3.69.78 $\int \frac{-4e^4+2x^2-4x\log (625)+2{\log }^2(625)}{x^2-2x\log (625)+{\log }^2(625)}dx$

Optimal. Leaf size=20 $$e^3+2x+\frac{4e^4}{x-\log (625)}$$

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Rubi [A]  time = 0.04, antiderivative size = 17, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 2, integrand size = 38, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.053, Rules used = {27, 683} $$\begin{array}{c}2x+\frac{4e^4}{x-\log (625)}\end{array}$$

Antiderivative was successfully verified.

[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 683

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-4e^4+2x^2-4x\log (625)+2{\log }^2(625)}{(x-\log (625){)}^2}dx\\ & =\int (2-\frac{4e^4}{(x-\log (625){)}^2})dx\\ & =2x+\frac{4e^4}{x-\log (625)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.85 $$\begin{array}{c}2x+\frac{4e^4}{x-\log (625)}\end{array}$$

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 1.92, size = 23, normalized size = 1.15 $$\begin{array}{c}\frac{2(x^2-4x\log (5)+2e^4)}{x-4\log (5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.14, size = 16, normalized size = 0.80 $$\begin{array}{c}2x+\frac{4e^4}{x-4\log (5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.13, size = 17, normalized size = 0.85

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.51, size = 16, normalized size = 0.80 $$\begin{array}{c}2x+\frac{4e^4}{x-4\log (5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 4.17, size = 59, normalized size = 2.95 $$\begin{array}{c}2x+\frac{4\mathrm{atanh}\left(\frac{2x-8\ln (5)}{2\sqrt{4\ln (5)+\ln \left(625\right)}\sqrt{4\ln (5)-\ln \left(625\right)}}\right){\mathrm{e}}^4}{\sqrt{4\ln (5)+\ln \left(625\right)}\sqrt{4\ln (5)-\ln \left(625\right)}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.13, size = 14, normalized size = 0.70 $$\begin{array}{c}2x+\frac{4e^4}{x-4\log (5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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))

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