3.17.68 $\int \frac{-4e-8e\log (\frac{1}{x})}{144x^2-24x\log (5)+{\log }^2(5)+(192x^2-16x\log (5))\log (\frac{1}{x})+64x^2{\log }^2(\frac{1}{x})}dx$

Optimal. Leaf size=22 $$\frac{e}{-\log (5)+4x(5+2(-1+\log \left(\frac{1}{x}\right)))}$$

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Rubi [A]  time = 0.14, antiderivative size = 19, normalized size of antiderivative = 0.86, number of steps used = 3, number of rules used = 3, integrand size = 56, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.054, Rules used = {6688, 12, 6686} $$\begin{array}{c}\frac{e}{12x+8x\log \left(\frac{1}{x}\right)-\log (5)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-4*E - 8*E*Log[x^(-1)])/(144*x^2 - 24*x*Log[5] + Log[5]^2 + (192*x^2 - 16*x*Log[5])*Log[x^(-1)] + 64*x^2*
Log[x^(-1)]^2),x]

[Out]

E/(12*x - Log[5] + 8*x*Log[x^(-1)])

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{4e(-1-2\log \left(\frac{1}{x}\right))}{{(12x-\log (5)+8x\log \left(\frac{1}{x}\right))}^2}dx\\ & =(4e)\int \frac{-1-2\log \left(\frac{1}{x}\right)}{{(12x-\log (5)+8x\log \left(\frac{1}{x}\right))}^2}dx\\ & =\frac{e}{12x-\log (5)+8x\log \left(\frac{1}{x}\right)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.82 $$\begin{array}{c}-\frac{e}{-12x+\log (5)-8x\log \left(\frac{1}{x}\right)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-4*E - 8*E*Log[x^(-1)])/(144*x^2 - 24*x*Log[5] + Log[5]^2 + (192*x^2 - 16*x*Log[5])*Log[x^(-1)] + 6
4*x^2*Log[x^(-1)]^2),x]

[Out]

-(E/(-12*x + Log[5] - 8*x*Log[x^(-1)]))

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fricas [A]  time = 0.91, size = 20, normalized size = 0.91 $$\begin{array}{c}\frac{e}{8x\log \left(\frac{1}{x}\right)+12x-\log (5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)*log(1/x)-4*exp(1))/(64*x^2*log(1/x)^2+(-16*x*log(5)+192*x^2)*log(1/x)+log(5)^2-24*x*log(5
)+144*x^2),x, algorithm="fricas")

[Out]

e/(8*x*log(1/x) + 12*x - log(5))

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giac [A]  time = 0.42, size = 17, normalized size = 0.77 $$\begin{array}{c}-\frac{e}{8x\log (x)-12x+\log (5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)*log(1/x)-4*exp(1))/(64*x^2*log(1/x)^2+(-16*x*log(5)+192*x^2)*log(1/x)+log(5)^2-24*x*log(5
)+144*x^2),x, algorithm="giac")

[Out]

-e/(8*x*log(x) - 12*x + log(5))

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maple [A]  time = 0.08, size = 20, normalized size = 0.91

method	result	size
norman	$-\frac{\mathrm{e}}{-8x\ln \left(\frac{1}{x}\right)+\ln (5)-12x}$	$20$
risch	$-\frac{\mathrm{e}}{-8x\ln \left(\frac{1}{x}\right)+\ln (5)-12x}$	$20$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*exp(1)*ln(1/x)-4*exp(1))/(64*x^2*ln(1/x)^2+(-16*x*ln(5)+192*x^2)*ln(1/x)+ln(5)^2-24*x*ln(5)+144*x^2),x
,method=_RETURNVERBOSE)

[Out]

-exp(1)/(-8*x*ln(1/x)+ln(5)-12*x)

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maxima [A]  time = 0.61, size = 17, normalized size = 0.77 $$\begin{array}{c}-\frac{e}{8x\log (x)-12x+\log (5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)*log(1/x)-4*exp(1))/(64*x^2*log(1/x)^2+(-16*x*log(5)+192*x^2)*log(1/x)+log(5)^2-24*x*log(5
)+144*x^2),x, algorithm="maxima")

[Out]

-e/(8*x*log(x) - 12*x + log(5))

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mupad [B]  time = 1.47, size = 20, normalized size = 0.91 $$\begin{array}{c}\frac{\mathrm{e}}{12x-\ln (5)+8x\ln \left(\frac{1}{x}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(1) + 8*log(1/x)*exp(1))/(log(5)^2 - log(1/x)*(16*x*log(5) - 192*x^2) - 24*x*log(5) + 144*x^2 + 64*
x^2*log(1/x)^2),x)

[Out]

exp(1)/(12*x - log(5) + 8*x*log(1/x))

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sympy [A]  time = 0.23, size = 17, normalized size = 0.77 $$\begin{array}{c}\frac{e}{8x\log \left(\frac{1}{x}\right)+12x-\log (5)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)*ln(1/x)-4*exp(1))/(64*x**2*ln(1/x)**2+(-16*x*ln(5)+192*x**2)*ln(1/x)+ln(5)**2-24*x*ln(5)+
144*x**2),x)

[Out]

E/(8*x*log(1/x) + 12*x - log(5))

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