3.28.81 \(\int \frac {-2 e^3+e^6 (-4 x+10 x^2)+e^6 (-2+20 x-50 x^2) \log (4)}{x^2+2 e^3 x^3+e^6 x^4+(2 x-10 x^2+e^3 (4 x^2-20 x^3)+e^6 (2 x^3-10 x^4)) \log (4)+(1-10 x+25 x^2+e^3 (2 x-20 x^2+50 x^3)+e^6 (x^2-10 x^3+25 x^4)) \log ^2(4)} \, dx\)

Optimal. Leaf size=23 \[ \frac {2}{\left (\frac {1}{e^3}+x\right ) \left (\frac {x}{1-5 x}+\log (4)\right )} \]

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Rubi [B]  time = 0.47, antiderivative size = 69, normalized size of antiderivative = 3.00, number of steps used = 4, number of rules used = 4, integrand size = 150, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {1680, 12, 1814, 8} \begin {gather*} \frac {2 e^6 (1-5 x) (1-5 \log (4))}{e^6 x^2 (1-5 \log (4))^2+e^3 x (1-5 \log (4)) \left (1-5 \log (4)+e^3 \log (4)\right )+e^3 (1-5 \log (4)) \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^3 + E^6*(-4*x + 10*x^2) + E^6*(-2 + 20*x - 50*x^2)*Log[4])/(x^2 + 2*E^3*x^3 + E^6*x^4 + (2*x - 10*x^
2 + E^3*(4*x^2 - 20*x^3) + E^6*(2*x^3 - 10*x^4))*Log[4] + (1 - 10*x + 25*x^2 + E^3*(2*x - 20*x^2 + 50*x^3) + E
^6*(x^2 - 10*x^3 + 25*x^4))*Log[4]^2),x]

[Out]

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

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 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 {8 e^6 (1-5 \log (4)) \left (20 e^6 x^2 (1-5 \log (4))^2-4 e^3 x (1-5 \log (4)) \left (5+2 e^3-25 \log (4)-5 e^3 \log (4)\right )+5 \left (1-5 \log (4)-e^3 \log (4)\right )^2\right )}{\left (4 e^6 x^2 (1-5 \log (4))^2-\left (-1+5 \log (4)+e^3 \log (4)\right )^2\right )^2} \, dx,x,x+\frac {2 e^3-20 e^3 \log (4)+2 e^6 \log (4)+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)}{4 \left (e^6-10 e^6 \log (4)+25 e^6 \log ^2(4)\right )}\right )\\ &=\left (8 e^6 (1-5 \log (4))\right ) \operatorname {Subst}\left (\int \frac {20 e^6 x^2 (1-5 \log (4))^2-4 e^3 x (1-5 \log (4)) \left (5+2 e^3-25 \log (4)-5 e^3 \log (4)\right )+5 \left (1-5 \log (4)-e^3 \log (4)\right )^2}{\left (4 e^6 x^2 (1-5 \log (4))^2-\left (-1+5 \log (4)+e^3 \log (4)\right )^2\right )^2} \, dx,x,x+\frac {2 e^3-20 e^3 \log (4)+2 e^6 \log (4)+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)}{4 \left (e^6-10 e^6 \log (4)+25 e^6 \log ^2(4)\right )}\right )\\ &=\frac {2 e^6 (1-5 x)}{e^6 x^2 (1-5 \log (4))+e^3 \log (4)+e^3 x \left (1-5 \log (4)+e^3 \log (4)\right )}+\frac {\left (4 e^6 (1-5 \log (4))\right ) \operatorname {Subst}\left (\int 0 \, dx,x,x+\frac {2 e^3-20 e^3 \log (4)+2 e^6 \log (4)+50 e^3 \log ^2(4)-10 e^6 \log ^2(4)}{4 \left (e^6-10 e^6 \log (4)+25 e^6 \log ^2(4)\right )}\right )}{\left (1-5 \log (4)-e^3 \log (4)\right )^2}\\ &=\frac {2 e^6 (1-5 x)}{e^6 x^2 (1-5 \log (4))+e^3 \log (4)+e^3 x \left (1-5 \log (4)+e^3 \log (4)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 34, normalized size = 1.48 \begin {gather*} -\frac {2 e^3 (1-5 x)}{\left (1+e^3 x\right ) (-\log (4)+x (-1+5 \log (4)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^3 + E^6*(-4*x + 10*x^2) + E^6*(-2 + 20*x - 50*x^2)*Log[4])/(x^2 + 2*E^3*x^3 + E^6*x^4 + (2*x -
10*x^2 + E^3*(4*x^2 - 20*x^3) + E^6*(2*x^3 - 10*x^4))*Log[4] + (1 - 10*x + 25*x^2 + E^3*(2*x - 20*x^2 + 50*x^
3) + E^6*(x^2 - 10*x^3 + 25*x^4))*Log[4]^2),x]

[Out]

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

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fricas [A]  time = 0.71, size = 40, normalized size = 1.74 \begin {gather*} -\frac {2 \, {\left (5 \, x - 1\right )} e^{3}}{x^{2} e^{3} - 2 \, {\left ({\left (5 \, x^{2} - x\right )} e^{3} + 5 \, x - 1\right )} \log \relax (2) + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-50*x^2+20*x-2)*exp(3)^2*log(2)+(10*x^2-4*x)*exp(3)^2-2*exp(3))/(4*((25*x^4-10*x^3+x^2)*exp(3)^2
+(50*x^3-20*x^2+2*x)*exp(3)+25*x^2-10*x+1)*log(2)^2+2*((-10*x^4+2*x^3)*exp(3)^2+(-20*x^3+4*x^2)*exp(3)-10*x^2+
2*x)*log(2)+x^4*exp(3)^2+2*x^3*exp(3)+x^2),x, algorithm="fricas")

[Out]

-2*(5*x - 1)*e^3/(x^2*e^3 - 2*((5*x^2 - x)*e^3 + 5*x - 1)*log(2) + x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-50*x^2+20*x-2)*exp(3)^2*log(2)+(10*x^2-4*x)*exp(3)^2-2*exp(3))/(4*((25*x^4-10*x^3+x^2)*exp(3)^2
+(50*x^3-20*x^2+2*x)*exp(3)+25*x^2-10*x+1)*log(2)^2+2*((-10*x^4+2*x^3)*exp(3)^2+(-20*x^3+4*x^2)*exp(3)-10*x^2+
2*x)*log(2)+x^4*exp(3)^2+2*x^3*exp(3)+x^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -2*((-100*exp(6)*ln(2)^2+20*exp(6)*ln(2)
-exp(6)+100*exp(3)^2*ln(2)^2-20*exp(3)^2*ln(2)+exp(3)^2)/(16*exp(6)^2*ln(2)^4+320*exp(6)*exp(3)*ln(2)^4-32*exp
(6)*exp(3)*ln(2)^3+80

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maple [A]  time = 0.24, size = 47, normalized size = 2.04

method	result	size
risch	\(\frac {x \,{\mathrm e}^{3}-\frac {{\mathrm e}^{3}}{5}}{{\mathrm e}^{3} \ln \relax (2) x^{2}-\frac {{\mathrm e}^{3} \ln \relax (2) x}{5}-\frac {x^{2} {\mathrm e}^{3}}{10}+x \ln \relax (2)-\frac {\ln \relax (2)}{5}-\frac {x}{10}}\)	\(47\)
gosper	\(\frac {2 \left (5 x -1\right ) {\mathrm e}^{3}}{10 \,{\mathrm e}^{3} \ln \relax (2) x^{2}-2 \,{\mathrm e}^{3} \ln \relax (2) x -x^{2} {\mathrm e}^{3}+10 x \ln \relax (2)-2 \ln \relax (2)-x}\)	\(48\)
norman	\(\frac {\frac {\left (4 \ln \relax (2) {\mathrm e}^{6}+2 \,{\mathrm e}^{3}\right ) x}{2 \ln \relax (2)}-\frac {\left (10 \ln \relax (2)-1\right ) {\mathrm e}^{6} x^{2}}{\ln \relax (2)}}{\left (10 x \ln \relax (2)-2 \ln \relax (2)-x \right ) \left (x \,{\mathrm e}^{3}+1\right )}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(-50*x^2+20*x-2)*exp(3)^2*ln(2)+(10*x^2-4*x)*exp(3)^2-2*exp(3))/(4*((25*x^4-10*x^3+x^2)*exp(3)^2+(50*x^
3-20*x^2+2*x)*exp(3)+25*x^2-10*x+1)*ln(2)^2+2*((-10*x^4+2*x^3)*exp(3)^2+(-20*x^3+4*x^2)*exp(3)-10*x^2+2*x)*ln(
2)+x^4*exp(3)^2+2*x^3*exp(3)+x^2),x,method=_RETURNVERBOSE)

[Out]

(x*exp(3)-1/5*exp(3))/(exp(3)*ln(2)*x^2-1/5*exp(3)*ln(2)*x-1/10*x^2*exp(3)+x*ln(2)-1/5*ln(2)-1/10*x)

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maxima [A]  time = 0.43, size = 47, normalized size = 2.04 \begin {gather*} \frac {2 \, {\left (5 \, x e^{3} - e^{3}\right )}}{{\left (10 \, e^{3} \log \relax (2) - e^{3}\right )} x^{2} - {\left (2 \, {\left (e^{3} - 5\right )} \log \relax (2) + 1\right )} x - 2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-50*x^2+20*x-2)*exp(3)^2*log(2)+(10*x^2-4*x)*exp(3)^2-2*exp(3))/(4*((25*x^4-10*x^3+x^2)*exp(3)^2
+(50*x^3-20*x^2+2*x)*exp(3)+25*x^2-10*x+1)*log(2)^2+2*((-10*x^4+2*x^3)*exp(3)^2+(-20*x^3+4*x^2)*exp(3)-10*x^2+
2*x)*log(2)+x^4*exp(3)^2+2*x^3*exp(3)+x^2),x, algorithm="maxima")

[Out]

2*(5*x*e^3 - e^3)/((10*e^3*log(2) - e^3)*x^2 - (2*(e^3 - 5)*log(2) + 1)*x - 2*log(2))

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*exp(3) + exp(6)*(4*x - 10*x^2) + 2*exp(6)*log(2)*(50*x^2 - 20*x + 2))/(2*log(2)*(2*x + exp(6)*(2*x^3 -
10*x^4) + exp(3)*(4*x^2 - 20*x^3) - 10*x^2) + 4*log(2)^2*(exp(3)*(2*x - 20*x^2 + 50*x^3) - 10*x + exp(6)*(x^2
- 10*x^3 + 25*x^4) + 25*x^2 + 1) + 2*x^3*exp(3) + x^4*exp(6) + x^2),x)

[Out]

\text{Hanged}

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sympy [B]  time = 4.55, size = 49, normalized size = 2.13 \begin {gather*} - \frac {- 10 x e^{3} + 2 e^{3}}{x^{2} \left (- e^{3} + 10 e^{3} \log {\relax (2 )}\right ) + x \left (- 2 e^{3} \log {\relax (2 )} - 1 + 10 \log {\relax (2 )}\right ) - 2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-50*x**2+20*x-2)*exp(3)**2*ln(2)+(10*x**2-4*x)*exp(3)**2-2*exp(3))/(4*((25*x**4-10*x**3+x**2)*ex
p(3)**2+(50*x**3-20*x**2+2*x)*exp(3)+25*x**2-10*x+1)*ln(2)**2+2*((-10*x**4+2*x**3)*exp(3)**2+(-20*x**3+4*x**2)
*exp(3)-10*x**2+2*x)*ln(2)+x**4*exp(3)**2+2*x**3*exp(3)+x**2),x)

[Out]

-(-10*x*exp(3) + 2*exp(3))/(x**2*(-exp(3) + 10*exp(3)*log(2)) + x*(-2*exp(3)*log(2) - 1 + 10*log(2)) - 2*log(2
))

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