3.42.22 \(\int \frac {4-2 x^4+(-2+5 x^4) \log (x^2)+25 \log ^2(x^2)}{25 \log ^2(x^2)} \, dx\)

Optimal. Leaf size=19 \[ 2+x+\frac {x \left (-2+x^4\right )}{25 \log \left (x^2\right )} \]

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Rubi [A]  time = 0.19, antiderivative size = 26, normalized size of antiderivative = 1.37, number of steps used = 17, number of rules used = 8, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {12, 6742, 2330, 2297, 2300, 2178, 2306, 2310} \begin {gather*} -\frac {2 x}{25 \log \left (x^2\right )}+\frac {x^5}{25 \log \left (x^2\right )}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 - 2*x^4 + (-2 + 5*x^4)*Log[x^2] + 25*Log[x^2]^2)/(25*Log[x^2]^2),x]

[Out]

x - (2*x)/(25*Log[x^2]) + x^5/(25*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 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 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 2330

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

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}{25} \int \frac {4-2 x^4+\left (-2+5 x^4\right ) \log \left (x^2\right )+25 \log ^2\left (x^2\right )}{\log ^2\left (x^2\right )} \, dx\\ &=\frac {1}{25} \int \left (25-\frac {2 \left (-2+x^4\right )}{\log ^2\left (x^2\right )}+\frac {-2+5 x^4}{\log \left (x^2\right )}\right ) \, dx\\ &=x+\frac {1}{25} \int \frac {-2+5 x^4}{\log \left (x^2\right )} \, dx-\frac {2}{25} \int \frac {-2+x^4}{\log ^2\left (x^2\right )} \, dx\\ &=x+\frac {1}{25} \int \left (-\frac {2}{\log \left (x^2\right )}+\frac {5 x^4}{\log \left (x^2\right )}\right ) \, dx-\frac {2}{25} \int \left (-\frac {2}{\log ^2\left (x^2\right )}+\frac {x^4}{\log ^2\left (x^2\right )}\right ) \, dx\\ &=x-\frac {2}{25} \int \frac {x^4}{\log ^2\left (x^2\right )} \, dx-\frac {2}{25} \int \frac {1}{\log \left (x^2\right )} \, dx+\frac {4}{25} \int \frac {1}{\log ^2\left (x^2\right )} \, dx+\frac {1}{5} \int \frac {x^4}{\log \left (x^2\right )} \, dx\\ &=x-\frac {2 x}{25 \log \left (x^2\right )}+\frac {x^5}{25 \log \left (x^2\right )}+\frac {2}{25} \int \frac {1}{\log \left (x^2\right )} \, dx-\frac {1}{5} \int \frac {x^4}{\log \left (x^2\right )} \, dx+\frac {x^5 \operatorname {Subst}\left (\int \frac {e^{5 x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{10 \left (x^2\right )^{5/2}}-\frac {x \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{25 \sqrt {x^2}}\\ &=x-\frac {x \text {Ei}\left (\frac {\log \left (x^2\right )}{2}\right )}{25 \sqrt {x^2}}+\frac {x^5 \text {Ei}\left (\frac {5 \log \left (x^2\right )}{2}\right )}{10 \left (x^2\right )^{5/2}}-\frac {2 x}{25 \log \left (x^2\right )}+\frac {x^5}{25 \log \left (x^2\right )}-\frac {x^5 \operatorname {Subst}\left (\int \frac {e^{5 x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{10 \left (x^2\right )^{5/2}}+\frac {x \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{25 \sqrt {x^2}}\\ &=x-\frac {2 x}{25 \log \left (x^2\right )}+\frac {x^5}{25 \log \left (x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 18, normalized size = 0.95 \begin {gather*} x+\frac {x \left (-2+x^4\right )}{25 \log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 2*x^4 + (-2 + 5*x^4)*Log[x^2] + 25*Log[x^2]^2)/(25*Log[x^2]^2),x]

[Out]

x + (x*(-2 + x^4))/(25*Log[x^2])

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fricas [A]  time = 0.65, size = 22, normalized size = 1.16 \begin {gather*} \frac {x^{5} + 25 \, x \log \left (x^{2}\right ) - 2 \, x}{25 \, \log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(25*log(x^2)^2+(5*x^4-2)*log(x^2)-2*x^4+4)/log(x^2)^2,x, algorithm="fricas")

[Out]

1/25*(x^5 + 25*x*log(x^2) - 2*x)/log(x^2)

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giac [A]  time = 0.24, size = 17, normalized size = 0.89 \begin {gather*} x + \frac {x^{5} - 2 \, x}{25 \, \log \left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(25*log(x^2)^2+(5*x^4-2)*log(x^2)-2*x^4+4)/log(x^2)^2,x, algorithm="giac")

[Out]

x + 1/25*(x^5 - 2*x)/log(x^2)

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maple [A]  time = 0.03, size = 17, normalized size = 0.89

method	result	size
risch	\(x +\frac {x \left (x^{4}-2\right )}{25 \ln \left (x^{2}\right )}\)	\(17\)
norman	\(\frac {x \ln \left (x^{2}\right )-\frac {2 x}{25}+\frac {x^{5}}{25}}{\ln \left (x^{2}\right )}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*(25*ln(x^2)^2+(5*x^4-2)*ln(x^2)-2*x^4+4)/ln(x^2)^2,x,method=_RETURNVERBOSE)

[Out]

x+1/25*x/ln(x^2)*(x^4-2)

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maxima [A]  time = 0.39, size = 15, normalized size = 0.79 \begin {gather*} x + \frac {x^{5} - 2 \, x}{50 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(25*log(x^2)^2+(5*x^4-2)*log(x^2)-2*x^4+4)/log(x^2)^2,x, algorithm="maxima")

[Out]

x + 1/50*(x^5 - 2*x)/log(x)

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mupad [B]  time = 3.75, size = 16, normalized size = 0.84 \begin {gather*} x+\frac {x\,\left (x^4-2\right )}{25\,\ln \left (x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(x^2)*(5*x^4 - 2))/25 + log(x^2)^2 - (2*x^4)/25 + 4/25)/log(x^2)^2,x)

[Out]

x + (x*(x^4 - 2))/(25*log(x^2))

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sympy [A]  time = 0.10, size = 14, normalized size = 0.74 \begin {gather*} x + \frac {x^{5} - 2 x}{25 \log {\left (x^{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(25*ln(x**2)**2+(5*x**4-2)*ln(x**2)-2*x**4+4)/ln(x**2)**2,x)

[Out]

x + (x**5 - 2*x)/(25*log(x**2))

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