\(\int \frac {-4 e^4+2 x^2-4 x \log (625)+2 \log ^2(625)}{x^2-2 x \log (625)+\log ^2(625)} \, dx\) [6878]

Optimal result

Integrand size = 38, antiderivative size = 20 \[ \int \frac {-4 e^4+2 x^2-4 x \log (625)+2 \log ^2(625)}{x^2-2 x \log (625)+\log ^2(625)} \, dx=e^3+2 x+\frac {4 e^4}{x-\log (625)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {27, 697} \[ \int \frac {-4 e^4+2 x^2-4 x \log (625)+2 \log ^2(625)}{x^2-2 x \log (625)+\log ^2(625)} \, dx=2 x+\frac {4 e^4}{x-\log (625)} \]

[In]

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

[Out]

2*x + (4*E^4)/(x - Log[625])

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 697

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 e^4+2 x^2-4 x \log (625)+2 \log ^2(625)}{(x-\log (625))^2} \, dx \\ & = \int \left (2-\frac {4 e^4}{(x-\log (625))^2}\right ) \, dx \\ & = 2 x+\frac {4 e^4}{x-\log (625)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-4 e^4+2 x^2-4 x \log (625)+2 \log ^2(625)}{x^2-2 x \log (625)+\log ^2(625)} \, dx=2 x+\frac {4 e^4}{x-\log (625)} \]

[In]

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

[Out]

2*x + (4*E^4)/(x - Log[625])

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
default	\(2 x +\frac {4 \,{\mathrm e}^{4}}{x -4 \ln \left (5\right )}\)	\(17\)
risch	\(2 x -\frac {{\mathrm e}^{4}}{\ln \left (5\right )-\frac {x}{4}}\)	\(17\)
norman	\(\frac {-2 x^{2}+32 \ln \left (5\right )^{2}-4 \,{\mathrm e}^{4}}{4 \ln \left (5\right )-x}\)	\(28\)
parallelrisch	\(\frac {-2 x^{2}+32 \ln \left (5\right )^{2}-4 \,{\mathrm e}^{4}}{4 \ln \left (5\right )-x}\)	\(28\)
gosper	\(\frac {-2 x^{2}+32 \ln \left (5\right )^{2}-4 \,{\mathrm e}^{4}}{4 \ln \left (5\right )-x}\)	\(29\)
meijerg	\(\frac {2 x}{1-\frac {x}{4 \ln \left (5\right )}}-16 \ln \left (5\right ) \left (\frac {x}{4 \ln \left (5\right ) \left (1-\frac {x}{4 \ln \left (5\right )}\right )}+\ln \left (1-\frac {x}{4 \ln \left (5\right )}\right )\right )-\frac {{\mathrm e}^{4} x}{4 \ln \left (5\right )^{2} \left (1-\frac {x}{4 \ln \left (5\right )}\right )}-8 \ln \left (5\right ) \left (-\frac {x \left (-\frac {3 x}{4 \ln \left (5\right )}+6\right )}{12 \ln \left (5\right ) \left (1-\frac {x}{4 \ln \left (5\right )}\right )}-2 \ln \left (1-\frac {x}{4 \ln \left (5\right )}\right )\right )\)	\(113\)

[In]

int((32*ln(5)^2-16*x*ln(5)-4*exp(4)+2*x^2)/(16*ln(5)^2-8*x*ln(5)+x^2),x,method=_RETURNVERBOSE)

[Out]

2*x+4*exp(4)/(x-4*ln(5))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-4 e^4+2 x^2-4 x \log (625)+2 \log ^2(625)}{x^2-2 x \log (625)+\log ^2(625)} \, dx=\frac {2 \, {\left (x^{2} - 4 \, x \log \left (5\right ) + 2 \, e^{4}\right )}}{x - 4 \, \log \left (5\right )} \]

[In]

integrate((32*log(5)^2-16*x*log(5)-4*exp(4)+2*x^2)/(16*log(5)^2-8*x*log(5)+x^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-4 e^4+2 x^2-4 x \log (625)+2 \log ^2(625)}{x^2-2 x \log (625)+\log ^2(625)} \, dx=2 x + \frac {4 e^{4}}{x - 4 \log {\left (5 \right )}} \]

[In]

integrate((32*ln(5)**2-16*x*ln(5)-4*exp(4)+2*x**2)/(16*ln(5)**2-8*x*ln(5)+x**2),x)

[Out]

2*x + 4*exp(4)/(x - 4*log(5))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {-4 e^4+2 x^2-4 x \log (625)+2 \log ^2(625)}{x^2-2 x \log (625)+\log ^2(625)} \, dx=2 \, x + \frac {4 \, e^{4}}{x - 4 \, \log \left (5\right )} \]

[In]

integrate((32*log(5)^2-16*x*log(5)-4*exp(4)+2*x^2)/(16*log(5)^2-8*x*log(5)+x^2),x, algorithm="maxima")

[Out]

2*x + 4*e^4/(x - 4*log(5))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {-4 e^4+2 x^2-4 x \log (625)+2 \log ^2(625)}{x^2-2 x \log (625)+\log ^2(625)} \, dx=2 \, x + \frac {4 \, e^{4}}{x - 4 \, \log \left (5\right )} \]

[In]

integrate((32*log(5)^2-16*x*log(5)-4*exp(4)+2*x^2)/(16*log(5)^2-8*x*log(5)+x^2),x, algorithm="giac")

[Out]

2*x + 4*e^4/(x - 4*log(5))

Mupad [B] (verification not implemented)

Time = 12.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.95 \[ \int \frac {-4 e^4+2 x^2-4 x \log (625)+2 \log ^2(625)}{x^2-2 x \log (625)+\log ^2(625)} \, dx=2\,x+\frac {4\,\mathrm {atanh}\left (\frac {2\,x-8\,\ln \left (5\right )}{2\,\sqrt {4\,\ln \left (5\right )+\ln \left (625\right )}\,\sqrt {4\,\ln \left (5\right )-\ln \left (625\right )}}\right )\,{\mathrm {e}}^4}{\sqrt {4\,\ln \left (5\right )+\ln \left (625\right )}\,\sqrt {4\,\ln \left (5\right )-\ln \left (625\right )}} \]

[In]

int(-(4*exp(4) + 16*x*log(5) - 32*log(5)^2 - 2*x^2)/(16*log(5)^2 - 8*x*log(5) + x^2),x)

[Out]

2*x + (4*atanh((2*x - 8*log(5))/(2*(4*log(5) + log(625))^(1/2)*(4*log(5) - log(625))^(1/2)))*exp(4))/((4*log(5
) + log(625))^(1/2)*(4*log(5) - log(625))^(1/2))