3.100.10 \(\int \frac {6+(-3+20 x^3) \log (x^2)+10 x^3 \log ^2(x^2)}{45 x^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{9} x^2 \log ^2\left (x^2\right )+\frac {1}{15} \left (-1+\frac {\log \left (x^2\right )}{x}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 44, normalized size of antiderivative = 1.57, number of steps used = 7, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {12, 14, 2334, 2305, 2304} \begin {gather*} \frac {1}{9} x^2 \log ^2\left (x^2\right )-\frac {2}{9} x^2 \log \left (x^2\right )+\frac {1}{45} \left (10 x^2+\frac {3}{x}\right ) \log \left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 + (-3 + 20*x^3)*Log[x^2] + 10*x^3*Log[x^2]^2)/(45*x^2),x]

[Out]

(-2*x^2*Log[x^2])/9 + ((3/x + 10*x^2)*Log[x^2])/45 + (x^2*Log[x^2]^2)/9

Rule 12

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

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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 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] && GtQ[p, 0]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{45} \int \frac {6+\left (-3+20 x^3\right ) \log \left (x^2\right )+10 x^3 \log ^2\left (x^2\right )}{x^2} \, dx\\ &=\frac {1}{45} \int \left (\frac {6}{x^2}+\frac {\left (-3+20 x^3\right ) \log \left (x^2\right )}{x^2}+10 x \log ^2\left (x^2\right )\right ) \, dx\\ &=-\frac {2}{15 x}+\frac {1}{45} \int \frac {\left (-3+20 x^3\right ) \log \left (x^2\right )}{x^2} \, dx+\frac {2}{9} \int x \log ^2\left (x^2\right ) \, dx\\ &=-\frac {2}{15 x}+\frac {1}{45} \left (\frac {3}{x}+10 x^2\right ) \log \left (x^2\right )+\frac {1}{9} x^2 \log ^2\left (x^2\right )-\frac {2}{45} \int \left (\frac {3}{x^2}+10 x\right ) \, dx-\frac {4}{9} \int x \log \left (x^2\right ) \, dx\\ &=-\frac {2}{9} x^2 \log \left (x^2\right )+\frac {1}{45} \left (\frac {3}{x}+10 x^2\right ) \log \left (x^2\right )+\frac {1}{9} x^2 \log ^2\left (x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 25, normalized size = 0.89 \begin {gather*} \frac {\log \left (x^2\right )}{15 x}+\frac {1}{9} x^2 \log ^2\left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 + (-3 + 20*x^3)*Log[x^2] + 10*x^3*Log[x^2]^2)/(45*x^2),x]

[Out]

Log[x^2]/(15*x) + (x^2*Log[x^2]^2)/9

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fricas [A]  time = 1.08, size = 23, normalized size = 0.82 \begin {gather*} \frac {5 \, x^{3} \log \left (x^{2}\right )^{2} + 3 \, \log \left (x^{2}\right )}{45 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/45*(10*x^3*log(x^2)^2+(20*x^3-3)*log(x^2)+6)/x^2,x, algorithm="fricas")

[Out]

1/45*(5*x^3*log(x^2)^2 + 3*log(x^2))/x

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giac [A]  time = 0.24, size = 21, normalized size = 0.75 \begin {gather*} \frac {1}{9} \, x^{2} \log \left (x^{2}\right )^{2} + \frac {\log \left (x^{2}\right )}{15 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/45*(10*x^3*log(x^2)^2+(20*x^3-3)*log(x^2)+6)/x^2,x, algorithm="giac")

[Out]

1/9*x^2*log(x^2)^2 + 1/15*log(x^2)/x

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maple [A]  time = 0.02, size = 22, normalized size = 0.79

method	result	size
default	\(\frac {\ln \left (x^{2}\right )}{15 x}+\frac {x^{2} \ln \left (x^{2}\right )^{2}}{9}\)	\(22\)
risch	\(\frac {\ln \left (x^{2}\right )}{15 x}+\frac {x^{2} \ln \left (x^{2}\right )^{2}}{9}\)	\(22\)
norman	\(\frac {\frac {x^{3} \ln \left (x^{2}\right )^{2}}{9}+\frac {\ln \left (x^{2}\right )}{15}}{x}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/45*(10*x^3*ln(x^2)^2+(20*x^3-3)*ln(x^2)+6)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/15*ln(x^2)/x+1/9*x^2*ln(x^2)^2

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maxima [A]  time = 0.35, size = 21, normalized size = 0.75 \begin {gather*} \frac {1}{9} \, x^{2} \log \left (x^{2}\right )^{2} + \frac {\log \left (x^{2}\right )}{15 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/45*(10*x^3*log(x^2)^2+(20*x^3-3)*log(x^2)+6)/x^2,x, algorithm="maxima")

[Out]

1/9*x^2*log(x^2)^2 + 1/15*log(x^2)/x

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mupad [B]  time = 5.70, size = 20, normalized size = 0.71 \begin {gather*} \frac {\ln \left (x^2\right )\,\left (5\,x^3\,\ln \left (x^2\right )+3\right )}{45\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(x^2)*(20*x^3 - 3))/45 + (2*x^3*log(x^2)^2)/9 + 2/15)/x^2,x)

[Out]

(log(x^2)*(5*x^3*log(x^2) + 3))/(45*x)

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sympy [A]  time = 0.13, size = 19, normalized size = 0.68 \begin {gather*} \frac {x^{2} \log {\left (x^{2} \right )}^{2}}{9} + \frac {\log {\left (x^{2} \right )}}{15 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/45*(10*x**3*ln(x**2)**2+(20*x**3-3)*ln(x**2)+6)/x**2,x)

[Out]

x**2*log(x**2)**2/9 + log(x**2)/(15*x)

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