3.100.23 \(\int \frac {(100+80 x+40 x^2) \log (2)-20 x \log (2) \log (5)}{25-20 x^2+4 x^4} \, dx\)

Optimal. Leaf size=31 \[ \frac {5 x \log (2) \left (x-\frac {(2+x)^2-\log (5)}{x}\right )}{-5+2 x^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 0.71, number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {28, 1814, 8} \begin {gather*} \frac {5 \log (2) (4 x+4-\log (5))}{5-2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((100 + 80*x + 40*x^2)*Log[2] - 20*x*Log[2]*Log[5])/(25 - 20*x^2 + 4*x^4),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=4 \int \frac {\left (100+80 x+40 x^2\right ) \log (2)-20 x \log (2) \log (5)}{\left (-10+4 x^2\right )^2} \, dx\\ &=\frac {5 \log (2) (4+4 x-\log (5))}{5-2 x^2}+\frac {\int 0 \, dx}{5}\\ &=\frac {5 \log (2) (4+4 x-\log (5))}{5-2 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.65 \begin {gather*} \frac {20 \log (2) (-4-4 x+\log (5))}{-20+8 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((100 + 80*x + 40*x^2)*Log[2] - 20*x*Log[2]*Log[5])/(25 - 20*x^2 + 4*x^4),x]

[Out]

(20*Log[2]*(-4 - 4*x + Log[5]))/(-20 + 8*x^2)

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fricas [A]  time = 1.05, size = 25, normalized size = 0.81 \begin {gather*} -\frac {5 \, {\left (4 \, {\left (x + 1\right )} \log \relax (2) - \log \relax (5) \log \relax (2)\right )}}{2 \, x^{2} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x*log(2)*log(5)+(40*x^2+80*x+100)*log(2))/(4*x^4-20*x^2+25),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 27, normalized size = 0.87 \begin {gather*} -\frac {5 \, {\left (4 \, x \log \relax (2) - \log \relax (5) \log \relax (2) + 4 \, \log \relax (2)\right )}}{2 \, x^{2} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x*log(2)*log(5)+(40*x^2+80*x+100)*log(2))/(4*x^4-20*x^2+25),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 21, normalized size = 0.68

method	result	size
gosper	\(\frac {5 \ln \relax (2) \left (\ln \relax (5)-4 x -4\right )}{2 x^{2}-5}\)	\(21\)
default	\(-\frac {20 \ln \relax (2) \left (\frac {x}{2}-\frac {\ln \relax (5)}{8}+\frac {1}{2}\right )}{x^{2}-\frac {5}{2}}\)	\(21\)
risch	\(\frac {-10 x \ln \relax (2)+\frac {5 \ln \relax (2) \ln \relax (5)}{2}-10 \ln \relax (2)}{x^{2}-\frac {5}{2}}\)	\(25\)
norman	\(\frac {-20 x \ln \relax (2)+5 \ln \relax (2) \ln \relax (5)-20 \ln \relax (2)}{2 x^{2}-5}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-20*x*ln(2)*ln(5)+(40*x^2+80*x+100)*ln(2))/(4*x^4-20*x^2+25),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 27, normalized size = 0.87 \begin {gather*} -\frac {5 \, {\left (4 \, x \log \relax (2) - \log \relax (5) \log \relax (2) + 4 \, \log \relax (2)\right )}}{2 \, x^{2} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x*log(2)*log(5)+(40*x^2+80*x+100)*log(2))/(4*x^4-20*x^2+25),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.13, size = 22, normalized size = 0.71 \begin {gather*} -\frac {5\,\ln \relax (2)\,\left (4\,x-\ln \relax (5)+4\right )}{2\,x^2-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(2)*(80*x + 40*x^2 + 100) - 20*x*log(2)*log(5))/(4*x^4 - 20*x^2 + 25),x)

[Out]

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

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sympy [A]  time = 0.39, size = 26, normalized size = 0.84 \begin {gather*} \frac {- 20 x \log {\relax (2 )} - 20 \log {\relax (2 )} + 5 \log {\relax (2 )} \log {\relax (5 )}}{2 x^{2} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x*ln(2)*ln(5)+(40*x**2+80*x+100)*ln(2))/(4*x**4-20*x**2+25),x)

[Out]

(-20*x*log(2) - 20*log(2) + 5*log(2)*log(5))/(2*x**2 - 5)

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