3.43.73 \(\int \frac {-250+(-4-100 x-625 x^2) \log (3)}{(4+100 x+625 x^2) \log (3)} \, dx\)

Optimal. Leaf size=26 \[ -e^4-x+5 \left (3+\frac {2}{(2+25 x) \log (3)}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 0.65, number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {12, 27, 1850} \begin {gather*} \frac {10}{(25 x+2) \log (3)}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-250 + (-4 - 100*x - 625*x^2)*Log[3])/((4 + 100*x + 625*x^2)*Log[3]),x]

[Out]

-x + 10/((2 + 25*x)*Log[3])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-250+\left (-4-100 x-625 x^2\right ) \log (3)}{4+100 x+625 x^2} \, dx}{\log (3)}\\ &=\frac {\int \frac {-250+\left (-4-100 x-625 x^2\right ) \log (3)}{(2+25 x)^2} \, dx}{\log (3)}\\ &=\frac {\int \left (-\frac {250}{(2+25 x)^2}-\log (3)\right ) \, dx}{\log (3)}\\ &=-x+\frac {10}{(2+25 x) \log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 1.12 \begin {gather*} -\frac {-250+100 x \log (3)+625 x^2 \log (3)+\log (81)}{(50+625 x) \log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-250 + (-4 - 100*x - 625*x^2)*Log[3])/((4 + 100*x + 625*x^2)*Log[3]),x]

[Out]

-((-250 + 100*x*Log[3] + 625*x^2*Log[3] + Log[81])/((50 + 625*x)*Log[3]))

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fricas [A]  time = 0.66, size = 27, normalized size = 1.04 \begin {gather*} -\frac {{\left (25 \, x^{2} + 2 \, x\right )} \log \relax (3) - 10}{{\left (25 \, x + 2\right )} \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-625*x^2-100*x-4)*log(3)-250)/(625*x^2+100*x+4)/log(3),x, algorithm="fricas")

[Out]

-((25*x^2 + 2*x)*log(3) - 10)/((25*x + 2)*log(3))

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giac [A]  time = 0.12, size = 20, normalized size = 0.77 \begin {gather*} -\frac {x \log \relax (3) - \frac {10}{25 \, x + 2}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-625*x^2-100*x-4)*log(3)-250)/(625*x^2+100*x+4)/log(3),x, algorithm="giac")

[Out]

-(x*log(3) - 10/(25*x + 2))/log(3)

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maple [A]  time = 0.10, size = 16, normalized size = 0.62

method	result	size
risch	\(-x +\frac {2}{5 \ln \relax (3) \left (x +\frac {2}{25}\right )}\)	\(16\)
default	\(\frac {-x \ln \relax (3)+\frac {10}{25 x +2}}{\ln \relax (3)}\)	\(21\)
gosper	\(-\frac {x \left (25 x \ln \relax (3)+2 \ln \relax (3)+125\right )}{\ln \relax (3) \left (25 x +2\right )}\)	\(26\)
meijerg	\(-\frac {125 x}{2 \ln \relax (3) \left (1+\frac {25 x}{2}\right )}-\frac {x \left (\frac {75 x}{2}+6\right )}{3 \left (1+\frac {25 x}{2}\right )}+\frac {x}{1+\frac {25 x}{2}}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-625*x^2-100*x-4)*ln(3)-250)/(625*x^2+100*x+4)/ln(3),x,method=_RETURNVERBOSE)

[Out]

-x+2/5/ln(3)/(x+2/25)

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maxima [A]  time = 0.34, size = 20, normalized size = 0.77 \begin {gather*} -\frac {x \log \relax (3) - \frac {10}{25 \, x + 2}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-625*x^2-100*x-4)*log(3)-250)/(625*x^2+100*x+4)/log(3),x, algorithm="maxima")

[Out]

-(x*log(3) - 10/(25*x + 2))/log(3)

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mupad [B]  time = 3.02, size = 17, normalized size = 0.65 \begin {gather*} \frac {10}{\ln \relax (3)\,\left (25\,x+2\right )}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(3)*(100*x + 625*x^2 + 4) + 250)/(log(3)*(100*x + 625*x^2 + 4)),x)

[Out]

10/(log(3)*(25*x + 2)) - x

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sympy [A]  time = 0.13, size = 14, normalized size = 0.54 \begin {gather*} - x + \frac {10}{25 x \log {\relax (3 )} + 2 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-625*x**2-100*x-4)*ln(3)-250)/(625*x**2+100*x+4)/ln(3),x)

[Out]

-x + 10/(25*x*log(3) + 2*log(3))

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