1}{36} e^4 x^2 \log \left (\frac {x^2}{5}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^4*x + E^4*x*Log[x^2/5])/18,x]

[Out]

(E^4*x^2*Log[x^2/5])/36

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fricas [A]  time = 0.51, size = 13, normalized size = 0.72 \begin {gather*} \frac {1}{36} \, x^{2} e^{4} \log \left (\frac {1}{5} \, x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*x*exp(4)*log(1/5*x^2)+1/18*x*exp(4),x, algorithm="fricas")

[Out]

1/36*x^2*e^4*log(1/5*x^2)

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giac [B]  time = 0.18, size = 28, normalized size = 1.56 \begin {gather*} \frac {1}{36} \, x^{2} e^{4} + \frac {1}{36} \, {\left (x^{2} \log \left (\frac {1}{5} \, x^{2}\right ) - x^{2}\right )} e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*x*exp(4)*log(1/5*x^2)+1/18*x*exp(4),x, algorithm="giac")

[Out]

1/36*x^2*e^4 + 1/36*(x^2*log(1/5*x^2) - x^2)*e^4

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maple [A]  time = 0.03, size = 14, normalized size = 0.78

method	result	size
default	\(\frac {x^{2} \ln \left (\frac {x^{2}}{5}\right ) {\mathrm e}^{4}}{36}\)	\(14\)
norman	\(\frac {x^{2} \ln \left (\frac {x^{2}}{5}\right ) {\mathrm e}^{4}}{36}\)	\(14\)
risch	\(\frac {x^{2} \ln \left (\frac {x^{2}}{5}\right ) {\mathrm e}^{4}}{36}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/18*x*exp(4)*ln(1/5*x^2)+1/18*x*exp(4),x,method=_RETURNVERBOSE)

[Out]

1/36*x^2*ln(1/5*x^2)*exp(4)

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maxima [B]  time = 0.34, size = 28, normalized size = 1.56 \begin {gather*} \frac {1}{36} \, x^{2} e^{4} + \frac {1}{36} \, {\left (x^{2} \log \left (\frac {1}{5} \, x^{2}\right ) - x^{2}\right )} e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*x*exp(4)*log(1/5*x^2)+1/18*x*exp(4),x, algorithm="maxima")

[Out]

1/36*x^2*e^4 + 1/36*(x^2*log(1/5*x^2) - x^2)*e^4

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mupad [B]  time = 6.67, size = 16, normalized size = 0.89 \begin {gather*} \frac {x^2\,{\mathrm {e}}^4\,\left (\ln \left (x^2\right )-\ln \relax (5)\right )}{36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(4))/18 + (x*exp(4)*log(x^2/5))/18,x)

[Out]

(x^2*exp(4)*(log(x^2) - log(5)))/36

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sympy [A]  time = 0.10, size = 14, normalized size = 0.78 \begin {gather*} \frac {x^{2} e^{4} \log {\left (\frac {x^{2}}{5} \right )}}{36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/18*x*exp(4)*ln(1/5*x**2)+1/18*x*exp(4),x)

[Out]

x**2*exp(4)*log(x**2/5)/36

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