3.360 \(\int \frac {2+x+3 x^2-x^3+5 x^4}{(3-x+2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=68 \[ \frac {219 x+89}{276 \left (2 x^2-x+3\right )^{3/2}}-\frac {2604 x+1465}{2116 \sqrt {2 x^2-x+3}}-\frac {5 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{4 \sqrt {2}} \]

[Out]

1/276*(89+219*x)/(2*x^2-x+3)^(3/2)-5/8*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)+1/2116*(-1465-2604*x)/(2*x^2-x+3
)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1660, 12, 619, 215} \[ \frac {219 x+89}{276 \left (2 x^2-x+3\right )^{3/2}}-\frac {2604 x+1465}{2116 \sqrt {2 x^2-x+3}}-\frac {5 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{4 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + x + 3*x^2 - x^3 + 5*x^4)/(3 - x + 2*x^2)^(5/2),x]

[Out]

(89 + 219*x)/(276*(3 - x + 2*x^2)^(3/2)) - (1465 + 2604*x)/(2116*Sqrt[3 - x + 2*x^2]) - (5*ArcSinh[(1 - 4*x)/S
qrt[23]])/(4*Sqrt[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 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rubi steps

\begin {align*} \int \frac {2+x+3 x^2-x^3+5 x^4}{\left (3-x+2 x^2\right )^{5/2}} \, dx &=\frac {89+219 x}{276 \left (3-x+2 x^2\right )^{3/2}}+\frac {2}{69} \int \frac {-\frac {159}{16}+\frac {207 x}{8}+\frac {345 x^2}{4}}{\left (3-x+2 x^2\right )^{3/2}} \, dx\\ &=\frac {89+219 x}{276 \left (3-x+2 x^2\right )^{3/2}}-\frac {1465+2604 x}{2116 \sqrt {3-x+2 x^2}}+\frac {4 \int \frac {7935}{16 \sqrt {3-x+2 x^2}} \, dx}{1587}\\ &=\frac {89+219 x}{276 \left (3-x+2 x^2\right )^{3/2}}-\frac {1465+2604 x}{2116 \sqrt {3-x+2 x^2}}+\frac {5}{4} \int \frac {1}{\sqrt {3-x+2 x^2}} \, dx\\ &=\frac {89+219 x}{276 \left (3-x+2 x^2\right )^{3/2}}-\frac {1465+2604 x}{2116 \sqrt {3-x+2 x^2}}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+4 x\right )}{4 \sqrt {46}}\\ &=\frac {89+219 x}{276 \left (3-x+2 x^2\right )^{3/2}}-\frac {1465+2604 x}{2116 \sqrt {3-x+2 x^2}}-\frac {5 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{4 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 55, normalized size = 0.81 \[ \frac {5 \sinh ^{-1}\left (\frac {4 x-1}{\sqrt {23}}\right )}{4 \sqrt {2}}-\frac {7812 x^3+489 x^2+7002 x+5569}{3174 \left (2 x^2-x+3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + x + 3*x^2 - x^3 + 5*x^4)/(3 - x + 2*x^2)^(5/2),x]

[Out]

-1/3174*(5569 + 7002*x + 489*x^2 + 7812*x^3)/(3 - x + 2*x^2)^(3/2) + (5*ArcSinh[(-1 + 4*x)/Sqrt[23]])/(4*Sqrt[
2])

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fricas [B]  time = 0.89, size = 112, normalized size = 1.65 \[ \frac {7935 \, \sqrt {2} {\left (4 \, x^{4} - 4 \, x^{3} + 13 \, x^{2} - 6 \, x + 9\right )} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) - 8 \, {\left (7812 \, x^{3} + 489 \, x^{2} + 7002 \, x + 5569\right )} \sqrt {2 \, x^{2} - x + 3}}{25392 \, {\left (4 \, x^{4} - 4 \, x^{3} + 13 \, x^{2} - 6 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^4-x^3+3*x^2+x+2)/(2*x^2-x+3)^(5/2),x, algorithm="fricas")

[Out]

1/25392*(7935*sqrt(2)*(4*x^4 - 4*x^3 + 13*x^2 - 6*x + 9)*log(-4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2
+ 16*x - 25) - 8*(7812*x^3 + 489*x^2 + 7002*x + 5569)*sqrt(2*x^2 - x + 3))/(4*x^4 - 4*x^3 + 13*x^2 - 6*x + 9)

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giac [A]  time = 0.30, size = 62, normalized size = 0.91 \[ -\frac {5}{8} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) - \frac {3 \, {\left ({\left (2604 \, x + 163\right )} x + 2334\right )} x + 5569}{3174 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^4-x^3+3*x^2+x+2)/(2*x^2-x+3)^(5/2),x, algorithm="giac")

[Out]

-5/8*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) - 1/3174*(3*((2604*x + 163)*x + 2334)*x + 5
569)/(2*x^2 - x + 3)^(3/2)

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maple [B]  time = 0.01, size = 146, normalized size = 2.15 \[ -\frac {5 x^{3}}{6 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}-\frac {x^{2}}{8 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}-\frac {47 x}{64 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}-\frac {5 x}{4 \sqrt {2 x^{2}-x +3}}+\frac {5 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{8}-\frac {271}{768 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}+\frac {\frac {2423 x}{4416}-\frac {2423}{17664}}{\left (2 x^{2}-x +3\right )^{\frac {3}{2}}}+\frac {\frac {692 x}{1587}-\frac {173}{1587}}{\sqrt {2 x^{2}-x +3}}-\frac {5}{16 \sqrt {2 x^{2}-x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^4-x^3+3*x^2+x+2)/(2*x^2-x+3)^(5/2),x)

[Out]

-5/6/(2*x^2-x+3)^(3/2)*x^3-1/8/(2*x^2-x+3)^(3/2)*x^2-47/64/(2*x^2-x+3)^(3/2)*x-271/768/(2*x^2-x+3)^(3/2)+2423/
17664*(4*x-1)/(2*x^2-x+3)^(3/2)+173/1587*(4*x-1)/(2*x^2-x+3)^(1/2)-5/4/(2*x^2-x+3)^(1/2)*x-5/16/(2*x^2-x+3)^(1
/2)+5/8*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))

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maxima [B]  time = 0.98, size = 185, normalized size = 2.72 \[ \frac {5}{6348} \, x {\left (\frac {284 \, x}{\sqrt {2 \, x^{2} - x + 3}} - \frac {3174 \, x^{2}}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {71}{\sqrt {2 \, x^{2} - x + 3}} + \frac {805 \, x}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {3243}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}\right )} + \frac {5}{8} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {355}{3174} \, \sqrt {2 \, x^{2} - x + 3} - \frac {58 \, x}{1587 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {x^{2}}{2 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {1897}{6348 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {95 \, x}{276 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {41}{276 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^4-x^3+3*x^2+x+2)/(2*x^2-x+3)^(5/2),x, algorithm="maxima")

[Out]

5/6348*x*(284*x/sqrt(2*x^2 - x + 3) - 3174*x^2/(2*x^2 - x + 3)^(3/2) - 71/sqrt(2*x^2 - x + 3) + 805*x/(2*x^2 -
x + 3)^(3/2) - 3243/(2*x^2 - x + 3)^(3/2)) + 5/8*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) - 355/3174*sqrt(2*x
^2 - x + 3) - 58/1587*x/sqrt(2*x^2 - x + 3) + 1/2*x^2/(2*x^2 - x + 3)^(3/2) - 1897/6348/sqrt(2*x^2 - x + 3) -
95/276*x/(2*x^2 - x + 3)^(3/2) + 41/276/(2*x^2 - x + 3)^(3/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {5\,x^4-x^3+3\,x^2+x+2}{{\left (2\,x^2-x+3\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 3*x^2 - x^3 + 5*x^4 + 2)/(2*x^2 - x + 3)^(5/2),x)

[Out]

int((x + 3*x^2 - x^3 + 5*x^4 + 2)/(2*x^2 - x + 3)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {5 x^{4} - x^{3} + 3 x^{2} + x + 2}{\left (2 x^{2} - x + 3\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**4-x**3+3*x**2+x+2)/(2*x**2-x+3)**(5/2),x)

[Out]

Integral((5*x**4 - x**3 + 3*x**2 + x + 2)/(2*x**2 - x + 3)**(5/2), x)

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