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

Optimal. Leaf size=143 \[ \frac {5}{112} \left (2 x^2-x+3\right )^{3/2} (2 x+5)^4-\frac {823 \left (2 x^2-x+3\right )^{3/2} (2 x+5)^3}{1344}+\frac {11433 \left (2 x^2-x+3\right )^{3/2} (2 x+5)^2}{4480}-\frac {(295276 x+1005757) \left (2 x^2-x+3\right )^{3/2}}{71680}-\frac {51435 (1-4 x) \sqrt {2 x^2-x+3}}{32768}-\frac {1183005 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{65536 \sqrt {2}} \]

[Out]

11433/4480*(5+2*x)^2*(2*x^2-x+3)^(3/2)-823/1344*(5+2*x)^3*(2*x^2-x+3)^(3/2)+5/112*(5+2*x)^4*(2*x^2-x+3)^(3/2)-
1/71680*(1005757+295276*x)*(2*x^2-x+3)^(3/2)-1183005/131072*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)-51435/32768
*(1-4*x)*(2*x^2-x+3)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {1653, 779, 612, 619, 215} \[ \frac {5}{112} \left (2 x^2-x+3\right )^{3/2} (2 x+5)^4-\frac {823 \left (2 x^2-x+3\right )^{3/2} (2 x+5)^3}{1344}+\frac {11433 \left (2 x^2-x+3\right )^{3/2} (2 x+5)^2}{4480}-\frac {(295276 x+1005757) \left (2 x^2-x+3\right )^{3/2}}{71680}-\frac {51435 (1-4 x) \sqrt {2 x^2-x+3}}{32768}-\frac {1183005 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{65536 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-51435*(1 - 4*x)*Sqrt[3 - x + 2*x^2])/32768 + (11433*(5 + 2*x)^2*(3 - x + 2*x^2)^(3/2))/4480 - (823*(5 + 2*x)
^3*(3 - x + 2*x^2)^(3/2))/1344 + (5*(5 + 2*x)^4*(3 - x + 2*x^2)^(3/2))/112 - ((1005757 + 295276*x)*(3 - x + 2*
x^2)^(3/2))/71680 - (1183005*ArcSinh[(1 - 4*x)/Sqrt[23]])/(65536*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 612

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

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 (5+2 x) \sqrt {3-x+2 x^2} \left (2+x+3 x^2-x^3+5 x^4\right ) \, dx &=\frac {5}{112} (5+2 x)^4 \left (3-x+2 x^2\right )^{3/2}+\frac {1}{224} \int (5+2 x) \sqrt {3-x+2 x^2} \left (-3677-7826 x-10788 x^2-6584 x^3\right ) \, dx\\ &=-\frac {823 (5+2 x)^3 \left (3-x+2 x^2\right )^{3/2}}{1344}+\frac {5}{112} (5+2 x)^4 \left (3-x+2 x^2\right )^{3/2}+\frac {\int (5+2 x) \sqrt {3-x+2 x^2} \left (338328+907872 x+1097568 x^2\right ) \, dx}{21504}\\ &=\frac {11433 (5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}}{4480}-\frac {823 (5+2 x)^3 \left (3-x+2 x^2\right )^{3/2}}{1344}+\frac {5}{112} (5+2 x)^4 \left (3-x+2 x^2\right )^{3/2}+\frac {\int (3655008-14173248 x) (5+2 x) \sqrt {3-x+2 x^2} \, dx}{860160}\\ &=\frac {11433 (5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}}{4480}-\frac {823 (5+2 x)^3 \left (3-x+2 x^2\right )^{3/2}}{1344}+\frac {5}{112} (5+2 x)^4 \left (3-x+2 x^2\right )^{3/2}-\frac {(1005757+295276 x) \left (3-x+2 x^2\right )^{3/2}}{71680}+\frac {51435 \int \sqrt {3-x+2 x^2} \, dx}{4096}\\ &=-\frac {51435 (1-4 x) \sqrt {3-x+2 x^2}}{32768}+\frac {11433 (5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}}{4480}-\frac {823 (5+2 x)^3 \left (3-x+2 x^2\right )^{3/2}}{1344}+\frac {5}{112} (5+2 x)^4 \left (3-x+2 x^2\right )^{3/2}-\frac {(1005757+295276 x) \left (3-x+2 x^2\right )^{3/2}}{71680}+\frac {1183005 \int \frac {1}{\sqrt {3-x+2 x^2}} \, dx}{65536}\\ &=-\frac {51435 (1-4 x) \sqrt {3-x+2 x^2}}{32768}+\frac {11433 (5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}}{4480}-\frac {823 (5+2 x)^3 \left (3-x+2 x^2\right )^{3/2}}{1344}+\frac {5}{112} (5+2 x)^4 \left (3-x+2 x^2\right )^{3/2}-\frac {(1005757+295276 x) \left (3-x+2 x^2\right )^{3/2}}{71680}+\frac {\left (51435 \sqrt {\frac {23}{2}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+4 x\right )}{65536}\\ &=-\frac {51435 (1-4 x) \sqrt {3-x+2 x^2}}{32768}+\frac {11433 (5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}}{4480}-\frac {823 (5+2 x)^3 \left (3-x+2 x^2\right )^{3/2}}{1344}+\frac {5}{112} (5+2 x)^4 \left (3-x+2 x^2\right )^{3/2}-\frac {(1005757+295276 x) \left (3-x+2 x^2\right )^{3/2}}{71680}-\frac {1183005 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{65536 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 70, normalized size = 0.49 \[ \frac {4 \sqrt {2 x^2-x+3} \left (4915200 x^6+12984320 x^5+1390592 x^4+20304768 x^3+11357024 x^2+14742332 x+6231117\right )-124215525 \sqrt {2} \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{13762560} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[3 - x + 2*x^2]*(6231117 + 14742332*x + 11357024*x^2 + 20304768*x^3 + 1390592*x^4 + 12984320*x^5 + 4915
200*x^6) - 124215525*Sqrt[2]*ArcSinh[(1 - 4*x)/Sqrt[23]])/13762560

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fricas [A]  time = 0.81, size = 83, normalized size = 0.58 \[ \frac {1}{3440640} \, {\left (4915200 \, x^{6} + 12984320 \, x^{5} + 1390592 \, x^{4} + 20304768 \, x^{3} + 11357024 \, x^{2} + 14742332 \, x + 6231117\right )} \sqrt {2 \, x^{2} - x + 3} + \frac {1183005}{262144} \, \sqrt {2} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) \]

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]

1/3440640*(4915200*x^6 + 12984320*x^5 + 1390592*x^4 + 20304768*x^3 + 11357024*x^2 + 14742332*x + 6231117)*sqrt
(2*x^2 - x + 3) + 1183005/262144*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 = 0.20, size = 78, normalized size = 0.55 \[ \frac {1}{3440640} \, {\left (4 \, {\left (8 \, {\left (4 \, {\left (16 \, {\left (20 \, {\left (120 \, x + 317\right )} x + 679\right )} x + 158631\right )} x + 354907\right )} x + 3685583\right )} x + 6231117\right )} \sqrt {2 \, x^{2} - x + 3} - \frac {1183005}{131072} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) \]

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]

1/3440640*(4*(8*(4*(16*(20*(120*x + 317)*x + 679)*x + 158631)*x + 354907)*x + 3685583)*x + 6231117)*sqrt(2*x^2
- x + 3) - 1183005/131072*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1)

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maple [A]  time = 0.02, size = 115, normalized size = 0.80 \[ \frac {5 \left (2 x^{2}-x +3\right )^{\frac {3}{2}} x^{4}}{7}+\frac {377 \left (2 x^{2}-x +3\right )^{\frac {3}{2}} x^{3}}{168}+\frac {283 \left (2 x^{2}-x +3\right )^{\frac {3}{2}} x^{2}}{1120}-\frac {5179 \left (2 x^{2}-x +3\right )^{\frac {3}{2}} x}{17920}+\frac {1183005 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{131072}+\frac {51435 \left (4 x -1\right ) \sqrt {2 x^{2}-x +3}}{32768}+\frac {242329 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}{215040} \]

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]

5/7*x^4*(2*x^2-x+3)^(3/2)+377/168*x^3*(2*x^2-x+3)^(3/2)+283/1120*x^2*(2*x^2-x+3)^(3/2)-5179/17920*x*(2*x^2-x+3
)^(3/2)+1183005/131072*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))+51435/32768*(4*x-1)*(2*x^2-x+3)^(1/2)+242329/215
040*(2*x^2-x+3)^(3/2)

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maxima [A]  time = 0.97, size = 126, normalized size = 0.88 \[ \frac {5}{7} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x^{4} + \frac {377}{168} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x^{3} + \frac {283}{1120} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x^{2} - \frac {5179}{17920} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + \frac {242329}{215040} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} + \frac {51435}{8192} \, \sqrt {2 \, x^{2} - x + 3} x + \frac {1183005}{131072} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {51435}{32768} \, \sqrt {2 \, x^{2} - x + 3} \]

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]

5/7*(2*x^2 - x + 3)^(3/2)*x^4 + 377/168*(2*x^2 - x + 3)^(3/2)*x^3 + 283/1120*(2*x^2 - x + 3)^(3/2)*x^2 - 5179/
17920*(2*x^2 - x + 3)^(3/2)*x + 242329/215040*(2*x^2 - x + 3)^(3/2) + 51435/8192*sqrt(2*x^2 - x + 3)*x + 11830
05/131072*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) - 51435/32768*sqrt(2*x^2 - x + 3)

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mupad [B]  time = 1.72, size = 170, normalized size = 1.19 \[ \frac {283\,x^2\,{\left (2\,x^2-x+3\right )}^{3/2}}{1120}+\frac {377\,x^3\,{\left (2\,x^2-x+3\right )}^{3/2}}{168}+\frac {5\,x^4\,{\left (2\,x^2-x+3\right )}^{3/2}}{7}+\frac {4478951\,\sqrt {2}\,\ln \left (\sqrt {2\,x^2-x+3}+\frac {\sqrt {2}\,\left (2\,x-\frac {1}{2}\right )}{2}\right )}{573440}+\frac {194737\,\left (\frac {x}{2}-\frac {1}{8}\right )\,\sqrt {2\,x^2-x+3}}{17920}+\frac {242329\,\sqrt {2\,x^2-x+3}\,\left (32\,x^2-4\,x+45\right )}{3440640}-\frac {5179\,x\,{\left (2\,x^2-x+3\right )}^{3/2}}{17920}+\frac {5573567\,\sqrt {2}\,\ln \left (2\,\sqrt {2\,x^2-x+3}+\frac {\sqrt {2}\,\left (4\,x-1\right )}{2}\right )}{4587520} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(283*x^2*(2*x^2 - x + 3)^(3/2))/1120 + (377*x^3*(2*x^2 - x + 3)^(3/2))/168 + (5*x^4*(2*x^2 - x + 3)^(3/2))/7 +
(4478951*2^(1/2)*log((2*x^2 - x + 3)^(1/2) + (2^(1/2)*(2*x - 1/2))/2))/573440 + (194737*(x/2 - 1/8)*(2*x^2 -
x + 3)^(1/2))/17920 + (242329*(2*x^2 - x + 3)^(1/2)*(32*x^2 - 4*x + 45))/3440640 - (5179*x*(2*x^2 - x + 3)^(3/
2))/17920 + (5573567*2^(1/2)*log(2*(2*x^2 - x + 3)^(1/2) + (2^(1/2)*(4*x - 1))/2))/4587520

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

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

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