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

Optimal. Leaf size=86 \[ \frac{6055-28981 x}{3174 \sqrt{2 x^2-x+3}}+\frac{5}{4} \sqrt{2 x^2-x+3}-\frac{53-373 x}{69 \left (2 x^2-x+3\right )^{3/2}}-\frac{71 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{8 \sqrt{2}} \]

[Out]

-(53 - 373*x)/(69*(3 - x + 2*x^2)^(3/2)) + (6055 - 28981*x)/(3174*Sqrt[3 - x + 2*x^2]) + (5*Sqrt[3 - x + 2*x^2
])/4 - (71*ArcSinh[(1 - 4*x)/Sqrt[23]])/(8*Sqrt[2])

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Rubi [A]  time = 0.0822592, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {1660, 640, 619, 215} \[ \frac{6055-28981 x}{3174 \sqrt{2 x^2-x+3}}+\frac{5}{4} \sqrt{2 x^2-x+3}-\frac{53-373 x}{69 \left (2 x^2-x+3\right )^{3/2}}-\frac{71 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{8 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(53 - 373*x)/(69*(3 - x + 2*x^2)^(3/2)) + (6055 - 28981*x)/(3174*Sqrt[3 - x + 2*x^2]) + (5*Sqrt[3 - x + 2*x^2
])/4 - (71*ArcSinh[(1 - 4*x)/Sqrt[23]])/(8*Sqrt[2])

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
]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

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 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]

Rubi steps

\begin{align*} \int \frac{(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx &=-\frac{53-373 x}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac{2}{69} \int \frac{-\frac{233}{4}+483 x^2+\frac{345 x^3}{2}}{\left (3-x+2 x^2\right )^{3/2}} \, dx\\ &=-\frac{53-373 x}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac{6055-28981 x}{3174 \sqrt{3-x+2 x^2}}+\frac{4 \int \frac{\frac{52371}{16}+\frac{7935 x}{8}}{\sqrt{3-x+2 x^2}} \, dx}{1587}\\ &=-\frac{53-373 x}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac{6055-28981 x}{3174 \sqrt{3-x+2 x^2}}+\frac{5}{4} \sqrt{3-x+2 x^2}+\frac{71}{8} \int \frac{1}{\sqrt{3-x+2 x^2}} \, dx\\ &=-\frac{53-373 x}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac{6055-28981 x}{3174 \sqrt{3-x+2 x^2}}+\frac{5}{4} \sqrt{3-x+2 x^2}+\frac{71 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+4 x\right )}{8 \sqrt{46}}\\ &=-\frac{53-373 x}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac{6055-28981 x}{3174 \sqrt{3-x+2 x^2}}+\frac{5}{4} \sqrt{3-x+2 x^2}-\frac{71 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{8 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.259808, size = 60, normalized size = 0.7 \[ \frac{31740 x^4-147664 x^3+185337 x^2-199290 x+102869}{6348 \left (2 x^2-x+3\right )^{3/2}}+\frac{71 \sinh ^{-1}\left (\frac{4 x-1}{\sqrt{23}}\right )}{8 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(102869 - 199290*x + 185337*x^2 - 147664*x^3 + 31740*x^4)/(6348*(3 - x + 2*x^2)^(3/2)) + (71*ArcSinh[(-1 + 4*x
)/Sqrt[23]])/(8*Sqrt[2])

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Maple [B]  time = 0.055, size = 163, normalized size = 1.9 \begin{align*} -{\frac{71\,{x}^{3}}{12} \left ( 2\,{x}^{2}-x+3 \right ) ^{-{\frac{3}{2}}}}+{\frac{401\,{x}^{2}}{16} \left ( 2\,{x}^{2}-x+3 \right ) ^{-{\frac{3}{2}}}}+{\frac{71\,\sqrt{2}}{16}{\it Arcsinh} \left ({\frac{4\,\sqrt{23}}{23} \left ( x-{\frac{1}{4}} \right ) } \right ) }+{\frac{-643+2572\,x}{12696}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}-{\frac{-2327+9308\,x}{35328} \left ( 2\,{x}^{2}-x+3 \right ) ^{-{\frac{3}{2}}}}-{\frac{945\,x}{128} \left ( 2\,{x}^{2}-x+3 \right ) ^{-{\frac{3}{2}}}}+5\,{\frac{{x}^{4}}{ \left ( 2\,{x}^{2}-x+3 \right ) ^{3/2}}}-{\frac{71\,x}{8}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}-{\frac{71}{32}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}+{\frac{11749}{512} \left ( 2\,{x}^{2}-x+3 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-71/12*x^3/(2*x^2-x+3)^(3/2)+401/16*x^2/(2*x^2-x+3)^(3/2)+71/16*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))+643/126
96*(-1+4*x)/(2*x^2-x+3)^(1/2)-2327/35328*(-1+4*x)/(2*x^2-x+3)^(3/2)-945/128*x/(2*x^2-x+3)^(3/2)+5*x^4/(2*x^2-x
+3)^(3/2)-71/8*x/(2*x^2-x+3)^(1/2)-71/32/(2*x^2-x+3)^(1/2)+11749/512/(2*x^2-x+3)^(3/2)

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Maxima [B]  time = 1.50182, size = 273, normalized size = 3.17 \begin{align*} \frac{5 \, x^{4}}{{\left (2 \, x^{2} - x + 3\right )}^{\frac{3}{2}}} + \frac{71}{12696} \, 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{71}{16} \, \sqrt{2} \operatorname{arsinh}\left (\frac{1}{23} \, \sqrt{23}{\left (4 \, x - 1\right )}\right ) - \frac{5041}{6348} \, \sqrt{2 \, x^{2} - x + 3} - \frac{10007 \, x}{3174 \, \sqrt{2 \, x^{2} - x + 3}} + \frac{59 \, x^{2}}{2 \,{\left (2 \, x^{2} - x + 3\right )}^{\frac{3}{2}}} - \frac{2959}{2116 \, \sqrt{2 \, x^{2} - x + 3}} - \frac{807 \, x}{92 \,{\left (2 \, x^{2} - x + 3\right )}^{\frac{3}{2}}} + \frac{7603}{276 \,{\left (2 \, x^{2} - x + 3\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*x^4/(2*x^2 - x + 3)^(3/2) + 71/12696*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)) + 71/16*sqrt(2)*arcsinh(1/23*sqrt(
23)*(4*x - 1)) - 5041/6348*sqrt(2*x^2 - x + 3) - 10007/3174*x/sqrt(2*x^2 - x + 3) + 59/2*x^2/(2*x^2 - x + 3)^(
3/2) - 2959/2116/sqrt(2*x^2 - x + 3) - 807/92*x/(2*x^2 - x + 3)^(3/2) + 7603/276/(2*x^2 - x + 3)^(3/2)

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Fricas [A]  time = 1.34327, size = 331, normalized size = 3.85 \begin{align*} \frac{112677 \, \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 (31740 \, x^{4} - 147664 \, x^{3} + 185337 \, x^{2} - 199290 \, x + 102869\right )} \sqrt{2 \, x^{2} - x + 3}}{50784 \,{\left (4 \, x^{4} - 4 \, x^{3} + 13 \, x^{2} - 6 \, x + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/50784*(112677*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*(31740*x^4 - 147664*x^3 + 185337*x^2 - 199290*x + 102869)*sqrt(2*x^2 - x + 3))/(4*x^4 - 4*
x^3 + 13*x^2 - 6*x + 9)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (2 x + 5\right ) \left (5 x^{4} - x^{3} + 3 x^{2} + x + 2\right )}{\left (2 x^{2} - x + 3\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17698, size = 89, normalized size = 1.03 \begin{align*} -\frac{71}{16} \, \sqrt{2} \log \left (-2 \, \sqrt{2}{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )} + 1\right ) + \frac{{\left ({\left (4 \,{\left (7935 \, x - 36916\right )} x + 185337\right )} x - 199290\right )} x + 102869}{6348 \,{\left (2 \, x^{2} - x + 3\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-71/16*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) + 1/6348*(((4*(7935*x - 36916)*x + 185337
)*x - 199290)*x + 102869)/(2*x^2 - x + 3)^(3/2)