3.5.92 \(\int \frac {(-48-16 x) \log (3)}{80 x^4+40 x^5+5 x^6} \, dx\)

Optimal. Leaf size=14 \[ \frac {4 \log (3)}{5 x^3 (4+x)} \]

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Rubi [A]  time = 0.03, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {12, 1594, 27, 74} \begin {gather*} \frac {4 \log (3)}{5 x^3 (x+4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-48 - 16*x)*Log[3])/(80*x^4 + 40*x^5 + 5*x^6),x]

[Out]

(4*Log[3])/(5*x^3*(4 + x))

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 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (3) \int \frac {-48-16 x}{80 x^4+40 x^5+5 x^6} \, dx\\ &=\log (3) \int \frac {-48-16 x}{x^4 \left (80+40 x+5 x^2\right )} \, dx\\ &=\log (3) \int \frac {-48-16 x}{5 x^4 (4+x)^2} \, dx\\ &=\frac {1}{5} \log (3) \int \frac {-48-16 x}{x^4 (4+x)^2} \, dx\\ &=\frac {4 \log (3)}{5 x^3 (4+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 1.00 \begin {gather*} \frac {4 \log (3)}{5 x^3 (4+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-48 - 16*x)*Log[3])/(80*x^4 + 40*x^5 + 5*x^6),x]

[Out]

(4*Log[3])/(5*x^3*(4 + x))

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fricas [A]  time = 0.77, size = 15, normalized size = 1.07 \begin {gather*} \frac {4 \, \log \relax (3)}{5 \, {\left (x^{4} + 4 \, x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x-48)*log(3)/(5*x^6+40*x^5+80*x^4),x, algorithm="fricas")

[Out]

4/5*log(3)/(x^4 + 4*x^3)

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giac [A]  time = 0.36, size = 23, normalized size = 1.64 \begin {gather*} -\frac {1}{80} \, {\left (\frac {1}{x + 4} - \frac {x^{2} - 4 \, x + 16}{x^{3}}\right )} \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x-48)*log(3)/(5*x^6+40*x^5+80*x^4),x, algorithm="giac")

[Out]

-1/80*(1/(x + 4) - (x^2 - 4*x + 16)/x^3)*log(3)

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maple [A]  time = 0.03, size = 13, normalized size = 0.93

method	result	size
gosper	\(\frac {4 \ln \relax (3)}{5 x^{3} \left (4+x \right )}\)	\(13\)
norman	\(\frac {4 \ln \relax (3)}{5 x^{3} \left (4+x \right )}\)	\(13\)
risch	\(\frac {4 \ln \relax (3)}{5 x^{3} \left (4+x \right )}\)	\(13\)
default	\(\frac {16 \ln \relax (3) \left (\frac {1}{16 x^{3}}-\frac {1}{64 x^{2}}+\frac {1}{256 x}-\frac {1}{256 \left (4+x \right )}\right )}{5}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16*x-48)*ln(3)/(5*x^6+40*x^5+80*x^4),x,method=_RETURNVERBOSE)

[Out]

4/5*ln(3)/x^3/(4+x)

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maxima [A]  time = 0.35, size = 15, normalized size = 1.07 \begin {gather*} \frac {4 \, \log \relax (3)}{5 \, {\left (x^{4} + 4 \, x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x-48)*log(3)/(5*x^6+40*x^5+80*x^4),x, algorithm="maxima")

[Out]

4/5*log(3)/(x^4 + 4*x^3)

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mupad [B]  time = 0.06, size = 17, normalized size = 1.21 \begin {gather*} \frac {4\,\ln \relax (3)}{5\,\left (x^4+4\,x^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(3)*(16*x + 48))/(80*x^4 + 40*x^5 + 5*x^6),x)

[Out]

(4*log(3))/(5*(4*x^3 + x^4))

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sympy [A]  time = 0.16, size = 14, normalized size = 1.00 \begin {gather*} \frac {4 \log {\relax (3 )}}{5 x^{4} + 20 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x-48)*ln(3)/(5*x**6+40*x**5+80*x**4),x)

[Out]

4*log(3)/(5*x**4 + 20*x**3)

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