3.80.17 \(\int \frac {270+90 e^{13/2}-90 x^2}{9+e^{13}+12 x-4 e^{39/4} x+10 x^2+4 x^3+x^4+e^{13/2} (6+4 x+6 x^2)+e^{13/4} (-12 x-8 x^2-4 x^3)} \, dx\)

Optimal. Leaf size=23 \[ 1+\frac {90 x}{3+2 x+\left (-e^{13/4}+x\right )^2} \]

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Rubi [A]  time = 0.12, antiderivative size = 27, normalized size of antiderivative = 1.17, number of steps used = 5, number of rules used = 5, integrand size = 82, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {6, 1680, 12, 1814, 8} \begin {gather*} \frac {90 x}{x^2+2 \left (1-e^{13/4}\right ) x+e^{13/2}+3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(270 + 90*E^(13/2) - 90*x^2)/(9 + E^13 + 12*x - 4*E^(39/4)*x + 10*x^2 + 4*x^3 + x^4 + E^(13/2)*(6 + 4*x +
6*x^2) + E^(13/4)*(-12*x - 8*x^2 - 4*x^3)),x]

[Out]

(90*x)/(3 + E^(13/2) + 2*(1 - E^(13/4))*x + 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 {270+90 e^{13/2}-90 x^2}{9+e^{13}+\left (12-4 e^{39/4}\right ) x+10 x^2+4 x^3+x^4+e^{13/2} \left (6+4 x+6 x^2\right )+e^{13/4} \left (-12 x-8 x^2-4 x^3\right )} \, dx\\ &=\operatorname {Subst}\left (\int \frac {90 \left (2 \left (1+e^{13/4}\right )+2 \left (1-e^{13/4}\right ) x-x^2\right )}{\left (2+2 e^{13/4}+x^2\right )^2} \, dx,x,\frac {1}{4} \left (4-4 e^{13/4}\right )+x\right )\\ &=90 \operatorname {Subst}\left (\int \frac {2 \left (1+e^{13/4}\right )+2 \left (1-e^{13/4}\right ) x-x^2}{\left (2+2 e^{13/4}+x^2\right )^2} \, dx,x,\frac {1}{4} \left (4-4 e^{13/4}\right )+x\right )\\ &=\frac {90 x}{3+e^{13/2}+2 \left (1-e^{13/4}\right ) x+x^2}-\frac {45 \operatorname {Subst}\left (\int 0 \, dx,x,\frac {1}{4} \left (4-4 e^{13/4}\right )+x\right )}{2 \left (1+e^{13/4}\right )}\\ &=\frac {90 x}{3+e^{13/2}+2 \left (1-e^{13/4}\right ) x+x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 26, normalized size = 1.13 \begin {gather*} \frac {90 x}{3+e^{13/2}+2 x-2 e^{13/4} x+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(270 + 90*E^(13/2) - 90*x^2)/(9 + E^13 + 12*x - 4*E^(39/4)*x + 10*x^2 + 4*x^3 + x^4 + E^(13/2)*(6 +
4*x + 6*x^2) + E^(13/4)*(-12*x - 8*x^2 - 4*x^3)),x]

[Out]

(90*x)/(3 + E^(13/2) + 2*x - 2*E^(13/4)*x + x^2)

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fricas [A]  time = 0.91, size = 20, normalized size = 0.87 \begin {gather*} \frac {90 \, x}{x^{2} - 2 \, x e^{\frac {13}{4}} + 2 \, x + e^{\frac {13}{2}} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*exp(13/4)^2-90*x^2+270)/(exp(13/4)^4-4*x*exp(13/4)^3+(6*x^2+4*x+6)*exp(13/4)^2+(-4*x^3-8*x^2-12*
x)*exp(13/4)+x^4+4*x^3+10*x^2+12*x+9),x, algorithm="fricas")

[Out]

90*x/(x^2 - 2*x*e^(13/4) + 2*x + e^(13/2) + 3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {90 \, {\left (x^{2} - e^{\frac {13}{2}} - 3\right )}}{x^{4} + 4 \, x^{3} + 10 \, x^{2} - 4 \, x e^{\frac {39}{4}} + 2 \, {\left (3 \, x^{2} + 2 \, x + 3\right )} e^{\frac {13}{2}} - 4 \, {\left (x^{3} + 2 \, x^{2} + 3 \, x\right )} e^{\frac {13}{4}} + 12 \, x + e^{13} + 9}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*exp(13/4)^2-90*x^2+270)/(exp(13/4)^4-4*x*exp(13/4)^3+(6*x^2+4*x+6)*exp(13/4)^2+(-4*x^3-8*x^2-12*
x)*exp(13/4)+x^4+4*x^3+10*x^2+12*x+9),x, algorithm="giac")

[Out]

integrate(-90*(x^2 - e^(13/2) - 3)/(x^4 + 4*x^3 + 10*x^2 - 4*x*e^(39/4) + 2*(3*x^2 + 2*x + 3)*e^(13/2) - 4*(x^
3 + 2*x^2 + 3*x)*e^(13/4) + 12*x + e^13 + 9), x)

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maple [A]  time = 0.15, size = 21, normalized size = 0.91

method	result	size
risch	\(\frac {90 x}{{\mathrm e}^{\frac {13}{2}}-2 \,{\mathrm e}^{\frac {13}{4}} x +x^{2}+2 x +3}\)	\(21\)
gosper	\(\frac {90 x}{{\mathrm e}^{\frac {13}{2}}-2 \,{\mathrm e}^{\frac {13}{4}} x +x^{2}+2 x +3}\)	\(23\)
norman	\(\frac {90 x}{{\mathrm e}^{\frac {13}{2}}-2 \,{\mathrm e}^{\frac {13}{4}} x +x^{2}+2 x +3}\)	\(23\)
default	\(-\frac {45 \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}+\left (-4 \,{\mathrm e}^{\frac {13}{4}}+4\right ) \textit {\_Z}^{3}+\left (-8 \,{\mathrm e}^{\frac {13}{4}}+6 \,{\mathrm e}^{\frac {13}{2}}+10\right ) \textit {\_Z}^{2}+\left (-12 \,{\mathrm e}^{\frac {13}{4}}+4 \,{\mathrm e}^{\frac {13}{2}}-4 \,{\mathrm e}^{\frac {39}{4}}+12\right ) \textit {\_Z} +6 \,{\mathrm e}^{\frac {13}{2}}+{\mathrm e}^{13}+9\right )}{\sum }\frac {\left ({\mathrm e}^{\frac {13}{2}}-\textit {\_R}^{2}+3\right ) \ln \left (x -\textit {\_R} \right )}{-3+3 \,{\mathrm e}^{\frac {13}{4}} \textit {\_R}^{2}-\textit {\_R}^{3}+4 \,{\mathrm e}^{\frac {13}{4}} \textit {\_R} -3 \,{\mathrm e}^{\frac {13}{2}} \textit {\_R} -3 \textit {\_R}^{2}+3 \,{\mathrm e}^{\frac {13}{4}}-{\mathrm e}^{\frac {13}{2}}+{\mathrm e}^{\frac {39}{4}}-5 \textit {\_R}}\right )}{2}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((90*exp(13/4)^2-90*x^2+270)/(exp(13/4)^4-4*x*exp(13/4)^3+(6*x^2+4*x+6)*exp(13/4)^2+(-4*x^3-8*x^2-12*x)*exp
(13/4)+x^4+4*x^3+10*x^2+12*x+9),x,method=_RETURNVERBOSE)

[Out]

90*x/(exp(13/2)-2*exp(13/4)*x+x^2+2*x+3)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -90 \, \int \frac {x^{2} - e^{\frac {13}{2}} - 3}{x^{4} + 4 \, x^{3} + 10 \, x^{2} - 4 \, x e^{\frac {39}{4}} + 2 \, {\left (3 \, x^{2} + 2 \, x + 3\right )} e^{\frac {13}{2}} - 4 \, {\left (x^{3} + 2 \, x^{2} + 3 \, x\right )} e^{\frac {13}{4}} + 12 \, x + e^{13} + 9}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*exp(13/4)^2-90*x^2+270)/(exp(13/4)^4-4*x*exp(13/4)^3+(6*x^2+4*x+6)*exp(13/4)^2+(-4*x^3-8*x^2-12*
x)*exp(13/4)+x^4+4*x^3+10*x^2+12*x+9),x, algorithm="maxima")

[Out]

-90*integrate((x^2 - e^(13/2) - 3)/(x^4 + 4*x^3 + 10*x^2 - 4*x*e^(39/4) + 2*(3*x^2 + 2*x + 3)*e^(13/2) - 4*(x^
3 + 2*x^2 + 3*x)*e^(13/4) + 12*x + e^13 + 9), x)

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mupad [B]  time = 4.89, size = 21, normalized size = 0.91 \begin {gather*} \frac {90\,x}{x^2+\left (2-2\,{\mathrm {e}}^{13/4}\right )\,x+{\mathrm {e}}^{13/2}+3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((90*exp(13/2) - 90*x^2 + 270)/(12*x + exp(13) + exp(13/2)*(4*x + 6*x^2 + 6) - 4*x*exp(39/4) - exp(13/4)*(1
2*x + 8*x^2 + 4*x^3) + 10*x^2 + 4*x^3 + x^4 + 9),x)

[Out]

(90*x)/(exp(13/2) + x^2 - x*(2*exp(13/4) - 2) + 3)

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sympy [B]  time = 1.17, size = 51, normalized size = 2.22 \begin {gather*} - \frac {x \left (- 90 e^{\frac {13}{4}} - 90\right )}{x^{2} \left (1 + e^{\frac {13}{4}}\right ) + x \left (2 - 2 e^{\frac {13}{2}}\right ) + 3 + 3 e^{\frac {13}{4}} + e^{\frac {13}{2}} + e^{\frac {39}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*exp(13/4)**2-90*x**2+270)/(exp(13/4)**4-4*x*exp(13/4)**3+(6*x**2+4*x+6)*exp(13/4)**2+(-4*x**3-8*
x**2-12*x)*exp(13/4)+x**4+4*x**3+10*x**2+12*x+9),x)

[Out]

-x*(-90*exp(13/4) - 90)/(x**2*(1 + exp(13/4)) + x*(2 - 2*exp(13/2)) + 3 + 3*exp(13/4) + exp(13/2) + exp(39/4))

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