3.94.60 \(\int \frac {-48 x^2+(-40-80 x+8 x^2) \log (5)}{25 x^2-10 x^3+x^4+(50 x-20 x^2+2 x^3) \log (5)+(25-10 x+x^2) \log ^2(5)} \, dx\)

Optimal. Leaf size=21 \[ -9+\frac {2 x (4+4 x)}{(-5+x) (x+\log (5))} \]

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Rubi [A]  time = 0.16, antiderivative size = 34, normalized size of antiderivative = 1.62, number of steps used = 4, number of rules used = 4, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {1680, 12, 1814, 8} \begin {gather*} \frac {8 (x (6-\log (5))+5 \log (5))}{x^2-x (5-\log (5))-5 \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-48*x^2 + (-40 - 80*x + 8*x^2)*Log[5])/(25*x^2 - 10*x^3 + x^4 + (50*x - 20*x^2 + 2*x^3)*Log[5] + (25 - 10
*x + x^2)*Log[5]^2),x]

[Out]

(8*(x*(6 - Log[5]) + 5*Log[5]))/(x^2 - x*(5 - Log[5]) - 5*Log[5])

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 {32 \left (-4 x^2 (6-\log (5))-(6-\log (5)) (5+\log (5))^2-4 x \left (30-\log (5)+\log ^2(5)\right )\right )}{\left (4 x^2-(5+\log (5))^2\right )^2} \, dx,x,x+\frac {1}{4} (-10+2 \log (5))\right )\\ &=32 \operatorname {Subst}\left (\int \frac {-4 x^2 (6-\log (5))-(6-\log (5)) (5+\log (5))^2-4 x \left (30-\log (5)+\log ^2(5)\right )}{\left (4 x^2-(5+\log (5))^2\right )^2} \, dx,x,x+\frac {1}{4} (-10+2 \log (5))\right )\\ &=\frac {8 (x (6-\log (5))+5 \log (5))}{x^2-x (5-\log (5))-5 \log (5)}+\frac {16 \operatorname {Subst}\left (\int 0 \, dx,x,x+\frac {1}{4} (-10+2 \log (5))\right )}{(5+\log (5))^2}\\ &=\frac {8 (x (6-\log (5))+5 \log (5))}{x^2-x (5-\log (5))-5 \log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 25, normalized size = 1.19 \begin {gather*} \frac {8 (-x (-6+\log (5))+5 \log (5))}{(-5+x) (x+\log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-48*x^2 + (-40 - 80*x + 8*x^2)*Log[5])/(25*x^2 - 10*x^3 + x^4 + (50*x - 20*x^2 + 2*x^3)*Log[5] + (2
5 - 10*x + x^2)*Log[5]^2),x]

[Out]

(8*(-(x*(-6 + Log[5])) + 5*Log[5]))/((-5 + x)*(x + Log[5]))

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fricas [A]  time = 0.83, size = 27, normalized size = 1.29 \begin {gather*} -\frac {8 \, {\left ({\left (x - 5\right )} \log \relax (5) - 6 \, x\right )}}{x^{2} + {\left (x - 5\right )} \log \relax (5) - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2-80*x-40)*log(5)-48*x^2)/((x^2-10*x+25)*log(5)^2+(2*x^3-20*x^2+50*x)*log(5)+x^4-10*x^3+25*x^2
),x, algorithm="fricas")

[Out]

-8*((x - 5)*log(5) - 6*x)/(x^2 + (x - 5)*log(5) - 5*x)

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giac [A]  time = 0.20, size = 31, normalized size = 1.48 \begin {gather*} -\frac {8 \, {\left (x \log \relax (5) - 6 \, x - 5 \, \log \relax (5)\right )}}{x^{2} + x \log \relax (5) - 5 \, x - 5 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2-80*x-40)*log(5)-48*x^2)/((x^2-10*x+25)*log(5)^2+(2*x^3-20*x^2+50*x)*log(5)+x^4-10*x^3+25*x^2
),x, algorithm="giac")

[Out]

-8*(x*log(5) - 6*x - 5*log(5))/(x^2 + x*log(5) - 5*x - 5*log(5))

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maple [A]  time = 0.09, size = 26, normalized size = 1.24




method result size



norman \(\frac {\left (-8 \ln \relax (5)+48\right ) x +40 \ln \relax (5)}{\left (x -5\right ) \left (\ln \relax (5)+x \right )}\) \(26\)
gosper \(-\frac {8 \left (x \ln \relax (5)-5 \ln \relax (5)-6 x \right )}{x \ln \relax (5)+x^{2}-5 \ln \relax (5)-5 x}\) \(32\)
risch \(\frac {\left (-8 \ln \relax (5)+48\right ) x +40 \ln \relax (5)}{x \ln \relax (5)+x^{2}-5 \ln \relax (5)-5 x}\) \(32\)
default \(-\frac {8 \ln \relax (5) \left (\ln \relax (5)-1\right )}{\left (5+\ln \relax (5)\right ) \left (\ln \relax (5)+x \right )}+\frac {240}{\left (5+\ln \relax (5)\right ) \left (x -5\right )}\) \(35\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^2-80*x-40)*ln(5)-48*x^2)/((x^2-10*x+25)*ln(5)^2+(2*x^3-20*x^2+50*x)*ln(5)+x^4-10*x^3+25*x^2),x,metho
d=_RETURNVERBOSE)

[Out]

((-8*ln(5)+48)*x+40*ln(5))/(x-5)/(ln(5)+x)

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maxima [A]  time = 0.42, size = 29, normalized size = 1.38 \begin {gather*} -\frac {8 \, {\left (x {\left (\log \relax (5) - 6\right )} - 5 \, \log \relax (5)\right )}}{x^{2} + x {\left (\log \relax (5) - 5\right )} - 5 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2-80*x-40)*log(5)-48*x^2)/((x^2-10*x+25)*log(5)^2+(2*x^3-20*x^2+50*x)*log(5)+x^4-10*x^3+25*x^2
),x, algorithm="maxima")

[Out]

-8*(x*(log(5) - 6) - 5*log(5))/(x^2 + x*(log(5) - 5) - 5*log(5))

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mupad [B]  time = 10.30, size = 684, normalized size = 32.57 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5)*(80*x - 8*x^2 + 40) + 48*x^2)/(log(5)^2*(x^2 - 10*x + 25) + log(5)*(50*x - 20*x^2 + 2*x^3) + 25*x
^2 - 10*x^3 + x^4),x)

[Out]

symsum(log(1843200*x*log(5) - 2880000*root(110592000*log(5) + 33177600*log(5)^2 - 32440320*log(5)^3 - 16367616
*log(5)^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*log(5) - 2880000*root(110592000*log(5) + 3317
7600*log(5)^2 - 32440320*log(5)^3 - 16367616*log(5)^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*x
 - 2304000*log(5) - 1136000*root(110592000*log(5) + 33177600*log(5)^2 - 32440320*log(5)^3 - 16367616*log(5)^4
- 2654208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*log(5)^2 - 140800*root(110592000*log(5) + 33177600*log(
5)^2 - 32440320*log(5)^3 - 16367616*log(5)^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*log(5)^3 +
 3840*root(110592000*log(5) + 33177600*log(5)^2 - 32440320*log(5)^3 - 16367616*log(5)^4 - 2654208*log(5)^5 - 1
47456*log(5)^6 - 92160000, z, k)*log(5)^4 + 5120*root(110592000*log(5) + 33177600*log(5)^2 - 32440320*log(5)^3
 - 16367616*log(5)^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*log(5)^5 + 640*root(110592000*log(
5) + 33177600*log(5)^2 - 32440320*log(5)^3 - 16367616*log(5)^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 92160000
, z, k)*log(5)^6 - 1904640*x*log(5)^2 + 122880*x*log(5)^3 - 61440*x*log(5)^4 + 3916800*log(5)^2 - 1858560*log(
5)^3 + 261120*log(5)^4 - 15360*log(5)^5 - 1152000*root(110592000*log(5) + 33177600*log(5)^2 - 32440320*log(5)^
3 - 16367616*log(5)^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*x*log(5) - 118400*root(110592000*
log(5) + 33177600*log(5)^2 - 32440320*log(5)^3 - 16367616*log(5)^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 9216
0000, z, k)*x*log(5)^2 + 5120*root(110592000*log(5) + 33177600*log(5)^2 - 32440320*log(5)^3 - 16367616*log(5)^
4 - 2654208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*x*log(5)^3 - 768*root(110592000*log(5) + 33177600*log
(5)^2 - 32440320*log(5)^3 - 16367616*log(5)^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*x*log(5)^
4 - 1024*root(110592000*log(5) + 33177600*log(5)^2 - 32440320*log(5)^3 - 16367616*log(5)^4 - 2654208*log(5)^5
- 147456*log(5)^6 - 92160000, z, k)*x*log(5)^5 - 128*root(110592000*log(5) + 33177600*log(5)^2 - 32440320*log(
5)^3 - 16367616*log(5)^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 92160000, z, k)*x*log(5)^6)*root(110592000*log
(5) + 33177600*log(5)^2 - 32440320*log(5)^3 - 16367616*log(5)^4 - 2654208*log(5)^5 - 147456*log(5)^6 - 9216000
0, z, k), k, 1, 4)

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sympy [A]  time = 0.66, size = 29, normalized size = 1.38 \begin {gather*} - \frac {x \left (-48 + 8 \log {\relax (5 )}\right ) - 40 \log {\relax (5 )}}{x^{2} + x \left (-5 + \log {\relax (5 )}\right ) - 5 \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**2-80*x-40)*ln(5)-48*x**2)/((x**2-10*x+25)*ln(5)**2+(2*x**3-20*x**2+50*x)*ln(5)+x**4-10*x**3+2
5*x**2),x)

[Out]

-(x*(-48 + 8*log(5)) - 40*log(5))/(x**2 + x*(-5 + log(5)) - 5*log(5))

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