3.98.19 \(\int \frac {20+x+2 \log (\frac {1}{5} (-80-4 x) \log (5))}{180+9 x} \, dx\)

Optimal. Leaf size=19 \[ \frac {1}{9} \left (x+\log ^2\left (\left (-16-\frac {4 x}{5}\right ) \log (5)\right )\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 22, normalized size of antiderivative = 1.16, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6742, 2390, 12, 2301} \begin {gather*} \frac {x}{9}+\frac {1}{9} \log ^2\left (-\frac {4}{5} (x+20) \log (5)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20 + x + 2*Log[((-80 - 4*x)*Log[5])/5])/(180 + 9*x),x]

[Out]

x/9 + Log[(-4*(20 + x)*Log[5])/5]^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 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]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

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}{9}+\frac {2 \log \left (-16 \log (5)-\frac {4}{5} x \log (5)\right )}{9 (20+x)}\right ) \, dx\\ &=\frac {x}{9}+\frac {2}{9} \int \frac {\log \left (-16 \log (5)-\frac {4}{5} x \log (5)\right )}{20+x} \, dx\\ &=\frac {x}{9}-\frac {5 \operatorname {Subst}\left (\int -\frac {4 \log (5) \log (x)}{5 x} \, dx,x,-16 \log (5)-\frac {4}{5} x \log (5)\right )}{18 \log (5)}\\ &=\frac {x}{9}+\frac {2}{9} \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,-16 \log (5)-\frac {4}{5} x \log (5)\right )\\ &=\frac {x}{9}+\frac {1}{9} \log ^2\left (-\frac {4}{5} (20+x) \log (5)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 1.11 \begin {gather*} \frac {1}{9} \left (x+\log ^2\left (-16 \log (5)-\frac {4}{5} x \log (5)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 + x + 2*Log[((-80 - 4*x)*Log[5])/5])/(180 + 9*x),x]

[Out]

(x + Log[-16*Log[5] - (4*x*Log[5])/5]^2)/9

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fricas [A]  time = 0.74, size = 16, normalized size = 0.84 \begin {gather*} \frac {1}{9} \, \log \left (-\frac {4}{5} \, {\left (x + 20\right )} \log \relax (5)\right )^{2} + \frac {1}{9} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(1/5*(-4*x-80)*log(5))+20+x)/(9*x+180),x, algorithm="fricas")

[Out]

1/9*log(-4/5*(x + 20)*log(5))^2 + 1/9*x

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giac [A]  time = 0.54, size = 27, normalized size = 1.42 \begin {gather*} \frac {1}{9} \, \log \left (-4 \, x \log \relax (5) - 80 \, \log \relax (5)\right )^{2} - \frac {2}{9} \, \log \relax (5) \log \left (x + 20\right ) + \frac {1}{9} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(1/5*(-4*x-80)*log(5))+20+x)/(9*x+180),x, algorithm="giac")

[Out]

1/9*log(-4*x*log(5) - 80*log(5))^2 - 2/9*log(5)*log(x + 20) + 1/9*x

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maple [A]  time = 0.08, size = 19, normalized size = 1.00

method	result	size
norman	\(\frac {x}{9}+\frac {\ln \left (\frac {\left (-4 x -80\right ) \ln \relax (5)}{5}\right )^{2}}{9}\)	\(19\)
risch	\(\frac {x}{9}+\frac {\ln \left (\frac {\left (-4 x -80\right ) \ln \relax (5)}{5}\right )^{2}}{9}\)	\(19\)
derivativedivides	\(-\frac {-4 \ln \relax (5) \ln \left (-\frac {4 x \ln \relax (5)}{5}-16 \ln \relax (5)\right )^{2}-4 x \ln \relax (5)-80 \ln \relax (5)}{36 \ln \relax (5)}\)	\(34\)
default	\(-\frac {-4 \ln \relax (5) \ln \left (-\frac {4 x \ln \relax (5)}{5}-16 \ln \relax (5)\right )^{2}-4 x \ln \relax (5)-80 \ln \relax (5)}{36 \ln \relax (5)}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(1/5*(-4*x-80)*ln(5))+20+x)/(9*x+180),x,method=_RETURNVERBOSE)

[Out]

1/9*x+1/9*ln(1/5*(-4*x-80)*ln(5))^2

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maxima [C]  time = 0.48, size = 33, normalized size = 1.74 \begin {gather*} -\frac {2}{9} \, {\left (-i \, \pi + \log \relax (5) - 2 \, \log \relax (2) - \log \left (\log \relax (5)\right )\right )} \log \left (x + 20\right ) + \frac {1}{9} \, \log \left (x + 20\right )^{2} + \frac {1}{9} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(1/5*(-4*x-80)*log(5))+20+x)/(9*x+180),x, algorithm="maxima")

[Out]

-2/9*(-I*pi + log(5) - 2*log(2) - log(log(5)))*log(x + 20) + 1/9*log(x + 20)^2 + 1/9*x

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mupad [B]  time = 0.24, size = 18, normalized size = 0.95 \begin {gather*} \frac {{\ln \left (-\frac {\ln \relax (5)\,\left (4\,x+80\right )}{5}\right )}^2}{9}+\frac {x}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 2*log(-(log(5)*(4*x + 80))/5) + 20)/(9*x + 180),x)

[Out]

x/9 + log(-(log(5)*(4*x + 80))/5)^2/9

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sympy [A]  time = 0.11, size = 19, normalized size = 1.00 \begin {gather*} \frac {x}{9} + \frac {\log {\left (\left (- \frac {4 x}{5} - 16\right ) \log {\relax (5 )} \right )}^{2}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(1/5*(-4*x-80)*ln(5))+20+x)/(9*x+180),x)

[Out]

x/9 + log((-4*x/5 - 16)*log(5))**2/9

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