3.17.2 $\int \frac{200+200\log (x)+x^2{\log }^2(x)}{x^2{\log }^2(x)}dx$

Optimal. Leaf size=12 $$2+x-\frac{200}{x\log (x)}$$

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Rubi [A]  time = 0.17, antiderivative size = 11, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 4, integrand size = 22, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.182, Rules used = {6742, 2306, 2309, 2178} $$\begin{array}{c}x-\frac{200}{x\log (x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(200 + 200*Log[x] + x^2*Log[x]^2)/(x^2*Log[x]^2),x]

[Out]

x - 200/(x*Log[x])

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2306

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

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (1+\frac{200}{x^2{\log }^2(x)}+\frac{200}{x^2\log (x)})dx\\ & =x+200\int \frac{1}{x^2{\log }^2(x)}dx+200\int \frac{1}{x^2\log (x)}dx\\ & =x-\frac{200}{x\log (x)}-200\int \frac{1}{x^2\log (x)}dx+200\mathrm{Subst}(\int \frac{e^{-x}}{x}dx,x,\log (x))\\ & =x+200\text{Ei}(-\log (x))-\frac{200}{x\log (x)}-200\mathrm{Subst}(\int \frac{e^{-x}}{x}dx,x,\log (x))\\ & =x-\frac{200}{x\log (x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.01, size = 11, normalized size = 0.92 $$\begin{array}{c}x-\frac{200}{x\log (x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(200 + 200*Log[x] + x^2*Log[x]^2)/(x^2*Log[x]^2),x]

[Out]

x - 200/(x*Log[x])

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fricas [A]  time = 0.61, size = 16, normalized size = 1.33 $$\begin{array}{c}\frac{x^2\log (x)-200}{x\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2+200*log(x)+200)/x^2/log(x)^2,x, algorithm="fricas")

[Out]

(x^2*log(x) - 200)/(x*log(x))

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giac [A]  time = 0.18, size = 11, normalized size = 0.92 $$\begin{array}{c}x-\frac{200}{x\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2+200*log(x)+200)/x^2/log(x)^2,x, algorithm="giac")

[Out]

x - 200/(x*log(x))

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maple [A]  time = 0.02, size = 12, normalized size = 1.00

method	result	size
default	$x-\frac{200}{x\ln (x)}$	$12$
risch	$x-\frac{200}{x\ln (x)}$	$12$
norman	$\frac{-200+x^2\ln (x)}{x\ln (x)}$	$17$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*ln(x)^2+200*ln(x)+200)/x^2/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

x-200/x/ln(x)

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maxima [C]  time = 0.45, size = 15, normalized size = 1.25 $$\begin{array}{c}x+200\mathrm{E}\mathrm{i}(-\log (x))-200\mathrm{\Gamma }(-1,\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2*log(x)^2+200*log(x)+200)/x^2/log(x)^2,x, algorithm="maxima")

[Out]

x + 200*Ei(-log(x)) - 200*gamma(-1, log(x))

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mupad [B]  time = 1.05, size = 11, normalized size = 0.92 $$\begin{array}{c}x-\frac{200}{x\ln (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((200*log(x) + x^2*log(x)^2 + 200)/(x^2*log(x)^2),x)

[Out]

x - 200/(x*log(x))

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sympy [A]  time = 0.08, size = 7, normalized size = 0.58 $$\begin{array}{c}x-\frac{200}{x\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2*ln(x)**2+200*ln(x)+200)/x**2/ln(x)**2,x)

[Out]

x - 200/(x*log(x))

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