3.3.30 \(\int \frac {-100+210 x-30 x^2+e^2 (20 x^2-24 x^3+4 x^4)}{e^2 (5 x-6 x^2+x^3)} \, dx\)

Optimal. Leaf size=25 \[ 2 \left (1+x^2+\frac {5 \log \left (\frac {-5+x}{(-1+x)^2 x^2}\right )}{e^2}\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 35, normalized size of antiderivative = 1.40, number of steps used = 4, number of rules used = 3, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 1594, 1628} \begin {gather*} 2 x^2-\frac {20 \log (1-x)}{e^2}+\frac {10 \log (5-x)}{e^2}-\frac {20 \log (x)}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-100 + 210*x - 30*x^2 + E^2*(20*x^2 - 24*x^3 + 4*x^4))/(E^2*(5*x - 6*x^2 + x^3)),x]

[Out]

2*x^2 - (20*Log[1 - x])/E^2 + (10*Log[5 - x])/E^2 - (20*Log[x])/E^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-100+210 x-30 x^2+e^2 \left (20 x^2-24 x^3+4 x^4\right )}{5 x-6 x^2+x^3} \, dx}{e^2}\\ &=\frac {\int \frac {-100+210 x-30 x^2+e^2 \left (20 x^2-24 x^3+4 x^4\right )}{x \left (5-6 x+x^2\right )} \, dx}{e^2}\\ &=\frac {\int \left (\frac {10}{-5+x}-\frac {20}{-1+x}-\frac {20}{x}+4 e^2 x\right ) \, dx}{e^2}\\ &=2 x^2-\frac {20 \log (1-x)}{e^2}+\frac {10 \log (5-x)}{e^2}-\frac {20 \log (x)}{e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 1.32 \begin {gather*} \frac {2 e^2 x^2-20 \log (1-x)+10 \log (5-x)-20 \log (x)}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-100 + 210*x - 30*x^2 + E^2*(20*x^2 - 24*x^3 + 4*x^4))/(E^2*(5*x - 6*x^2 + x^3)),x]

[Out]

(2*E^2*x^2 - 20*Log[1 - x] + 10*Log[5 - x] - 20*Log[x])/E^2

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fricas [A]  time = 0.48, size = 27, normalized size = 1.08 \begin {gather*} 2 \, {\left (x^{2} e^{2} - 10 \, \log \left (x^{2} - x\right ) + 5 \, \log \left (x - 5\right )\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4-24*x^3+20*x^2)*exp(2)-30*x^2+210*x-100)/(x^3-6*x^2+5*x)/exp(2),x, algorithm="fricas")

[Out]

2*(x^2*e^2 - 10*log(x^2 - x) + 5*log(x - 5))*e^(-2)

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giac [A]  time = 0.37, size = 30, normalized size = 1.20 \begin {gather*} 2 \, {\left (x^{2} e^{2} - 10 \, \log \left ({\left | x - 1 \right |}\right ) + 5 \, \log \left ({\left | x - 5 \right |}\right ) - 10 \, \log \left ({\left | x \right |}\right )\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4-24*x^3+20*x^2)*exp(2)-30*x^2+210*x-100)/(x^3-6*x^2+5*x)/exp(2),x, algorithm="giac")

[Out]

2*(x^2*e^2 - 10*log(abs(x - 1)) + 5*log(abs(x - 5)) - 10*log(abs(x)))*e^(-2)

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maple [A]  time = 0.06, size = 27, normalized size = 1.08

method	result	size
risch	\(2 x^{2}+10 \,{\mathrm e}^{-2} \ln \left (x -5\right )-20 \,{\mathrm e}^{-2} \ln \left (x^{2}-x \right )\)	\(27\)
default	\({\mathrm e}^{-2} \left (2 x^{2} {\mathrm e}^{2}+10 \ln \left (x -5\right )-20 \ln \relax (x )-20 \ln \left (x -1\right )\right )\)	\(30\)
norman	\(2 x^{2}-20 \ln \relax (x ) {\mathrm e}^{-2}+10 \,{\mathrm e}^{-2} \ln \left (x -5\right )-20 \,{\mathrm e}^{-2} \ln \left (x -1\right )\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^4-24*x^3+20*x^2)*exp(2)-30*x^2+210*x-100)/(x^3-6*x^2+5*x)/exp(2),x,method=_RETURNVERBOSE)

[Out]

2*x^2+10*exp(-2)*ln(x-5)-20*exp(-2)*ln(x^2-x)

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maxima [A]  time = 0.42, size = 27, normalized size = 1.08 \begin {gather*} 2 \, {\left (x^{2} e^{2} - 10 \, \log \left (x - 1\right ) + 5 \, \log \left (x - 5\right ) - 10 \, \log \relax (x)\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4-24*x^3+20*x^2)*exp(2)-30*x^2+210*x-100)/(x^3-6*x^2+5*x)/exp(2),x, algorithm="maxima")

[Out]

2*(x^2*e^2 - 10*log(x - 1) + 5*log(x - 5) - 10*log(x))*e^(-2)

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mupad [B]  time = 0.12, size = 24, normalized size = 0.96 \begin {gather*} 10\,\ln \left (x-5\right )\,{\mathrm {e}}^{-2}+2\,x^2-20\,\ln \left (x\,\left (x-1\right )\right )\,{\mathrm {e}}^{-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2)*(210*x + exp(2)*(20*x^2 - 24*x^3 + 4*x^4) - 30*x^2 - 100))/(5*x - 6*x^2 + x^3),x)

[Out]

10*log(x - 5)*exp(-2) + 2*x^2 - 20*log(x*(x - 1))*exp(-2)

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sympy [A]  time = 0.27, size = 26, normalized size = 1.04 \begin {gather*} 2 x^{2} + \frac {10 \log {\left (x - 5 \right )}}{e^{2}} - \frac {20 \log {\left (x^{2} - x \right )}}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**4-24*x**3+20*x**2)*exp(2)-30*x**2+210*x-100)/(x**3-6*x**2+5*x)/exp(2),x)

[Out]

2*x**2 + 10*exp(-2)*log(x - 5) - 20*exp(-2)*log(x**2 - x)

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