3.100.3 \(\int \frac {90 e^{53}-120 x^2}{9 e^{106}+100 x^2+4 e^2 x^2+80 x^3+16 x^4+e^{53} (60 x+12 e x+24 x^2)+e (40 x^2+16 x^3)} \, dx\)

Optimal. Leaf size=20 \[ \frac {15}{5+e+\frac {3 e^{53}}{2 x}+2 x} \]

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Rubi [A]  time = 0.17, antiderivative size = 22, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6, 1680, 12, 1814, 8} \begin {gather*} \frac {30 x}{4 x^2+2 (5+e) x+3 e^{53}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(90*E^53 - 120*x^2)/(9*E^106 + 100*x^2 + 4*E^2*x^2 + 80*x^3 + 16*x^4 + E^53*(60*x + 12*E*x + 24*x^2) + E*(
40*x^2 + 16*x^3)),x]

[Out]

(30*x)/(3*E^53 + 2*(5 + E)*x + 4*x^2)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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 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} &=\int \frac {90 e^{53}-120 x^2}{9 e^{106}+\left (100+4 e^2\right ) x^2+80 x^3+16 x^4+e^{53} \left (60 x+12 e x+24 x^2\right )+e \left (40 x^2+16 x^3\right )} \, dx\\ &=\operatorname {Subst}\left (\int \frac {120 \left (-25-10 e-e^2+12 e^{53}+8 (5+e) x-16 x^2\right )}{\left (25+10 e+e^2-12 e^{53}-16 x^2\right )^2} \, dx,x,\frac {1}{64} (80+16 e)+x\right )\\ &=120 \operatorname {Subst}\left (\int \frac {-25-10 e-e^2+12 e^{53}+8 (5+e) x-16 x^2}{\left (25+10 e+e^2-12 e^{53}-16 x^2\right )^2} \, dx,x,\frac {1}{64} (80+16 e)+x\right )\\ &=\frac {30 x}{3 e^{53}+2 (5+e) x+4 x^2}-\frac {60 \operatorname {Subst}\left (\int 0 \, dx,x,\frac {1}{64} (80+16 e)+x\right )}{25+10 e+e^2-12 e^{53}}\\ &=\frac {30 x}{3 e^{53}+2 (5+e) x+4 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 23, normalized size = 1.15 \begin {gather*} \frac {30 x}{3 e^{53}+10 x+2 e x+4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(90*E^53 - 120*x^2)/(9*E^106 + 100*x^2 + 4*E^2*x^2 + 80*x^3 + 16*x^4 + E^53*(60*x + 12*E*x + 24*x^2)
+ E*(40*x^2 + 16*x^3)),x]

[Out]

(30*x)/(3*E^53 + 10*x + 2*E*x + 4*x^2)

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fricas [A]  time = 0.68, size = 23, normalized size = 1.15 \begin {gather*} \frac {30 \, x}{4 \, x^{2} + 2 \, x e + 10 \, x + 3 \, e^{53}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*exp(3)*exp(25)^2-120*x^2)/(9*exp(3)^2*exp(25)^4+(12*x*exp(1)+24*x^2+60*x)*exp(3)*exp(25)^2+4*x^2
*exp(1)^2+(16*x^3+40*x^2)*exp(1)+16*x^4+80*x^3+100*x^2),x, algorithm="fricas")

[Out]

30*x/(4*x^2 + 2*x*e + 10*x + 3*e^53)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {30 \, {\left (4 \, x^{2} - 3 \, e^{53}\right )}}{16 \, x^{4} + 80 \, x^{3} + 4 \, x^{2} e^{2} + 100 \, x^{2} + 12 \, {\left (2 \, x^{2} + x e + 5 \, x\right )} e^{53} + 8 \, {\left (2 \, x^{3} + 5 \, x^{2}\right )} e + 9 \, e^{106}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*exp(3)*exp(25)^2-120*x^2)/(9*exp(3)^2*exp(25)^4+(12*x*exp(1)+24*x^2+60*x)*exp(3)*exp(25)^2+4*x^2
*exp(1)^2+(16*x^3+40*x^2)*exp(1)+16*x^4+80*x^3+100*x^2),x, algorithm="giac")

[Out]

integrate(-30*(4*x^2 - 3*e^53)/(16*x^4 + 80*x^3 + 4*x^2*e^2 + 100*x^2 + 12*(2*x^2 + x*e + 5*x)*e^53 + 8*(2*x^3
+ 5*x^2)*e + 9*e^106), x)

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maple [A]  time = 0.18, size = 22, normalized size = 1.10

method	result	size
risch	\(\frac {10 x}{{\mathrm e}^{53}+\frac {2 x \,{\mathrm e}}{3}+\frac {4 x^{2}}{3}+\frac {10 x}{3}}\)	\(22\)
gosper	\(\frac {30 x}{3 \,{\mathrm e}^{3} {\mathrm e}^{50}+2 x \,{\mathrm e}+4 x^{2}+10 x}\)	\(28\)
norman	\(\frac {30 x}{3 \,{\mathrm e}^{3} {\mathrm e}^{50}+2 x \,{\mathrm e}+4 x^{2}+10 x}\)	\(28\)
default	\(\frac {15 \left (\munderset {\textit {\_R} =\RootOf \left (16 \textit {\_Z}^{4}+\left (16 \,{\mathrm e}+80\right ) \textit {\_Z}^{3}+\left (24 \,{\mathrm e}^{53}+4 \,{\mathrm e}^{2}+40 \,{\mathrm e}+100\right ) \textit {\_Z}^{2}+\left (12 \,{\mathrm e}^{54}+60 \,{\mathrm e}^{53}\right ) \textit {\_Z} +9 \,{\mathrm e}^{106}\right )}{\sum }\frac {\left (3 \,{\mathrm e}^{53}-4 \textit {\_R}^{2}\right ) \ln \left (x -\textit {\_R} \right )}{3 \,{\mathrm e}^{54}+12 \textit {\_R} \,{\mathrm e}^{53}+2 \,{\mathrm e}^{2} \textit {\_R} +12 \textit {\_R}^{2} {\mathrm e}+15 \,{\mathrm e}^{53}+16 \textit {\_R}^{3}+20 \textit {\_R} \,{\mathrm e}+60 \textit {\_R}^{2}+50 \textit {\_R}}\right )}{2}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((90*exp(3)*exp(25)^2-120*x^2)/(9*exp(3)^2*exp(25)^4+(12*x*exp(1)+24*x^2+60*x)*exp(3)*exp(25)^2+4*x^2*exp(1
)^2+(16*x^3+40*x^2)*exp(1)+16*x^4+80*x^3+100*x^2),x,method=_RETURNVERBOSE)

[Out]

10*x/(exp(53)+2/3*x*exp(1)+4/3*x^2+10/3*x)

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maxima [A]  time = 0.36, size = 22, normalized size = 1.10 \begin {gather*} \frac {30 \, x}{4 \, x^{2} + 2 \, x {\left (e + 5\right )} + 3 \, e^{53}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*exp(3)*exp(25)^2-120*x^2)/(9*exp(3)^2*exp(25)^4+(12*x*exp(1)+24*x^2+60*x)*exp(3)*exp(25)^2+4*x^2
*exp(1)^2+(16*x^3+40*x^2)*exp(1)+16*x^4+80*x^3+100*x^2),x, algorithm="maxima")

[Out]

30*x/(4*x^2 + 2*x*(e + 5) + 3*e^53)

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mupad [B]  time = 0.12, size = 23, normalized size = 1.15 \begin {gather*} \frac {30\,x}{4\,x^2+\left (2\,\mathrm {e}+10\right )\,x+3\,{\mathrm {e}}^{53}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((90*exp(53) - 120*x^2)/(9*exp(106) + exp(53)*(60*x + 12*x*exp(1) + 24*x^2) + exp(1)*(40*x^2 + 16*x^3) + 4*
x^2*exp(2) + 100*x^2 + 80*x^3 + 16*x^4),x)

[Out]

(30*x)/(3*exp(53) + 4*x^2 + x*(2*exp(1) + 10))

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sympy [A]  time = 0.62, size = 20, normalized size = 1.00 \begin {gather*} \frac {30 x}{4 x^{2} + x \left (2 e + 10\right ) + 3 e^{53}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*exp(3)*exp(25)**2-120*x**2)/(9*exp(3)**2*exp(25)**4+(12*x*exp(1)+24*x**2+60*x)*exp(3)*exp(25)**2
+4*x**2*exp(1)**2+(16*x**3+40*x**2)*exp(1)+16*x**4+80*x**3+100*x**2),x)

[Out]

30*x/(4*x**2 + x*(2*E + 10) + 3*exp(53))

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