3.53.12 \(\int \frac {8 x^2+2 x^6+(-57 x-12 x^5) \log ^2(2)+(90+18 x^4) \log ^4(2)}{x^5-6 x^4 \log ^2(2)+9 x^3 \log ^4(2)} \, dx\)

Optimal. Leaf size=23 \[ x^2-\frac {4-\frac {3}{-3+\frac {x}{\log ^2(2)}}}{x^2} \]

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Rubi [A]  time = 0.10, antiderivative size = 38, normalized size of antiderivative = 1.65, number of steps used = 4, number of rules used = 3, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {1594, 27, 1620} \begin {gather*} x^2-\frac {5}{x^2}+\frac {1}{3 \log ^2(2) \left (x-3 \log ^2(2)\right )}-\frac {1}{3 x \log ^2(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8*x^2 + 2*x^6 + (-57*x - 12*x^5)*Log[2]^2 + (90 + 18*x^4)*Log[2]^4)/(x^5 - 6*x^4*Log[2]^2 + 9*x^3*Log[2]^
4),x]

[Out]

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

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 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 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8 x^2+2 x^6+\left (-57 x-12 x^5\right ) \log ^2(2)+\left (90+18 x^4\right ) \log ^4(2)}{x^3 \left (x^2-6 x \log ^2(2)+9 \log ^4(2)\right )} \, dx\\ &=\int \frac {8 x^2+2 x^6+\left (-57 x-12 x^5\right ) \log ^2(2)+\left (90+18 x^4\right ) \log ^4(2)}{x^3 \left (x-3 \log ^2(2)\right )^2} \, dx\\ &=\int \left (\frac {10}{x^3}+2 x+\frac {1}{3 x^2 \log ^2(2)}-\frac {1}{3 \log ^2(2) \left (-x+3 \log ^2(2)\right )^2}\right ) \, dx\\ &=-\frac {5}{x^2}+x^2-\frac {1}{3 x \log ^2(2)}+\frac {1}{3 \log ^2(2) \left (x-3 \log ^2(2)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 38, normalized size = 1.65 \begin {gather*} -\frac {5}{x^2}+x^2-\frac {1}{3 x \log ^2(2)}+\frac {1}{3 \log ^2(2) \left (x-3 \log ^2(2)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*x^2 + 2*x^6 + (-57*x - 12*x^5)*Log[2]^2 + (90 + 18*x^4)*Log[2]^4)/(x^5 - 6*x^4*Log[2]^2 + 9*x^3*L
og[2]^4),x]

[Out]

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

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fricas [A]  time = 0.67, size = 37, normalized size = 1.61 \begin {gather*} -\frac {x^{5} - 3 \, {\left (x^{4} - 5\right )} \log \relax (2)^{2} - 4 \, x}{3 \, x^{2} \log \relax (2)^{2} - x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x^4+90)*log(2)^4+(-12*x^5-57*x)*log(2)^2+2*x^6+8*x^2)/(9*x^3*log(2)^4-6*x^4*log(2)^2+x^5),x, al
gorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 39, normalized size = 1.70 \begin {gather*} x^{2} - \frac {1}{3 \, {\left (3 \, \log \relax (2)^{2} - x\right )} \log \relax (2)^{2}} - \frac {15 \, \log \relax (2)^{2} + x}{3 \, x^{2} \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x^4+90)*log(2)^4+(-12*x^5-57*x)*log(2)^2+2*x^6+8*x^2)/(9*x^3*log(2)^4-6*x^4*log(2)^2+x^5),x, al
gorithm="giac")

[Out]

x^2 - 1/3/((3*log(2)^2 - x)*log(2)^2) - 1/3*(15*log(2)^2 + x)/(x^2*log(2)^2)

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maple [A]  time = 0.10, size = 32, normalized size = 1.39

method	result	size
risch	\(x^{2}+\frac {-15 \ln \relax (2)^{2}+4 x}{x^{2} \left (3 \ln \relax (2)^{2}-x \right )}\)	\(32\)
default	\(x^{2}+\frac {1}{3 \ln \relax (2)^{2} \left (-3 \ln \relax (2)^{2}+x \right )}-\frac {5}{x^{2}}-\frac {1}{3 \ln \relax (2)^{2} x}\)	\(35\)
gosper	\(\frac {-x^{5}+3 x^{4} \ln \relax (2)^{2}+4 x -15 \ln \relax (2)^{2}}{x^{2} \left (3 \ln \relax (2)^{2}-x \right )}\)	\(41\)
norman	\(\frac {-x^{5}+3 x^{4} \ln \relax (2)^{2}+4 x -15 \ln \relax (2)^{2}}{x^{2} \left (3 \ln \relax (2)^{2}-x \right )}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((18*x^4+90)*ln(2)^4+(-12*x^5-57*x)*ln(2)^2+2*x^6+8*x^2)/(9*x^3*ln(2)^4-6*x^4*ln(2)^2+x^5),x,method=_RETUR
NVERBOSE)

[Out]

x^2+3*(-5*ln(2)^2+4/3*x)/x^2/(3*ln(2)^2-x)

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maxima [A]  time = 0.34, size = 33, normalized size = 1.43 \begin {gather*} x^{2} - \frac {15 \, \log \relax (2)^{2} - 4 \, x}{3 \, x^{2} \log \relax (2)^{2} - x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x^4+90)*log(2)^4+(-12*x^5-57*x)*log(2)^2+2*x^6+8*x^2)/(9*x^3*log(2)^4-6*x^4*log(2)^2+x^5),x, al
gorithm="maxima")

[Out]

x^2 - (15*log(2)^2 - 4*x)/(3*x^2*log(2)^2 - x^3)

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mupad [B]  time = 3.39, size = 29, normalized size = 1.26 \begin {gather*} x^2-\frac {4\,x-15\,{\ln \relax (2)}^2}{x^2\,\left (x-3\,{\ln \relax (2)}^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(2)^4*(18*x^4 + 90) - log(2)^2*(57*x + 12*x^5) + 8*x^2 + 2*x^6)/(9*x^3*log(2)^4 - 6*x^4*log(2)^2 + x^5
),x)

[Out]

x^2 - (4*x - 15*log(2)^2)/(x^2*(x - 3*log(2)^2))

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sympy [A]  time = 0.21, size = 26, normalized size = 1.13 \begin {gather*} x^{2} + \frac {- 4 x + 15 \log {\relax (2 )}^{2}}{x^{3} - 3 x^{2} \log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x**4+90)*ln(2)**4+(-12*x**5-57*x)*ln(2)**2+2*x**6+8*x**2)/(9*x**3*ln(2)**4-6*x**4*ln(2)**2+x**5
),x)

[Out]

x**2 + (-4*x + 15*log(2)**2)/(x**3 - 3*x**2*log(2)**2)

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