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

Optimal. Leaf size=25 \[ -x+\log \left (4 e^{20+e^{\log (5) \log (x)}-x} \log \left (x^2\right )\right ) \]

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Rubi [A]  time = 0.22, antiderivative size = 13, normalized size of antiderivative = 0.52, number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {6688, 2274, 30, 2302, 29} \begin {gather*} x^{\log (5)}+\log \left (\log \left (x^2\right )\right )-2 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*x + x^Log[5] + Log[Log[x^2]]

Rule 29

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, 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 6688

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 13, normalized size = 0.52 \begin {gather*} 5^{\log (x)}-2 x+\log \left (\log \left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

5^Log[x] - 2*x + Log[Log[x^2]]

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fricas [A]  time = 0.75, size = 13, normalized size = 0.52 \begin {gather*} -2 \, x + e^{\left (\log \relax (5) \log \relax (x)\right )} + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x + e^(log(5)*log(x)) + log(log(x))

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giac [A]  time = 0.39, size = 15, normalized size = 0.60 \begin {gather*} -2 \, x + e^{\left (\log \relax (5) \log \relax (x)\right )} + \log \left (\log \left (x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x + e^(log(5)*log(x)) + log(log(x^2))

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maple [A]  time = 0.10, size = 16, normalized size = 0.64

method	result	size
default	\(-2 x +{\mathrm e}^{\ln \relax (5) \ln \relax (x )}+\ln \left (\ln \left (x^{2}\right )\right )\)	\(16\)
norman	\(-2 x +{\mathrm e}^{\ln \relax (5) \ln \relax (x )}+\ln \left (\ln \left (x^{2}\right )\right )\)	\(16\)
risch	\(-2 x +\ln \left (\ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (\mathrm {csgn}\left (i x \right )^{2}-2 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right )^{2}\right )}{4}\right )+x^{\ln \relax (5)}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.54, size = 15, normalized size = 0.60 \begin {gather*} -2 \, x + e^{\left (\log \relax (5) \log \relax (x)\right )} + \log \left (\log \left (x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x + e^(log(5)*log(x)) + log(log(x^2))

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mupad [B]  time = 0.81, size = 13, normalized size = 0.52 \begin {gather*} \ln \left (\ln \left (x^2\right )\right )-2\,x+x^{\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(x^2)) - 2*x + x^log(5)

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sympy [A]  time = 1.05, size = 15, normalized size = 0.60 \begin {gather*} - 2 x + e^{\log {\relax (5 )} \log {\relax (x )}} + \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x + exp(log(5)*log(x)) + log(log(x))

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