3.46.15 \(\int \frac {-45+30 x-48 x^2+(-45+96 x^2) \log (x^2)-48 x^2 \log ^2(x^2)}{5 x^2-10 x^2 \log (x^2)+5 x^2 \log ^2(x^2)} \, dx\)

Optimal. Leaf size=24 \[ -\frac {48 x}{5}+\frac {3 (3-x)}{x \left (-1+\log \left (x^2\right )\right )} \]

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Rubi [A]  time = 0.45, antiderivative size = 33, normalized size of antiderivative = 1.38, number of steps used = 13, number of rules used = 9, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.145, Rules used = {6688, 12, 6742, 2353, 2306, 2310, 2178, 2302, 30} \begin {gather*} \frac {3}{1-\log \left (x^2\right )}-\frac {9}{x \left (1-\log \left (x^2\right )\right )}-\frac {48 x}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-45 + 30*x - 48*x^2 + (-45 + 96*x^2)*Log[x^2] - 48*x^2*Log[x^2]^2)/(5*x^2 - 10*x^2*Log[x^2] + 5*x^2*Log[x
^2]^2),x]

[Out]

(-48*x)/5 + 3/(1 - Log[x^2]) - 9/(x*(1 - Log[x^2]))

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 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 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 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 2310

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)*x)/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, 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 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 {gather*} \begin {aligned} \text {integral} &=\int \frac {-45+30 x-48 x^2+\left (-45+96 x^2\right ) \log \left (x^2\right )-48 x^2 \log ^2\left (x^2\right )}{5 x^2 \left (1-\log \left (x^2\right )\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {-45+30 x-48 x^2+\left (-45+96 x^2\right ) \log \left (x^2\right )-48 x^2 \log ^2\left (x^2\right )}{x^2 \left (1-\log \left (x^2\right )\right )^2} \, dx\\ &=\frac {1}{5} \int \left (-48+\frac {30 (-3+x)}{x^2 \left (-1+\log \left (x^2\right )\right )^2}-\frac {45}{x^2 \left (-1+\log \left (x^2\right )\right )}\right ) \, dx\\ &=-\frac {48 x}{5}+6 \int \frac {-3+x}{x^2 \left (-1+\log \left (x^2\right )\right )^2} \, dx-9 \int \frac {1}{x^2 \left (-1+\log \left (x^2\right )\right )} \, dx\\ &=-\frac {48 x}{5}+6 \int \left (-\frac {3}{x^2 \left (-1+\log \left (x^2\right )\right )^2}+\frac {1}{x \left (-1+\log \left (x^2\right )\right )^2}\right ) \, dx-\frac {\left (9 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{-x/2}}{-1+x} \, dx,x,\log \left (x^2\right )\right )}{2 x}\\ &=-\frac {48 x}{5}-\frac {9 \sqrt {x^2} \text {Ei}\left (\frac {1}{2} \left (1-\log \left (x^2\right )\right )\right )}{2 \sqrt {e} x}+6 \int \frac {1}{x \left (-1+\log \left (x^2\right )\right )^2} \, dx-18 \int \frac {1}{x^2 \left (-1+\log \left (x^2\right )\right )^2} \, dx\\ &=-\frac {48 x}{5}-\frac {9 \sqrt {x^2} \text {Ei}\left (\frac {1}{2} \left (1-\log \left (x^2\right )\right )\right )}{2 \sqrt {e} x}-\frac {9}{x \left (1-\log \left (x^2\right )\right )}+3 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,-1+\log \left (x^2\right )\right )+9 \int \frac {1}{x^2 \left (-1+\log \left (x^2\right )\right )} \, dx\\ &=-\frac {48 x}{5}-\frac {9 \sqrt {x^2} \text {Ei}\left (\frac {1}{2} \left (1-\log \left (x^2\right )\right )\right )}{2 \sqrt {e} x}+\frac {3}{1-\log \left (x^2\right )}-\frac {9}{x \left (1-\log \left (x^2\right )\right )}+\frac {\left (9 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{-x/2}}{-1+x} \, dx,x,\log \left (x^2\right )\right )}{2 x}\\ &=-\frac {48 x}{5}+\frac {3}{1-\log \left (x^2\right )}-\frac {9}{x \left (1-\log \left (x^2\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 24, normalized size = 1.00 \begin {gather*} \frac {1}{5} \left (-48 x-\frac {15 (-3+x)}{x \left (-1+\log \left (x^2\right )\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-45 + 30*x - 48*x^2 + (-45 + 96*x^2)*Log[x^2] - 48*x^2*Log[x^2]^2)/(5*x^2 - 10*x^2*Log[x^2] + 5*x^2
*Log[x^2]^2),x]

[Out]

(-48*x - (15*(-3 + x))/(x*(-1 + Log[x^2])))/5

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fricas [A]  time = 0.57, size = 33, normalized size = 1.38 \begin {gather*} -\frac {3 \, {\left (16 \, x^{2} \log \left (x^{2}\right ) - 16 \, x^{2} + 5 \, x - 15\right )}}{5 \, {\left (x \log \left (x^{2}\right ) - x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x^2*log(x^2)^2+(96*x^2-45)*log(x^2)-48*x^2+30*x-45)/(5*x^2*log(x^2)^2-10*x^2*log(x^2)+5*x^2),x,
algorithm="fricas")

[Out]

-3/5*(16*x^2*log(x^2) - 16*x^2 + 5*x - 15)/(x*log(x^2) - x)

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giac [A]  time = 0.20, size = 21, normalized size = 0.88 \begin {gather*} -\frac {48}{5} \, x - \frac {3 \, {\left (x - 3\right )}}{x \log \left (x^{2}\right ) - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x^2*log(x^2)^2+(96*x^2-45)*log(x^2)-48*x^2+30*x-45)/(5*x^2*log(x^2)^2-10*x^2*log(x^2)+5*x^2),x,
algorithm="giac")

[Out]

-48/5*x - 3*(x - 3)/(x*log(x^2) - x)

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maple [A]  time = 0.04, size = 21, normalized size = 0.88

method	result	size
risch	\(-\frac {48 x}{5}-\frac {3 \left (x -3\right )}{x \left (\ln \left (x^{2}\right )-1\right )}\)	\(21\)
norman	\(\frac {9-3 x +\frac {48 x^{2}}{5}-\frac {48 x^{2} \ln \left (x^{2}\right )}{5}}{x \left (\ln \left (x^{2}\right )-1\right )}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-48*x^2*ln(x^2)^2+(96*x^2-45)*ln(x^2)-48*x^2+30*x-45)/(5*x^2*ln(x^2)^2-10*x^2*ln(x^2)+5*x^2),x,method=_RE
TURNVERBOSE)

[Out]

-48/5*x-3*(x-3)/x/(ln(x^2)-1)

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maxima [A]  time = 0.39, size = 30, normalized size = 1.25 \begin {gather*} -\frac {3 \, {\left (32 \, x^{2} \log \relax (x) - 16 \, x^{2} + 5 \, x - 15\right )}}{5 \, {\left (2 \, x \log \relax (x) - x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x^2*log(x^2)^2+(96*x^2-45)*log(x^2)-48*x^2+30*x-45)/(5*x^2*log(x^2)^2-10*x^2*log(x^2)+5*x^2),x,
algorithm="maxima")

[Out]

-3/5*(32*x^2*log(x) - 16*x^2 + 5*x - 15)/(2*x*log(x) - x)

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mupad [B]  time = 3.07, size = 33, normalized size = 1.38 \begin {gather*} -\frac {\frac {48\,x^2}{5}+3\,x}{x}-\frac {3\,x-9}{x\,\left (\ln \left (x^2\right )-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(48*x^2 - log(x^2)*(96*x^2 - 45) - 30*x + 48*x^2*log(x^2)^2 + 45)/(5*x^2 - 10*x^2*log(x^2) + 5*x^2*log(x^
2)^2),x)

[Out]

- (3*x + (48*x^2)/5)/x - (3*x - 9)/(x*(log(x^2) - 1))

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sympy [A]  time = 0.11, size = 17, normalized size = 0.71 \begin {gather*} - \frac {48 x}{5} + \frac {9 - 3 x}{x \log {\left (x^{2} \right )} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*x**2*ln(x**2)**2+(96*x**2-45)*ln(x**2)-48*x**2+30*x-45)/(5*x**2*ln(x**2)**2-10*x**2*ln(x**2)+5*
x**2),x)

[Out]

-48*x/5 + (9 - 3*x)/(x*log(x**2) - x)

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