3.76.77 \(\int \frac {612+840 x-519 x^2+12 x^3+(-420+336 x-6 x^2) \log (4)}{2000-1600 x-680 x^2+800 x^3-35 x^4-100 x^5+20 x^6+(1000 x-800 x^2-90 x^3+200 x^4-40 x^5) \log (4)+(125 x^2-100 x^3+20 x^4) \log ^2(4)} \, dx\)

Optimal. Leaf size=32 \[ \frac {3 \left (5+\frac {3-x}{5}\right )}{(-5+2 x) (-4+x (x-\log (4)))} \]

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Rubi [A]  time = 0.27, antiderivative size = 57, normalized size of antiderivative = 1.78, number of steps used = 7, number of rules used = 4, integrand size = 109, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {2074, 618, 206, 638} \begin {gather*} \frac {3 (51 x+4 (33-14 \log (4)))}{5 (9-10 \log (4)) \left (-x^2+x \log (4)+4\right )}-\frac {306}{5 (5-2 x) (9-10 \log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(612 + 840*x - 519*x^2 + 12*x^3 + (-420 + 336*x - 6*x^2)*Log[4])/(2000 - 1600*x - 680*x^2 + 800*x^3 - 35*x
^4 - 100*x^5 + 20*x^6 + (1000*x - 800*x^2 - 90*x^3 + 200*x^4 - 40*x^5)*Log[4] + (125*x^2 - 100*x^3 + 20*x^4)*L
og[4]^2),x]

[Out]

-306/(5*(5 - 2*x)*(9 - 10*Log[4])) + (3*(51*x + 4*(33 - 14*Log[4])))/(5*(9 - 10*Log[4])*(4 - x^2 + x*Log[4]))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

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 \left (\frac {612}{5 (-5+2 x)^2 (-9+10 \log (4))}+\frac {153}{5 (-9+10 \log (4)) \left (4-x^2+x \log (4)\right )}+\frac {3 \left (x (264-61 \log (4))+4 \left (102-33 \log (4)+14 \log ^2(4)\right )\right )}{5 (9-10 \log (4)) \left (4-x^2+x \log (4)\right )^2}\right ) \, dx\\ &=-\frac {306}{5 (5-2 x) (9-10 \log (4))}+\frac {3 \int \frac {x (264-61 \log (4))+4 \left (102-33 \log (4)+14 \log ^2(4)\right )}{\left (4-x^2+x \log (4)\right )^2} \, dx}{5 (9-10 \log (4))}-\frac {153 \int \frac {1}{4-x^2+x \log (4)} \, dx}{5 (9-10 \log (4))}\\ &=-\frac {306}{5 (5-2 x) (9-10 \log (4))}+\frac {3 (51 x+4 (33-14 \log (4)))}{5 (9-10 \log (4)) \left (4-x^2+x \log (4)\right )}+\frac {153 \int \frac {1}{4-x^2+x \log (4)} \, dx}{5 (9-10 \log (4))}+\frac {306 \operatorname {Subst}\left (\int \frac {1}{16-x^2+\log ^2(4)} \, dx,x,-2 x+\log (4)\right )}{5 (9-10 \log (4))}\\ &=-\frac {306}{5 (5-2 x) (9-10 \log (4))}+\frac {3 (51 x+4 (33-14 \log (4)))}{5 (9-10 \log (4)) \left (4-x^2+x \log (4)\right )}-\frac {306 \tanh ^{-1}\left (\frac {2 x-\log (4)}{\sqrt {16+\log ^2(4)}}\right )}{5 (9-10 \log (4)) \sqrt {16+\log ^2(4)}}-\frac {306 \operatorname {Subst}\left (\int \frac {1}{16-x^2+\log ^2(4)} \, dx,x,-2 x+\log (4)\right )}{5 (9-10 \log (4))}\\ &=-\frac {306}{5 (5-2 x) (9-10 \log (4))}+\frac {3 (51 x+4 (33-14 \log (4)))}{5 (9-10 \log (4)) \left (4-x^2+x \log (4)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.11, size = 132, normalized size = 4.12 \begin {gather*} \frac {3 \left (x \left (-2240 \log ^4(4)+72 (-18+107 \log (16))+2 \log ^3(4) (2007+535 \log (16))+48 \log (4) (-261+589 \log (16))-\log ^2(4) (58225+1917 \log (16))\right )+4 \left (9072-5640 \log ^3(4)+700 \log ^4(4)+39280 \log (16)-5 \log (4) (19744+1129 \log (16))+\log ^2(4) (23057+2190 \log (16))\right )\right )}{5 (-5+2 x) (9-10 \log (4))^2 \left (-4+x^2-x \log (4)\right ) \left (16+\log ^2(4)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(612 + 840*x - 519*x^2 + 12*x^3 + (-420 + 336*x - 6*x^2)*Log[4])/(2000 - 1600*x - 680*x^2 + 800*x^3
- 35*x^4 - 100*x^5 + 20*x^6 + (1000*x - 800*x^2 - 90*x^3 + 200*x^4 - 40*x^5)*Log[4] + (125*x^2 - 100*x^3 + 20*
x^4)*Log[4]^2),x]

[Out]

(3*(x*(-2240*Log[4]^4 + 72*(-18 + 107*Log[16]) + 2*Log[4]^3*(2007 + 535*Log[16]) + 48*Log[4]*(-261 + 589*Log[1
6]) - Log[4]^2*(58225 + 1917*Log[16])) + 4*(9072 - 5640*Log[4]^3 + 700*Log[4]^4 + 39280*Log[16] - 5*Log[4]*(19
744 + 1129*Log[16]) + Log[4]^2*(23057 + 2190*Log[16]))))/(5*(-5 + 2*x)*(9 - 10*Log[4])^2*(-4 + x^2 - x*Log[4])
*(16 + Log[4]^2))

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fricas [A]  time = 0.54, size = 35, normalized size = 1.09 \begin {gather*} -\frac {3 \, {\left (x - 28\right )}}{5 \, {\left (2 \, x^{3} - 5 \, x^{2} - 2 \, {\left (2 \, x^{2} - 5 \, x\right )} \log \relax (2) - 8 \, x + 20\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-6*x^2+336*x-420)*log(2)+12*x^3-519*x^2+840*x+612)/(4*(20*x^4-100*x^3+125*x^2)*log(2)^2+2*(-40*x
^5+200*x^4-90*x^3-800*x^2+1000*x)*log(2)+20*x^6-100*x^5-35*x^4+800*x^3-680*x^2-1600*x+2000),x, algorithm="fric
as")

[Out]

-3/5*(x - 28)/(2*x^3 - 5*x^2 - 2*(2*x^2 - 5*x)*log(2) - 8*x + 20)

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giac [A]  time = 0.17, size = 34, normalized size = 1.06 \begin {gather*} -\frac {3 \, {\left (x - 28\right )}}{5 \, {\left (2 \, x^{3} - 4 \, x^{2} \log \relax (2) - 5 \, x^{2} + 10 \, x \log \relax (2) - 8 \, x + 20\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-6*x^2+336*x-420)*log(2)+12*x^3-519*x^2+840*x+612)/(4*(20*x^4-100*x^3+125*x^2)*log(2)^2+2*(-40*x
^5+200*x^4-90*x^3-800*x^2+1000*x)*log(2)+20*x^6-100*x^5-35*x^4+800*x^3-680*x^2-1600*x+2000),x, algorithm="giac
")

[Out]

-3/5*(x - 28)/(2*x^3 - 4*x^2*log(2) - 5*x^2 + 10*x*log(2) - 8*x + 20)

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maple [A]  time = 0.11, size = 28, normalized size = 0.88

method	result	size
norman	\(\frac {\frac {3 x}{5}-\frac {84}{5}}{\left (2 x -5\right ) \left (2 x \ln \relax (2)-x^{2}+4\right )}\)	\(28\)
gosper	\(\frac {\frac {3 x}{5}-\frac {84}{5}}{4 x^{2} \ln \relax (2)-2 x^{3}-10 x \ln \relax (2)+5 x^{2}+8 x -20}\)	\(35\)
risch	\(\frac {\frac {3 x}{20}-\frac {21}{5}}{x^{2} \ln \relax (2)-\frac {x^{3}}{2}-\frac {5 x \ln \relax (2)}{2}+\frac {5 x^{2}}{4}+2 x -5}\)	\(35\)
default	\(-\frac {3 \left (\frac {51 x}{2}-56 \ln \relax (2)+66\right )}{5 \left (20 \ln \relax (2)-9\right ) \left (x \ln \relax (2)-\frac {x^{2}}{2}+2\right )}-\frac {306}{5 \left (20 \ln \relax (2)-9\right ) \left (2 x -5\right )}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(-6*x^2+336*x-420)*ln(2)+12*x^3-519*x^2+840*x+612)/(4*(20*x^4-100*x^3+125*x^2)*ln(2)^2+2*(-40*x^5+200*x
^4-90*x^3-800*x^2+1000*x)*ln(2)+20*x^6-100*x^5-35*x^4+800*x^3-680*x^2-1600*x+2000),x,method=_RETURNVERBOSE)

[Out]

(3/5*x-84/5)/(2*x-5)/(2*x*ln(2)-x^2+4)

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maxima [A]  time = 0.36, size = 34, normalized size = 1.06 \begin {gather*} -\frac {3 \, {\left (x - 28\right )}}{5 \, {\left (2 \, x^{3} - x^{2} {\left (4 \, \log \relax (2) + 5\right )} + 2 \, x {\left (5 \, \log \relax (2) - 4\right )} + 20\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-6*x^2+336*x-420)*log(2)+12*x^3-519*x^2+840*x+612)/(4*(20*x^4-100*x^3+125*x^2)*log(2)^2+2*(-40*x
^5+200*x^4-90*x^3-800*x^2+1000*x)*log(2)+20*x^6-100*x^5-35*x^4+800*x^3-680*x^2-1600*x+2000),x, algorithm="maxi
ma")

[Out]

-3/5*(x - 28)/(2*x^3 - x^2*(4*log(2) + 5) + 2*x*(5*log(2) - 4) + 20)

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mupad [B]  time = 0.22, size = 26, normalized size = 0.81 \begin {gather*} \frac {3\,\left (x-28\right )}{5\,\left (2\,x-5\right )\,\left (-x^2+2\,\ln \relax (2)\,x+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(840*x - 2*log(2)*(6*x^2 - 336*x + 420) - 519*x^2 + 12*x^3 + 612)/(1600*x - 4*log(2)^2*(125*x^2 - 100*x^3
+ 20*x^4) + 680*x^2 - 800*x^3 + 35*x^4 + 100*x^5 - 20*x^6 + 2*log(2)*(800*x^2 - 1000*x + 90*x^3 - 200*x^4 + 4
0*x^5) - 2000),x)

[Out]

(3*(x - 28))/(5*(2*x - 5)*(2*x*log(2) - x^2 + 4))

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sympy [A]  time = 1.60, size = 31, normalized size = 0.97 \begin {gather*} \frac {84 - 3 x}{10 x^{3} + x^{2} \left (-25 - 20 \log {\relax (2 )}\right ) + x \left (-40 + 50 \log {\relax (2 )}\right ) + 100} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-6*x**2+336*x-420)*ln(2)+12*x**3-519*x**2+840*x+612)/(4*(20*x**4-100*x**3+125*x**2)*ln(2)**2+2*(
-40*x**5+200*x**4-90*x**3-800*x**2+1000*x)*ln(2)+20*x**6-100*x**5-35*x**4+800*x**3-680*x**2-1600*x+2000),x)

[Out]

(84 - 3*x)/(10*x**3 + x**2*(-25 - 20*log(2)) + x*(-40 + 50*log(2)) + 100)

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