3.71.98 \(\int \frac {-1000-1000 x-250 x^2+(-80-40 x) \log ^3(x)+(-2000-1000 x) \log (2 x)-1000 \log ^2(2 x)}{(4 x+4 x^2+x^3) \log ^3(x)+(8 x+4 x^2) \log ^3(x) \log (2 x)+4 x \log ^3(x) \log ^2(2 x)} \, dx\)

Optimal. Leaf size=24 \[ 5 \left (\frac {25}{\log ^2(x)}+\frac {4}{1+\frac {x}{2}+\log (2 x)}\right ) \]

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Rubi [A]  time = 0.20, antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 7, number of rules used = 6, integrand size = 90, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {6688, 12, 14, 2302, 30, 6686} \begin {gather*} \frac {125}{\log ^2(x)}+\frac {40}{x+2 \log (2 x)+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1000 - 1000*x - 250*x^2 + (-80 - 40*x)*Log[x]^3 + (-2000 - 1000*x)*Log[2*x] - 1000*Log[2*x]^2)/((4*x + 4
*x^2 + x^3)*Log[x]^3 + (8*x + 4*x^2)*Log[x]^3*Log[2*x] + 4*x*Log[x]^3*Log[2*x]^2),x]

[Out]

125/Log[x]^2 + 40/(2 + x + 2*Log[2*x])

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, 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 {gather*} \begin {aligned} \text {integral} &=\int \frac {10 \left (-\frac {25}{\log ^3(x)}-\frac {4 (2+x)}{(2+x+2 \log (2 x))^2}\right )}{x} \, dx\\ &=10 \int \frac {-\frac {25}{\log ^3(x)}-\frac {4 (2+x)}{(2+x+2 \log (2 x))^2}}{x} \, dx\\ &=10 \int \left (-\frac {25}{x \log ^3(x)}-\frac {4 (2+x)}{x (2+x+2 \log (2 x))^2}\right ) \, dx\\ &=-\left (40 \int \frac {2+x}{x (2+x+2 \log (2 x))^2} \, dx\right )-250 \int \frac {1}{x \log ^3(x)} \, dx\\ &=\frac {40}{2+x+2 \log (2 x)}-250 \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right )\\ &=\frac {125}{\log ^2(x)}+\frac {40}{2+x+2 \log (2 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 24, normalized size = 1.00 \begin {gather*} 10 \left (\frac {25}{2 \log ^2(x)}+\frac {4}{2+x+\log (4)+2 \log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1000 - 1000*x - 250*x^2 + (-80 - 40*x)*Log[x]^3 + (-2000 - 1000*x)*Log[2*x] - 1000*Log[2*x]^2)/((4
*x + 4*x^2 + x^3)*Log[x]^3 + (8*x + 4*x^2)*Log[x]^3*Log[2*x] + 4*x*Log[x]^3*Log[2*x]^2),x]

[Out]

10*(25/(2*Log[x]^2) + 4/(2 + x + Log[4] + 2*Log[x]))

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fricas [B]  time = 0.96, size = 42, normalized size = 1.75 \begin {gather*} \frac {5 \, {\left (8 \, \log \relax (x)^{2} + 25 \, x + 50 \, \log \relax (2) + 50 \, \log \relax (x) + 50\right )}}{{\left (x + 2 \, \log \relax (2) + 2\right )} \log \relax (x)^{2} + 2 \, \log \relax (x)^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1000*log(2*x)^2+(-1000*x-2000)*log(2*x)+(-40*x-80)*log(x)^3-250*x^2-1000*x-1000)/(4*x*log(x)^3*log
(2*x)^2+(4*x^2+8*x)*log(x)^3*log(2*x)+(x^3+4*x^2+4*x)*log(x)^3),x, algorithm="fricas")

[Out]

5*(8*log(x)^2 + 25*x + 50*log(2) + 50*log(x) + 50)/((x + 2*log(2) + 2)*log(x)^2 + 2*log(x)^3)

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giac [A]  time = 0.22, size = 22, normalized size = 0.92 \begin {gather*} \frac {40}{x + 2 \, \log \relax (2) + 2 \, \log \relax (x) + 2} + \frac {125}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1000*log(2*x)^2+(-1000*x-2000)*log(2*x)+(-40*x-80)*log(x)^3-250*x^2-1000*x-1000)/(4*x*log(x)^3*log
(2*x)^2+(4*x^2+8*x)*log(x)^3*log(2*x)+(x^3+4*x^2+4*x)*log(x)^3),x, algorithm="giac")

[Out]

40/(x + 2*log(2) + 2*log(x) + 2) + 125/log(x)^2

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maple [A]  time = 0.09, size = 39, normalized size = 1.62

method	result	size
risch	\(\frac {250+250 \ln \relax (2)+125 x +250 \ln \relax (x )+40 \ln \relax (x )^{2}}{\ln \relax (x )^{2} \left (2+2 \ln \relax (2)+x +2 \ln \relax (x )\right )}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1000*ln(2*x)^2+(-1000*x-2000)*ln(2*x)+(-40*x-80)*ln(x)^3-250*x^2-1000*x-1000)/(4*x*ln(x)^3*ln(2*x)^2+(4*
x^2+8*x)*ln(x)^3*ln(2*x)+(x^3+4*x^2+4*x)*ln(x)^3),x,method=_RETURNVERBOSE)

[Out]

5*(50+50*ln(2)+25*x+50*ln(x)+8*ln(x)^2)/ln(x)^2/(2+2*ln(2)+x+2*ln(x))

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maxima [B]  time = 0.51, size = 42, normalized size = 1.75 \begin {gather*} \frac {5 \, {\left (8 \, \log \relax (x)^{2} + 25 \, x + 50 \, \log \relax (2) + 50 \, \log \relax (x) + 50\right )}}{{\left (x + 2 \, \log \relax (2) + 2\right )} \log \relax (x)^{2} + 2 \, \log \relax (x)^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1000*log(2*x)^2+(-1000*x-2000)*log(2*x)+(-40*x-80)*log(x)^3-250*x^2-1000*x-1000)/(4*x*log(x)^3*log
(2*x)^2+(4*x^2+8*x)*log(x)^3*log(2*x)+(x^3+4*x^2+4*x)*log(x)^3),x, algorithm="maxima")

[Out]

5*(8*log(x)^2 + 25*x + 50*log(2) + 50*log(x) + 50)/((x + 2*log(2) + 2)*log(x)^2 + 2*log(x)^3)

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mupad [B]  time = 4.36, size = 20, normalized size = 0.83 \begin {gather*} \frac {125}{{\ln \relax (x)}^2}+\frac {40}{x+\ln \relax (4)+2\,\ln \relax (x)+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(1000*x + 1000*log(2*x)^2 + 250*x^2 + log(2*x)*(1000*x + 2000) + log(x)^3*(40*x + 80) + 1000)/(log(x)^3*(
4*x + 4*x^2 + x^3) + log(2*x)*log(x)^3*(8*x + 4*x^2) + 4*x*log(2*x)^2*log(x)^3),x)

[Out]

125/log(x)^2 + 40/(x + log(4) + 2*log(x) + 2)

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sympy [B]  time = 0.47, size = 41, normalized size = 1.71 \begin {gather*} \frac {125 x + 40 \log {\relax (x )}^{2} + 250 \log {\relax (x )} + 250 \log {\relax (2 )} + 250}{\left (x + 2 \log {\relax (2 )} + 2\right ) \log {\relax (x )}^{2} + 2 \log {\relax (x )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1000*ln(2*x)**2+(-1000*x-2000)*ln(2*x)+(-40*x-80)*ln(x)**3-250*x**2-1000*x-1000)/(4*x*ln(x)**3*ln(
2*x)**2+(4*x**2+8*x)*ln(x)**3*ln(2*x)+(x**3+4*x**2+4*x)*ln(x)**3),x)

[Out]

(125*x + 40*log(x)**2 + 250*log(x) + 250*log(2) + 250)/((x + 2*log(2) + 2)*log(x)**2 + 2*log(x)**3)

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