\(\int \frac {(-80-256 e^2-192 x) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 (19200 x^2+15360 x^3)+e^2 (6000 x^2+9600 x^3+3840 x^4)} \, dx\) [3389]

Optimal result

Integrand size = 79, antiderivative size = 25 \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=e^2+\frac {\log (2)}{5 x \left (\frac {5}{4}+4 e^2+x\right )^2} \]

[Out]

exp(2)+1/5*ln(2)/(x+5/4+4*exp(2))^2/x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(73\) vs. \(2(25)=50\).

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.92, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6, 12, 2099} \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=-\frac {64 \log (2)}{5 \left (5+16 e^2\right )^2 \left (4 x+16 e^2+5\right )}-\frac {64 \log (2)}{5 \left (5+16 e^2\right ) \left (4 x+16 e^2+5\right )^2}+\frac {16 \log (2)}{5 \left (5+16 e^2\right )^2 x} \]

[In]

Int[((-80 - 256*E^2 - 192*x)*Log[2])/(625*x^2 + 20480*E^6*x^2 + 1500*x^3 + 1200*x^4 + 320*x^5 + E^4*(19200*x^2
+ 15360*x^3) + E^2*(6000*x^2 + 9600*x^3 + 3840*x^4)),x]

[Out]

(16*Log[2])/(5*(5 + 16*E^2)^2*x) - (64*Log[2])/(5*(5 + 16*E^2)*(5 + 16*E^2 + 4*x)^2) - (64*Log[2])/(5*(5 + 16*
E^2)^2*(5 + 16*E^2 + 4*x))

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 12

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

Rule 2099

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{\left (625+20480 e^6\right ) x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx \\ & = \log (2) \int \frac {-80-256 e^2-192 x}{\left (625+20480 e^6\right ) x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx \\ & = \log (2) \int \left (-\frac {16}{5 \left (5+16 e^2\right )^2 x^2}+\frac {512}{5 \left (5+16 e^2\right ) \left (5+16 e^2+4 x\right )^3}+\frac {256}{5 \left (5+16 e^2\right )^2 \left (5+16 e^2+4 x\right )^2}\right ) \, dx \\ & = \frac {16 \log (2)}{5 \left (5+16 e^2\right )^2 x}-\frac {64 \log (2)}{5 \left (5+16 e^2\right ) \left (5+16 e^2+4 x\right )^2}-\frac {64 \log (2)}{5 \left (5+16 e^2\right )^2 \left (5+16 e^2+4 x\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=\frac {16 \log (2)}{5 x \left (5+16 e^2+4 x\right )^2} \]

[In]

Integrate[((-80 - 256*E^2 - 192*x)*Log[2])/(625*x^2 + 20480*E^6*x^2 + 1500*x^3 + 1200*x^4 + 320*x^5 + E^4*(192
00*x^2 + 15360*x^3) + E^2*(6000*x^2 + 9600*x^3 + 3840*x^4)),x]

[Out]

(16*Log[2])/(5*x*(5 + 16*E^2 + 4*x)^2)

Maple [A] (verified)

Time = 0.93 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76

method	result	size
norman	\(\frac {16 \ln \left (2\right )}{5 x \left (16 \,{\mathrm e}^{2}+4 x +5\right )^{2}}\)	\(19\)
risch	\(\frac {16 \ln \left (2\right )}{5 x \left (256 \,{\mathrm e}^{4}+128 \,{\mathrm e}^{2} x +16 x^{2}+160 \,{\mathrm e}^{2}+40 x +25\right )}\)	\(33\)
gosper	\(\frac {16 \ln \left (2\right )}{5 x \left (256 \,{\mathrm e}^{4}+128 \,{\mathrm e}^{2} x +16 x^{2}+160 \,{\mathrm e}^{2}+40 x +25\right )}\)	\(35\)
parallelrisch	\(\frac {16 \ln \left (2\right )}{5 x \left (256 \,{\mathrm e}^{4}+128 \,{\mathrm e}^{2} x +16 x^{2}+160 \,{\mathrm e}^{2}+40 x +25\right )}\)	\(35\)
default	\(16 \ln \left (2\right ) \left (\frac {4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (64 \textit {\_Z}^{3}+\left (768 \,{\mathrm e}^{2}+240\right ) \textit {\_Z}^{2}+\left (1920 \,{\mathrm e}^{2}+3072 \,{\mathrm e}^{4}+300\right ) \textit {\_Z} +1200 \,{\mathrm e}^{2}+3840 \,{\mathrm e}^{4}+4096 \,{\mathrm e}^{6}+125\right )}{\sum }\frac {\left (9375+32000 \,{\mathrm e}^{2} \textit {\_R} +153600 \textit {\_R} \,{\mathrm e}^{4}+327680 \textit {\_R} \,{\mathrm e}^{6}+262144 \textit {\_R} \,{\mathrm e}^{8}+150000 \,{\mathrm e}^{2}+960000 \,{\mathrm e}^{4}+3072000 \,{\mathrm e}^{6}+4915200 \,{\mathrm e}^{8}+3145728 \,{\mathrm e}^{10}+2500 \textit {\_R} \right ) \ln \left (x -\textit {\_R} \right )}{25+256 \,{\mathrm e}^{4}+128 \,{\mathrm e}^{2} \textit {\_R} +16 \textit {\_R}^{2}+160 \,{\mathrm e}^{2}+40 \textit {\_R}}\right )}{15 \left (1200 \,{\mathrm e}^{2}+3840 \,{\mathrm e}^{4}+4096 \,{\mathrm e}^{6}+125\right )^{2}}-\frac {-38400 \,{\mathrm e}^{4}-81920 \,{\mathrm e}^{6}-65536 \,{\mathrm e}^{8}-8000 \,{\mathrm e}^{2}-625}{5 \left (1200 \,{\mathrm e}^{2}+3840 \,{\mathrm e}^{4}+4096 \,{\mathrm e}^{6}+125\right )^{2} x}\right )\)	\(185\)

[In]

int((-256*exp(2)-192*x-80)*ln(2)/(20480*x^2*exp(2)^3+(15360*x^3+19200*x^2)*exp(2)^2+(3840*x^4+9600*x^3+6000*x^
2)*exp(2)+320*x^5+1200*x^4+1500*x^3+625*x^2),x,method=_RETURNVERBOSE)

[Out]

16/5*ln(2)/x/(16*exp(2)+4*x+5)^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=\frac {16 \, \log \left (2\right )}{5 \, {\left (16 \, x^{3} + 40 \, x^{2} + 256 \, x e^{4} + 32 \, {\left (4 \, x^{2} + 5 \, x\right )} e^{2} + 25 \, x\right )}} \]

[In]

integrate((-256*exp(2)-192*x-80)*log(2)/(20480*x^2*exp(2)^3+(15360*x^3+19200*x^2)*exp(2)^2+(3840*x^4+9600*x^3+
6000*x^2)*exp(2)+320*x^5+1200*x^4+1500*x^3+625*x^2),x, algorithm="fricas")

[Out]

16/5*log(2)/(16*x^3 + 40*x^2 + 256*x*e^4 + 32*(4*x^2 + 5*x)*e^2 + 25*x)

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=\frac {16 \log {\left (2 \right )}}{80 x^{3} + x^{2} \cdot \left (200 + 640 e^{2}\right ) + x \left (125 + 800 e^{2} + 1280 e^{4}\right )} \]

[In]

integrate((-256*exp(2)-192*x-80)*ln(2)/(20480*x**2*exp(2)**3+(15360*x**3+19200*x**2)*exp(2)**2+(3840*x**4+9600
*x**3+6000*x**2)*exp(2)+320*x**5+1200*x**4+1500*x**3+625*x**2),x)

[Out]

16*log(2)/(80*x**3 + x**2*(200 + 640*exp(2)) + x*(125 + 800*exp(2) + 1280*exp(4)))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=\frac {16 \, \log \left (2\right )}{5 \, {\left (16 \, x^{3} + 8 \, x^{2} {\left (16 \, e^{2} + 5\right )} + x {\left (256 \, e^{4} + 160 \, e^{2} + 25\right )}\right )}} \]

[In]

integrate((-256*exp(2)-192*x-80)*log(2)/(20480*x^2*exp(2)^3+(15360*x^3+19200*x^2)*exp(2)^2+(3840*x^4+9600*x^3+
6000*x^2)*exp(2)+320*x^5+1200*x^4+1500*x^3+625*x^2),x, algorithm="maxima")

[Out]

16/5*log(2)/(16*x^3 + 8*x^2*(16*e^2 + 5) + x*(256*e^4 + 160*e^2 + 25))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (21) = 42\).

Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.24 \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=-\frac {16}{5} \, {\left (\frac {8 \, {\left (2 \, x + 16 \, e^{2} + 5\right )}}{{\left (4 \, x + 16 \, e^{2} + 5\right )}^{2} {\left (256 \, e^{4} + 160 \, e^{2} + 25\right )}} - \frac {1}{x {\left (256 \, e^{4} + 160 \, e^{2} + 25\right )}}\right )} \log \left (2\right ) \]

[In]

integrate((-256*exp(2)-192*x-80)*log(2)/(20480*x^2*exp(2)^3+(15360*x^3+19200*x^2)*exp(2)^2+(3840*x^4+9600*x^3+
6000*x^2)*exp(2)+320*x^5+1200*x^4+1500*x^3+625*x^2),x, algorithm="giac")

[Out]

-16/5*(8*(2*x + 16*e^2 + 5)/((4*x + 16*e^2 + 5)^2*(256*e^4 + 160*e^2 + 25)) - 1/(x*(256*e^4 + 160*e^2 + 25)))*
log(2)

Mupad [B] (verification not implemented)

Time = 8.58 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {\left (-80-256 e^2-192 x\right ) \log (2)}{625 x^2+20480 e^6 x^2+1500 x^3+1200 x^4+320 x^5+e^4 \left (19200 x^2+15360 x^3\right )+e^2 \left (6000 x^2+9600 x^3+3840 x^4\right )} \, dx=\frac {16\,\ln \left (2\right )}{5\,x\,{\left (4\,x+16\,{\mathrm {e}}^2+5\right )}^2} \]

[In]

int(-(log(2)*(192*x + 256*exp(2) + 80))/(exp(4)*(19200*x^2 + 15360*x^3) + 20480*x^2*exp(6) + exp(2)*(6000*x^2
+ 9600*x^3 + 3840*x^4) + 625*x^2 + 1500*x^3 + 1200*x^4 + 320*x^5),x)

[Out]

(16*log(2))/(5*x*(4*x + 16*exp(2) + 5)^2)