3.70.11 \(\int \frac {50 x^2+20 x^3-10 x^4+e^4 (-100 x-30 x^2)+e^2 (-100 x^2-40 x^3)+(-10 x^2-4 x^3+2 x^4+e^4 (20 x+6 x^2)+e^2 (20 x^2+8 x^3)) \log (9)}{e^8+4 e^6 x+x^2-2 x^3+x^4+e^4 (-2 x+6 x^2)+e^2 (-4 x^2+4 x^3)} \, dx\)

Optimal. Leaf size=25 \[ \frac {2 x^2 (5+x) (-5+\log (9))}{-x+\left (e^2+x\right )^2} \]

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Rubi [B]  time = 0.57, antiderivative size = 86, normalized size of antiderivative = 3.44, number of steps used = 5, number of rules used = 5, integrand size = 143, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {1680, 12, 1814, 21, 8} \begin {gather*} \frac {2 \left (\left (6-38 e^2+59 e^4-12 e^6\right ) x-2 e^4 \left (3-13 e^2+4 e^4\right )\right ) (5-\log (9))}{\left (1-4 e^2\right ) \left (-x^2+\left (1-2 e^2\right ) x-e^4\right )}-2 x (5-\log (9)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(50*x^2 + 20*x^3 - 10*x^4 + E^4*(-100*x - 30*x^2) + E^2*(-100*x^2 - 40*x^3) + (-10*x^2 - 4*x^3 + 2*x^4 + E
^4*(20*x + 6*x^2) + E^2*(20*x^2 + 8*x^3))*Log[9])/(E^8 + 4*E^6*x + x^2 - 2*x^3 + x^4 + E^4*(-2*x + 6*x^2) + E^
2*(-4*x^2 + 4*x^3)),x]

[Out]

-2*x*(5 - Log[9]) + (2*(-2*E^4*(3 - 13*E^2 + 4*E^4) + (6 - 38*E^2 + 59*E^4 - 12*E^6)*x)*(5 - Log[9]))/((1 - 4*
E^2)*(-E^4 + (1 - 2*E^2)*x - x^2))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {2 \left (23-144 e^2+220 e^4-48 e^6+16 \left (6-26 e^2+19 e^4-2 e^6\right ) x+8 \left (13-32 e^2+6 e^4\right ) x^2-16 x^4\right ) (5-\log (9))}{\left (1-4 e^2-4 x^2\right )^2} \, dx,x,\frac {1}{4} \left (-2+4 e^2\right )+x\right )\\ &=(2 (5-\log (9))) \operatorname {Subst}\left (\int \frac {23-144 e^2+220 e^4-48 e^6+16 \left (6-26 e^2+19 e^4-2 e^6\right ) x+8 \left (13-32 e^2+6 e^4\right ) x^2-16 x^4}{\left (1-4 e^2-4 x^2\right )^2} \, dx,x,\frac {1}{4} \left (-2+4 e^2\right )+x\right )\\ &=\frac {2 \left (2 e^4 \left (3-13 e^2+4 e^4\right )-\left (6-38 e^2+59 e^4-12 e^6\right ) x\right ) (5-\log (9))}{\left (1-4 e^2\right ) \left (e^4-\left (1-2 e^2\right ) x+x^2\right )}-\frac {(5-\log (9)) \operatorname {Subst}\left (\int \frac {2 \left (1-4 e^2\right )^2-8 \left (1-4 e^2\right ) x^2}{1-4 e^2-4 x^2} \, dx,x,\frac {1}{4} \left (-2+4 e^2\right )+x\right )}{1-4 e^2}\\ &=\frac {2 \left (2 e^4 \left (3-13 e^2+4 e^4\right )-\left (6-38 e^2+59 e^4-12 e^6\right ) x\right ) (5-\log (9))}{\left (1-4 e^2\right ) \left (e^4-\left (1-2 e^2\right ) x+x^2\right )}-(2 (5-\log (9))) \operatorname {Subst}\left (\int 1 \, dx,x,\frac {1}{4} \left (-2+4 e^2\right )+x\right )\\ &=-2 x (5-\log (9))+\frac {2 \left (2 e^4 \left (3-13 e^2+4 e^4\right )-\left (6-38 e^2+59 e^4-12 e^6\right ) x\right ) (5-\log (9))}{\left (1-4 e^2\right ) \left (e^4-\left (1-2 e^2\right ) x+x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 49, normalized size = 1.96 \begin {gather*} 2 \left (x+\frac {2 e^6+3 e^4 (-2+x)+6 x-14 e^2 x}{e^4+2 e^2 x+(-1+x) x}\right ) (-5+\log (9)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(50*x^2 + 20*x^3 - 10*x^4 + E^4*(-100*x - 30*x^2) + E^2*(-100*x^2 - 40*x^3) + (-10*x^2 - 4*x^3 + 2*x
^4 + E^4*(20*x + 6*x^2) + E^2*(20*x^2 + 8*x^3))*Log[9])/(E^8 + 4*E^6*x + x^2 - 2*x^3 + x^4 + E^4*(-2*x + 6*x^2
) + E^2*(-4*x^2 + 4*x^3)),x]

[Out]

2*(x + (2*E^6 + 3*E^4*(-2 + x) + 6*x - 14*E^2*x)/(E^4 + 2*E^2*x + (-1 + x)*x))*(-5 + Log[9])

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fricas [B]  time = 1.14, size = 96, normalized size = 3.84 \begin {gather*} -\frac {2 \, {\left (5 \, x^{3} - 5 \, x^{2} + 10 \, {\left (2 \, x - 3\right )} e^{4} + 10 \, {\left (x^{2} - 7 \, x\right )} e^{2} - 2 \, {\left (x^{3} - x^{2} + 2 \, {\left (2 \, x - 3\right )} e^{4} + 2 \, {\left (x^{2} - 7 \, x\right )} e^{2} + 6 \, x + 2 \, e^{6}\right )} \log \relax (3) + 30 \, x + 10 \, e^{6}\right )}}{x^{2} + 2 \, x e^{2} - x + e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*((6*x^2+20*x)*exp(2)^2+(8*x^3+20*x^2)*exp(2)+2*x^4-4*x^3-10*x^2)*log(3)+(-30*x^2-100*x)*exp(2)^2+
(-40*x^3-100*x^2)*exp(2)-10*x^4+20*x^3+50*x^2)/(exp(2)^4+4*x*exp(2)^3+(6*x^2-2*x)*exp(2)^2+(4*x^3-4*x^2)*exp(2
)+x^4-2*x^3+x^2),x, algorithm="fricas")

[Out]

-2*(5*x^3 - 5*x^2 + 10*(2*x - 3)*e^4 + 10*(x^2 - 7*x)*e^2 - 2*(x^3 - x^2 + 2*(2*x - 3)*e^4 + 2*(x^2 - 7*x)*e^2
+ 6*x + 2*e^6)*log(3) + 30*x + 10*e^6)/(x^2 + 2*x*e^2 - x + e^4)

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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((2*((6*x^2+20*x)*exp(2)^2+(8*x^3+20*x^2)*exp(2)+2*x^4-4*x^3-10*x^2)*log(3)+(-30*x^2-100*x)*exp(2)^2+
(-40*x^3-100*x^2)*exp(2)-10*x^4+20*x^3+50*x^2)/(exp(2)^4+4*x*exp(2)^3+(6*x^2-2*x)*exp(2)^2+(4*x^3-4*x^2)*exp(2
)+x^4-2*x^3+x^2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.14, size = 47, normalized size = 1.88

method	result	size
gosper	\(-\frac {2 \left (-x^{3}+5 \,{\mathrm e}^{4}+10 \,{\mathrm e}^{2} x -5 x \right ) \left (2 \ln \relax (3)-5\right )}{{\mathrm e}^{4}+2 \,{\mathrm e}^{2} x +x^{2}-x}\)	\(47\)
norman	\(\frac {-10 \left (2 \,{\mathrm e}^{2}-1\right ) \left (2 \ln \relax (3)-5\right ) x +\left (4 \ln \relax (3)-10\right ) x^{3}-10 \,{\mathrm e}^{4} \left (2 \ln \relax (3)-5\right )}{{\mathrm e}^{4}+2 \,{\mathrm e}^{2} x +x^{2}-x}\)	\(58\)
risch	\(4 x \ln \relax (3)-10 x +\frac {\left (12 \,{\mathrm e}^{4} \ln \relax (3)-30 \,{\mathrm e}^{4}-56 \,{\mathrm e}^{2} \ln \relax (3)+140 \,{\mathrm e}^{2}+24 \ln \relax (3)-60\right ) x +4 \left (2 \,{\mathrm e}^{2} \ln \relax (3)-5 \,{\mathrm e}^{2}-6 \ln \relax (3)+15\right ) {\mathrm e}^{4}}{{\mathrm e}^{4}+2 \,{\mathrm e}^{2} x +x^{2}-x}\)	\(76\)
default	\(\left (4 \ln \relax (3)-10\right ) \left (x +\frac {\left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}+\left (4 \,{\mathrm e}^{2}-2\right ) \textit {\_Z}^{3}+\left (-4 \,{\mathrm e}^{2}+6 \,{\mathrm e}^{4}+1\right ) \textit {\_Z}^{2}+\left (-2 \,{\mathrm e}^{4}+4 \,{\mathrm e}^{6}\right ) \textit {\_Z} +{\mathrm e}^{8}\right )}{\sum }\frac {\left (\left (14 \,{\mathrm e}^{2}-3 \,{\mathrm e}^{4}-6\right ) \textit {\_R}^{2}+4 \left (3 \,{\mathrm e}^{4}-{\mathrm e}^{6}\right ) \textit {\_R} -{\mathrm e}^{8}\right ) \ln \left (x -\textit {\_R} \right )}{2 \,{\mathrm e}^{6}+6 \textit {\_R} \,{\mathrm e}^{4}+6 \textit {\_R}^{2} {\mathrm e}^{2}+2 \textit {\_R}^{3}-{\mathrm e}^{4}-4 \,{\mathrm e}^{2} \textit {\_R} -3 \textit {\_R}^{2}+\textit {\_R}}\right )}{2}\right )\)	\(134\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*((6*x^2+20*x)*exp(2)^2+(8*x^3+20*x^2)*exp(2)+2*x^4-4*x^3-10*x^2)*ln(3)+(-30*x^2-100*x)*exp(2)^2+(-40*x^
3-100*x^2)*exp(2)-10*x^4+20*x^3+50*x^2)/(exp(2)^4+4*x*exp(2)^3+(6*x^2-2*x)*exp(2)^2+(4*x^3-4*x^2)*exp(2)+x^4-2
*x^3+x^2),x,method=_RETURNVERBOSE)

[Out]

-2*(-x^3+5*exp(2)^2+10*exp(2)*x-5*x)*(2*ln(3)-5)/(exp(2)^2+2*exp(2)*x+x^2-x)

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maxima [B]  time = 0.36, size = 74, normalized size = 2.96 \begin {gather*} 2 \, x {\left (2 \, \log \relax (3) - 5\right )} + \frac {2 \, {\left ({\left (2 \, {\left (3 \, e^{4} - 14 \, e^{2} + 6\right )} \log \relax (3) - 15 \, e^{4} + 70 \, e^{2} - 30\right )} x + 4 \, {\left (e^{6} - 3 \, e^{4}\right )} \log \relax (3) - 10 \, e^{6} + 30 \, e^{4}\right )}}{x^{2} + x {\left (2 \, e^{2} - 1\right )} + e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*((6*x^2+20*x)*exp(2)^2+(8*x^3+20*x^2)*exp(2)+2*x^4-4*x^3-10*x^2)*log(3)+(-30*x^2-100*x)*exp(2)^2+
(-40*x^3-100*x^2)*exp(2)-10*x^4+20*x^3+50*x^2)/(exp(2)^4+4*x*exp(2)^3+(6*x^2-2*x)*exp(2)^2+(4*x^3-4*x^2)*exp(2
)+x^4-2*x^3+x^2),x, algorithm="maxima")

[Out]

2*x*(2*log(3) - 5) + 2*((2*(3*e^4 - 14*e^2 + 6)*log(3) - 15*e^4 + 70*e^2 - 30)*x + 4*(e^6 - 3*e^4)*log(3) - 10
*e^6 + 30*e^4)/(x^2 + x*(2*e^2 - 1) + e^4)

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mupad [B]  time = 4.62, size = 187, normalized size = 7.48 \begin {gather*} x\,\left (\ln \left (81\right )-10\right )-\frac {\frac {60\,{\mathrm {e}}^4-260\,{\mathrm {e}}^6+80\,{\mathrm {e}}^8+\ln \left (\frac {9^{14\,{\mathrm {e}}^8}\,9^{40\,{\mathrm {e}}^6}\,{81}^{6\,{\mathrm {e}}^6}\,{81}^{14\,{\mathrm {e}}^{10}}}{9^{10\,{\mathrm {e}}^4}\,9^{28\,{\mathrm {e}}^{10}}\,{81}^{{\mathrm {e}}^4}\,{81}^{15\,{\mathrm {e}}^8}}\right )}{4\,{\mathrm {e}}^2-1}+\frac {x\,\left (380\,{\mathrm {e}}^2-590\,{\mathrm {e}}^4+120\,{\mathrm {e}}^6+\ln \left (\frac {282429536481\,9^{40\,{\mathrm {e}}^6}\,9^{70\,{\mathrm {e}}^4}\,{81}^{14\,{\mathrm {e}}^8}\,{81}^{24\,{\mathrm {e}}^4}}{9^{28\,{\mathrm {e}}^8}\,9^{60\,{\mathrm {e}}^2}\,{81}^{8\,{\mathrm {e}}^2}\,{81}^{32\,{\mathrm {e}}^6}}\right )-60\right )}{4\,{\mathrm {e}}^2-1}}{x^2+\left (2\,{\mathrm {e}}^2-1\right )\,x+{\mathrm {e}}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(4)*(100*x + 30*x^2) + exp(2)*(100*x^2 + 40*x^3) - 2*log(3)*(exp(4)*(20*x + 6*x^2) + exp(2)*(20*x^2 +
8*x^3) - 10*x^2 - 4*x^3 + 2*x^4) - 50*x^2 - 20*x^3 + 10*x^4)/(exp(8) - exp(4)*(2*x - 6*x^2) + 4*x*exp(6) - ex
p(2)*(4*x^2 - 4*x^3) + x^2 - 2*x^3 + x^4),x)

[Out]

x*(log(81) - 10) - ((60*exp(4) - 260*exp(6) + 80*exp(8) + log((9^(14*exp(8))*9^(40*exp(6))*81^(6*exp(6))*81^(1
4*exp(10)))/(9^(10*exp(4))*9^(28*exp(10))*81^exp(4)*81^(15*exp(8)))))/(4*exp(2) - 1) + (x*(380*exp(2) - 590*ex
p(4) + 120*exp(6) + log((282429536481*9^(40*exp(6))*9^(70*exp(4))*81^(14*exp(8))*81^(24*exp(4)))/(9^(28*exp(8)
)*9^(60*exp(2))*81^(8*exp(2))*81^(32*exp(6)))) - 60))/(4*exp(2) - 1))/(exp(4) + x^2 + x*(2*exp(2) - 1))

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sympy [B]  time = 4.83, size = 87, normalized size = 3.48 \begin {gather*} - x \left (10 - 4 \log {\relax (3 )}\right ) - \frac {x \left (- 140 e^{2} - 12 e^{4} \log {\relax (3 )} - 24 \log {\relax (3 )} + 60 + 56 e^{2} \log {\relax (3 )} + 30 e^{4}\right ) - 8 e^{6} \log {\relax (3 )} - 60 e^{4} + 24 e^{4} \log {\relax (3 )} + 20 e^{6}}{x^{2} + x \left (-1 + 2 e^{2}\right ) + e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*((6*x**2+20*x)*exp(2)**2+(8*x**3+20*x**2)*exp(2)+2*x**4-4*x**3-10*x**2)*ln(3)+(-30*x**2-100*x)*ex
p(2)**2+(-40*x**3-100*x**2)*exp(2)-10*x**4+20*x**3+50*x**2)/(exp(2)**4+4*x*exp(2)**3+(6*x**2-2*x)*exp(2)**2+(4
*x**3-4*x**2)*exp(2)+x**4-2*x**3+x**2),x)

[Out]

-x*(10 - 4*log(3)) - (x*(-140*exp(2) - 12*exp(4)*log(3) - 24*log(3) + 60 + 56*exp(2)*log(3) + 30*exp(4)) - 8*e
xp(6)*log(3) - 60*exp(4) + 24*exp(4)*log(3) + 20*exp(6))/(x**2 + x*(-1 + 2*exp(2)) + exp(4))

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