3.12.43 \(\int \frac {2-10 x^2+e^5 (1-5 x^2)}{(4 x-8 x^2+20 x^3) \log ^2(\frac {1-2 x+5 x^2}{x})} \, dx\)

Optimal. Leaf size=20 \[ \frac {2+e^5}{4 \log \left (-2+\frac {1}{x}+5 x\right )} \]

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Rubi [A]  time = 0.16, antiderivative size = 26, normalized size of antiderivative = 1.30, number of steps used = 2, number of rules used = 2, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {1594, 6686} \begin {gather*} \frac {2+e^5}{4 \log \left (\frac {5 x^2-2 x+1}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - 10*x^2 + E^5*(1 - 5*x^2))/((4*x - 8*x^2 + 20*x^3)*Log[(1 - 2*x + 5*x^2)/x]^2),x]

[Out]

(2 + E^5)/(4*Log[(1 - 2*x + 5*x^2)/x])

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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2-10 x^2+e^5 \left (1-5 x^2\right )}{x \left (4-8 x+20 x^2\right ) \log ^2\left (\frac {1-2 x+5 x^2}{x}\right )} \, dx\\ &=\frac {2+e^5}{4 \log \left (\frac {1-2 x+5 x^2}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 1.00 \begin {gather*} \frac {2+e^5}{4 \log \left (-2+\frac {1}{x}+5 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - 10*x^2 + E^5*(1 - 5*x^2))/((4*x - 8*x^2 + 20*x^3)*Log[(1 - 2*x + 5*x^2)/x]^2),x]

[Out]

(2 + E^5)/(4*Log[-2 + x^(-1) + 5*x])

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fricas [A]  time = 0.58, size = 23, normalized size = 1.15 \begin {gather*} \frac {e^{5} + 2}{4 \, \log \left (\frac {5 \, x^{2} - 2 \, x + 1}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2+1)*exp(5)-10*x^2+2)/(20*x^3-8*x^2+4*x)/log((5*x^2-2*x+1)/x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.30, size = 23, normalized size = 1.15 \begin {gather*} \frac {e^{5} + 2}{4 \, \log \left (\frac {5 \, x^{2} - 2 \, x + 1}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2+1)*exp(5)-10*x^2+2)/(20*x^3-8*x^2+4*x)/log((5*x^2-2*x+1)/x)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 25, normalized size = 1.25

method	result	size
norman	\(\frac {\frac {{\mathrm e}^{5}}{4}+\frac {1}{2}}{\ln \left (\frac {5 x^{2}-2 x +1}{x}\right )}\)	\(25\)
risch	\(\frac {{\mathrm e}^{5}}{4 \ln \left (\frac {5 x^{2}-2 x +1}{x}\right )}+\frac {1}{2 \ln \left (\frac {5 x^{2}-2 x +1}{x}\right )}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x^2+1)*exp(5)-10*x^2+2)/(20*x^3-8*x^2+4*x)/ln((5*x^2-2*x+1)/x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.68, size = 24, normalized size = 1.20 \begin {gather*} \frac {e^{5} + 2}{4 \, {\left (\log \left (5 \, x^{2} - 2 \, x + 1\right ) - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^2+1)*exp(5)-10*x^2+2)/(20*x^3-8*x^2+4*x)/log((5*x^2-2*x+1)/x)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.93, size = 24, normalized size = 1.20 \begin {gather*} \frac {\frac {{\mathrm {e}}^5}{4}+\frac {1}{2}}{\ln \left (\frac {5\,x^2-2\,x+1}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5)*(5*x^2 - 1) + 10*x^2 - 2)/(log((5*x^2 - 2*x + 1)/x)^2*(4*x - 8*x^2 + 20*x^3)),x)

[Out]

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

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sympy [A]  time = 0.15, size = 19, normalized size = 0.95 \begin {gather*} \frac {2 + e^{5}}{4 \log {\left (\frac {5 x^{2} - 2 x + 1}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x**2+1)*exp(5)-10*x**2+2)/(20*x**3-8*x**2+4*x)/ln((5*x**2-2*x+1)/x)**2,x)

[Out]

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

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