3.33.92 $\int \frac{-92e-56e\log (\frac{x}{15})-8e{\log }^2(\frac{x}{15})}{16x^3+8x^3\log (\frac{x}{15})+x^3{\log }^2(\frac{x}{15})}dx$

Optimal. Leaf size=19 $$\frac{e(4-\frac{4}{4+\log \left(\frac{x}{15}\right)})}{x^2}$$

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Rubi [A]  time = 0.33, antiderivative size = 23, normalized size of antiderivative = 1.21, number of steps used = 9, number of rules used = 6, integrand size = 56, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.107, Rules used = {6688, 12, 6742, 2306, 2309, 2178} $$\begin{array}{c}\frac{4e}{x^2}-\frac{4e}{x^2(\log \left(\frac{x}{15}\right)+4)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-92*E - 56*E*Log[x/15] - 8*E*Log[x/15]^2)/(16*x^3 + 8*x^3*Log[x/15] + x^3*Log[x/15]^2),x]

[Out]

(4*E)/x^2 - (4*E)/(x^2*(4 + Log[x/15]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 6688

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

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{4e(-23-14\log \left(\frac{x}{15}\right)-2{\log }^2\left(\frac{x}{15}\right))}{x^3{(4+\log \left(\frac{x}{15}\right))}^2}dx\\ & =(4e)\int \frac{-23-14\log \left(\frac{x}{15}\right)-2{\log }^2\left(\frac{x}{15}\right)}{x^3{(4+\log \left(\frac{x}{15}\right))}^2}dx\\ & =(4e)\int (-\frac{2}{x^3}+\frac{1}{x^3{(4+\log \left(\frac{x}{15}\right))}^2}+\frac{2}{x^3(4+\log \left(\frac{x}{15}\right))})dx\\ & =\frac{4e}{x^2}+(4e)\int \frac{1}{x^3{(4+\log \left(\frac{x}{15}\right))}^2}dx+(8e)\int \frac{1}{x^3(4+\log \left(\frac{x}{15}\right))}dx\\ & =\frac{4e}{x^2}-\frac{4e}{x^2(4+\log \left(\frac{x}{15}\right))}+\frac{1}{225}(8e)\mathrm{Subst}(\int \frac{e^{-2x}}{4+x}dx,x,\log \left(\frac{x}{15}\right))-(8e)\int \frac{1}{x^3(4+\log \left(\frac{x}{15}\right))}dx\\ & =\frac{4e}{x^2}+\frac{8}{225}{e}^9\text{Ei}(-2(4+\log \left(\frac{x}{15}\right)))-\frac{4e}{x^2(4+\log \left(\frac{x}{15}\right))}-\frac{1}{225}(8e)\mathrm{Subst}(\int \frac{e^{-2x}}{4+x}dx,x,\log \left(\frac{x}{15}\right))\\ & =\frac{4e}{x^2}-\frac{4e}{x^2(4+\log \left(\frac{x}{15}\right))}\end{array}\end{array}$$

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Mathematica [A]  time = 0.13, size = 18, normalized size = 0.95 $$\begin{array}{c}-\frac{4e(-1+\frac{1}{4+\log \left(\frac{x}{15}\right)})}{x^2}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-92*E - 56*E*Log[x/15] - 8*E*Log[x/15]^2)/(16*x^3 + 8*x^3*Log[x/15] + x^3*Log[x/15]^2),x]

[Out]

(-4*E*(-1 + (4 + Log[x/15])^(-1)))/x^2

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fricas [A]  time = 0.50, size = 30, normalized size = 1.58 $$\begin{array}{c}\frac{4(e\log \left(\frac{1}{15}x\right)+3e)}{x^2\log \left(\frac{1}{15}x\right)+4x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)*log(1/15*x)^2-56*exp(1)*log(1/15*x)-92*exp(1))/(x^3*log(1/15*x)^2+8*x^3*log(1/15*x)+16*x^
3),x, algorithm="fricas")

[Out]

4*(e*log(1/15*x) + 3*e)/(x^2*log(1/15*x) + 4*x^2)

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giac [A]  time = 0.28, size = 30, normalized size = 1.58 $$\begin{array}{c}\frac{4(e\log \left(\frac{1}{15}x\right)+3e)}{x^2\log \left(\frac{1}{15}x\right)+4x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)*log(1/15*x)^2-56*exp(1)*log(1/15*x)-92*exp(1))/(x^3*log(1/15*x)^2+8*x^3*log(1/15*x)+16*x^
3),x, algorithm="giac")

[Out]

4*(e*log(1/15*x) + 3*e)/(x^2*log(1/15*x) + 4*x^2)

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maple [A]  time = 0.05, size = 24, normalized size = 1.26

method	result	size
risch	$\frac{4\mathrm{e}}{x^2}-\frac{4\mathrm{e}}{x^2(4+\ln \left(\frac{x}{15}\right))}$	$24$
norman	$\frac{4\mathrm{e}\ln \left(\frac{x}{15}\right)+12\mathrm{e}}{x^2(4+\ln \left(\frac{x}{15}\right))}$	$26$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*exp(1)*ln(1/15*x)^2-56*exp(1)*ln(1/15*x)-92*exp(1))/(x^3*ln(1/15*x)^2+8*x^3*ln(1/15*x)+16*x^3),x,metho
d=_RETURNVERBOSE)

[Out]

4*exp(1)/x^2-4/x^2*exp(1)/(4+ln(1/15*x))

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maxima [B]  time = 0.64, size = 38, normalized size = 2.00 $$\begin{array}{c}\frac{4((\log (5)+\log (3)-3)e-e\log (x))}{x^2(\log (5)+\log (3)-4)-x^2\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)*log(1/15*x)^2-56*exp(1)*log(1/15*x)-92*exp(1))/(x^3*log(1/15*x)^2+8*x^3*log(1/15*x)+16*x^
3),x, algorithm="maxima")

[Out]

4*((log(5) + log(3) - 3)*e - e*log(x))/(x^2*(log(5) + log(3) - 4) - x^2*log(x))

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mupad [B]  time = 2.18, size = 21, normalized size = 1.11 $$\begin{array}{c}\frac{4\mathrm{e}(\ln \left(\frac{x}{15}\right)+3)}{x^2(\ln \left(\frac{x}{15}\right)+4)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(92*exp(1) + 56*log(x/15)*exp(1) + 8*log(x/15)^2*exp(1))/(8*x^3*log(x/15) + 16*x^3 + x^3*log(x/15)^2),x)

[Out]

(4*exp(1)*(log(x/15) + 3))/(x^2*(log(x/15) + 4))

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sympy [A]  time = 0.23, size = 26, normalized size = 1.37 $$\begin{array}{c}-\frac{4e}{x^2\log \left(\frac{x}{15}\right)+4x^2}+\frac{4e}{x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)*ln(1/15*x)**2-56*exp(1)*ln(1/15*x)-92*exp(1))/(x**3*ln(1/15*x)**2+8*x**3*ln(1/15*x)+16*x*
*3),x)

[Out]

-4*E/(x**2*log(x/15) + 4*x**2) + 4*E/x**2

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