3.72.43 \(\int \frac {18 e^5 x^2+54 e^5 x^2 \log (x)}{-1+e^5 (18 e+e^2)} \, dx\)

Optimal. Leaf size=23 \[ \frac {x^3 \log (x)}{e+\frac {1}{18} \left (-\frac {1}{e^5}+e^2\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {12, 2304} \begin {gather*} -\frac {18 e^5 x^3 \log (x)}{1-18 e^6-e^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18*E^5*x^2 + 54*E^5*x^2*Log[x])/(-1 + E^5*(18*E + E^2)),x]

[Out]

(-18*E^5*x^3*Log[x])/(1 - 18*E^6 - E^7)

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 {\int \left (18 e^5 x^2+54 e^5 x^2 \log (x)\right ) \, dx}{-1+e^5 \left (18 e+e^2\right )}\\ &=-\frac {6 e^5 x^3}{1-18 e^6-e^7}-\frac {\left (54 e^5\right ) \int x^2 \log (x) \, dx}{1-18 e^6-e^7}\\ &=-\frac {18 e^5 x^3 \log (x)}{1-18 e^6-e^7}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 22, normalized size = 0.96 \begin {gather*} \frac {18 e^5 x^3 \log (x)}{-1+18 e^6+e^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18*E^5*x^2 + 54*E^5*x^2*Log[x])/(-1 + E^5*(18*E + E^2)),x]

[Out]

(18*E^5*x^3*Log[x])/(-1 + 18*E^6 + E^7)

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fricas [A]  time = 1.57, size = 19, normalized size = 0.83 \begin {gather*} \frac {18 \, x^{3} e^{5} \log \relax (x)}{e^{7} + 18 \, e^{6} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((54*x^2*exp(5)*log(x)+18*x^2*exp(5))/((exp(2)+18*exp(1))*exp(5)-1),x, algorithm="fricas")

[Out]

18*x^3*e^5*log(x)/(e^7 + 18*e^6 - 1)

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giac [A]  time = 1.05, size = 39, normalized size = 1.70 \begin {gather*} \frac {6 \, {\left (x^{3} e^{5} + {\left (3 \, x^{3} \log \relax (x) - x^{3}\right )} e^{5}\right )}}{{\left (e^{2} + 18 \, e\right )} e^{5} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((54*x^2*exp(5)*log(x)+18*x^2*exp(5))/((exp(2)+18*exp(1))*exp(5)-1),x, algorithm="giac")

[Out]

6*(x^3*e^5 + (3*x^3*log(x) - x^3)*e^5)/((e^2 + 18*e)*e^5 - 1)

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maple [A]  time = 0.04, size = 24, normalized size = 1.04

method	result	size
default	\(\frac {18 x^{3} {\mathrm e}^{5} \ln \relax (x )}{\left ({\mathrm e}^{2}+18 \,{\mathrm e}\right ) {\mathrm e}^{5}-1}\)	\(24\)
risch	\(\frac {18 x^{3} {\mathrm e}^{5} \ln \relax (x )}{\left ({\mathrm e}^{2}+18 \,{\mathrm e}\right ) {\mathrm e}^{5}-1}\)	\(24\)
norman	\(\frac {18 \,{\mathrm e}^{5} x^{3} \ln \relax (x )}{{\mathrm e}^{2} {\mathrm e}^{5}+18 \,{\mathrm e} \,{\mathrm e}^{5}-1}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((54*x^2*exp(5)*ln(x)+18*x^2*exp(5))/((exp(2)+18*exp(1))*exp(5)-1),x,method=_RETURNVERBOSE)

[Out]

18/((exp(2)+18*exp(1))*exp(5)-1)*x^3*exp(5)*ln(x)

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maxima [A]  time = 0.37, size = 39, normalized size = 1.70 \begin {gather*} \frac {6 \, {\left (x^{3} e^{5} + {\left (3 \, x^{3} \log \relax (x) - x^{3}\right )} e^{5}\right )}}{{\left (e^{2} + 18 \, e\right )} e^{5} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((54*x^2*exp(5)*log(x)+18*x^2*exp(5))/((exp(2)+18*exp(1))*exp(5)-1),x, algorithm="maxima")

[Out]

6*(x^3*e^5 + (3*x^3*log(x) - x^3)*e^5)/((e^2 + 18*e)*e^5 - 1)

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mupad [B]  time = 4.13, size = 19, normalized size = 0.83 \begin {gather*} \frac {18\,x^3\,{\mathrm {e}}^5\,\ln \relax (x)}{18\,{\mathrm {e}}^6+{\mathrm {e}}^7-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((18*x^2*exp(5) + 54*x^2*exp(5)*log(x))/(exp(5)*(18*exp(1) + exp(2)) - 1),x)

[Out]

(18*x^3*exp(5)*log(x))/(18*exp(6) + exp(7) - 1)

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sympy [A]  time = 0.15, size = 20, normalized size = 0.87 \begin {gather*} \frac {18 x^{3} e^{5} \log {\relax (x )}}{-1 + e^{7} + 18 e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((54*x**2*exp(5)*ln(x)+18*x**2*exp(5))/((exp(2)+18*exp(1))*exp(5)-1),x)

[Out]

18*x**3*exp(5)*log(x)/(-1 + exp(7) + 18*exp(6))

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