\(\int \frac {-4 x^2+(5-e^5+3 x) \log (15)}{(10 x^2-2 e^5 x^2+2 x^3+(-5 x+e^5 x-x^2) \log (15)) \log (\frac {-2 x+\log (15)}{e^{15} x+e^{10} (-10 x-2 x^2)+e^5 (25 x+10 x^2+x^3)})} \, dx\) [5363]

Optimal result

Integrand size = 105, antiderivative size = 24 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\log \left (\log \left (\frac {-2+\frac {\log (15)}{x}}{e^5 \left (-5+e^5-x\right )^2}\right )\right ) \]

[Out]

ln(ln((ln(15)/x-2)/(exp(5)-x-5)^2/exp(5)))

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {6, 6820, 6816} \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\log \left (\log \left (-\frac {2 x-\log (15)}{e^5 x \left (x-e^5+5\right )^2}\right )\right ) \]

[In]

Int[(-4*x^2 + (5 - E^5 + 3*x)*Log[15])/((10*x^2 - 2*E^5*x^2 + 2*x^3 + (-5*x + E^5*x - x^2)*Log[15])*Log[(-2*x
+ Log[15])/(E^15*x + E^10*(-10*x - 2*x^2) + E^5*(25*x + 10*x^2 + x^3))]),x]

[Out]

Log[Log[-((2*x - Log[15])/(E^5*x*(5 - E^5 + x)^2))]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (\left (10-2 e^5\right ) x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx \\ & = \int \frac {-4 x^2+\left (5-e^5\right ) \log (15)+3 x \log (15)}{x \left (5-e^5+x\right ) (2 x-\log (15)) \log \left (\frac {-2 x+\log (15)}{e^5 x \left (5-e^5+x\right )^2}\right )} \, dx \\ & = \log \left (\log \left (-\frac {2 x-\log (15)}{e^5 x \left (5-e^5+x\right )^2}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\log \left (\log \left (\frac {-2 x+\log (15)}{e^5 x \left (5-e^5+x\right )^2}\right )\right ) \]

[In]

Integrate[(-4*x^2 + (5 - E^5 + 3*x)*Log[15])/((10*x^2 - 2*E^5*x^2 + 2*x^3 + (-5*x + E^5*x - x^2)*Log[15])*Log[
(-2*x + Log[15])/(E^15*x + E^10*(-10*x - 2*x^2) + E^5*(25*x + 10*x^2 + x^3))]),x]

[Out]

Log[Log[(-2*x + Log[15])/(E^5*x*(5 - E^5 + x)^2)]]

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58

method	result	size
default	\(\ln \left (-5+\ln \left (\frac {\ln \left (15\right )-2 x}{x \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}-10 \,{\mathrm e}^{5}+10 x +25\right )}\right )\right )\)	\(38\)
parallelrisch	\(\ln \left (\ln \left (\frac {\left (\ln \left (15\right )-2 x \right ) {\mathrm e}^{-5}}{x \left ({\mathrm e}^{10}-2 x \,{\mathrm e}^{5}+x^{2}-10 \,{\mathrm e}^{5}+10 x +25\right )}\right )\right )\)	\(40\)
risch	\(\ln \left (\ln \left (\frac {\ln \left (3\right )+\ln \left (5\right )-2 x}{x \,{\mathrm e}^{15}+\left (-2 x^{2}-10 x \right ) {\mathrm e}^{10}+\left (x^{3}+10 x^{2}+25 x \right ) {\mathrm e}^{5}}\right )\right )\)	\(46\)
norman	\(\ln \left (\ln \left (\frac {\ln \left (15\right )-2 x}{x \,{\mathrm e}^{15}+\left (-2 x^{2}-10 x \right ) {\mathrm e}^{10}+\left (x^{3}+10 x^{2}+25 x \right ) {\mathrm e}^{5}}\right )\right )\)	\(48\)

[In]

int(((-exp(5)+3*x+5)*ln(15)-4*x^2)/((x*exp(5)-x^2-5*x)*ln(15)-2*x^2*exp(5)+2*x^3+10*x^2)/ln((ln(15)-2*x)/(x*ex
p(5)^3+(-2*x^2-10*x)*exp(5)^2+(x^3+10*x^2+25*x)*exp(5))),x,method=_RETURNVERBOSE)

[Out]

ln(-5+ln((ln(15)-2*x)/x/(exp(5)^2-2*x*exp(5)+x^2-10*exp(5)+10*x+25)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).

Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\log \left (\log \left (-\frac {2 \, x - \log \left (15\right )}{x e^{15} - 2 \, {\left (x^{2} + 5 \, x\right )} e^{10} + {\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} e^{5}}\right )\right ) \]

[In]

integrate(((-exp(5)+3*x+5)*log(15)-4*x^2)/((x*exp(5)-x^2-5*x)*log(15)-2*x^2*exp(5)+2*x^3+10*x^2)/log((log(15)-
2*x)/(x*exp(5)^3+(-2*x^2-10*x)*exp(5)^2+(x^3+10*x^2+25*x)*exp(5))),x, algorithm="fricas")

[Out]

log(log(-(2*x - log(15))/(x*e^15 - 2*(x^2 + 5*x)*e^10 + (x^3 + 10*x^2 + 25*x)*e^5)))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).

Time = 0.34 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\log {\left (\log {\left (\frac {- 2 x + \log {\left (15 \right )}}{x e^{15} + \left (- 2 x^{2} - 10 x\right ) e^{10} + \left (x^{3} + 10 x^{2} + 25 x\right ) e^{5}} \right )} \right )} \]

[In]

integrate(((-exp(5)+3*x+5)*ln(15)-4*x**2)/((x*exp(5)-x**2-5*x)*ln(15)-2*x**2*exp(5)+2*x**3+10*x**2)/ln((ln(15)
-2*x)/(x*exp(5)**3+(-2*x**2-10*x)*exp(5)**2+(x**3+10*x**2+25*x)*exp(5))),x)

[Out]

log(log((-2*x + log(15))/(x*exp(15) + (-2*x**2 - 10*x)*exp(10) + (x**3 + 10*x**2 + 25*x)*exp(5))))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\log \left (-2 \, \log \left (x - e^{5} + 5\right ) - \log \left (x\right ) + \log \left (-2 \, x + \log \left (5\right ) + \log \left (3\right )\right ) - 5\right ) \]

[In]

integrate(((-exp(5)+3*x+5)*log(15)-4*x^2)/((x*exp(5)-x^2-5*x)*log(15)-2*x^2*exp(5)+2*x^3+10*x^2)/log((log(15)-
2*x)/(x*exp(5)^3+(-2*x^2-10*x)*exp(5)^2+(x^3+10*x^2+25*x)*exp(5))),x, algorithm="maxima")

[Out]

log(-2*log(x - e^5 + 5) - log(x) + log(-2*x + log(5) + log(3)) - 5)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (22) = 44\).

Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\log \left (-\log \left (x^{3} e^{5} - 2 \, x^{2} e^{10} + 10 \, x^{2} e^{5} + x e^{15} - 10 \, x e^{10} + 25 \, x e^{5}\right ) + \log \left (-2 \, x + \log \left (15\right )\right )\right ) \]

[In]

integrate(((-exp(5)+3*x+5)*log(15)-4*x^2)/((x*exp(5)-x^2-5*x)*log(15)-2*x^2*exp(5)+2*x^3+10*x^2)/log((log(15)-
2*x)/(x*exp(5)^3+(-2*x^2-10*x)*exp(5)^2+(x^3+10*x^2+25*x)*exp(5))),x, algorithm="giac")

[Out]

log(-log(x^3*e^5 - 2*x^2*e^10 + 10*x^2*e^5 + x*e^15 - 10*x*e^10 + 25*x*e^5) + log(-2*x + log(15)))

Mupad [B] (verification not implemented)

Time = 14.69 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {-4 x^2+\left (5-e^5+3 x\right ) \log (15)}{\left (10 x^2-2 e^5 x^2+2 x^3+\left (-5 x+e^5 x-x^2\right ) \log (15)\right ) \log \left (\frac {-2 x+\log (15)}{e^{15} x+e^{10} \left (-10 x-2 x^2\right )+e^5 \left (25 x+10 x^2+x^3\right )}\right )} \, dx=\ln \left (\ln \left (-\frac {2\,x-\ln \left (15\right )}{{\mathrm {e}}^5\,\left (x^3+10\,x^2+25\,x\right )-{\mathrm {e}}^{10}\,\left (2\,x^2+10\,x\right )+x\,{\mathrm {e}}^{15}}\right )\right ) \]

[In]

int(-(log(15)*(3*x - exp(5) + 5) - 4*x^2)/(log(-(2*x - log(15))/(exp(5)*(25*x + 10*x^2 + x^3) - exp(10)*(10*x
+ 2*x^2) + x*exp(15)))*(log(15)*(5*x - x*exp(5) + x^2) + 2*x^2*exp(5) - 10*x^2 - 2*x^3)),x)

[Out]

log(log(-(2*x - log(15))/(exp(5)*(25*x + 10*x^2 + x^3) - exp(10)*(10*x + 2*x^2) + x*exp(15))))