3.32.45 \(\int \frac {16+(-16+8 x+8 x^2) \log (17)+4 \log (17) \log (x^2)}{x \log (17)} \, dx\)

Optimal. Leaf size=23 \[ -4+(2+2 x)^2+\left (-4+\frac {4}{\log (17)}+\log \left (x^2\right )\right )^2 \]

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Rubi [A]  time = 0.04, antiderivative size = 27, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 14, 2301} \begin {gather*} 4 x^2+\log ^2\left (x^2\right )+8 x-16 \left (1-\frac {1}{\log (17)}\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16 + (-16 + 8*x + 8*x^2)*Log[17] + 4*Log[17]*Log[x^2])/(x*Log[17]),x]

[Out]

8*x + 4*x^2 - 16*(1 - Log[17]^(-1))*Log[x] + 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 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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {16+\left (-16+8 x+8 x^2\right ) \log (17)+4 \log (17) \log \left (x^2\right )}{x} \, dx}{\log (17)}\\ &=\frac {\int \left (\frac {8 \left (2 (1-\log (17))+x \log (17)+x^2 \log (17)\right )}{x}+\frac {4 \log (17) \log \left (x^2\right )}{x}\right ) \, dx}{\log (17)}\\ &=4 \int \frac {\log \left (x^2\right )}{x} \, dx+\frac {8 \int \frac {2 (1-\log (17))+x \log (17)+x^2 \log (17)}{x} \, dx}{\log (17)}\\ &=\log ^2\left (x^2\right )+\frac {8 \int \left (-\frac {2 (-1+\log (17))}{x}+\log (17)+x \log (17)\right ) \, dx}{\log (17)}\\ &=8 x+4 x^2-16 \left (1-\frac {1}{\log (17)}\right ) \log (x)+\log ^2\left (x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 31, normalized size = 1.35 \begin {gather*} 4 \left (2 x+x^2-4 \log (x)+\frac {4 \log (x)}{\log (17)}+\frac {1}{4} \log ^2\left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 + (-16 + 8*x + 8*x^2)*Log[17] + 4*Log[17]*Log[x^2])/(x*Log[17]),x]

[Out]

4*(2*x + x^2 - 4*Log[x] + (4*Log[x])/Log[17] + Log[x^2]^2/4)

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fricas [A]  time = 0.65, size = 36, normalized size = 1.57 \begin {gather*} \frac {\log \left (17\right ) \log \left (x^{2}\right )^{2} + 4 \, {\left (x^{2} + 2 \, x\right )} \log \left (17\right ) - 8 \, {\left (\log \left (17\right ) - 1\right )} \log \left (x^{2}\right )}{\log \left (17\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(17)*log(x^2)+(8*x^2+8*x-16)*log(17)+16)/x/log(17),x, algorithm="fricas")

[Out]

(log(17)*log(x^2)^2 + 4*(x^2 + 2*x)*log(17) - 8*(log(17) - 1)*log(x^2))/log(17)

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giac [A]  time = 0.20, size = 35, normalized size = 1.52 \begin {gather*} \frac {4 \, x^{2} \log \left (17\right ) + \log \left (17\right ) \log \left (x^{2}\right )^{2} + 8 \, x \log \left (17\right ) - 16 \, {\left (\log \left (17\right ) - 1\right )} \log \relax (x)}{\log \left (17\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(17)*log(x^2)+(8*x^2+8*x-16)*log(17)+16)/x/log(17),x, algorithm="giac")

[Out]

(4*x^2*log(17) + log(17)*log(x^2)^2 + 8*x*log(17) - 16*(log(17) - 1)*log(x))/log(17)

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maple [A]  time = 0.05, size = 28, normalized size = 1.22

method	result	size
norman	\(\ln \left (x^{2}\right )^{2}+8 x +4 x^{2}-\frac {16 \left (\ln \left (17\right )-1\right ) \ln \relax (x )}{\ln \left (17\right )}\)	\(28\)
risch	\(\ln \left (x^{2}\right )^{2}+4 x^{2}-16 \ln \relax (x )+8 x +\frac {16 \ln \relax (x )}{\ln \left (17\right )}\)	\(28\)
default	\(\frac {\ln \left (17\right ) \ln \left (x^{2}\right )^{2}+4 x^{2} \ln \left (17\right )-16 \ln \left (17\right ) \ln \relax (x )+8 \ln \left (17\right ) x +16 \ln \relax (x )}{\ln \left (17\right )}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*ln(17)*ln(x^2)+(8*x^2+8*x-16)*ln(17)+16)/x/ln(17),x,method=_RETURNVERBOSE)

[Out]

ln(x^2)^2+8*x+4*x^2-16*(ln(17)-1)/ln(17)*ln(x)

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maxima [A]  time = 0.85, size = 37, normalized size = 1.61 \begin {gather*} \frac {4 \, x^{2} \log \left (17\right ) + \log \left (17\right ) \log \left (x^{2}\right )^{2} + 8 \, x \log \left (17\right ) - 16 \, \log \left (17\right ) \log \relax (x) + 16 \, \log \relax (x)}{\log \left (17\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(17)*log(x^2)+(8*x^2+8*x-16)*log(17)+16)/x/log(17),x, algorithm="maxima")

[Out]

(4*x^2*log(17) + log(17)*log(x^2)^2 + 8*x*log(17) - 16*log(17)*log(x) + 16*log(x))/log(17)

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mupad [B]  time = 1.86, size = 31, normalized size = 1.35 \begin {gather*} 8\,x-8\,\ln \left (x^2\right )+\frac {8\,\ln \left (x^2\right )}{\ln \left (17\right )}+{\ln \left (x^2\right )}^2+4\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*log(x^2)*log(17) + log(17)*(8*x + 8*x^2 - 16) + 16)/(x*log(17)),x)

[Out]

8*x - 8*log(x^2) + (8*log(x^2))/log(17) + log(x^2)^2 + 4*x^2

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sympy [A]  time = 0.15, size = 34, normalized size = 1.48 \begin {gather*} \frac {4 x^{2} \log {\left (17 \right )} + 8 x \log {\left (17 \right )} - 16 \left (-1 + \log {\left (17 \right )}\right ) \log {\relax (x )}}{\log {\left (17 \right )}} + \log {\left (x^{2} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*ln(17)*ln(x**2)+(8*x**2+8*x-16)*ln(17)+16)/x/ln(17),x)

[Out]

(4*x**2*log(17) + 8*x*log(17) - 16*(-1 + log(17))*log(x))/log(17) + log(x**2)**2

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