3.95.16 \(\int \frac {-128 x-256 x^2-128 x^3+(128 x+384 x^2+256 x^3) \log (x^2)+(512 x+1536 x^2+1024 x^3) \log ^2(x^2)}{\log ^2(x^2)} \, dx\)

Optimal. Leaf size=22 \[ 256 x^2 (1+x)^2 \left (1+\frac {1}{4 \log \left (x^2\right )}\right ) \]

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Rubi [B]  time = 0.34, antiderivative size = 49, normalized size of antiderivative = 2.23, number of steps used = 23, number of rules used = 9, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6742, 14, 2353, 2306, 2307, 2298, 2310, 2178, 2356} \begin {gather*} 256 x^4+512 x^3+256 x^2+\frac {64 x^2}{\log \left (x^2\right )}+\frac {64 x^4}{\log \left (x^2\right )}+\frac {128 x^3}{\log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-128*x - 256*x^2 - 128*x^3 + (128*x + 384*x^2 + 256*x^3)*Log[x^2] + (512*x + 1536*x^2 + 1024*x^3)*Log[x^2
]^2)/Log[x^2]^2,x]

[Out]

256*x^2 + 512*x^3 + 256*x^4 + (64*x^2)/Log[x^2] + (128*x^3)/Log[x^2] + (64*x^4)/Log[x^2]

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 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 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, 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 2307

Int[(x_)^(m_.)/Log[(c_.)*(x_)^(n_)], x_Symbol] :> Dist[1/n, Subst[Int[1/Log[c*x], x], x, x^n], x] /; FreeQ[{c,
m, n}, x] && EqQ[m, n - 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 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Polyx*(a + b*Log[c*
x^n])^p, x], x] /; FreeQ[{a, b, c, n, p}, x] && PolynomialQ[Polyx, 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 \left (512 x \left (1+3 x+2 x^2\right )-\frac {128 x (1+x)^2}{\log ^2\left (x^2\right )}+\frac {128 x (1+x) (1+2 x)}{\log \left (x^2\right )}\right ) \, dx\\ &=-\left (128 \int \frac {x (1+x)^2}{\log ^2\left (x^2\right )} \, dx\right )+128 \int \frac {x (1+x) (1+2 x)}{\log \left (x^2\right )} \, dx+512 \int x \left (1+3 x+2 x^2\right ) \, dx\\ &=-\left (128 \int \left (\frac {x}{\log ^2\left (x^2\right )}+\frac {2 x^2}{\log ^2\left (x^2\right )}+\frac {x^3}{\log ^2\left (x^2\right )}\right ) \, dx\right )+128 \int \left (\frac {x}{\log \left (x^2\right )}+\frac {3 x^2}{\log \left (x^2\right )}+\frac {2 x^3}{\log \left (x^2\right )}\right ) \, dx+512 \int \left (x+3 x^2+2 x^3\right ) \, dx\\ &=256 x^2+512 x^3+256 x^4-128 \int \frac {x}{\log ^2\left (x^2\right )} \, dx-128 \int \frac {x^3}{\log ^2\left (x^2\right )} \, dx+128 \int \frac {x}{\log \left (x^2\right )} \, dx-256 \int \frac {x^2}{\log ^2\left (x^2\right )} \, dx+256 \int \frac {x^3}{\log \left (x^2\right )} \, dx+384 \int \frac {x^2}{\log \left (x^2\right )} \, dx\\ &=256 x^2+512 x^3+256 x^4+\frac {64 x^2}{\log \left (x^2\right )}+\frac {128 x^3}{\log \left (x^2\right )}+\frac {64 x^4}{\log \left (x^2\right )}+64 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,x^2\right )-128 \int \frac {x}{\log \left (x^2\right )} \, dx+128 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log \left (x^2\right )\right )-256 \int \frac {x^3}{\log \left (x^2\right )} \, dx-384 \int \frac {x^2}{\log \left (x^2\right )} \, dx+\frac {\left (192 x^3\right ) \operatorname {Subst}\left (\int \frac {e^{3 x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{\left (x^2\right )^{3/2}}\\ &=256 x^2+512 x^3+256 x^4+\frac {192 x^3 \text {Ei}\left (\frac {3 \log \left (x^2\right )}{2}\right )}{\left (x^2\right )^{3/2}}+128 \text {Ei}\left (2 \log \left (x^2\right )\right )+\frac {64 x^2}{\log \left (x^2\right )}+\frac {128 x^3}{\log \left (x^2\right )}+\frac {64 x^4}{\log \left (x^2\right )}+64 \text {li}\left (x^2\right )-64 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,x^2\right )-128 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log \left (x^2\right )\right )-\frac {\left (192 x^3\right ) \operatorname {Subst}\left (\int \frac {e^{3 x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{\left (x^2\right )^{3/2}}\\ &=256 x^2+512 x^3+256 x^4+\frac {64 x^2}{\log \left (x^2\right )}+\frac {128 x^3}{\log \left (x^2\right )}+\frac {64 x^4}{\log \left (x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.07, size = 49, normalized size = 2.23 \begin {gather*} 256 x^2+512 x^3+256 x^4+\frac {64 x^2}{\log \left (x^2\right )}+\frac {128 x^3}{\log \left (x^2\right )}+\frac {64 x^4}{\log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-128*x - 256*x^2 - 128*x^3 + (128*x + 384*x^2 + 256*x^3)*Log[x^2] + (512*x + 1536*x^2 + 1024*x^3)*L
og[x^2]^2)/Log[x^2]^2,x]

[Out]

256*x^2 + 512*x^3 + 256*x^4 + (64*x^2)/Log[x^2] + (128*x^3)/Log[x^2] + (64*x^4)/Log[x^2]

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fricas [B]  time = 0.70, size = 38, normalized size = 1.73 \begin {gather*} \frac {64 \, {\left (x^{4} + 2 \, x^{3} + x^{2} + 4 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )} \log \left (x^{2}\right )\right )}}{\log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1024*x^3+1536*x^2+512*x)*log(x^2)^2+(256*x^3+384*x^2+128*x)*log(x^2)-128*x^3-256*x^2-128*x)/log(x^
2)^2,x, algorithm="fricas")

[Out]

64*(x^4 + 2*x^3 + x^2 + 4*(x^4 + 2*x^3 + x^2)*log(x^2))/log(x^2)

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giac [A]  time = 0.16, size = 36, normalized size = 1.64 \begin {gather*} 256 \, x^{4} + 512 \, x^{3} + 256 \, x^{2} + \frac {64 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )}}{\log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1024*x^3+1536*x^2+512*x)*log(x^2)^2+(256*x^3+384*x^2+128*x)*log(x^2)-128*x^3-256*x^2-128*x)/log(x^
2)^2,x, algorithm="giac")

[Out]

256*x^4 + 512*x^3 + 256*x^2 + 64*(x^4 + 2*x^3 + x^2)/log(x^2)

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maple [A]  time = 0.03, size = 36, normalized size = 1.64

method	result	size
risch	\(256 x^{4}+512 x^{3}+256 x^{2}+\frac {64 x^{2} \left (x^{2}+2 x +1\right )}{\ln \left (x^{2}\right )}\)	\(36\)
norman	\(\frac {64 x^{2}+128 x^{3}+64 x^{4}+256 x^{2} \ln \left (x^{2}\right )+512 x^{3} \ln \left (x^{2}\right )+256 x^{4} \ln \left (x^{2}\right )}{\ln \left (x^{2}\right )}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1024*x^3+1536*x^2+512*x)*ln(x^2)^2+(256*x^3+384*x^2+128*x)*ln(x^2)-128*x^3-256*x^2-128*x)/ln(x^2)^2,x,me
thod=_RETURNVERBOSE)

[Out]

256*x^4+512*x^3+256*x^2+64*x^2*(x^2+2*x+1)/ln(x^2)

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maxima [A]  time = 0.36, size = 34, normalized size = 1.55 \begin {gather*} 256 \, x^{4} + 512 \, x^{3} + 256 \, x^{2} + \frac {32 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1024*x^3+1536*x^2+512*x)*log(x^2)^2+(256*x^3+384*x^2+128*x)*log(x^2)-128*x^3-256*x^2-128*x)/log(x^
2)^2,x, algorithm="maxima")

[Out]

256*x^4 + 512*x^3 + 256*x^2 + 32*(x^4 + 2*x^3 + x^2)/log(x)

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mupad [B]  time = 7.44, size = 27, normalized size = 1.23 \begin {gather*} 256\,x^2\,{\left (x+1\right )}^2+\frac {64\,x^2\,{\left (x+1\right )}^2}{\ln \left (x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(128*x - log(x^2)*(128*x + 384*x^2 + 256*x^3) - log(x^2)^2*(512*x + 1536*x^2 + 1024*x^3) + 256*x^2 + 128*
x^3)/log(x^2)^2,x)

[Out]

256*x^2*(x + 1)^2 + (64*x^2*(x + 1)^2)/log(x^2)

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sympy [A]  time = 0.11, size = 34, normalized size = 1.55 \begin {gather*} 256 x^{4} + 512 x^{3} + 256 x^{2} + \frac {64 x^{4} + 128 x^{3} + 64 x^{2}}{\log {\left (x^{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((1024*x**3+1536*x**2+512*x)*ln(x**2)**2+(256*x**3+384*x**2+128*x)*ln(x**2)-128*x**3-256*x**2-128*x)
/ln(x**2)**2,x)

[Out]

256*x**4 + 512*x**3 + 256*x**2 + (64*x**4 + 128*x**3 + 64*x**2)/log(x**2)

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