\(\int \frac {e^{\frac {9}{e^{10} \log ^2(e^5+x)}} (-18-18 \log (3))}{(e^{15}+e^{10} x) \log ^3(e^5+x)} \, dx\) [149]

Optimal result

Integrand size = 41, antiderivative size = 20 \[ \int \frac {e^{\frac {9}{e^{10} \log ^2\left (e^5+x\right )}} (-18-18 \log (3))}{\left (e^{15}+e^{10} x\right ) \log ^3\left (e^5+x\right )} \, dx=e^{\frac {9}{e^{10} \log ^2\left (e^5+x\right )}} (1+\log (3)) \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {12, 6838} \[ \int \frac {e^{\frac {9}{e^{10} \log ^2\left (e^5+x\right )}} (-18-18 \log (3))}{\left (e^{15}+e^{10} x\right ) \log ^3\left (e^5+x\right )} \, dx=(1+\log (3)) e^{\frac {9}{e^{10} \log ^2\left (x+e^5\right )}} \]

[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 6838

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {9}{e^{10} \log ^2\left (e^5+x\right )}} (-18-18 \log (3))}{\left (e^{15}+e^{10} x\right ) \log ^3\left (e^5+x\right )} \, dx=e^{\frac {9}{e^{10} \log ^2\left (e^5+x\right )}} (1+\log (3)) \]

[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])

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

method	result	size
norman	\(\left (\ln \left (3\right )+1\right ) {\mathrm e}^{\frac {9 \,{\mathrm e}^{-10}}{\ln \left ({\mathrm e}^{5}+x \right )^{2}}}\)	\(20\)
parallelrisch	\(-\frac {\left (-18 \ln \left (3\right )-18\right ) {\mathrm e}^{\frac {9 \,{\mathrm e}^{-10}}{\ln \left ({\mathrm e}^{5}+x \right )^{2}}}}{18}\)	\(23\)
risch	\({\mathrm e}^{\frac {9 \,{\mathrm e}^{-10}}{\ln \left ({\mathrm e}^{5}+x \right )^{2}}} \ln \left (3\right )+{\mathrm e}^{\frac {9 \,{\mathrm e}^{-10}}{\ln \left ({\mathrm e}^{5}+x \right )^{2}}}\)	\(29\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {9}{e^{10} \log ^2\left (e^5+x\right )}} (-18-18 \log (3))}{\left (e^{15}+e^{10} x\right ) \log ^3\left (e^5+x\right )} \, dx={\left (\log \left (3\right ) + 1\right )} e^{\left (\frac {9 \, e^{\left (-10\right )}}{\log \left (x + e^{5}\right )^{2}}\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\frac {9}{e^{10} \log ^2\left (e^5+x\right )}} (-18-18 \log (3))}{\left (e^{15}+e^{10} x\right ) \log ^3\left (e^5+x\right )} \, dx=\left (1 + \log {\left (3 \right )}\right ) e^{\frac {9}{e^{10} \log {\left (x + e^{5} \right )}^{2}}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {9}{e^{10} \log ^2\left (e^5+x\right )}} (-18-18 \log (3))}{\left (e^{15}+e^{10} x\right ) \log ^3\left (e^5+x\right )} \, dx={\left (\log \left (3\right ) + 1\right )} e^{\left (\frac {9 \, e^{\left (-10\right )}}{\log \left (x + e^{5}\right )^{2}}\right )} \]

[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)

Giac [F]

\[ \int \frac {e^{\frac {9}{e^{10} \log ^2\left (e^5+x\right )}} (-18-18 \log (3))}{\left (e^{15}+e^{10} x\right ) \log ^3\left (e^5+x\right )} \, dx=\int { -\frac {18 \, {\left (\log \left (3\right ) + 1\right )} e^{\left (\frac {9 \, e^{\left (-10\right )}}{\log \left (x + e^{5}\right )^{2}}\right )}}{{\left (x e^{10} + e^{15}\right )} \log \left (x + e^{5}\right )^{3}} \,d x } \]

[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

Mupad [B] (verification not implemented)

Time = 8.90 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {9}{e^{10} \log ^2\left (e^5+x\right )}} (-18-18 \log (3))}{\left (e^{15}+e^{10} x\right ) \log ^3\left (e^5+x\right )} \, dx={\mathrm {e}}^{\frac {9\,{\mathrm {e}}^{-10}}{{\ln \left (x+{\mathrm {e}}^5\right )}^2}}\,\left (\ln \left (3\right )+1\right ) \]

[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)