3.8.32 \(\int \frac {(200-75 x+3 x^3) \log ^2(5)}{64 x^3-32 x^4+4 x^5} \, dx\)

Optimal. Leaf size=34 \[ 3+\frac {5 \left (\frac {5}{x}+\frac {1}{5} (1-x) x\right ) \log ^2(5)}{4 (-4+x) x} \]

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Rubi [A]  time = 0.06, antiderivative size = 38, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 1594, 27, 1620} \begin {gather*} -\frac {25 \log ^2(5)}{16 x^2}-\frac {25 \log ^2(5)}{64 x}+\frac {23 \log ^2(5)}{64 (4-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((200 - 75*x + 3*x^3)*Log[5]^2)/(64*x^3 - 32*x^4 + 4*x^5),x]

[Out]

(23*Log[5]^2)/(64*(4 - x)) - (25*Log[5]^2)/(16*x^2) - (25*Log[5]^2)/(64*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 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} &=\log ^2(5) \int \frac {200-75 x+3 x^3}{64 x^3-32 x^4+4 x^5} \, dx\\ &=\log ^2(5) \int \frac {200-75 x+3 x^3}{x^3 \left (64-32 x+4 x^2\right )} \, dx\\ &=\log ^2(5) \int \frac {200-75 x+3 x^3}{4 (-4+x)^2 x^3} \, dx\\ &=\frac {1}{4} \log ^2(5) \int \frac {200-75 x+3 x^3}{(-4+x)^2 x^3} \, dx\\ &=\frac {1}{4} \log ^2(5) \int \left (\frac {23}{16 (-4+x)^2}+\frac {25}{2 x^3}+\frac {25}{16 x^2}\right ) \, dx\\ &=\frac {23 \log ^2(5)}{64 (4-x)}-\frac {25 \log ^2(5)}{16 x^2}-\frac {25 \log ^2(5)}{64 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 0.68 \begin {gather*} \frac {\left (25-3 x^2\right ) \log ^2(5)}{4 (-4+x) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((200 - 75*x + 3*x^3)*Log[5]^2)/(64*x^3 - 32*x^4 + 4*x^5),x]

[Out]

((25 - 3*x^2)*Log[5]^2)/(4*(-4 + x)*x^2)

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fricas [A]  time = 0.48, size = 24, normalized size = 0.71 \begin {gather*} -\frac {{\left (3 \, x^{2} - 25\right )} \log \relax (5)^{2}}{4 \, {\left (x^{3} - 4 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^3-75*x+200)*log(5)^2/(4*x^5-32*x^4+64*x^3),x, algorithm="fricas")

[Out]

-1/4*(3*x^2 - 25)*log(5)^2/(x^3 - 4*x^2)

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giac [A]  time = 0.26, size = 22, normalized size = 0.65 \begin {gather*} -\frac {1}{64} \, {\left (\frac {23}{x - 4} + \frac {25 \, {\left (x + 4\right )}}{x^{2}}\right )} \log \relax (5)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^3-75*x+200)*log(5)^2/(4*x^5-32*x^4+64*x^3),x, algorithm="giac")

[Out]

-1/64*(23/(x - 4) + 25*(x + 4)/x^2)*log(5)^2

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maple [A]  time = 0.04, size = 21, normalized size = 0.62

method	result	size
risch	\(\frac {\ln \relax (5)^{2} \left (-\frac {3 x^{2}}{4}+\frac {25}{4}\right )}{\left (x -4\right ) x^{2}}\)	\(21\)
gosper	\(-\frac {\ln \relax (5)^{2} \left (3 x^{2}-25\right )}{4 x^{2} \left (x -4\right )}\)	\(22\)
default	\(\frac {\ln \relax (5)^{2} \left (-\frac {25}{4 x^{2}}-\frac {25}{16 x}-\frac {23}{16 \left (x -4\right )}\right )}{4}\)	\(25\)
norman	\(\frac {-\frac {3 x^{2} \ln \relax (5)^{2}}{4}+\frac {25 \ln \relax (5)^{2}}{4}}{\left (x -4\right ) x^{2}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^3-75*x+200)*ln(5)^2/(4*x^5-32*x^4+64*x^3),x,method=_RETURNVERBOSE)

[Out]

ln(5)^2*(-3/4*x^2+25/4)/(x-4)/x^2

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maxima [A]  time = 0.42, size = 24, normalized size = 0.71 \begin {gather*} -\frac {{\left (3 \, x^{2} - 25\right )} \log \relax (5)^{2}}{4 \, {\left (x^{3} - 4 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^3-75*x+200)*log(5)^2/(4*x^5-32*x^4+64*x^3),x, algorithm="maxima")

[Out]

-1/4*(3*x^2 - 25)*log(5)^2/(x^3 - 4*x^2)

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mupad [B]  time = 0.54, size = 30, normalized size = 0.88 \begin {gather*} \frac {3\,x^2\,{\ln \relax (5)}^2-25\,{\ln \relax (5)}^2}{16\,x^2-4\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(5)^2*(3*x^3 - 75*x + 200))/(64*x^3 - 32*x^4 + 4*x^5),x)

[Out]

(3*x^2*log(5)^2 - 25*log(5)^2)/(16*x^2 - 4*x^3)

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sympy [A]  time = 0.19, size = 26, normalized size = 0.76 \begin {gather*} \frac {- 3 x^{2} \log {\relax (5 )}^{2} + 25 \log {\relax (5 )}^{2}}{4 x^{3} - 16 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**3-75*x+200)*ln(5)**2/(4*x**5-32*x**4+64*x**3),x)

[Out]

(-3*x**2*log(5)**2 + 25*log(5)**2)/(4*x**3 - 16*x**2)

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