3.2.59 \(\int \frac {(-120 x^2+240 x^3-40 x^4+e (24-24 x^2-16 x^3)) \log (2)}{225 x^2-150 x^3+25 x^4+e^2 (1+2 x+x^2)+e (-30 x-20 x^2+10 x^3)} \, dx\)

Optimal. Leaf size=30 \[ \frac {8 x \left (3-x^2\right ) \log (2)}{e-(-e+5 (3-x)) x} \]

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Rubi [A]  time = 0.23, antiderivative size = 52, normalized size of antiderivative = 1.73, number of steps used = 6, number of rules used = 5, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {12, 1680, 1814, 21, 8} \begin {gather*} \frac {8 ((15-e) e-(5-e) (30-e) x) \log (2)}{25 \left (5 x^2-(15-e) x+e\right )}-\frac {8}{5} x \log (2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-120*x^2 + 240*x^3 - 40*x^4 + E*(24 - 24*x^2 - 16*x^3))*Log[2])/(225*x^2 - 150*x^3 + 25*x^4 + E^2*(1 + 2
*x + x^2) + E*(-30*x - 20*x^2 + 10*x^3)),x]

[Out]

(-8*x*Log[2])/5 + (8*((15 - E)*E - (5 - E)*(30 - E)*x)*Log[2])/(25*(E - (15 - E)*x + 5*x^2))

Rule 8

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

Rule 12

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 1680

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]

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} &=\log (2) \int \frac {-120 x^2+240 x^3-40 x^4+e \left (24-24 x^2-16 x^3\right )}{225 x^2-150 x^3+25 x^4+e^2 \left (1+2 x+x^2\right )+e \left (-30 x-20 x^2+10 x^3\right )} \, dx\\ &=\log (2) \operatorname {Subst}\left (\int \frac {8 \left (3 (5-e)^2 (25-e) (45-e)+80 (15-e) \left (150-45 e+e^2\right ) x+600 (5-e) (35-e) x^2-10000 x^4\right )}{5 \left ((5-e) (45-e)-100 x^2\right )^2} \, dx,x,\frac {1}{100} (-150+10 e)+x\right )\\ &=\frac {1}{5} (8 \log (2)) \operatorname {Subst}\left (\int \frac {3 (5-e)^2 (25-e) (45-e)+80 (15-e) \left (150-45 e+e^2\right ) x+600 (5-e) (35-e) x^2-10000 x^4}{\left ((5-e) (45-e)-100 x^2\right )^2} \, dx,x,\frac {1}{100} (-150+10 e)+x\right )\\ &=\frac {8 ((15-e) e-(5-e) (30-e) x) \log (2)}{25 \left (e-(15-e) x+5 x^2\right )}-\frac {(4 \log (2)) \operatorname {Subst}\left (\int \frac {2 \left (225-50 e+e^2\right )^2-200 (5-e) (45-e) x^2}{(5-e) (45-e)-100 x^2} \, dx,x,\frac {1}{100} (-150+10 e)+x\right )}{5 \left (225-50 e+e^2\right )}\\ &=\frac {8 ((15-e) e-(5-e) (30-e) x) \log (2)}{25 \left (e-(15-e) x+5 x^2\right )}-\frac {1}{5} (8 \log (2)) \operatorname {Subst}\left (\int 1 \, dx,x,\frac {1}{100} (-150+10 e)+x\right )\\ &=-\frac {8}{5} x \log (2)+\frac {8 ((15-e) e-(5-e) (30-e) x) \log (2)}{25 \left (e-(15-e) x+5 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 48, normalized size = 1.60 \begin {gather*} -8 \left (\frac {x}{5}+\frac {-15 e+e^2+150 x-35 e x+e^2 x}{25 \left (e-15 x+e x+5 x^2\right )}\right ) \log (2) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-120*x^2 + 240*x^3 - 40*x^4 + E*(24 - 24*x^2 - 16*x^3))*Log[2])/(225*x^2 - 150*x^3 + 25*x^4 + E^2*
(1 + 2*x + x^2) + E*(-30*x - 20*x^2 + 10*x^3)),x]

[Out]

-8*(x/5 + (-15*E + E^2 + 150*x - 35*E*x + E^2*x)/(25*(E - 15*x + E*x + 5*x^2)))*Log[2]

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fricas [B]  time = 0.78, size = 53, normalized size = 1.77 \begin {gather*} -\frac {8 \, {\left (25 \, x^{3} - 75 \, x^{2} + {\left (x + 1\right )} e^{2} + 5 \, {\left (x^{2} - 6 \, x - 3\right )} e + 150 \, x\right )} \log \relax (2)}{25 \, {\left (5 \, x^{2} + {\left (x + 1\right )} e - 15 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^3-24*x^2+24)*exp(1)-40*x^4+240*x^3-120*x^2)*log(2)/((x^2+2*x+1)*exp(1)^2+(10*x^3-20*x^2-30*x
)*exp(1)+25*x^4-150*x^3+225*x^2),x, algorithm="fricas")

[Out]

-8/25*(25*x^3 - 75*x^2 + (x + 1)*e^2 + 5*(x^2 - 6*x - 3)*e + 150*x)*log(2)/(5*x^2 + (x + 1)*e - 15*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {8 \, {\left (5 \, x^{4} - 30 \, x^{3} + 15 \, x^{2} + {\left (2 \, x^{3} + 3 \, x^{2} - 3\right )} e\right )} \log \relax (2)}{25 \, x^{4} - 150 \, x^{3} + 225 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{2} + 10 \, {\left (x^{3} - 2 \, x^{2} - 3 \, x\right )} e}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^3-24*x^2+24)*exp(1)-40*x^4+240*x^3-120*x^2)*log(2)/((x^2+2*x+1)*exp(1)^2+(10*x^3-20*x^2-30*x
)*exp(1)+25*x^4-150*x^3+225*x^2),x, algorithm="giac")

[Out]

integrate(-8*(5*x^4 - 30*x^3 + 15*x^2 + (2*x^3 + 3*x^2 - 3)*e)*log(2)/(25*x^4 - 150*x^3 + 225*x^2 + (x^2 + 2*x
+ 1)*e^2 + 10*(x^3 - 2*x^2 - 3*x)*e), x)

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maple [A]  time = 0.14, size = 28, normalized size = 0.93

method	result	size
gosper	\(-\frac {8 x \left (x^{2}-3\right ) \ln \relax (2)}{x \,{\mathrm e}+5 x^{2}+{\mathrm e}-15 x}\)	\(28\)
norman	\(\frac {24 x \ln \relax (2)-8 x^{3} \ln \relax (2)}{x \,{\mathrm e}+5 x^{2}+{\mathrm e}-15 x}\)	\(32\)
risch	\(-\frac {8 x \ln \relax (2)}{5}+\frac {\ln \relax (2) \left (\frac {\left (-\frac {8 \,{\mathrm e}^{2}}{5}+56 \,{\mathrm e}-240\right ) x}{5}-\frac {8 \,{\mathrm e}^{2}}{25}+\frac {24 \,{\mathrm e}}{5}\right )}{x \,{\mathrm e}+5 x^{2}+{\mathrm e}-15 x}\)	\(49\)
default	\(\ln \relax (2) \left (-\frac {8 x}{5}+\frac {4 \left (\munderset {\textit {\_R} =\RootOf \left (25 \textit {\_Z}^{4}+\left (10 \,{\mathrm e}-150\right ) \textit {\_Z}^{3}+\left ({\mathrm e}^{2}-20 \,{\mathrm e}+225\right ) \textit {\_Z}^{2}+\left (2 \,{\mathrm e}^{2}-30 \,{\mathrm e}\right ) \textit {\_Z} +{\mathrm e}^{2}\right )}{\sum }\frac {\left (\left ({\mathrm e}^{2}-35 \,{\mathrm e}+150\right ) \textit {\_R}^{2}+2 \left ({\mathrm e}^{2}-15 \,{\mathrm e}\right ) \textit {\_R} +{\mathrm e}^{2}+15 \,{\mathrm e}\right ) \ln \left (x -\textit {\_R} \right )}{{\mathrm e}^{2} \textit {\_R} +15 \textit {\_R}^{2} {\mathrm e}+50 \textit {\_R}^{3}+{\mathrm e}^{2}-20 \textit {\_R} \,{\mathrm e}-225 \textit {\_R}^{2}-15 \,{\mathrm e}+225 \textit {\_R}}\right )}{5}\right )\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-16*x^3-24*x^2+24)*exp(1)-40*x^4+240*x^3-120*x^2)*ln(2)/((x^2+2*x+1)*exp(1)^2+(10*x^3-20*x^2-30*x)*exp(1
)+25*x^4-150*x^3+225*x^2),x,method=_RETURNVERBOSE)

[Out]

-8*x*(x^2-3)*ln(2)/(x*exp(1)+5*x^2+exp(1)-15*x)

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maxima [A]  time = 0.36, size = 42, normalized size = 1.40 \begin {gather*} -\frac {8}{25} \, {\left (5 \, x + \frac {x {\left (e^{2} - 35 \, e + 150\right )} + e^{2} - 15 \, e}{5 \, x^{2} + x {\left (e - 15\right )} + e}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^3-24*x^2+24)*exp(1)-40*x^4+240*x^3-120*x^2)*log(2)/((x^2+2*x+1)*exp(1)^2+(10*x^3-20*x^2-30*x
)*exp(1)+25*x^4-150*x^3+225*x^2),x, algorithm="maxima")

[Out]

-8/25*(5*x + (x*(e^2 - 35*e + 150) + e^2 - 15*e)/(5*x^2 + x*(e - 15) + e))*log(2)

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mupad [B]  time = 0.24, size = 60, normalized size = 2.00 \begin {gather*} -\frac {\frac {8\,{\mathrm {e}}^2\,\ln \relax (2)}{5}-24\,\mathrm {e}\,\ln \relax (2)+x\,\left (240\,\ln \relax (2)-56\,\mathrm {e}\,\ln \relax (2)+\frac {8\,{\mathrm {e}}^2\,\ln \relax (2)}{5}\right )}{25\,x^2+\left (5\,\mathrm {e}-75\right )\,x+5\,\mathrm {e}}-\frac {8\,x\,\ln \relax (2)}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2)*(exp(1)*(24*x^2 + 16*x^3 - 24) + 120*x^2 - 240*x^3 + 40*x^4))/(exp(2)*(2*x + x^2 + 1) - exp(1)*(3
0*x + 20*x^2 - 10*x^3) + 225*x^2 - 150*x^3 + 25*x^4),x)

[Out]

- ((8*exp(2)*log(2))/5 - 24*exp(1)*log(2) + x*(240*log(2) - 56*exp(1)*log(2) + (8*exp(2)*log(2))/5))/(5*exp(1)
+ 25*x^2 + x*(5*exp(1) - 75)) - (8*x*log(2))/5

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sympy [B]  time = 1.06, size = 68, normalized size = 2.27 \begin {gather*} - \frac {8 x \log {\relax (2 )}}{5} - \frac {x \left (- 280 e \log {\relax (2 )} + 8 e^{2} \log {\relax (2 )} + 1200 \log {\relax (2 )}\right ) - 120 e \log {\relax (2 )} + 8 e^{2} \log {\relax (2 )}}{125 x^{2} + x \left (-375 + 25 e\right ) + 25 e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x**3-24*x**2+24)*exp(1)-40*x**4+240*x**3-120*x**2)*ln(2)/((x**2+2*x+1)*exp(1)**2+(10*x**3-20*x
**2-30*x)*exp(1)+25*x**4-150*x**3+225*x**2),x)

[Out]

-8*x*log(2)/5 - (x*(-280*E*log(2) + 8*exp(2)*log(2) + 1200*log(2)) - 120*E*log(2) + 8*exp(2)*log(2))/(125*x**2
+ x*(-375 + 25*E) + 25*E)

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