\(\int \frac {50-2 x^2-2 \log (5)}{625+50 x^2+x^4+(-50-2 x^2) \log (5)+\log ^2(5)} \, dx\) [9559]

Optimal result

Integrand size = 38, antiderivative size = 14 \[ \int \frac {50-2 x^2-2 \log (5)}{625+50 x^2+x^4+\left (-50-2 x^2\right ) \log (5)+\log ^2(5)} \, dx=\frac {2 x}{25+x^2-\log (5)} \]

[Out]

4*x/(50+2*x^2-2*ln(5))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2015, 28, 391} \[ \int \frac {50-2 x^2-2 \log (5)}{625+50 x^2+x^4+\left (-50-2 x^2\right ) \log (5)+\log ^2(5)} \, dx=\frac {2 x}{x^2+25-\log (5)} \]

[In]

Int[(50 - 2*x^2 - 2*Log[5])/(625 + 50*x^2 + x^4 + (-50 - 2*x^2)*Log[5] + Log[5]^2),x]

[Out]

(2*x)/(25 + x^2 - Log[5])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 391

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*x*((a + b*x^n)^(p + 1)/a), x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a*d - b*c*(n*(p + 1) + 1), 0]

Rule 2015

Int[(u_)^(q_.)*(v_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*ExpandToSum[v, x]^p, x] /; FreeQ[{p, q}, x] &&
BinomialQ[u, x] && TrinomialQ[v, x] &&  !(BinomialMatchQ[u, x] && TrinomialMatchQ[v, x])

Rubi steps \begin{align*} \text {integral}& = \int \frac {-2 x^2+2 (25-\log (5))}{x^4+2 x^2 (25-\log (5))+(-25+\log (5))^2} \, dx \\ & = \int \frac {-2 x^2+2 (25-\log (5))}{\left (25+x^2-\log (5)\right )^2} \, dx \\ & = \frac {2 x}{25+x^2-\log (5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {50-2 x^2-2 \log (5)}{625+50 x^2+x^4+\left (-50-2 x^2\right ) \log (5)+\log ^2(5)} \, dx=\frac {2 x}{25+x^2-\log (5)} \]

[In]

Integrate[(50 - 2*x^2 - 2*Log[5])/(625 + 50*x^2 + x^4 + (-50 - 2*x^2)*Log[5] + Log[5]^2),x]

[Out]

(2*x)/(25 + x^2 - Log[5])

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07

method	result	size
gosper	\(-\frac {2 x}{-x^{2}+\ln \left (5\right )-25}\)	\(15\)
default	\(-\frac {2 x}{-x^{2}+\ln \left (5\right )-25}\)	\(15\)
norman	\(-\frac {2 x}{-x^{2}+\ln \left (5\right )-25}\)	\(15\)
risch	\(-\frac {2 x}{-x^{2}+\ln \left (5\right )-25}\)	\(15\)
parallelrisch	\(-\frac {2 x}{-x^{2}+\ln \left (5\right )-25}\)	\(15\)

[In]

int((-2*ln(5)-2*x^2+50)/(ln(5)^2+(-2*x^2-50)*ln(5)+x^4+50*x^2+625),x,method=_RETURNVERBOSE)

[Out]

-2*x/(-x^2+ln(5)-25)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {50-2 x^2-2 \log (5)}{625+50 x^2+x^4+\left (-50-2 x^2\right ) \log (5)+\log ^2(5)} \, dx=\frac {2 \, x}{x^{2} - \log \left (5\right ) + 25} \]

[In]

integrate((-2*log(5)-2*x^2+50)/(log(5)^2+(-2*x^2-50)*log(5)+x^4+50*x^2+625),x, algorithm="fricas")

[Out]

2*x/(x^2 - log(5) + 25)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {50-2 x^2-2 \log (5)}{625+50 x^2+x^4+\left (-50-2 x^2\right ) \log (5)+\log ^2(5)} \, dx=\frac {2 x}{x^{2} - \log {\left (5 \right )} + 25} \]

[In]

integrate((-2*ln(5)-2*x**2+50)/(ln(5)**2+(-2*x**2-50)*ln(5)+x**4+50*x**2+625),x)

[Out]

2*x/(x**2 - log(5) + 25)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {50-2 x^2-2 \log (5)}{625+50 x^2+x^4+\left (-50-2 x^2\right ) \log (5)+\log ^2(5)} \, dx=\frac {2 \, x}{x^{2} - \log \left (5\right ) + 25} \]

[In]

integrate((-2*log(5)-2*x^2+50)/(log(5)^2+(-2*x^2-50)*log(5)+x^4+50*x^2+625),x, algorithm="maxima")

[Out]

2*x/(x^2 - log(5) + 25)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {50-2 x^2-2 \log (5)}{625+50 x^2+x^4+\left (-50-2 x^2\right ) \log (5)+\log ^2(5)} \, dx=\frac {2 \, x}{x^{2} - \log \left (5\right ) + 25} \]

[In]

integrate((-2*log(5)-2*x^2+50)/(log(5)^2+(-2*x^2-50)*log(5)+x^4+50*x^2+625),x, algorithm="giac")

[Out]

2*x/(x^2 - log(5) + 25)

Mupad [B] (verification not implemented)

Time = 17.70 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.07 \[ \int \frac {50-2 x^2-2 \log (5)}{625+50 x^2+x^4+\left (-50-2 x^2\right ) \log (5)+\log ^2(5)} \, dx=0 \]

[In]

int(-(2*log(5) + 2*x^2 - 50)/(log(5)^2 - log(5)*(2*x^2 + 50) + 50*x^2 + x^4 + 625),x)

[Out]

0