3.60.15 \(\int \frac {76-6 x+40 \log (2)}{361 x^3-38 x^4+x^5+(380 x^3-20 x^4) \log (2)+100 x^3 \log ^2(2)} \, dx\)

Optimal. Leaf size=20 \[ \frac {2}{x^2+x^3-10 x^2 (2+\log (2))} \]

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Rubi [A]  time = 0.08, antiderivative size = 14, normalized size of antiderivative = 0.70, number of steps used = 5, number of rules used = 5, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {6, 1998, 1594, 27, 74} \begin {gather*} -\frac {2}{x^2 (-x+19+\log (1024))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(76 - 6*x + 40*Log[2])/(361*x^3 - 38*x^4 + x^5 + (380*x^3 - 20*x^4)*Log[2] + 100*x^3*Log[2]^2),x]

[Out]

-2/(x^2*(19 - x + Log[1024]))

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 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]

Rule 1998

Int[(u_)^(p_.)*(z_), x_Symbol] :> Int[ExpandToSum[z, x]*ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && BinomialQ[z,
x] && GeneralizedTrinomialQ[u, x] && EqQ[BinomialDegree[z, x] - GeneralizedTrinomialDegree[u, x], 0] &&  !(Bi
nomialMatchQ[z, x] && GeneralizedTrinomialMatchQ[u, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {76-6 x+40 \log (2)}{-38 x^4+x^5+\left (380 x^3-20 x^4\right ) \log (2)+x^3 \left (361+100 \log ^2(2)\right )} \, dx\\ &=\int \frac {-6 x+4 (19+10 \log (2))}{x^5-2 x^4 (19+\log (1024))+x^3 (19+\log (1024))^2} \, dx\\ &=\int \frac {-6 x+4 (19+10 \log (2))}{x^3 \left (x^2-2 x (19+\log (1024))+(19+\log (1024))^2\right )} \, dx\\ &=\int \frac {-6 x+4 (19+10 \log (2))}{x^3 (-19+x-\log (1024))^2} \, dx\\ &=-\frac {2}{x^2 (19-x+\log (1024))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 14, normalized size = 0.70 \begin {gather*} \frac {2}{x^2 (-19+x-10 \log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(76 - 6*x + 40*Log[2])/(361*x^3 - 38*x^4 + x^5 + (380*x^3 - 20*x^4)*Log[2] + 100*x^3*Log[2]^2),x]

[Out]

2/(x^2*(-19 + x - 10*Log[2]))

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fricas [A]  time = 0.57, size = 20, normalized size = 1.00 \begin {gather*} \frac {2}{x^{3} - 10 \, x^{2} \log \relax (2) - 19 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((40*log(2)-6*x+76)/(100*x^3*log(2)^2+(-20*x^4+380*x^3)*log(2)+x^5-38*x^4+361*x^3),x, algorithm="fric
as")

[Out]

2/(x^3 - 10*x^2*log(2) - 19*x^2)

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giac [B]  time = 0.19, size = 52, normalized size = 2.60 \begin {gather*} \frac {2}{{\left (100 \, \log \relax (2)^{2} + 380 \, \log \relax (2) + 361\right )} {\left (x - 10 \, \log \relax (2) - 19\right )}} - \frac {2 \, {\left (x + 10 \, \log \relax (2) + 19\right )}}{{\left (100 \, \log \relax (2)^{2} + 380 \, \log \relax (2) + 361\right )} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((40*log(2)-6*x+76)/(100*x^3*log(2)^2+(-20*x^4+380*x^3)*log(2)+x^5-38*x^4+361*x^3),x, algorithm="giac
")

[Out]

2/((100*log(2)^2 + 380*log(2) + 361)*(x - 10*log(2) - 19)) - 2*(x + 10*log(2) + 19)/((100*log(2)^2 + 380*log(2
) + 361)*x^2)

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maple [A]  time = 0.08, size = 17, normalized size = 0.85

method	result	size
gosper	\(-\frac {2}{x^{2} \left (19+10 \ln \relax (2)-x \right )}\)	\(17\)
norman	\(-\frac {2}{x^{2} \left (19+10 \ln \relax (2)-x \right )}\)	\(17\)
risch	\(-\frac {2}{x^{2} \left (19+10 \ln \relax (2)-x \right )}\)	\(17\)
default	\(\frac {2}{\left (19+10 \ln \relax (2)\right )^{2} \left (-19-10 \ln \relax (2)+x \right )}-\frac {2}{\left (19+10 \ln \relax (2)\right )^{2} x}-\frac {2}{\left (19+10 \ln \relax (2)\right ) x^{2}}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((40*ln(2)-6*x+76)/(100*x^3*ln(2)^2+(-20*x^4+380*x^3)*ln(2)+x^5-38*x^4+361*x^3),x,method=_RETURNVERBOSE)

[Out]

-2/x^2/(19+10*ln(2)-x)

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maxima [A]  time = 0.36, size = 19, normalized size = 0.95 \begin {gather*} \frac {2}{x^{3} - x^{2} {\left (10 \, \log \relax (2) + 19\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((40*log(2)-6*x+76)/(100*x^3*log(2)^2+(-20*x^4+380*x^3)*log(2)+x^5-38*x^4+361*x^3),x, algorithm="maxi
ma")

[Out]

2/(x^3 - x^2*(10*log(2) + 19))

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mupad [B]  time = 0.15, size = 20, normalized size = 1.00 \begin {gather*} -\frac {2}{x^2\,\left (10\,\ln \relax (2)+19\right )-x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((40*log(2) - 6*x + 76)/(100*x^3*log(2)^2 + log(2)*(380*x^3 - 20*x^4) + 361*x^3 - 38*x^4 + x^5),x)

[Out]

-2/(x^2*(10*log(2) + 19) - x^3)

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sympy [A]  time = 0.35, size = 15, normalized size = 0.75 \begin {gather*} \frac {2}{x^{3} + x^{2} \left (-19 - 10 \log {\relax (2 )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((40*ln(2)-6*x+76)/(100*x**3*ln(2)**2+(-20*x**4+380*x**3)*ln(2)+x**5-38*x**4+361*x**3),x)

[Out]

2/(x**3 + x**2*(-19 - 10*log(2)))

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