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

Optimal. Leaf size=25 \[ e^2+\frac {\log (2)}{5 x \left (\frac {5}{4}+4 e^2+x\right )^2} \]

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Rubi [B]  time = 0.12, antiderivative size = 73, normalized size of antiderivative = 2.92, number of steps used = 4, number of rules used = 3, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6, 12, 2074} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[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 2074

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 0.84 \begin {gather*} \frac {16 \log (2)}{5 x \left (5+16 e^2+4 x\right )^2} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.57, size = 38, normalized size = 1.52 \begin {gather*} \frac {16 \, \log \relax (2)}{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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

Timed out

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maple [A]  time = 0.11, size = 19, normalized size = 0.76

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.35, size = 35, normalized size = 1.40 \begin {gather*} \frac {16 \, \log \relax (2)}{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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 2.21, size = 18, normalized size = 0.72 \begin {gather*} \frac {16\,\ln \relax (2)}{5\,x\,{\left (4\,x+16\,{\mathrm {e}}^2+5\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.65, size = 32, normalized size = 1.28 \begin {gather*} \frac {16 \log {\relax (2 )}}{80 x^{3} + x^{2} \left (200 + 640 e^{2}\right ) + x \left (125 + 800 e^{2} + 1280 e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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