3.99.25 \(\int \frac {20+(16 x+8 x^2+x^3) \log (19)-5 x \log (\frac {x}{\log (5)})}{(16 x+8 x^2+x^3) \log (19)} \, dx\)

Optimal. Leaf size=25 \[ x+\frac {5 \left (-3-x+\log \left (\frac {x}{\log (5)}\right )\right )}{(4+x) \log (19)} \]

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Rubi [A]  time = 0.26, antiderivative size = 44, normalized size of antiderivative = 1.76, number of steps used = 9, number of rules used = 7, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 1594, 27, 6742, 1620, 2314, 31} \begin {gather*} x-\frac {5 x \log \left (\frac {x}{\log (5)}\right )}{4 (x+4) \log (19)}+\frac {5 \log (x)}{4 \log (19)}+\frac {5}{(x+4) \log (19)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20 + (16*x + 8*x^2 + x^3)*Log[19] - 5*x*Log[x/Log[5]])/((16*x + 8*x^2 + x^3)*Log[19]),x]

[Out]

x + 5/((4 + x)*Log[19]) + (5*Log[x])/(4*Log[19]) - (5*x*Log[x/Log[5]])/(4*(4 + x)*Log[19])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 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} &=\frac {\int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{16 x+8 x^2+x^3} \, dx}{\log (19)}\\ &=\frac {\int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{x \left (16+8 x+x^2\right )} \, dx}{\log (19)}\\ &=\frac {\int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{x (4+x)^2} \, dx}{\log (19)}\\ &=\frac {\int \left (\frac {20+16 x \log (19)+8 x^2 \log (19)+x^3 \log (19)}{x (4+x)^2}-\frac {5 \log \left (\frac {x}{\log (5)}\right )}{(4+x)^2}\right ) \, dx}{\log (19)}\\ &=\frac {\int \frac {20+16 x \log (19)+8 x^2 \log (19)+x^3 \log (19)}{x (4+x)^2} \, dx}{\log (19)}-\frac {5 \int \frac {\log \left (\frac {x}{\log (5)}\right )}{(4+x)^2} \, dx}{\log (19)}\\ &=-\frac {5 x \log \left (\frac {x}{\log (5)}\right )}{4 (4+x) \log (19)}+\frac {\int \left (\frac {5}{4 x}-\frac {5}{(4+x)^2}-\frac {5}{4 (4+x)}+\log (19)\right ) \, dx}{\log (19)}+\frac {5 \int \frac {1}{4+x} \, dx}{4 \log (19)}\\ &=x+\frac {5}{(4+x) \log (19)}+\frac {5 \log (x)}{4 \log (19)}-\frac {5 x \log \left (\frac {x}{\log (5)}\right )}{4 (4+x) \log (19)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 26, normalized size = 1.04 \begin {gather*} \frac {x \log (19)+\frac {5 \left (1+\log \left (\frac {x}{\log (5)}\right )\right )}{4+x}}{\log (19)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 + (16*x + 8*x^2 + x^3)*Log[19] - 5*x*Log[x/Log[5]])/((16*x + 8*x^2 + x^3)*Log[19]),x]

[Out]

(x*Log[19] + (5*(1 + Log[x/Log[5]]))/(4 + x))/Log[19]

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fricas [A]  time = 0.57, size = 31, normalized size = 1.24 \begin {gather*} \frac {{\left (x^{2} + 4 \, x\right )} \log \left (19\right ) + 5 \, \log \left (\frac {x}{\log \relax (5)}\right ) + 5}{{\left (x + 4\right )} \log \left (19\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x*log(x/log(5))+(x^3+8*x^2+16*x)*log(19)+20)/(x^3+8*x^2+16*x)/log(19),x, algorithm="fricas")

[Out]

((x^2 + 4*x)*log(19) + 5*log(x/log(5)) + 5)/((x + 4)*log(19))

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giac [A]  time = 0.14, size = 31, normalized size = 1.24 \begin {gather*} \frac {x \log \left (19\right ) - \frac {5 \, {\left (\log \left (\log \relax (5)\right ) - 1\right )}}{x + 4} + \frac {5 \, \log \relax (x)}{x + 4}}{\log \left (19\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x*log(x/log(5))+(x^3+8*x^2+16*x)*log(19)+20)/(x^3+8*x^2+16*x)/log(19),x, algorithm="giac")

[Out]

(x*log(19) - 5*(log(log(5)) - 1)/(x + 4) + 5*log(x)/(x + 4))/log(19)

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maple [A]  time = 0.10, size = 36, normalized size = 1.44

method	result	size
norman	\(\frac {x^{2}+\frac {5 \ln \left (\frac {x}{\ln \relax (5)}\right )}{\ln \left (19\right )}-\frac {16 \ln \left (19\right )-5}{\ln \left (19\right )}}{4+x}\)	\(36\)
risch	\(\frac {5 \ln \left (\frac {x}{\ln \relax (5)}\right )}{\ln \left (19\right ) \left (4+x \right )}+\frac {\ln \left (19\right ) x^{2}+4 x \ln \left (19\right )+5}{\ln \left (19\right ) \left (4+x \right )}\)	\(43\)
derivativedivides	\(\frac {\ln \relax (5) \left (-\frac {5 \ln \left (\frac {x}{\ln \relax (5)}\right ) x}{4 \ln \relax (5) \left (4+x \right )}+\frac {5 \ln \left (\frac {x}{\ln \relax (5)}\right )}{4 \ln \relax (5)}+\frac {5}{\ln \relax (5) \left (4+x \right )}+\frac {\ln \left (19\right ) x}{\ln \relax (5)}\right )}{\ln \left (19\right )}\)	\(60\)
default	\(\frac {\ln \relax (5) \left (-\frac {5 \ln \left (\frac {x}{\ln \relax (5)}\right ) x}{4 \ln \relax (5) \left (4+x \right )}+\frac {5 \ln \left (\frac {x}{\ln \relax (5)}\right )}{4 \ln \relax (5)}+\frac {5}{\ln \relax (5) \left (4+x \right )}+\frac {\ln \left (19\right ) x}{\ln \relax (5)}\right )}{\ln \left (19\right )}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*x*ln(x/ln(5))+(x^3+8*x^2+16*x)*ln(19)+20)/(x^3+8*x^2+16*x)/ln(19),x,method=_RETURNVERBOSE)

[Out]

(x^2+5/ln(19)*ln(x/ln(5))-(16*ln(19)-5)/ln(19))/(4+x)

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maxima [B]  time = 0.47, size = 70, normalized size = 2.80 \begin {gather*} \frac {{\left (x - \frac {16}{x + 4} - 8 \, \log \left (x + 4\right )\right )} \log \left (19\right ) + 8 \, {\left (\frac {4}{x + 4} + \log \left (x + 4\right )\right )} \log \left (19\right ) - \frac {16 \, \log \left (19\right )}{x + 4} + \frac {5 \, \log \left (\frac {x}{\log \relax (5)}\right )}{x + 4} + \frac {5}{x + 4}}{\log \left (19\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x*log(x/log(5))+(x^3+8*x^2+16*x)*log(19)+20)/(x^3+8*x^2+16*x)/log(19),x, algorithm="maxima")

[Out]

((x - 16/(x + 4) - 8*log(x + 4))*log(19) + 8*(4/(x + 4) + log(x + 4))*log(19) - 16*log(19)/(x + 4) + 5*log(x/l
og(5))/(x + 4) + 5/(x + 4))/log(19)

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mupad [B]  time = 5.89, size = 23, normalized size = 0.92 \begin {gather*} x+\frac {5\,\ln \left (\frac {x}{\ln \relax (5)}\right )+5}{\ln \left (19\right )\,\left (x+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(19)*(16*x + 8*x^2 + x^3) - 5*x*log(x/log(5)) + 20)/(log(19)*(16*x + 8*x^2 + x^3)),x)

[Out]

x + (5*log(x/log(5)) + 5)/(log(19)*(x + 4))

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sympy [A]  time = 0.24, size = 31, normalized size = 1.24 \begin {gather*} x + \frac {5 \log {\left (\frac {x}{\log {\relax (5 )}} \right )}}{x \log {\left (19 \right )} + 4 \log {\left (19 \right )}} + \frac {5}{x \log {\left (19 \right )} + 4 \log {\left (19 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*x*ln(x/ln(5))+(x**3+8*x**2+16*x)*ln(19)+20)/(x**3+8*x**2+16*x)/ln(19),x)

[Out]

x + 5*log(x/log(5))/(x*log(19) + 4*log(19)) + 5/(x*log(19) + 4*log(19))

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