3.63.77 \(\int -\frac {9 e^{\frac {16}{e^2}}}{81 e^8 x+18 e^4 x \log (x)+x \log ^2(x)} \, dx\)

Optimal. Leaf size=20 \[ \frac {e^{\frac {16}{e^2}}}{e^4+\frac {\log (x)}{9}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 32} \begin {gather*} \frac {9 e^{\frac {16}{e^2}}}{\log (x)+9 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-9*E^(16/E^2))/(81*E^8*x + 18*E^4*x*Log[x] + x*Log[x]^2),x]

[Out]

(9*E^(16/E^2))/(9*E^4 + 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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (\left (9 e^{\frac {16}{e^2}}\right ) \int \frac {1}{81 e^8 x+18 e^4 x \log (x)+x \log ^2(x)} \, dx\right )\\ &=-\left (\left (9 e^{\frac {16}{e^2}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (9 e^4+x\right )^2} \, dx,x,\log (x)\right )\right )\\ &=\frac {9 e^{\frac {16}{e^2}}}{9 e^4+\log (x)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 19, normalized size = 0.95 \begin {gather*} \frac {9 e^{\frac {16}{e^2}}}{9 e^4+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9*E^(16/E^2))/(81*E^8*x + 18*E^4*x*Log[x] + x*Log[x]^2),x]

[Out]

(9*E^(16/E^2))/(9*E^4 + Log[x])

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 16, normalized size = 0.80 \begin {gather*} \frac {9 \, e^{\left (16 \, e^{\left (-2\right )}\right )}}{9 \, e^{4} + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9*exp(16/exp(1)^2)/(x*log(x)^2+18*x*exp(4)*log(x)+81*x*exp(4)^2),x, algorithm="fricas")

[Out]

9*e^(16*e^(-2))/(9*e^4 + log(x))

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9*exp(16/exp(1)^2)/(x*log(x)^2+18*x*exp(4)*log(x)+81*x*exp(4)^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -9*exp(16/exp(2))/18/sqrt(-exp(4)^2+exp(
8))*(-i)*ln(sqrt((2*ln(abs(sageVARx))+18*exp(4)+18*sqrt(exp(4)^2-exp(8)))^2+(2*(1-sign(sageVARx))*pi/2)^2)/sqr
t((2*ln(abs(sageVARx)

________________________________________________________________________________________

maple [A]  time = 0.18, size = 17, normalized size = 0.85

method	result	size
risch	\(\frac {9 \,{\mathrm e}^{16 \,{\mathrm e}^{-2}}}{\ln \relax (x )+9 \,{\mathrm e}^{4}}\)	\(17\)
norman	\(\frac {9 \,{\mathrm e}^{16 \,{\mathrm e}^{-2}}}{\ln \relax (x )+9 \,{\mathrm e}^{4}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-9*exp(16/exp(1)^2)/(x*ln(x)^2+18*x*exp(4)*ln(x)+81*x*exp(4)^2),x,method=_RETURNVERBOSE)

[Out]

9*exp(16*exp(-2))/(ln(x)+9*exp(4))

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 16, normalized size = 0.80 \begin {gather*} \frac {9 \, e^{\left (16 \, e^{\left (-2\right )}\right )}}{9 \, e^{4} + \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9*exp(16/exp(1)^2)/(x*log(x)^2+18*x*exp(4)*log(x)+81*x*exp(4)^2),x, algorithm="maxima")

[Out]

9*e^(16*e^(-2))/(9*e^4 + log(x))

________________________________________________________________________________________

mupad [B]  time = 4.29, size = 16, normalized size = 0.80 \begin {gather*} \frac {9\,{\mathrm {e}}^{16\,{\mathrm {e}}^{-2}}}{9\,{\mathrm {e}}^4+\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(9*exp(16*exp(-2)))/(x*log(x)^2 + 81*x*exp(8) + 18*x*exp(4)*log(x)),x)

[Out]

(9*exp(16*exp(-2)))/(9*exp(4) + log(x))

________________________________________________________________________________________

sympy [A]  time = 0.09, size = 15, normalized size = 0.75 \begin {gather*} \frac {9 e^{\frac {16}{e^{2}}}}{\log {\relax (x )} + 9 e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-9*exp(16/exp(1)**2)/(x*ln(x)**2+18*x*exp(4)*ln(x)+81*x*exp(4)**2),x)

[Out]

9*exp(16*exp(-2))/(log(x) + 9*exp(4))

________________________________________________________________________________________