3.45.96 \(\int \frac {-8-5 x+(2+x) \log (2+x)}{800 x^2+400 x^3} \, dx\)

Optimal. Leaf size=15 \[ \frac {4-\log (2+x)}{400 x} \]

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Rubi [A]  time = 0.15, antiderivative size = 19, normalized size of antiderivative = 1.27, number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1593, 6742, 77, 2395, 36, 29, 31} \begin {gather*} \frac {1}{100 x}-\frac {\log (x+2)}{400 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8 - 5*x + (2 + x)*Log[2 + x])/(800*x^2 + 400*x^3),x]

[Out]

1/(100*x) - Log[2 + x]/(400*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 {-8-5 x+(2+x) \log (2+x)}{x^2 (800+400 x)} \, dx\\ &=\int \left (\frac {-8-5 x}{400 x^2 (2+x)}+\frac {\log (2+x)}{400 x^2}\right ) \, dx\\ &=\frac {1}{400} \int \frac {-8-5 x}{x^2 (2+x)} \, dx+\frac {1}{400} \int \frac {\log (2+x)}{x^2} \, dx\\ &=-\frac {\log (2+x)}{400 x}+\frac {1}{400} \int \frac {1}{x (2+x)} \, dx+\frac {1}{400} \int \left (-\frac {4}{x^2}-\frac {1}{2 x}+\frac {1}{2 (2+x)}\right ) \, dx\\ &=\frac {1}{100 x}-\frac {\log (x)}{800}+\frac {1}{800} \log (2+x)-\frac {\log (2+x)}{400 x}+\frac {1}{800} \int \frac {1}{x} \, dx-\frac {1}{800} \int \frac {1}{2+x} \, dx\\ &=\frac {1}{100 x}-\frac {\log (2+x)}{400 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 15, normalized size = 1.00 \begin {gather*} \frac {4-\log (2+x)}{400 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 - 5*x + (2 + x)*Log[2 + x])/(800*x^2 + 400*x^3),x]

[Out]

(4 - Log[2 + x])/(400*x)

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fricas [A]  time = 0.72, size = 11, normalized size = 0.73 \begin {gather*} -\frac {\log \left (x + 2\right ) - 4}{400 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2+x)*log(2+x)-5*x-8)/(400*x^3+800*x^2),x, algorithm="fricas")

[Out]

-1/400*(log(x + 2) - 4)/x

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giac [A]  time = 0.20, size = 15, normalized size = 1.00 \begin {gather*} -\frac {\log \left (x + 2\right )}{400 \, x} + \frac {1}{100 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2+x)*log(2+x)-5*x-8)/(400*x^3+800*x^2),x, algorithm="giac")

[Out]

-1/400*log(x + 2)/x + 1/100/x

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maple [A]  time = 0.14, size = 13, normalized size = 0.87

method	result	size
norman	\(\frac {\frac {1}{100}-\frac {\ln \left (2+x \right )}{400}}{x}\)	\(13\)
risch	\(-\frac {\ln \left (2+x \right )}{400 x}+\frac {1}{100 x}\)	\(16\)
derivativedivides	\(-\frac {\ln \left (2+x \right ) \left (2+x \right )}{800 x}+\frac {\ln \left (2+x \right )}{800}+\frac {1}{100 x}\)	\(25\)
default	\(-\frac {\ln \left (2+x \right ) \left (2+x \right )}{800 x}+\frac {\ln \left (2+x \right )}{800}+\frac {1}{100 x}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2+x)*ln(2+x)-5*x-8)/(400*x^3+800*x^2),x,method=_RETURNVERBOSE)

[Out]

(1/100-1/400*ln(2+x))/x

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maxima [B]  time = 0.41, size = 24, normalized size = 1.60 \begin {gather*} -\frac {{\left (x + 2\right )} \log \left (x + 2\right )}{800 \, x} + \frac {1}{100 \, x} + \frac {1}{800} \, \log \left (x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2+x)*log(2+x)-5*x-8)/(400*x^3+800*x^2),x, algorithm="maxima")

[Out]

-1/800*(x + 2)*log(x + 2)/x + 1/100/x + 1/800*log(x + 2)

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mupad [B]  time = 0.10, size = 13, normalized size = 0.87 \begin {gather*} -\frac {\frac {\ln \left (x+2\right )}{400}-\frac {1}{100}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x - log(x + 2)*(x + 2) + 8)/(800*x^2 + 400*x^3),x)

[Out]

-(log(x + 2)/400 - 1/100)/x

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sympy [A]  time = 0.12, size = 12, normalized size = 0.80 \begin {gather*} - \frac {\log {\left (x + 2 \right )}}{400 x} + \frac {1}{100 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2+x)*ln(2+x)-5*x-8)/(400*x**3+800*x**2),x)

[Out]

-log(x + 2)/(400*x) + 1/(100*x)

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