3.103.5 \(\int \frac {e^{-2 x} (180 e^5 \log (\log (x))+e^5 (-180-180 x) \log (x) \log ^2(\log (x)))}{x^3 \log (x)} \, dx\)

Optimal. Leaf size=17 \[ \frac {90 e^{5-2 x} \log ^2(\log (x))}{x^2} \]

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Rubi [A]  time = 0.47, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {6741, 12, 2288} \begin {gather*} \frac {90 e^{5-2 x} \log ^2(\log (x))}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(180*E^5*Log[Log[x]] + E^5*(-180 - 180*x)*Log[x]*Log[Log[x]]^2)/(E^(2*x)*x^3*Log[x]),x]

[Out]

(90*E^(5 - 2*x)*Log[Log[x]]^2)/x^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {180 e^{5-2 x} \log (\log (x)) (1-\log (x) \log (\log (x))-x \log (x) \log (\log (x)))}{x^3 \log (x)} \, dx\\ &=180 \int \frac {e^{5-2 x} \log (\log (x)) (1-\log (x) \log (\log (x))-x \log (x) \log (\log (x)))}{x^3 \log (x)} \, dx\\ &=\frac {90 e^{5-2 x} \log ^2(\log (x))}{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 17, normalized size = 1.00 \begin {gather*} \frac {90 e^{5-2 x} \log ^2(\log (x))}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(180*E^5*Log[Log[x]] + E^5*(-180 - 180*x)*Log[x]*Log[Log[x]]^2)/(E^(2*x)*x^3*Log[x]),x]

[Out]

(90*E^(5 - 2*x)*Log[Log[x]]^2)/x^2

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fricas [A]  time = 0.91, size = 16, normalized size = 0.94 \begin {gather*} \frac {90 \, e^{\left (-2 \, x + 5\right )} \log \left (\log \relax (x)\right )^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-180*x-180)*exp(5)*log(x)*log(log(x))^2+180*exp(5)*log(log(x)))/x^3/exp(x)^2/log(x),x, algorithm="
fricas")

[Out]

90*e^(-2*x + 5)*log(log(x))^2/x^2

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giac [A]  time = 0.21, size = 16, normalized size = 0.94 \begin {gather*} \frac {90 \, e^{\left (-2 \, x + 5\right )} \log \left (\log \relax (x)\right )^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-180*x-180)*exp(5)*log(x)*log(log(x))^2+180*exp(5)*log(log(x)))/x^3/exp(x)^2/log(x),x, algorithm="
giac")

[Out]

90*e^(-2*x + 5)*log(log(x))^2/x^2

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maple [A]  time = 0.04, size = 17, normalized size = 1.00

method	result	size
risch	\(\frac {90 \ln \left (\ln \relax (x )\right )^{2} {\mathrm e}^{-2 x +5}}{x^{2}}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-180*x-180)*exp(5)*ln(x)*ln(ln(x))^2+180*exp(5)*ln(ln(x)))/x^3/exp(x)^2/ln(x),x,method=_RETURNVERBOSE)

[Out]

90/x^2*ln(ln(x))^2*exp(-2*x+5)

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maxima [A]  time = 0.43, size = 16, normalized size = 0.94 \begin {gather*} \frac {90 \, e^{\left (-2 \, x + 5\right )} \log \left (\log \relax (x)\right )^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-180*x-180)*exp(5)*log(x)*log(log(x))^2+180*exp(5)*log(log(x)))/x^3/exp(x)^2/log(x),x, algorithm="
maxima")

[Out]

90*e^(-2*x + 5)*log(log(x))^2/x^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int \frac {{\mathrm {e}}^{-2\,x}\,\left (180\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^5-{\ln \left (\ln \relax (x)\right )}^2\,{\mathrm {e}}^5\,\ln \relax (x)\,\left (180\,x+180\right )\right )}{x^3\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2*x)*(180*log(log(x))*exp(5) - log(log(x))^2*exp(5)*log(x)*(180*x + 180)))/(x^3*log(x)),x)

[Out]

int((exp(-2*x)*(180*log(log(x))*exp(5) - log(log(x))^2*exp(5)*log(x)*(180*x + 180)))/(x^3*log(x)), x)

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sympy [A]  time = 0.36, size = 19, normalized size = 1.12 \begin {gather*} \frac {90 e^{5} e^{- 2 x} \log {\left (\log {\relax (x )} \right )}^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-180*x-180)*exp(5)*ln(x)*ln(ln(x))**2+180*exp(5)*ln(ln(x)))/x**3/exp(x)**2/ln(x),x)

[Out]

90*exp(5)*exp(-2*x)*log(log(x))**2/x**2

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