3.11.58 \(\int \frac {-40 x+60 x^2-10 x^4+e (-6+8 x)}{5 x^2-10 x^3+5 x^5+e (x-2 x^2+x^4)} \, dx\)

Optimal. Leaf size=29 \[ 2 \log \left (\frac {3 \left (-\frac {2-\frac {1}{x}}{x}+x\right )}{x (e+5 x)}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 34, normalized size of antiderivative = 1.17, number of steps used = 3, number of rules used = 2, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {2074, 628} \begin {gather*} 2 \log \left (-x^2-x+1\right )+2 \log (1-x)-6 \log (x)-2 \log (5 x+e) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-40*x + 60*x^2 - 10*x^4 + E*(-6 + 8*x))/(5*x^2 - 10*x^3 + 5*x^5 + E*(x - 2*x^2 + x^4)),x]

[Out]

2*Log[1 - x] - 6*Log[x] - 2*Log[E + 5*x] + 2*Log[1 - x - x^2]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2}{-1+x}-\frac {6}{x}-\frac {10}{e+5 x}+\frac {2 (1+2 x)}{-1+x+x^2}\right ) \, dx\\ &=2 \log (1-x)-6 \log (x)-2 \log (e+5 x)+2 \int \frac {1+2 x}{-1+x+x^2} \, dx\\ &=2 \log (1-x)-6 \log (x)-2 \log (e+5 x)+2 \log \left (1-x-x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 24, normalized size = 0.83 \begin {gather*} -6 \log (x)-2 \log (e+5 x)+2 \log \left (1-2 x+x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-40*x + 60*x^2 - 10*x^4 + E*(-6 + 8*x))/(5*x^2 - 10*x^3 + 5*x^5 + E*(x - 2*x^2 + x^4)),x]

[Out]

-6*Log[x] - 2*Log[E + 5*x] + 2*Log[1 - 2*x + x^3]

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fricas [A]  time = 0.58, size = 25, normalized size = 0.86 \begin {gather*} 2 \, \log \left (x^{3} - 2 \, x + 1\right ) - 2 \, \log \left (5 \, x + e\right ) - 6 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-6)*exp(1)-10*x^4+60*x^2-40*x)/((x^4-2*x^2+x)*exp(1)+5*x^5-10*x^3+5*x^2),x, algorithm="fricas")

[Out]

2*log(x^3 - 2*x + 1) - 2*log(5*x + e) - 6*log(x)

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giac [A]  time = 0.31, size = 33, normalized size = 1.14 \begin {gather*} 2 \, \log \left ({\left | x^{2} + x - 1 \right |}\right ) - 2 \, \log \left ({\left | 5 \, x + e \right |}\right ) + 2 \, \log \left ({\left | x - 1 \right |}\right ) - 6 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-6)*exp(1)-10*x^4+60*x^2-40*x)/((x^4-2*x^2+x)*exp(1)+5*x^5-10*x^3+5*x^2),x, algorithm="giac")

[Out]

2*log(abs(x^2 + x - 1)) - 2*log(abs(5*x + e)) + 2*log(abs(x - 1)) - 6*log(abs(x))

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maple [A]  time = 0.08, size = 30, normalized size = 1.03

method	result	size
default	\(2 \ln \left (x -1\right )-6 \ln \relax (x )+2 \ln \left (x^{2}+x -1\right )-2 \ln \left (5 x +{\mathrm e}\right )\)	\(30\)
norman	\(2 \ln \left (x -1\right )-6 \ln \relax (x )+2 \ln \left (x^{2}+x -1\right )-2 \ln \left (5 x +{\mathrm e}\right )\)	\(30\)
risch	\(-2 \ln \left (-{\mathrm e}-5 x \right )-6 \ln \relax (x )+2 \ln \left (-x^{3}+2 x -1\right )\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x-6)*exp(1)-10*x^4+60*x^2-40*x)/((x^4-2*x^2+x)*exp(1)+5*x^5-10*x^3+5*x^2),x,method=_RETURNVERBOSE)

[Out]

2*ln(x-1)-6*ln(x)+2*ln(x^2+x-1)-2*ln(5*x+exp(1))

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maxima [A]  time = 0.44, size = 29, normalized size = 1.00 \begin {gather*} 2 \, \log \left (x^{2} + x - 1\right ) - 2 \, \log \left (5 \, x + e\right ) + 2 \, \log \left (x - 1\right ) - 6 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-6)*exp(1)-10*x^4+60*x^2-40*x)/((x^4-2*x^2+x)*exp(1)+5*x^5-10*x^3+5*x^2),x, algorithm="maxima")

[Out]

2*log(x^2 + x - 1) - 2*log(5*x + e) + 2*log(x - 1) - 6*log(x)

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mupad [B]  time = 0.85, size = 25, normalized size = 0.86 \begin {gather*} 2\,\ln \left (x^3-2\,x+1\right )-2\,\ln \left (x+\frac {\mathrm {e}}{5}\right )-6\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(40*x - 60*x^2 + 10*x^4 - exp(1)*(8*x - 6))/(exp(1)*(x - 2*x^2 + x^4) + 5*x^2 - 10*x^3 + 5*x^5),x)

[Out]

2*log(x^3 - 2*x + 1) - 2*log(x + exp(1)/5) - 6*log(x)

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sympy [A]  time = 9.40, size = 26, normalized size = 0.90 \begin {gather*} - 6 \log {\relax (x )} - 2 \log {\left (x + \frac {e}{5} \right )} + 2 \log {\left (x^{3} - 2 x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x-6)*exp(1)-10*x**4+60*x**2-40*x)/((x**4-2*x**2+x)*exp(1)+5*x**5-10*x**3+5*x**2),x)

[Out]

-6*log(x) - 2*log(x + E/5) + 2*log(x**3 - 2*x + 1)

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