3.97.3 \(\int \frac {-375+300 x-64 x^2+(-125+100 x-20 x^2) \log (4)}{25 x^2-20 x^3+4 x^4} \, dx\)

Optimal. Leaf size=24 \[ -2+\frac {1+x+\frac {5}{-5+2 x}+5 (3+\log (4))}{x} \]

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Rubi [A]  time = 0.07, antiderivative size = 19, normalized size of antiderivative = 0.79, number of steps used = 4, number of rules used = 3, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1594, 27, 1820} \begin {gather*} \frac {5 (3+\log (4))}{x}-\frac {2}{5-2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-375 + 300*x - 64*x^2 + (-125 + 100*x - 20*x^2)*Log[4])/(25*x^2 - 20*x^3 + 4*x^4),x]

[Out]

-2/(5 - 2*x) + (5*(3 + Log[4]))/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 1820

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-375+300 x-64 x^2+\left (-125+100 x-20 x^2\right ) \log (4)}{x^2 \left (25-20 x+4 x^2\right )} \, dx\\ &=\int \frac {-375+300 x-64 x^2+\left (-125+100 x-20 x^2\right ) \log (4)}{x^2 (-5+2 x)^2} \, dx\\ &=\int \left (-\frac {4}{(-5+2 x)^2}-\frac {5 (3+\log (4))}{x^2}\right ) \, dx\\ &=-\frac {2}{5-2 x}+\frac {5 (3+\log (4))}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 0.79 \begin {gather*} \frac {2}{-5+2 x}+\frac {5 (3+\log (4))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-375 + 300*x - 64*x^2 + (-125 + 100*x - 20*x^2)*Log[4])/(25*x^2 - 20*x^3 + 4*x^4),x]

[Out]

2/(-5 + 2*x) + (5*(3 + Log[4]))/x

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fricas [A]  time = 0.58, size = 26, normalized size = 1.08 \begin {gather*} \frac {10 \, {\left (2 \, x - 5\right )} \log \relax (2) + 32 \, x - 75}{2 \, x^{2} - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-20*x^2+100*x-125)*log(2)-64*x^2+300*x-375)/(4*x^4-20*x^3+25*x^2),x, algorithm="fricas")

[Out]

(10*(2*x - 5)*log(2) + 32*x - 75)/(2*x^2 - 5*x)

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giac [A]  time = 0.18, size = 26, normalized size = 1.08 \begin {gather*} \frac {20 \, x \log \relax (2) + 32 \, x - 50 \, \log \relax (2) - 75}{2 \, x^{2} - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-20*x^2+100*x-125)*log(2)-64*x^2+300*x-375)/(4*x^4-20*x^3+25*x^2),x, algorithm="giac")

[Out]

(20*x*log(2) + 32*x - 50*log(2) - 75)/(2*x^2 - 5*x)

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maple [A]  time = 0.06, size = 22, normalized size = 0.92

method	result	size
default	\(\frac {2}{2 x -5}-\frac {-10 \ln \relax (2)-15}{x}\)	\(22\)
gosper	\(\frac {20 x \ln \relax (2)-50 \ln \relax (2)+32 x -75}{x \left (2 x -5\right )}\)	\(26\)
norman	\(\frac {\left (20 \ln \relax (2)+32\right ) x -50 \ln \relax (2)-75}{x \left (2 x -5\right )}\)	\(26\)
risch	\(\frac {2 \left (10 \ln \relax (2)+16\right ) x -50 \ln \relax (2)-75}{x \left (2 x -5\right )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(-20*x^2+100*x-125)*ln(2)-64*x^2+300*x-375)/(4*x^4-20*x^3+25*x^2),x,method=_RETURNVERBOSE)

[Out]

2/(2*x-5)-(-10*ln(2)-15)/x

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maxima [A]  time = 0.35, size = 27, normalized size = 1.12 \begin {gather*} \frac {4 \, x {\left (5 \, \log \relax (2) + 8\right )} - 50 \, \log \relax (2) - 75}{2 \, x^{2} - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-20*x^2+100*x-125)*log(2)-64*x^2+300*x-375)/(4*x^4-20*x^3+25*x^2),x, algorithm="maxima")

[Out]

(4*x*(5*log(2) + 8) - 50*log(2) - 75)/(2*x^2 - 5*x)

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mupad [B]  time = 0.09, size = 20, normalized size = 0.83 \begin {gather*} \frac {10\,\ln \relax (2)+15}{x}+\frac {2}{2\,x-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(2)*(20*x^2 - 100*x + 125) - 300*x + 64*x^2 + 375)/(25*x^2 - 20*x^3 + 4*x^4),x)

[Out]

(10*log(2) + 15)/x + 2/(2*x - 5)

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sympy [A]  time = 0.31, size = 26, normalized size = 1.08 \begin {gather*} - \frac {x \left (-32 - 20 \log {\relax (2 )}\right ) + 50 \log {\relax (2 )} + 75}{2 x^{2} - 5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(-20*x**2+100*x-125)*ln(2)-64*x**2+300*x-375)/(4*x**4-20*x**3+25*x**2),x)

[Out]

-(x*(-32 - 20*log(2)) + 50*log(2) + 75)/(2*x**2 - 5*x)

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