3.32.87 \(\int \frac {1280 e^6 x^4+1024 x^5+2048 e^3 x^5+768 x^6}{(e^{12}+4 e^9 x+x^2+2 x^3+x^4+e^6 (2 x+6 x^2)+e^3 (4 x^2+4 x^3)) \log ^4(2)} \, dx\)

Optimal. Leaf size=20 \[ \frac {256 x^5}{\left (x+\left (e^3+x\right )^2\right ) \log ^4(2)} \]

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Rubi [A]  time = 0.26, antiderivative size = 27, normalized size of antiderivative = 1.35, number of steps used = 6, number of rules used = 5, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6, 12, 1594, 1680, 1590} \begin {gather*} \frac {256 x^5}{\left (x^2+\left (1+2 e^3\right ) x+e^6\right ) \log ^4(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1280*E^6*x^4 + 1024*x^5 + 2048*E^3*x^5 + 768*x^6)/((E^12 + 4*E^9*x + x^2 + 2*x^3 + x^4 + E^6*(2*x + 6*x^2
) + E^3*(4*x^2 + 4*x^3))*Log[2]^4),x]

[Out]

(256*x^5)/((E^6 + (1 + 2*E^3)*x + x^2)*Log[2]^4)

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 1590

Int[(Pp_)*(Qq_)^(m_.)*(Rr_)^(n_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x], r = Expon[Rr, x]}, S
imp[(Coeff[Pp, x, p]*x^(p - q - r + 1)*Qq^(m + 1)*Rr^(n + 1))/((p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x
, r]), x] /; NeQ[p + m*q + n*r + 1, 0] && EqQ[(p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x, r]*Pp, Coeff[Pp
, x, p]*x^(p - q - r)*((p - q - r + 1)*Qq*Rr + (m + 1)*x*Rr*D[Qq, x] + (n + 1)*x*Qq*D[Rr, x])]] /; FreeQ[{m, n
}, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && PolyQ[Rr, x] && NeQ[m, -1] && NeQ[n, -1]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1280 e^6 x^4+\left (1024+2048 e^3\right ) x^5+768 x^6}{\left (e^{12}+4 e^9 x+x^2+2 x^3+x^4+e^6 \left (2 x+6 x^2\right )+e^3 \left (4 x^2+4 x^3\right )\right ) \log ^4(2)} \, dx\\ &=\frac {\int \frac {1280 e^6 x^4+\left (1024+2048 e^3\right ) x^5+768 x^6}{e^{12}+4 e^9 x+x^2+2 x^3+x^4+e^6 \left (2 x+6 x^2\right )+e^3 \left (4 x^2+4 x^3\right )} \, dx}{\log ^4(2)}\\ &=\frac {\int \frac {x^4 \left (1280 e^6+\left (1024+2048 e^3\right ) x+768 x^2\right )}{e^{12}+4 e^9 x+x^2+2 x^3+x^4+e^6 \left (2 x+6 x^2\right )+e^3 \left (4 x^2+4 x^3\right )} \, dx}{\log ^4(2)}\\ &=\frac {\operatorname {Subst}\left (\int \frac {64 \left (1+2 e^3-2 x\right )^4 \left (-5 \left (1+4 e^3\right )+4 \left (1+2 e^3\right ) x+12 x^2\right )}{\left (1+4 e^3-4 x^2\right )^2} \, dx,x,\frac {1}{4} \left (2+4 e^3\right )+x\right )}{\log ^4(2)}\\ &=\frac {64 \operatorname {Subst}\left (\int \frac {\left (1+2 e^3-2 x\right )^4 \left (-5 \left (1+4 e^3\right )+4 \left (1+2 e^3\right ) x+12 x^2\right )}{\left (1+4 e^3-4 x^2\right )^2} \, dx,x,\frac {1}{4} \left (2+4 e^3\right )+x\right )}{\log ^4(2)}\\ &=\frac {256 x^5}{\left (e^6+\left (1+2 e^3\right ) x+x^2\right ) \log ^4(2)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 85, normalized size = 4.25 \begin {gather*} \frac {256 \left (4 e^{15}+2 e^3 x (4+3 x)+2 e^{12} (5+4 x)+e^9 \left (6+24 x+4 x^2\right )+e^6 \left (1+22 x+10 x^2\right )+x \left (1+x+x^4\right )\right )}{\left (e^6+x+2 e^3 x+x^2\right ) \log ^4(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1280*E^6*x^4 + 1024*x^5 + 2048*E^3*x^5 + 768*x^6)/((E^12 + 4*E^9*x + x^2 + 2*x^3 + x^4 + E^6*(2*x +
6*x^2) + E^3*(4*x^2 + 4*x^3))*Log[2]^4),x]

[Out]

(256*(4*E^15 + 2*E^3*x*(4 + 3*x) + 2*E^12*(5 + 4*x) + E^9*(6 + 24*x + 4*x^2) + E^6*(1 + 22*x + 10*x^2) + x*(1
+ x + x^4)))/((E^6 + x + 2*E^3*x + x^2)*Log[2]^4)

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fricas [B]  time = 0.62, size = 81, normalized size = 4.05 \begin {gather*} \frac {256 \, {\left (x^{5} + x^{2} + 2 \, {\left (4 \, x + 5\right )} e^{12} + 2 \, {\left (2 \, x^{2} + 12 \, x + 3\right )} e^{9} + {\left (10 \, x^{2} + 22 \, x + 1\right )} e^{6} + 2 \, {\left (3 \, x^{2} + 4 \, x\right )} e^{3} + x + 4 \, e^{15}\right )}}{{\left (x^{2} + 2 \, x e^{3} + x + e^{6}\right )} \log \relax (2)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1280*x^4*exp(3)^2+2048*x^5*exp(3)+768*x^6+1024*x^5)/(exp(3)^4+4*x*exp(3)^3+(6*x^2+2*x)*exp(3)^2+(4*
x^3+4*x^2)*exp(3)+x^4+2*x^3+x^2)/log(2)^4,x, algorithm="fricas")

[Out]

256*(x^5 + x^2 + 2*(4*x + 5)*e^12 + 2*(2*x^2 + 12*x + 3)*e^9 + (10*x^2 + 22*x + 1)*e^6 + 2*(3*x^2 + 4*x)*e^3 +
x + 4*e^15)/((x^2 + 2*x*e^3 + x + e^6)*log(2)^4)

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giac [B]  time = 0.45, size = 61, normalized size = 3.05 \begin {gather*} \frac {64 \, {\left (4 \, x^{3} - 8 \, x^{2} e^{3} - 4 \, x^{2} + 12 \, x e^{6} + 16 \, x e^{3} + 4 \, x + 297749522100 \, \log \left (x + 25.0950384229000\right ) - 297749578061 \, \log \left (x + 25.0950235980000\right ) - 4.43056261903000 \times 10^{10} \, \log \left (x + 16.0760482101000\right ) + 44305625602 \, \log \left (x + 16.0760374618000\right )\right )}}{\log \relax (2)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1280*x^4*exp(3)^2+2048*x^5*exp(3)+768*x^6+1024*x^5)/(exp(3)^4+4*x*exp(3)^3+(6*x^2+2*x)*exp(3)^2+(4*
x^3+4*x^2)*exp(3)+x^4+2*x^3+x^2)/log(2)^4,x, algorithm="giac")

[Out]

64*(4*x^3 - 8*x^2*e^3 - 4*x^2 + 12*x*e^6 + 16*x*e^3 + 4*x + 297749522100*log(x + 25.0950384229000) - 297749578
061*log(x + 25.0950235980000) - 4.43056261903000e10*log(x + 16.0760482101000) + 44305625602*log(x + 16.0760374
618000))/log(2)^4

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maple [A]  time = 0.20, size = 26, normalized size = 1.30

method	result	size
gosper	\(\frac {256 x^{5}}{\ln \relax (2)^{4} \left ({\mathrm e}^{6}+2 x \,{\mathrm e}^{3}+x^{2}+x \right )}\)	\(26\)
norman	\(\frac {256 x^{5}}{\ln \relax (2)^{4} \left ({\mathrm e}^{6}+2 x \,{\mathrm e}^{3}+x^{2}+x \right )}\)	\(26\)
risch	\(\frac {768 x \,{\mathrm e}^{6}}{\ln \relax (2)^{4}}-\frac {512 x^{2} {\mathrm e}^{3}}{\ln \relax (2)^{4}}+\frac {256 x^{3}}{\ln \relax (2)^{4}}+\frac {1024 x \,{\mathrm e}^{3}}{\ln \relax (2)^{4}}-\frac {256 x^{2}}{\ln \relax (2)^{4}}+\frac {256 x}{\ln \relax (2)^{4}}+\frac {\left (1280 \,{\mathrm e}^{12}+5120 \,{\mathrm e}^{9}+5376 \,{\mathrm e}^{6}+2048 \,{\mathrm e}^{3}+256\right ) x +1024 \,{\mathrm e}^{15}+2560 \,{\mathrm e}^{12}+1536 \,{\mathrm e}^{9}+256 \,{\mathrm e}^{6}}{\ln \relax (2)^{4} \left ({\mathrm e}^{6}+2 x \,{\mathrm e}^{3}+x^{2}+x \right )}\)	\(112\)
default	\(\frac {768 x \,{\mathrm e}^{6}-512 x^{2} {\mathrm e}^{3}+256 x^{3}+1024 x \,{\mathrm e}^{3}-256 x^{2}+256 x +128 \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}+\left (4 \,{\mathrm e}^{3}+2\right ) \textit {\_Z}^{3}+\left (6 \,{\mathrm e}^{6}+4 \,{\mathrm e}^{3}+1\right ) \textit {\_Z}^{2}+\left (2 \,{\mathrm e}^{6}+4 \,{\mathrm e}^{9}\right ) \textit {\_Z} +{\mathrm e}^{12}\right )}{\sum }\frac {\left (\left (-5 \,{\mathrm e}^{12}-20 \,{\mathrm e}^{9}-21 \,{\mathrm e}^{6}-8 \,{\mathrm e}^{3}-1\right ) \textit {\_R}^{2}+2 \left (-10 \,{\mathrm e}^{12}-4 \,{\mathrm e}^{15}-6 \,{\mathrm e}^{9}-{\mathrm e}^{6}\right ) \textit {\_R} -3 \,{\mathrm e}^{12} {\mathrm e}^{6}-4 \,{\mathrm e}^{12} {\mathrm e}^{3}-{\mathrm e}^{12}\right ) \ln \left (x -\textit {\_R} \right )}{2 \,{\mathrm e}^{9}+6 \textit {\_R} \,{\mathrm e}^{6}+6 \textit {\_R}^{2} {\mathrm e}^{3}+2 \textit {\_R}^{3}+{\mathrm e}^{6}+4 \textit {\_R} \,{\mathrm e}^{3}+3 \textit {\_R}^{2}+\textit {\_R}}\right )}{\ln \relax (2)^{4}}\)	\(191\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1280*x^4*exp(3)^2+2048*x^5*exp(3)+768*x^6+1024*x^5)/(exp(3)^4+4*x*exp(3)^3+(6*x^2+2*x)*exp(3)^2+(4*x^3+4*
x^2)*exp(3)+x^4+2*x^3+x^2)/ln(2)^4,x,method=_RETURNVERBOSE)

[Out]

256*x^5/ln(2)^4/(exp(3)^2+2*x*exp(3)+x^2+x)

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maxima [B]  time = 0.40, size = 85, normalized size = 4.25 \begin {gather*} \frac {256 \, {\left (x^{3} - x^{2} {\left (2 \, e^{3} + 1\right )} + x {\left (3 \, e^{6} + 4 \, e^{3} + 1\right )} + \frac {x {\left (5 \, e^{12} + 20 \, e^{9} + 21 \, e^{6} + 8 \, e^{3} + 1\right )} + 4 \, e^{15} + 10 \, e^{12} + 6 \, e^{9} + e^{6}}{x^{2} + x {\left (2 \, e^{3} + 1\right )} + e^{6}}\right )}}{\log \relax (2)^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1280*x^4*exp(3)^2+2048*x^5*exp(3)+768*x^6+1024*x^5)/(exp(3)^4+4*x*exp(3)^3+(6*x^2+2*x)*exp(3)^2+(4*
x^3+4*x^2)*exp(3)+x^4+2*x^3+x^2)/log(2)^4,x, algorithm="maxima")

[Out]

256*(x^3 - x^2*(2*e^3 + 1) + x*(3*e^6 + 4*e^3 + 1) + (x*(5*e^12 + 20*e^9 + 21*e^6 + 8*e^3 + 1) + 4*e^15 + 10*e
^12 + 6*e^9 + e^6)/(x^2 + x*(2*e^3 + 1) + e^6))/log(2)^4

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mupad [B]  time = 0.32, size = 170, normalized size = 8.50 \begin {gather*} \frac {256\,x^3}{{\ln \relax (2)}^4}+\frac {256\,{\mathrm {e}}^6+1536\,{\mathrm {e}}^9+2560\,{\mathrm {e}}^{12}+1024\,{\mathrm {e}}^{15}+x\,\left (2048\,{\mathrm {e}}^3+5376\,{\mathrm {e}}^6+5120\,{\mathrm {e}}^9+1280\,{\mathrm {e}}^{12}+256\right )}{{\ln \relax (2)}^4\,x^2+\left (2\,{\mathrm {e}}^3\,{\ln \relax (2)}^4+{\ln \relax (2)}^4\right )\,x+{\mathrm {e}}^6\,{\ln \relax (2)}^4}-x^2\,\left (\frac {384\,\left (4\,{\mathrm {e}}^3+2\right )}{{\ln \relax (2)}^4}-\frac {2048\,{\mathrm {e}}^3+1024}{2\,{\ln \relax (2)}^4}\right )+x\,\left (\left (4\,{\mathrm {e}}^3+2\right )\,\left (\frac {768\,\left (4\,{\mathrm {e}}^3+2\right )}{{\ln \relax (2)}^4}-\frac {2048\,{\mathrm {e}}^3+1024}{{\ln \relax (2)}^4}\right )-\frac {768\,\left (4\,{\mathrm {e}}^3+6\,{\mathrm {e}}^6+1\right )}{{\ln \relax (2)}^4}+\frac {1280\,{\mathrm {e}}^6}{{\ln \relax (2)}^4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2048*x^5*exp(3) + 1280*x^4*exp(6) + 1024*x^5 + 768*x^6)/(log(2)^4*(exp(12) + exp(6)*(2*x + 6*x^2) + 4*x*e
xp(9) + exp(3)*(4*x^2 + 4*x^3) + x^2 + 2*x^3 + x^4)),x)

[Out]

(256*x^3)/log(2)^4 + (256*exp(6) + 1536*exp(9) + 2560*exp(12) + 1024*exp(15) + x*(2048*exp(3) + 5376*exp(6) +
5120*exp(9) + 1280*exp(12) + 256))/(x^2*log(2)^4 + exp(6)*log(2)^4 + x*(2*exp(3)*log(2)^4 + log(2)^4)) - x^2*(
(384*(4*exp(3) + 2))/log(2)^4 - (2048*exp(3) + 1024)/(2*log(2)^4)) + x*((4*exp(3) + 2)*((768*(4*exp(3) + 2))/l
og(2)^4 - (2048*exp(3) + 1024)/log(2)^4) - (768*(4*exp(3) + 6*exp(6) + 1))/log(2)^4 + (1280*exp(6))/log(2)^4)

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sympy [B]  time = 1.32, size = 138, normalized size = 6.90 \begin {gather*} \frac {256 x^{3}}{\log {\relax (2 )}^{4}} + x^{2} \left (- \frac {512 e^{3}}{\log {\relax (2 )}^{4}} - \frac {256}{\log {\relax (2 )}^{4}}\right ) + x \left (\frac {256}{\log {\relax (2 )}^{4}} + \frac {1024 e^{3}}{\log {\relax (2 )}^{4}} + \frac {768 e^{6}}{\log {\relax (2 )}^{4}}\right ) + \frac {x \left (256 + 2048 e^{3} + 5376 e^{6} + 5120 e^{9} + 1280 e^{12}\right ) + 256 e^{6} + 1536 e^{9} + 2560 e^{12} + 1024 e^{15}}{x^{2} \log {\relax (2 )}^{4} + x \left (\log {\relax (2 )}^{4} + 2 e^{3} \log {\relax (2 )}^{4}\right ) + e^{6} \log {\relax (2 )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1280*x**4*exp(3)**2+2048*x**5*exp(3)+768*x**6+1024*x**5)/(exp(3)**4+4*x*exp(3)**3+(6*x**2+2*x)*exp(
3)**2+(4*x**3+4*x**2)*exp(3)+x**4+2*x**3+x**2)/ln(2)**4,x)

[Out]

256*x**3/log(2)**4 + x**2*(-512*exp(3)/log(2)**4 - 256/log(2)**4) + x*(256/log(2)**4 + 1024*exp(3)/log(2)**4 +
768*exp(6)/log(2)**4) + (x*(256 + 2048*exp(3) + 5376*exp(6) + 5120*exp(9) + 1280*exp(12)) + 256*exp(6) + 1536
*exp(9) + 2560*exp(12) + 1024*exp(15))/(x**2*log(2)**4 + x*(log(2)**4 + 2*exp(3)*log(2)**4) + exp(6)*log(2)**4
)

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