3.55.50 \(\int \frac {-20-4 e^2+8 x}{(5 e^5 x^2+e^3 (50 x^2-10 x^3)+e (125 x^2-50 x^3+5 x^4)) \log (6)} \, dx\) [5450]

Optimal. Leaf size=24 \[ \frac {4}{5 e \left (5+e^2-x\right ) x \log (6)} \]

[Out]

4/5/x/(exp(2)+5-x)/exp(1)/ln(6)

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Rubi [A]
time = 0.06, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 1694, 267} \begin {gather*} \frac {4}{5 e \left (-x+e^2+5\right ) x \log (6)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20 - 4*E^2 + 8*x)/((5*E^5*x^2 + E^3*(50*x^2 - 10*x^3) + E*(125*x^2 - 50*x^3 + 5*x^4))*Log[6]),x]

[Out]

4/(5*E*(5 + E^2 - x)*x*Log[6])

Rule 12

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

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 1694

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -d/(4*e) + x)*(a + d^4/(256*e^3)
- b*(d/(8*e)) + (c - 3*(d^2/(8*e)))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0]
&& NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-20-4 e^2+8 x}{5 e^5 x^2+e^3 \left (50 x^2-10 x^3\right )+e \left (125 x^2-50 x^3+5 x^4\right )} \, dx}{\log (6)}\\ &=\frac {\text {Subst}\left (\int \frac {128 x}{5 e \left (25+10 e^2+e^4-4 x^2\right )^2} \, dx,x,\frac {-50 e-10 e^3}{20 e}+x\right )}{\log (6)}\\ &=\frac {128 \text {Subst}\left (\int \frac {x}{\left (25+10 e^2+e^4-4 x^2\right )^2} \, dx,x,\frac {-50 e-10 e^3}{20 e}+x\right )}{5 e \log (6)}\\ &=\frac {4}{5 e \left (5+e^2-x\right ) x \log (6)}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 24, normalized size = 1.00 \begin {gather*} \frac {4}{5 e \left (5+e^2-x\right ) x \log (6)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 - 4*E^2 + 8*x)/((5*E^5*x^2 + E^3*(50*x^2 - 10*x^3) + E*(125*x^2 - 50*x^3 + 5*x^4))*Log[6]),x]

[Out]

4/(5*E*(5 + E^2 - x)*x*Log[6])

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Maple [A]
time = 0.37, size = 23, normalized size = 0.96

method	result	size
gosper	\(\frac {4 \,{\mathrm e}^{-1}}{5 x \left ({\mathrm e}^{2}+5-x \right ) \ln \left (6\right )}\)	\(23\)
norman	\(\frac {4 \,{\mathrm e}^{-1}}{5 x \left ({\mathrm e}^{2}+5-x \right ) \ln \left (6\right )}\)	\(23\)
risch	\(\frac {4 \,{\mathrm e}^{-1}}{5 \left (\ln \left (2\right )+\ln \left (3\right )\right ) \left ({\mathrm e}^{2}+5-x \right ) x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*exp(2)+8*x-20)/(5*x^2*exp(1)*exp(2)^2+(-10*x^3+50*x^2)*exp(1)*exp(2)+(5*x^4-50*x^3+125*x^2)*exp(1))/ln
(6),x,method=_RETURNVERBOSE)

[Out]

4/5/x/(exp(2)+5-x)/exp(1)/ln(6)

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Maxima [A]
time = 0.28, size = 25, normalized size = 1.04 \begin {gather*} -\frac {4}{5 \, {\left (x^{2} e - x {\left (e^{3} + 5 \, e\right )}\right )} \log \left (6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2)+8*x-20)/(5*x^2*exp(1)*exp(2)^2+(-10*x^3+50*x^2)*exp(1)*exp(2)+(5*x^4-50*x^3+125*x^2)*exp(
1))/log(6),x, algorithm="maxima")

[Out]

-4/5/((x^2*e - x*(e^3 + 5*e))*log(6))

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Fricas [A]
time = 0.38, size = 24, normalized size = 1.00 \begin {gather*} \frac {4}{5 \, {\left (x e^{3} - {\left (x^{2} - 5 \, x\right )} e\right )} \log \left (6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2)+8*x-20)/(5*x^2*exp(1)*exp(2)^2+(-10*x^3+50*x^2)*exp(1)*exp(2)+(5*x^4-50*x^3+125*x^2)*exp(
1))/log(6),x, algorithm="fricas")

[Out]

4/5/((x*e^3 - (x^2 - 5*x)*e)*log(6))

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Sympy [A]
time = 0.30, size = 34, normalized size = 1.42 \begin {gather*} - \frac {4}{5 e x^{2} \log {\left (6 \right )} + x \left (- 5 e^{3} \log {\left (6 \right )} - 25 e \log {\left (6 \right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2)+8*x-20)/(5*x**2*exp(1)*exp(2)**2+(-10*x**3+50*x**2)*exp(1)*exp(2)+(5*x**4-50*x**3+125*x**
2)*exp(1))/ln(6),x)

[Out]

-4/(5*E*x**2*log(6) + x*(-5*exp(3)*log(6) - 25*E*log(6)))

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Giac [A]
time = 0.38, size = 22, normalized size = 0.92 \begin {gather*} -\frac {4 \, e^{\left (-1\right )}}{5 \, {\left (x^{2} - x e^{2} - 5 \, x\right )} \log \left (6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2)+8*x-20)/(5*x^2*exp(1)*exp(2)^2+(-10*x^3+50*x^2)*exp(1)*exp(2)+(5*x^4-50*x^3+125*x^2)*exp(
1))/log(6),x, algorithm="giac")

[Out]

-4/5*e^(-1)/((x^2 - x*e^2 - 5*x)*log(6))

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Mupad [B]
time = 0.20, size = 20, normalized size = 0.83 \begin {gather*} \frac {4\,{\mathrm {e}}^{-1}}{5\,x\,\ln \left (6\right )\,\left ({\mathrm {e}}^2-x+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(2) - 8*x + 20)/(log(6)*(exp(3)*(50*x^2 - 10*x^3) + 5*x^2*exp(5) + exp(1)*(125*x^2 - 50*x^3 + 5*x^4
))),x)

[Out]

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

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