3.98.58 \(\int \frac {-24-5 x+(-8-2 x) \log (4+x)}{4 x^3+x^4} \, dx\)

Optimal. Leaf size=18 \[ e^{2 e^5}+\frac {3+\log (4+x)}{x^2} \]

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Rubi [A]  time = 0.16, antiderivative size = 14, normalized size of antiderivative = 0.78, number of steps used = 8, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1593, 6742, 77, 2395, 44} \begin {gather*} \frac {3}{x^2}+\frac {\log (x+4)}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-24 - 5*x + (-8 - 2*x)*Log[4 + x])/(4*x^3 + x^4),x]

[Out]

3/x^2 + Log[4 + x]/x^2

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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 {-24-5 x+(-8-2 x) \log (4+x)}{x^3 (4+x)} \, dx\\ &=\int \left (\frac {-24-5 x}{x^3 (4+x)}-\frac {2 \log (4+x)}{x^3}\right ) \, dx\\ &=-\left (2 \int \frac {\log (4+x)}{x^3} \, dx\right )+\int \frac {-24-5 x}{x^3 (4+x)} \, dx\\ &=\frac {\log (4+x)}{x^2}-\int \frac {1}{x^2 (4+x)} \, dx+\int \left (-\frac {6}{x^3}+\frac {1}{4 x^2}-\frac {1}{16 x}+\frac {1}{16 (4+x)}\right ) \, dx\\ &=\frac {3}{x^2}-\frac {1}{4 x}-\frac {\log (x)}{16}+\frac {1}{16} \log (4+x)+\frac {\log (4+x)}{x^2}-\int \left (\frac {1}{4 x^2}-\frac {1}{16 x}+\frac {1}{16 (4+x)}\right ) \, dx\\ &=\frac {3}{x^2}+\frac {\log (4+x)}{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 10, normalized size = 0.56 \begin {gather*} \frac {3+\log (4+x)}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-24 - 5*x + (-8 - 2*x)*Log[4 + x])/(4*x^3 + x^4),x]

[Out]

(3 + Log[4 + x])/x^2

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fricas [A]  time = 0.75, size = 10, normalized size = 0.56 \begin {gather*} \frac {\log \left (x + 4\right ) + 3}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-8)*log(4+x)-5*x-24)/(x^4+4*x^3),x, algorithm="fricas")

[Out]

(log(x + 4) + 3)/x^2

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giac [A]  time = 0.20, size = 14, normalized size = 0.78 \begin {gather*} \frac {\log \left (x + 4\right )}{x^{2}} + \frac {3}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-8)*log(4+x)-5*x-24)/(x^4+4*x^3),x, algorithm="giac")

[Out]

log(x + 4)/x^2 + 3/x^2

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maple [A]  time = 0.10, size = 11, normalized size = 0.61

method	result	size
norman	\(\frac {3+\ln \left (4+x \right )}{x^{2}}\)	\(11\)
risch	\(\frac {\ln \left (4+x \right )}{x^{2}}+\frac {3}{x^{2}}\)	\(15\)
derivativedivides	\(-\frac {\ln \left (4+x \right ) \left (4+x \right ) \left (x -4\right )}{16 x^{2}}+\frac {3}{x^{2}}+\frac {\ln \left (4+x \right )}{16}\)	\(28\)
default	\(-\frac {\ln \left (4+x \right ) \left (4+x \right ) \left (x -4\right )}{16 x^{2}}+\frac {3}{x^{2}}+\frac {\ln \left (4+x \right )}{16}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x-8)*ln(4+x)-5*x-24)/(x^4+4*x^3),x,method=_RETURNVERBOSE)

[Out]

(3+ln(4+x))/x^2

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maxima [B]  time = 0.39, size = 39, normalized size = 2.17 \begin {gather*} -\frac {{\left (x^{2} - 16\right )} \log \left (x + 4\right ) - 4 \, x}{16 \, x^{2}} - \frac {3 \, {\left (x - 2\right )}}{2 \, x^{2}} + \frac {5}{4 \, x} + \frac {1}{16} \, \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-8)*log(4+x)-5*x-24)/(x^4+4*x^3),x, algorithm="maxima")

[Out]

-1/16*((x^2 - 16)*log(x + 4) - 4*x)/x^2 - 3/2*(x - 2)/x^2 + 5/4/x + 1/16*log(x + 4)

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mupad [B]  time = 5.94, size = 10, normalized size = 0.56 \begin {gather*} \frac {\ln \left (x+4\right )+3}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x + log(x + 4)*(2*x + 8) + 24)/(4*x^3 + x^4),x)

[Out]

(log(x + 4) + 3)/x^2

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sympy [A]  time = 0.13, size = 12, normalized size = 0.67 \begin {gather*} \frac {\log {\left (x + 4 \right )}}{x^{2}} + \frac {3}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-8)*ln(4+x)-5*x-24)/(x**4+4*x**3),x)

[Out]

log(x + 4)/x**2 + 3/x**2

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