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

Optimal. Leaf size=120 \[ \frac {1}{16} \sqrt {2 x^2-x+3} (2 x+5)^4-\frac {105}{128} \sqrt {2 x^2-x+3} (2 x+5)^3+\frac {761}{256} \sqrt {2 x^2-x+3} (2 x+5)^2-\frac {(4676 x+19227) \sqrt {2 x^2-x+3}}{2048}-\frac {85429 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{4096 \sqrt {2}} \]

[Out]

-85429/8192*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)+761/256*(5+2*x)^2*(2*x^2-x+3)^(1/2)-105/128*(5+2*x)^3*(2*x^
2-x+3)^(1/2)+1/16*(5+2*x)^4*(2*x^2-x+3)^(1/2)-1/2048*(19227+4676*x)*(2*x^2-x+3)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1653, 779, 619, 215} \[ \frac {1}{16} \sqrt {2 x^2-x+3} (2 x+5)^4-\frac {105}{128} \sqrt {2 x^2-x+3} (2 x+5)^3+\frac {761}{256} \sqrt {2 x^2-x+3} (2 x+5)^2-\frac {(4676 x+19227) \sqrt {2 x^2-x+3}}{2048}-\frac {85429 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{4096 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(761*(5 + 2*x)^2*Sqrt[3 - x + 2*x^2])/256 - (105*(5 + 2*x)^3*Sqrt[3 - x + 2*x^2])/128 + ((5 + 2*x)^4*Sqrt[3 -
x + 2*x^2])/16 - ((19227 + 4676*x)*Sqrt[3 - x + 2*x^2])/2048 - (85429*ArcSinh[(1 - 4*x)/Sqrt[23]])/(4096*Sqrt[
2])

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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rubi steps

\begin {align*} \int \frac {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\sqrt {3-x+2 x^2}} \, dx &=\frac {1}{16} (5+2 x)^4 \sqrt {3-x+2 x^2}+\frac {1}{160} \int \frac {(5+2 x) \left (-5055-4390 x-5580 x^2-4200 x^3\right )}{\sqrt {3-x+2 x^2}} \, dx\\ &=-\frac {105}{128} (5+2 x)^3 \sqrt {3-x+2 x^2}+\frac {1}{16} (5+2 x)^4 \sqrt {3-x+2 x^2}+\frac {\int \frac {(5+2 x) \left (327480+105440 x+365280 x^2\right )}{\sqrt {3-x+2 x^2}} \, dx}{10240}\\ &=\frac {761}{256} (5+2 x)^2 \sqrt {3-x+2 x^2}-\frac {105}{128} (5+2 x)^3 \sqrt {3-x+2 x^2}+\frac {1}{16} (5+2 x)^4 \sqrt {3-x+2 x^2}+\frac {\int \frac {(919200-1122240 x) (5+2 x)}{\sqrt {3-x+2 x^2}} \, dx}{245760}\\ &=\frac {761}{256} (5+2 x)^2 \sqrt {3-x+2 x^2}-\frac {105}{128} (5+2 x)^3 \sqrt {3-x+2 x^2}+\frac {1}{16} (5+2 x)^4 \sqrt {3-x+2 x^2}-\frac {(19227+4676 x) \sqrt {3-x+2 x^2}}{2048}+\frac {85429 \int \frac {1}{\sqrt {3-x+2 x^2}} \, dx}{4096}\\ &=\frac {761}{256} (5+2 x)^2 \sqrt {3-x+2 x^2}-\frac {105}{128} (5+2 x)^3 \sqrt {3-x+2 x^2}+\frac {1}{16} (5+2 x)^4 \sqrt {3-x+2 x^2}-\frac {(19227+4676 x) \sqrt {3-x+2 x^2}}{2048}+\frac {85429 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+4 x\right )}{4096 \sqrt {46}}\\ &=\frac {761}{256} (5+2 x)^2 \sqrt {3-x+2 x^2}-\frac {105}{128} (5+2 x)^3 \sqrt {3-x+2 x^2}+\frac {1}{16} (5+2 x)^4 \sqrt {3-x+2 x^2}-\frac {(19227+4676 x) \sqrt {3-x+2 x^2}}{2048}-\frac {85429 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{4096 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 60, normalized size = 0.50 \[ \frac {4 \sqrt {2 x^2-x+3} \left (2048 x^4+7040 x^3+352 x^2-6916 x+2973\right )-85429 \sqrt {2} \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{8192} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[3 - x + 2*x^2]*(2973 - 6916*x + 352*x^2 + 7040*x^3 + 2048*x^4) - 85429*Sqrt[2]*ArcSinh[(1 - 4*x)/Sqrt[
23]])/8192

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fricas [A]  time = 0.88, size = 73, normalized size = 0.61 \[ \frac {1}{2048} \, {\left (2048 \, x^{4} + 7040 \, x^{3} + 352 \, x^{2} - 6916 \, x + 2973\right )} \sqrt {2 \, x^{2} - x + 3} + \frac {85429}{16384} \, \sqrt {2} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) \]

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)^(1/2),x, algorithm="fricas")

[Out]

1/2048*(2048*x^4 + 7040*x^3 + 352*x^2 - 6916*x + 2973)*sqrt(2*x^2 - x + 3) + 85429/16384*sqrt(2)*log(-4*sqrt(2
)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25)

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giac [A]  time = 0.19, size = 68, normalized size = 0.57 \[ \frac {1}{2048} \, {\left (4 \, {\left (8 \, {\left (4 \, {\left (16 \, x + 55\right )} x + 11\right )} x - 1729\right )} x + 2973\right )} \sqrt {2 \, x^{2} - x + 3} - \frac {85429}{8192} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) \]

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)^(1/2),x, algorithm="giac")

[Out]

1/2048*(4*(8*(4*(16*x + 55)*x + 11)*x - 1729)*x + 2973)*sqrt(2*x^2 - x + 3) - 85429/8192*sqrt(2)*log(-2*sqrt(2
)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1)

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maple [A]  time = 0.01, size = 95, normalized size = 0.79 \[ \sqrt {2 x^{2}-x +3}\, x^{4}+\frac {55 \sqrt {2 x^{2}-x +3}\, x^{3}}{16}+\frac {11 \sqrt {2 x^{2}-x +3}\, x^{2}}{64}-\frac {1729 \sqrt {2 x^{2}-x +3}\, x}{512}+\frac {85429 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{8192}+\frac {2973 \sqrt {2 x^{2}-x +3}}{2048} \]

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)^(1/2),x)

[Out]

x^4*(2*x^2-x+3)^(1/2)+55/16*x^3*(2*x^2-x+3)^(1/2)+11/64*x^2*(2*x^2-x+3)^(1/2)-1729/512*x*(2*x^2-x+3)^(1/2)+297
3/2048*(2*x^2-x+3)^(1/2)+85429/8192*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))

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maxima [A]  time = 0.97, size = 96, normalized size = 0.80 \[ \sqrt {2 \, x^{2} - x + 3} x^{4} + \frac {55}{16} \, \sqrt {2 \, x^{2} - x + 3} x^{3} + \frac {11}{64} \, \sqrt {2 \, x^{2} - x + 3} x^{2} - \frac {1729}{512} \, \sqrt {2 \, x^{2} - x + 3} x + \frac {85429}{8192} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {2973}{2048} \, \sqrt {2 \, x^{2} - x + 3} \]

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)^(1/2),x, algorithm="maxima")

[Out]

sqrt(2*x^2 - x + 3)*x^4 + 55/16*sqrt(2*x^2 - x + 3)*x^3 + 11/64*sqrt(2*x^2 - x + 3)*x^2 - 1729/512*sqrt(2*x^2
- x + 3)*x + 85429/8192*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) + 2973/2048*sqrt(2*x^2 - x + 3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x + 5)*(x + 3*x^2 - x^3 + 5*x^4 + 2))/(2*x^2 - x + 3)^(1/2),x)

[Out]

int(((2*x + 5)*(x + 3*x^2 - x^3 + 5*x^4 + 2))/(2*x^2 - x + 3)^(1/2), x)

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

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)**(1/2),x)

[Out]

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

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