3.23.17 \(\int -\frac {5}{50 x+20 e^{25} x \log (x)+2 e^{50} x \log ^2(x)} \, dx\)

Optimal. Leaf size=17 \[ \frac {\log (x)}{2 \left (-5-e^{25} \log (x)\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 32} \begin {gather*} \frac {5}{2 e^{25} \left (e^{25} \log (x)+5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-5/(50*x + 20*E^25*x*Log[x] + 2*E^50*x*Log[x]^2),x]

[Out]

5/(2*E^25*(5 + E^25*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 (5 \int \frac {1}{50 x+20 e^{25} x \log (x)+2 e^{50} x \log ^2(x)} \, dx\right )\\ &=-\left (5 \operatorname {Subst}\left (\int \frac {1}{2 \left (5+e^{25} x\right )^2} \, dx,x,\log (x)\right )\right )\\ &=-\left (\frac {5}{2} \operatorname {Subst}\left (\int \frac {1}{\left (5+e^{25} x\right )^2} \, dx,x,\log (x)\right )\right )\\ &=\frac {5}{2 e^{25} \left (5+e^{25} \log (x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 17, normalized size = 1.00 \begin {gather*} \frac {5}{2 e^{25} \left (5+e^{25} \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-5/(50*x + 20*E^25*x*Log[x] + 2*E^50*x*Log[x]^2),x]

[Out]

5/(2*E^25*(5 + E^25*Log[x]))

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fricas [A]  time = 0.87, size = 14, normalized size = 0.82 \begin {gather*} \frac {5}{2 \, {\left (e^{50} \log \relax (x) + 5 \, e^{25}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/(2*x*exp(25)^2*log(x)^2+20*x*exp(25)*log(x)+50*x),x, algorithm="fricas")

[Out]

5/2/(e^50*log(x) + 5*e^25)

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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(-5/(2*x*exp(25)^2*log(x)^2+20*x*exp(25)*log(x)+50*x),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -1/2/sqrt(-exp(25)^2+exp(50))*atan((ln(s
ageVARx)*exp(50)+5*exp(25))/5/sqrt(-exp(25)^2+exp(50)))

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maple [A]  time = 0.13, size = 14, normalized size = 0.82

method	result	size
norman	\(-\frac {\ln \relax (x )}{2 \left ({\mathrm e}^{25} \ln \relax (x )+5\right )}\)	\(14\)
risch	\(\frac {5 \,{\mathrm e}^{-25}}{2 \left ({\mathrm e}^{25} \ln \relax (x )+5\right )}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5/(2*x*exp(25)^2*ln(x)^2+20*x*exp(25)*ln(x)+50*x),x,method=_RETURNVERBOSE)

[Out]

-1/2*ln(x)/(exp(25)*ln(x)+5)

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maxima [A]  time = 0.45, size = 14, normalized size = 0.82 \begin {gather*} \frac {5}{2 \, {\left (e^{50} \log \relax (x) + 5 \, e^{25}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/(2*x*exp(25)^2*log(x)^2+20*x*exp(25)*log(x)+50*x),x, algorithm="maxima")

[Out]

5/2/(e^50*log(x) + 5*e^25)

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mupad [B]  time = 1.41, size = 14, normalized size = 0.82 \begin {gather*} \frac {5\,{\mathrm {e}}^{-25}}{2\,{\mathrm {e}}^{25}\,\ln \relax (x)+10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5/(50*x + 20*x*exp(25)*log(x) + 2*x*exp(50)*log(x)^2),x)

[Out]

(5*exp(-25))/(2*exp(25)*log(x) + 10)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.82 \begin {gather*} \frac {5}{2 e^{50} \log {\relax (x )} + 10 e^{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5/(2*x*exp(25)**2*ln(x)**2+20*x*exp(25)*ln(x)+50*x),x)

[Out]

5/(2*exp(50)*log(x) + 10*exp(25))

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