3.48.18 \(\int \frac {-256-512 x-256 x^2+(-64+64 x^2) \log (x^2)}{5 x^2 \log ^3(x^2)} \, dx\)

Optimal. Leaf size=22 \[ -5+\frac {16 (2+2 x)^2}{5 x \log ^2\left (x^2\right )} \]

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Rubi [A]  time = 0.48, antiderivative size = 35, normalized size of antiderivative = 1.59, number of steps used = 25, number of rules used = 11, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {12, 6741, 6742, 2353, 2297, 2300, 2178, 2306, 2310, 2302, 30} \begin {gather*} \frac {64 x}{5 \log ^2\left (x^2\right )}+\frac {128}{5 \log ^2\left (x^2\right )}+\frac {64}{5 x \log ^2\left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-256 - 512*x - 256*x^2 + (-64 + 64*x^2)*Log[x^2])/(5*x^2*Log[x^2]^3),x]

[Out]

128/(5*Log[x^2]^2) + 64/(5*x*Log[x^2]^2) + (64*x)/(5*Log[x^2]^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 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 2300

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, 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 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 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {-256-512 x-256 x^2+\left (-64+64 x^2\right ) \log \left (x^2\right )}{x^2 \log ^3\left (x^2\right )} \, dx\\ &=\frac {1}{5} \int \frac {64 (1+x) \left (-4-4 x-\log \left (x^2\right )+x \log \left (x^2\right )\right )}{x^2 \log ^3\left (x^2\right )} \, dx\\ &=\frac {64}{5} \int \frac {(1+x) \left (-4-4 x-\log \left (x^2\right )+x \log \left (x^2\right )\right )}{x^2 \log ^3\left (x^2\right )} \, dx\\ &=\frac {64}{5} \int \left (-\frac {4 (1+x)^2}{x^2 \log ^3\left (x^2\right )}+\frac {-1+x^2}{x^2 \log ^2\left (x^2\right )}\right ) \, dx\\ &=\frac {64}{5} \int \frac {-1+x^2}{x^2 \log ^2\left (x^2\right )} \, dx-\frac {256}{5} \int \frac {(1+x)^2}{x^2 \log ^3\left (x^2\right )} \, dx\\ &=\frac {64}{5} \int \left (\frac {1}{\log ^2\left (x^2\right )}-\frac {1}{x^2 \log ^2\left (x^2\right )}\right ) \, dx-\frac {256}{5} \int \left (\frac {1}{\log ^3\left (x^2\right )}+\frac {1}{x^2 \log ^3\left (x^2\right )}+\frac {2}{x \log ^3\left (x^2\right )}\right ) \, dx\\ &=\frac {64}{5} \int \frac {1}{\log ^2\left (x^2\right )} \, dx-\frac {64}{5} \int \frac {1}{x^2 \log ^2\left (x^2\right )} \, dx-\frac {256}{5} \int \frac {1}{\log ^3\left (x^2\right )} \, dx-\frac {256}{5} \int \frac {1}{x^2 \log ^3\left (x^2\right )} \, dx-\frac {512}{5} \int \frac {1}{x \log ^3\left (x^2\right )} \, dx\\ &=\frac {64}{5 x \log ^2\left (x^2\right )}+\frac {64 x}{5 \log ^2\left (x^2\right )}+\frac {32}{5 x \log \left (x^2\right )}-\frac {32 x}{5 \log \left (x^2\right )}+\frac {32}{5} \int \frac {1}{\log \left (x^2\right )} \, dx+\frac {32}{5} \int \frac {1}{x^2 \log \left (x^2\right )} \, dx-\frac {64}{5} \int \frac {1}{\log ^2\left (x^2\right )} \, dx+\frac {64}{5} \int \frac {1}{x^2 \log ^2\left (x^2\right )} \, dx-\frac {256}{5} \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log \left (x^2\right )\right )\\ &=\frac {128}{5 \log ^2\left (x^2\right )}+\frac {64}{5 x \log ^2\left (x^2\right )}+\frac {64 x}{5 \log ^2\left (x^2\right )}-\frac {32}{5} \int \frac {1}{\log \left (x^2\right )} \, dx-\frac {32}{5} \int \frac {1}{x^2 \log \left (x^2\right )} \, dx+\frac {(16 x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{5 \sqrt {x^2}}+\frac {\left (16 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{-x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{5 x}\\ &=\frac {16 \sqrt {x^2} \text {Ei}\left (-\frac {1}{2} \log \left (x^2\right )\right )}{5 x}+\frac {16 x \text {Ei}\left (\frac {\log \left (x^2\right )}{2}\right )}{5 \sqrt {x^2}}+\frac {128}{5 \log ^2\left (x^2\right )}+\frac {64}{5 x \log ^2\left (x^2\right )}+\frac {64 x}{5 \log ^2\left (x^2\right )}-\frac {(16 x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{5 \sqrt {x^2}}-\frac {\left (16 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{-x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{5 x}\\ &=\frac {128}{5 \log ^2\left (x^2\right )}+\frac {64}{5 x \log ^2\left (x^2\right )}+\frac {64 x}{5 \log ^2\left (x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 18, normalized size = 0.82 \begin {gather*} \frac {64 (1+x)^2}{5 x \log ^2\left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-256 - 512*x - 256*x^2 + (-64 + 64*x^2)*Log[x^2])/(5*x^2*Log[x^2]^3),x]

[Out]

(64*(1 + x)^2)/(5*x*Log[x^2]^2)

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fricas [A]  time = 0.55, size = 19, normalized size = 0.86 \begin {gather*} \frac {64 \, {\left (x^{2} + 2 \, x + 1\right )}}{5 \, x \log \left (x^{2}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((64*x^2-64)*log(x^2)-256*x^2-512*x-256)/x^2/log(x^2)^3,x, algorithm="fricas")

[Out]

64/5*(x^2 + 2*x + 1)/(x*log(x^2)^2)

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giac [A]  time = 0.13, size = 19, normalized size = 0.86 \begin {gather*} \frac {64 \, {\left (x^{2} + 2 \, x + 1\right )}}{5 \, x \log \left (x^{2}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((64*x^2-64)*log(x^2)-256*x^2-512*x-256)/x^2/log(x^2)^3,x, algorithm="giac")

[Out]

64/5*(x^2 + 2*x + 1)/(x*log(x^2)^2)

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maple [A]  time = 0.03, size = 20, normalized size = 0.91

method	result	size
risch	\(\frac {\frac {64}{5}+\frac {128}{5} x +\frac {64}{5} x^{2}}{x \ln \left (x^{2}\right )^{2}}\)	\(20\)
norman	\(\frac {\frac {64}{5}+\frac {128}{5} x +\frac {64}{5} x^{2}}{x \ln \left (x^{2}\right )^{2}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((64*x^2-64)*ln(x^2)-256*x^2-512*x-256)/x^2/ln(x^2)^3,x,method=_RETURNVERBOSE)

[Out]

64/5/x*(x^2+2*x+1)/ln(x^2)^2

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maxima [A]  time = 0.40, size = 17, normalized size = 0.77 \begin {gather*} \frac {16 \, {\left (x^{2} + 2 \, x + 1\right )}}{5 \, x \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((64*x^2-64)*log(x^2)-256*x^2-512*x-256)/x^2/log(x^2)^3,x, algorithm="maxima")

[Out]

16/5*(x^2 + 2*x + 1)/(x*log(x)^2)

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mupad [B]  time = 3.55, size = 21, normalized size = 0.95 \begin {gather*} \frac {64\,x^2+128\,x+64}{5\,x\,{\ln \left (x^2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((512*x)/5 - (log(x^2)*(64*x^2 - 64))/5 + (256*x^2)/5 + 256/5)/(x^2*log(x^2)^3),x)

[Out]

(128*x + 64*x^2 + 64)/(5*x*log(x^2)^2)

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sympy [A]  time = 0.11, size = 19, normalized size = 0.86 \begin {gather*} \frac {64 x^{2} + 128 x + 64}{5 x \log {\left (x^{2} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((64*x**2-64)*ln(x**2)-256*x**2-512*x-256)/x**2/ln(x**2)**3,x)

[Out]

(64*x**2 + 128*x + 64)/(5*x*log(x**2)**2)

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