∫ −5−x−2x log(3)+(10+2x) log(3) log(5+x)_____________________________

3.86.80 \(\int \frac {-5-x-2 x \log (3)+(10+2 x) \log (3) \log (5+x)}{10 x^2+2 x^3} \, dx\)

Optimal. Leaf size=22 \[ \frac {1+x}{2 x}-\frac {\log (3) \log (5+x)}{x} \]

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Rubi [B] time = 0.22, antiderivative size = 55, normalized size of antiderivative = 2.50, number of steps used = 10, number of rules used = 8, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6, 1593, 6742, 77, 2395, 36, 29, 31} \begin {gather*} \frac {1}{2 x}-\frac {1}{10} \log (9) \log (x)+\frac {1}{5} \log (3) \log (x)+\frac {1}{10} \log (9) \log (x+5)-\frac {\log (3) \log (x+5)}{x}-\frac {1}{5} \log (3) \log (x+5) \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[(-5 - x - 2*x*Log[3] + (10 + 2*x)*Log[3]*Log[5 + x])/(10*x^2 + 2*x^3),x]

[Out]

1/(2*x) + (Log[3]*Log[x])/5 - (Log[9]*Log[x])/10 - (Log[3]*Log[5 + x])/5 - (Log[3]*Log[5 + x])/x + (Log[9]*Log

[5 + x])/10

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 29

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

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -

Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran

d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[

n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1

, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F

reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g

*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +

e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

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 \frac {-5+x (-1-2 \log (3))+(10+2 x) \log (3) \log (5+x)}{10 x^2+2 x^3} \, dx\\ &=\int \frac {-5+x (-1-2 \log (3))+(10+2 x) \log (3) \log (5+x)}{x^2 (10+2 x)} \, dx\\ &=\int \left (\frac {-5-x (1+\log (9))}{2 x^2 (5+x)}+\frac {\log (3) \log (5+x)}{x^2}\right ) \, dx\\ &=\frac {1}{2} \int \frac {-5-x (1+\log (9))}{x^2 (5+x)} \, dx+\log (3) \int \frac {\log (5+x)}{x^2} \, dx\\ &=-\frac {\log (3) \log (5+x)}{x}+\frac {1}{2} \int \left (-\frac {1}{x^2}-\frac {\log (9)}{5 x}+\frac {\log (9)}{5 (5+x)}\right ) \, dx+\log (3) \int \frac {1}{x (5+x)} \, dx\\ &=\frac {1}{2 x}-\frac {1}{10} \log (9) \log (x)-\frac {\log (3) \log (5+x)}{x}+\frac {1}{10} \log (9) \log (5+x)+\frac {1}{5} \log (3) \int \frac {1}{x} \, dx-\frac {1}{5} \log (3) \int \frac {1}{5+x} \, dx\\ &=\frac {1}{2 x}+\frac {1}{5} \log (3) \log (x)-\frac {1}{10} \log (9) \log (x)-\frac {1}{5} \log (3) \log (5+x)-\frac {\log (3) \log (5+x)}{x}+\frac {1}{10} \log (9) \log (5+x)\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.05, size = 19, normalized size = 0.86 \begin {gather*} \frac {1}{2} \left (\frac {1}{x}-\frac {\log (9) \log (5+x)}{x}\right ) \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(-5 - x - 2*x*Log[3] + (10 + 2*x)*Log[3]*Log[5 + x])/(10*x^2 + 2*x^3),x]

[Out]

(x^(-1) - (Log[9]*Log[5 + x])/x)/2

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fricas [A] time = 0.65, size = 15, normalized size = 0.68 \begin {gather*} -\frac {2 \, \log \relax (3) \log \left (x + 5\right ) - 1}{2 \, x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+10)*log(3)*log(5+x)-2*x*log(3)-x-5)/(2*x^3+10*x^2),x, algorithm="fricas")

[Out]

-1/2*(2*log(3)*log(x + 5) - 1)/x

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giac [A] time = 0.15, size = 17, normalized size = 0.77 \begin {gather*} -\frac {\log \relax (3) \log \left (x + 5\right )}{x} + \frac {1}{2 \, x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+10)*log(3)*log(5+x)-2*x*log(3)-x-5)/(2*x^3+10*x^2),x, algorithm="giac")

[Out]

-log(3)*log(x + 5)/x + 1/2/x

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maple [A] time = 0.38, size = 15, normalized size = 0.68


method	result	size

norman	\(\frac {\frac {1}{2}-\ln \relax (3) \ln \left (5+x \right )}{x}\)	\(15\)
risch	\(-\frac {\ln \left (5+x \right ) \ln \relax (3)}{x}+\frac {1}{2 x}\)	\(18\)
derivativedivides	\(-\frac {\ln \relax (3) \ln \left (5+x \right ) \left (5+x \right )}{5 x}+\frac {\ln \relax (3) \ln \left (5+x \right )}{5}+\frac {1}{2 x}\)	\(29\)
default	\(-\frac {\ln \relax (3) \ln \left (5+x \right ) \left (5+x \right )}{5 x}+\frac {\ln \relax (3) \ln \left (5+x \right )}{5}+\frac {1}{2 x}\)	\(29\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x+10)*ln(3)*ln(5+x)-2*x*ln(3)-x-5)/(2*x^3+10*x^2),x,method=_RETURNVERBOSE)

[Out]

(1/2-ln(3)*ln(5+x))/x

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maxima [B] time = 0.48, size = 43, normalized size = 1.95 \begin {gather*} \frac {1}{5} \, {\left (\log \left (x + 5\right ) - \log \relax (x)\right )} \log \relax (3) + \frac {1}{5} \, \log \relax (3) \log \relax (x) - \frac {{\left (x \log \relax (3) + 5 \, \log \relax (3)\right )} \log \left (x + 5\right )}{5 \, x} + \frac {1}{2 \, x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+10)*log(3)*log(5+x)-2*x*log(3)-x-5)/(2*x^3+10*x^2),x, algorithm="maxima")

[Out]

1/5*(log(x + 5) - log(x))*log(3) + 1/5*log(3)*log(x) - 1/5*(x*log(3) + 5*log(3))*log(x + 5)/x + 1/2/x

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mupad [B] time = 5.49, size = 46, normalized size = 2.09 \begin {gather*} -\frac {25\,\ln \left (x+5\right )\,\ln \relax (3)+x\,\left (10\,\ln \left (x+5\right )\,\ln \relax (3)-5\right )+x^2\,\left (\frac {\ln \left (x+5\right )\,\ln \relax (9)}{2}-\frac {1}{2}\right )-\frac {25}{2}}{x\,{\left (x+5\right )}^2} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + 2*x*log(3) - log(x + 5)*log(3)*(2*x + 10) + 5)/(10*x^2 + 2*x^3),x)

[Out]

-(25*log(x + 5)*log(3) + x*(10*log(x + 5)*log(3) - 5) + x^2*((log(x + 5)*log(9))/2 - 1/2) - 25/2)/(x*(x + 5)^2

)

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sympy [A] time = 0.15, size = 14, normalized size = 0.64 \begin {gather*} - \frac {\log {\relax (3 )} \log {\left (x + 5 \right )}}{x} + \frac {1}{2 x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+10)*ln(3)*ln(5+x)-2*x*ln(3)-x-5)/(2*x**3+10*x**2),x)

[Out]

-log(3)*log(x + 5)/x + 1/(2*x)

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