3.2.49 \(\int \frac {e^{\frac {9}{e^{10} \log ^2(e^5+x)}} (-18-18 \log (3))}{(e^{15}+e^{10} x) \log ^3(e^5+x)} \, dx\)

Optimal. Leaf size=20 \[ e^{\frac {9}{e^{10} \log ^2\left (e^5+x\right )}} (1+\log (3)) \]

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Rubi [A]  time = 0.19, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {12, 6706} \begin {gather*} (1+\log (3)) e^{\frac {9}{e^{10} \log ^2\left (x+e^5\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(9/(E^10*Log[E^5 + x]^2))*(-18 - 18*Log[3]))/((E^15 + E^10*x)*Log[E^5 + x]^3),x]

[Out]

E^(9/(E^10*Log[E^5 + x]^2))*(1 + Log[3])

Rule 12

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left ((18 (1+\log (3))) \int \frac {e^{\frac {9}{e^{10} \log ^2\left (e^5+x\right )}}}{\left (e^{15}+e^{10} x\right ) \log ^3\left (e^5+x\right )} \, dx\right )\\ &=e^{\frac {9}{e^{10} \log ^2\left (e^5+x\right )}} (1+\log (3))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 20, normalized size = 1.00 \begin {gather*} e^{\frac {9}{e^{10} \log ^2\left (e^5+x\right )}} (1+\log (3)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(9/(E^10*Log[E^5 + x]^2))*(-18 - 18*Log[3]))/((E^15 + E^10*x)*Log[E^5 + x]^3),x]

[Out]

E^(9/(E^10*Log[E^5 + x]^2))*(1 + Log[3])

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fricas [A]  time = 1.10, size = 17, normalized size = 0.85 \begin {gather*} {\left (\log \relax (3) + 1\right )} e^{\left (\frac {9 \, e^{\left (-10\right )}}{\log \left (x + e^{5}\right )^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*log(3)-18)*exp(9/exp(5)^2/log(exp(5)+x)^2)/(exp(5)^3+x*exp(5)^2)/log(exp(5)+x)^3,x, algorithm="
fricas")

[Out]

(log(3) + 1)*e^(9*e^(-10)/log(x + e^5)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*log(3)-18)*exp(9/exp(5)^2/log(exp(5)+x)^2)/(exp(5)^3+x*exp(5)^2)/log(exp(5)+x)^3,x, algorithm="
giac")

[Out]

undef

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maple [A]  time = 0.09, size = 20, normalized size = 1.00

method	result	size
norman	\(\left (\ln \relax (3)+1\right ) {\mathrm e}^{\frac {9 \,{\mathrm e}^{-10}}{\ln \left ({\mathrm e}^{5}+x \right )^{2}}}\)	\(20\)
risch	\({\mathrm e}^{\frac {9 \,{\mathrm e}^{-10}}{\ln \left ({\mathrm e}^{5}+x \right )^{2}}} \ln \relax (3)+{\mathrm e}^{\frac {9 \,{\mathrm e}^{-10}}{\ln \left ({\mathrm e}^{5}+x \right )^{2}}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-18*ln(3)-18)*exp(9/exp(5)^2/ln(exp(5)+x)^2)/(exp(5)^3+x*exp(5)^2)/ln(exp(5)+x)^3,x,method=_RETURNVERBOSE
)

[Out]

(ln(3)+1)*exp(9/exp(5)^2/ln(exp(5)+x)^2)

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maxima [A]  time = 0.35, size = 17, normalized size = 0.85 \begin {gather*} {\left (\log \relax (3) + 1\right )} e^{\left (\frac {9 \, e^{\left (-10\right )}}{\log \left (x + e^{5}\right )^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*log(3)-18)*exp(9/exp(5)^2/log(exp(5)+x)^2)/(exp(5)^3+x*exp(5)^2)/log(exp(5)+x)^3,x, algorithm="
maxima")

[Out]

(log(3) + 1)*e^(9*e^(-10)/log(x + e^5)^2)

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mupad [B]  time = 0.67, size = 17, normalized size = 0.85 \begin {gather*} {\mathrm {e}}^{\frac {9\,{\mathrm {e}}^{-10}}{{\ln \left (x+{\mathrm {e}}^5\right )}^2}}\,\left (\ln \relax (3)+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((9*exp(-10))/log(x + exp(5))^2)*(18*log(3) + 18))/(log(x + exp(5))^3*(exp(15) + x*exp(10))),x)

[Out]

exp((9*exp(-10))/log(x + exp(5))^2)*(log(3) + 1)

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sympy [A]  time = 0.38, size = 19, normalized size = 0.95 \begin {gather*} \left (1 + \log {\relax (3 )}\right ) e^{\frac {9}{e^{10} \log {\left (x + e^{5} \right )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-18*ln(3)-18)*exp(9/exp(5)**2/ln(exp(5)+x)**2)/(exp(5)**3+x*exp(5)**2)/ln(exp(5)+x)**3,x)

[Out]

(1 + log(3))*exp(9*exp(-10)/log(x + exp(5))**2)

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