3.26.81 \(\int \frac {10-2 x+16 x \log (2)+16 x \log (2) \log (x)}{x} \, dx\)

Optimal. Leaf size=18 \[ -4+2 (-x+(5+8 x \log (2)) \log (x)) \]

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Rubi [A]  time = 0.03, antiderivative size = 26, normalized size of antiderivative = 1.44, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6, 14, 43, 2295} \begin {gather*} 16 x \log (2) \log (x)-2 x (1-\log (256))-16 x \log (2)+10 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 - 2*x + 16*x*Log[2] + 16*x*Log[2]*Log[x])/x,x]

[Out]

-16*x*Log[2] - 2*x*(1 - Log[256]) + 10*Log[x] + 16*x*Log[2]*Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10+x (-2+16 \log (2))+16 x \log (2) \log (x)}{x} \, dx\\ &=\int \left (\frac {2 (5-x (1-\log (256)))}{x}+16 \log (2) \log (x)\right ) \, dx\\ &=2 \int \frac {5-x (1-\log (256))}{x} \, dx+(16 \log (2)) \int \log (x) \, dx\\ &=-16 x \log (2)+16 x \log (2) \log (x)+2 \int \left (-1+\frac {5}{x}+\log (256)\right ) \, dx\\ &=-16 x \log (2)-2 x (1-\log (256))+10 \log (x)+16 x \log (2) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.39 \begin {gather*} -2 x-16 x \log (2)+2 x \log (256)+10 \log (x)+16 x \log (2) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 - 2*x + 16*x*Log[2] + 16*x*Log[2]*Log[x])/x,x]

[Out]

-2*x - 16*x*Log[2] + 2*x*Log[256] + 10*Log[x] + 16*x*Log[2]*Log[x]

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fricas [A]  time = 0.55, size = 15, normalized size = 0.83 \begin {gather*} 2 \, {\left (8 \, x \log \relax (2) + 5\right )} \log \relax (x) - 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x*log(2)*log(x)+16*x*log(2)-2*x+10)/x,x, algorithm="fricas")

[Out]

2*(8*x*log(2) + 5)*log(x) - 2*x

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giac [A]  time = 0.45, size = 15, normalized size = 0.83 \begin {gather*} 16 \, x \log \relax (2) \log \relax (x) - 2 \, x + 10 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x*log(2)*log(x)+16*x*log(2)-2*x+10)/x,x, algorithm="giac")

[Out]

16*x*log(2)*log(x) - 2*x + 10*log(x)

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maple [A]  time = 0.02, size = 16, normalized size = 0.89

method	result	size
norman	\(10 \ln \relax (x )-2 x +16 x \ln \relax (2) \ln \relax (x )\)	\(16\)
risch	\(10 \ln \relax (x )-2 x +16 x \ln \relax (2) \ln \relax (x )\)	\(16\)
default	\(16 \ln \relax (2) \left (x \ln \relax (x )-x \right )+16 x \ln \relax (2)-2 x +10 \ln \relax (x )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x*ln(2)*ln(x)+16*x*ln(2)-2*x+10)/x,x,method=_RETURNVERBOSE)

[Out]

10*ln(x)-2*x+16*x*ln(2)*ln(x)

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maxima [A]  time = 0.40, size = 25, normalized size = 1.39 \begin {gather*} 16 \, {\left (x \log \relax (x) - x\right )} \log \relax (2) + 16 \, x \log \relax (2) - 2 \, x + 10 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x*log(2)*log(x)+16*x*log(2)-2*x+10)/x,x, algorithm="maxima")

[Out]

16*(x*log(x) - x)*log(2) + 16*x*log(2) - 2*x + 10*log(x)

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mupad [B]  time = 1.45, size = 15, normalized size = 0.83 \begin {gather*} 10\,\ln \relax (x)-2\,x+16\,x\,\ln \relax (2)\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x*log(2) - 2*x + 16*x*log(2)*log(x) + 10)/x,x)

[Out]

10*log(x) - 2*x + 16*x*log(2)*log(x)

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sympy [A]  time = 0.11, size = 17, normalized size = 0.94 \begin {gather*} 16 x \log {\relax (2 )} \log {\relax (x )} - 2 x + 10 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x*ln(2)*ln(x)+16*x*ln(2)-2*x+10)/x,x)

[Out]

16*x*log(2)*log(x) - 2*x + 10*log(x)

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