3.96.24 \(\int \frac {-2 x^3+(-2500 x+2 x^3) \log (\frac {1}{125} (-1250+x^2))}{(-1250+x^2) \log ^2(\frac {1}{125} (-1250+x^2))} \, dx\)

Optimal. Leaf size=20 \[ \frac {x^2}{\log \left (-15+5 \left (1+\frac {x^2}{625}\right )\right )} \]

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Rubi [A]  time = 0.32, antiderivative size = 36, normalized size of antiderivative = 1.80, number of steps used = 14, number of rules used = 11, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.239, Rules used = {6725, 2475, 2411, 12, 2353, 2297, 2298, 2302, 30, 2454, 2389} \begin {gather*} \frac {1250}{\log \left (\frac {x^2}{125}-10\right )}-\frac {1250-x^2}{\log \left (\frac {x^2}{125}-10\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x^3 + (-2500*x + 2*x^3)*Log[(-1250 + x^2)/125])/((-1250 + x^2)*Log[(-1250 + x^2)/125]^2),x]

[Out]

1250/Log[-10 + x^2/125] - (1250 - x^2)/Log[-10 + x^2/125]

Rule 12

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

Rule 30

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

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

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 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {2 x^3}{\left (-1250+x^2\right ) \log ^2\left (-10+\frac {x^2}{125}\right )}+\frac {2 x}{\log \left (-10+\frac {x^2}{125}\right )}\right ) \, dx\\ &=-\left (2 \int \frac {x^3}{\left (-1250+x^2\right ) \log ^2\left (-10+\frac {x^2}{125}\right )} \, dx\right )+2 \int \frac {x}{\log \left (-10+\frac {x^2}{125}\right )} \, dx\\ &=-\operatorname {Subst}\left (\int \frac {x}{(-1250+x) \log ^2\left (-10+\frac {x}{125}\right )} \, dx,x,x^2\right )+\operatorname {Subst}\left (\int \frac {1}{\log \left (-10+\frac {x}{125}\right )} \, dx,x,x^2\right )\\ &=-\left (125 \operatorname {Subst}\left (\int \frac {1250+125 x}{125 x \log ^2(x)} \, dx,x,\frac {1}{125} \left (-1250+x^2\right )\right )\right )+125 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,\frac {1}{125} \left (-1250+x^2\right )\right )\\ &=125 \text {li}\left (-10+\frac {x^2}{125}\right )-\operatorname {Subst}\left (\int \frac {1250+125 x}{x \log ^2(x)} \, dx,x,\frac {1}{125} \left (-1250+x^2\right )\right )\\ &=125 \text {li}\left (-10+\frac {x^2}{125}\right )-\operatorname {Subst}\left (\int \left (\frac {125}{\log ^2(x)}+\frac {1250}{x \log ^2(x)}\right ) \, dx,x,\frac {1}{125} \left (-1250+x^2\right )\right )\\ &=125 \text {li}\left (-10+\frac {x^2}{125}\right )-125 \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,\frac {1}{125} \left (-1250+x^2\right )\right )-1250 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,\frac {1}{125} \left (-1250+x^2\right )\right )\\ &=-\frac {1250-x^2}{\log \left (-10+\frac {x^2}{125}\right )}+125 \text {li}\left (-10+\frac {x^2}{125}\right )-125 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,\frac {1}{125} \left (-1250+x^2\right )\right )-1250 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (-10+\frac {x^2}{125}\right )\right )\\ &=\frac {1250}{\log \left (-10+\frac {x^2}{125}\right )}-\frac {1250-x^2}{\log \left (-10+\frac {x^2}{125}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 16, normalized size = 0.80 \begin {gather*} \frac {x^2}{\log \left (-10+\frac {x^2}{125}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x^3 + (-2500*x + 2*x^3)*Log[(-1250 + x^2)/125])/((-1250 + x^2)*Log[(-1250 + x^2)/125]^2),x]

[Out]

x^2/Log[-10 + x^2/125]

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fricas [A]  time = 0.55, size = 14, normalized size = 0.70 \begin {gather*} \frac {x^{2}}{\log \left (\frac {1}{125} \, x^{2} - 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-2500*x)*log(1/125*x^2-10)-2*x^3)/(x^2-1250)/log(1/125*x^2-10)^2,x, algorithm="fricas")

[Out]

x^2/log(1/125*x^2 - 10)

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giac [A]  time = 0.19, size = 14, normalized size = 0.70 \begin {gather*} \frac {x^{2}}{\log \left (\frac {1}{125} \, x^{2} - 10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-2500*x)*log(1/125*x^2-10)-2*x^3)/(x^2-1250)/log(1/125*x^2-10)^2,x, algorithm="giac")

[Out]

x^2/log(1/125*x^2 - 10)

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maple [A]  time = 0.06, size = 15, normalized size = 0.75

method	result	size
norman	\(\frac {x^{2}}{\ln \left (\frac {x^{2}}{125}-10\right )}\)	\(15\)
risch	\(\frac {x^{2}}{\ln \left (\frac {x^{2}}{125}-10\right )}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3-2500*x)*ln(1/125*x^2-10)-2*x^3)/(x^2-1250)/ln(1/125*x^2-10)^2,x,method=_RETURNVERBOSE)

[Out]

x^2/ln(1/125*x^2-10)

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maxima [A]  time = 0.48, size = 20, normalized size = 1.00 \begin {gather*} -\frac {x^{2}}{3 \, \log \relax (5) - \log \left (x^{2} - 1250\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-2500*x)*log(1/125*x^2-10)-2*x^3)/(x^2-1250)/log(1/125*x^2-10)^2,x, algorithm="maxima")

[Out]

-x^2/(3*log(5) - log(x^2 - 1250))

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mupad [B]  time = 9.93, size = 14, normalized size = 0.70 \begin {gather*} \frac {x^2}{\ln \left (\frac {x^2}{125}-10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x^2/125 - 10)*(2500*x - 2*x^3) + 2*x^3)/(log(x^2/125 - 10)^2*(x^2 - 1250)),x)

[Out]

x^2/log(x^2/125 - 10)

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sympy [A]  time = 0.10, size = 10, normalized size = 0.50 \begin {gather*} \frac {x^{2}}{\log {\left (\frac {x^{2}}{125} - 10 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3-2500*x)*ln(1/125*x**2-10)-2*x**3)/(x**2-1250)/ln(1/125*x**2-10)**2,x)

[Out]

x**2/log(x**2/125 - 10)

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