3.30.5 \(\int \frac {-1+(1+x-2 x^2) \log (5 x)}{x \log (5 x)} \, dx\)

Optimal. Leaf size=17 \[ -11+x-x^2+\log (x)-\log (\log (5 x)) \]

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Rubi [A]  time = 0.14, antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6742, 14, 2302, 29} \begin {gather*} -x^2+x+\log (x)-\log (\log (5 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + (1 + x - 2*x^2)*Log[5*x])/(x*Log[5*x]),x]

[Out]

x - x^2 + Log[x] - Log[Log[5*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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.94 \begin {gather*} x-x^2+\log (x)-\log (\log (5 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + (1 + x - 2*x^2)*Log[5*x])/(x*Log[5*x]),x]

[Out]

x - x^2 + Log[x] - Log[Log[5*x]]

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fricas [A]  time = 0.69, size = 18, normalized size = 1.06 \begin {gather*} -x^{2} + x + \log \left (5 \, x\right ) - \log \left (\log \left (5 \, x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2 + x + log(5*x) - log(log(5*x))

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giac [A]  time = 0.16, size = 16, normalized size = 0.94 \begin {gather*} -x^{2} + x + \log \relax (x) - \log \left (\log \left (5 \, x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2 + x + log(x) - log(log(5*x))

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maple [A]  time = 0.03, size = 17, normalized size = 1.00

method	result	size
risch	\(\ln \relax (x )+x -x^{2}-\ln \left (\ln \left (5 x \right )\right )\)	\(17\)
derivativedivides	\(-x^{2}+x +\ln \left (5 x \right )-\ln \left (\ln \left (5 x \right )\right )\)	\(19\)
default	\(-x^{2}+x +\ln \left (5 x \right )-\ln \left (\ln \left (5 x \right )\right )\)	\(19\)
norman	\(-x^{2}+x +\ln \left (5 x \right )-\ln \left (\ln \left (5 x \right )\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(x)+x-x^2-ln(ln(5*x))

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maxima [A]  time = 0.41, size = 16, normalized size = 0.94 \begin {gather*} -x^{2} + x + \log \relax (x) - \log \left (\log \left (5 \, x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2 + x + log(x) - log(log(5*x))

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mupad [B]  time = 1.69, size = 16, normalized size = 0.94 \begin {gather*} x-\ln \left (\ln \left (5\,x\right )\right )+\ln \relax (x)-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - log(log(5*x)) + log(x) - x^2

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sympy [A]  time = 0.12, size = 14, normalized size = 0.82 \begin {gather*} - x^{2} + x + \log {\relax (x )} - \log {\left (\log {\left (5 x \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2 + x + log(x) - log(log(5*x))

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