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

Optimal. Leaf size=82 \[ \frac {5}{8} x \left (2 x^2-x+3\right )^{3/2}+\frac {73}{96} \left (2 x^2-x+3\right )^{3/2}-\frac {81}{512} (1-4 x) \sqrt {2 x^2-x+3}-\frac {1863 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{1024 \sqrt {2}} \]

[Out]

73/96*(2*x^2-x+3)^(3/2)+5/8*x*(2*x^2-x+3)^(3/2)-1863/2048*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)-81/512*(1-4*x
)*(2*x^2-x+3)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1661, 640, 612, 619, 215} \[ \frac {5}{8} x \left (2 x^2-x+3\right )^{3/2}+\frac {73}{96} \left (2 x^2-x+3\right )^{3/2}-\frac {81}{512} (1-4 x) \sqrt {2 x^2-x+3}-\frac {1863 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{1024 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-81*(1 - 4*x)*Sqrt[3 - x + 2*x^2])/512 + (73*(3 - x + 2*x^2)^(3/2))/96 + (5*x*(3 - x + 2*x^2)^(3/2))/8 - (186
3*ArcSinh[(1 - 4*x)/Sqrt[23]])/(1024*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 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 1661

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[(e*x^(q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*(q + 2*p + 1)), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right ) \, dx &=\frac {5}{8} x \left (3-x+2 x^2\right )^{3/2}+\frac {1}{8} \int \left (1+\frac {73 x}{2}\right ) \sqrt {3-x+2 x^2} \, dx\\ &=\frac {73}{96} \left (3-x+2 x^2\right )^{3/2}+\frac {5}{8} x \left (3-x+2 x^2\right )^{3/2}+\frac {81}{64} \int \sqrt {3-x+2 x^2} \, dx\\ &=-\frac {81}{512} (1-4 x) \sqrt {3-x+2 x^2}+\frac {73}{96} \left (3-x+2 x^2\right )^{3/2}+\frac {5}{8} x \left (3-x+2 x^2\right )^{3/2}+\frac {1863 \int \frac {1}{\sqrt {3-x+2 x^2}} \, dx}{1024}\\ &=-\frac {81}{512} (1-4 x) \sqrt {3-x+2 x^2}+\frac {73}{96} \left (3-x+2 x^2\right )^{3/2}+\frac {5}{8} x \left (3-x+2 x^2\right )^{3/2}+\frac {\left (81 \sqrt {\frac {23}{2}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+4 x\right )}{1024}\\ &=-\frac {81}{512} (1-4 x) \sqrt {3-x+2 x^2}+\frac {73}{96} \left (3-x+2 x^2\right )^{3/2}+\frac {5}{8} x \left (3-x+2 x^2\right )^{3/2}-\frac {1863 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{1024 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 55, normalized size = 0.67 \[ \frac {4 \sqrt {2 x^2-x+3} \left (1920 x^3+1376 x^2+2684 x+3261\right )-5589 \sqrt {2} \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{6144} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[3 - x + 2*x^2]*(3261 + 2684*x + 1376*x^2 + 1920*x^3) - 5589*Sqrt[2]*ArcSinh[(1 - 4*x)/Sqrt[23]])/6144

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fricas [A]  time = 1.01, size = 68, normalized size = 0.83 \[ \frac {1}{1536} \, {\left (1920 \, x^{3} + 1376 \, x^{2} + 2684 \, x + 3261\right )} \sqrt {2 \, x^{2} - x + 3} + \frac {1863}{4096} \, \sqrt {2} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)*(2*x^2-x+3)^(1/2),x, algorithm="fricas")

[Out]

1/1536*(1920*x^3 + 1376*x^2 + 2684*x + 3261)*sqrt(2*x^2 - x + 3) + 1863/4096*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 = 0.24, size = 63, normalized size = 0.77 \[ \frac {1}{1536} \, {\left (4 \, {\left (8 \, {\left (60 \, x + 43\right )} x + 671\right )} x + 3261\right )} \sqrt {2 \, x^{2} - x + 3} - \frac {1863}{2048} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)*(2*x^2-x+3)^(1/2),x, algorithm="giac")

[Out]

1/1536*(4*(8*(60*x + 43)*x + 671)*x + 3261)*sqrt(2*x^2 - x + 3) - 1863/2048*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 = 64, normalized size = 0.78 \[ \frac {5 \left (2 x^{2}-x +3\right )^{\frac {3}{2}} x}{8}+\frac {1863 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{2048}+\frac {73 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}{96}+\frac {81 \left (4 x -1\right ) \sqrt {2 x^{2}-x +3}}{512} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/8*(2*x^2-x+3)^(3/2)*x+73/96*(2*x^2-x+3)^(3/2)+81/512*(4*x-1)*(2*x^2-x+3)^(1/2)+1863/2048*2^(1/2)*arcsinh(4/2
3*23^(1/2)*(x-1/4))

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maxima [A]  time = 0.96, size = 75, normalized size = 0.91 \[ \frac {5}{8} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + \frac {73}{96} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} + \frac {81}{128} \, \sqrt {2 \, x^{2} - x + 3} x + \frac {1863}{2048} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {81}{512} \, \sqrt {2 \, x^{2} - x + 3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)*(2*x^2-x+3)^(1/2),x, algorithm="maxima")

[Out]

5/8*(2*x^2 - x + 3)^(3/2)*x + 73/96*(2*x^2 - x + 3)^(3/2) + 81/128*sqrt(2*x^2 - x + 3)*x + 1863/2048*sqrt(2)*a
rcsinh(1/23*sqrt(23)*(4*x - 1)) - 81/512*sqrt(2*x^2 - x + 3)

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mupad [B]  time = 3.84, size = 119, normalized size = 1.45 \[ \frac {23\,\sqrt {2}\,\ln \left (\sqrt {2\,x^2-x+3}+\frac {\sqrt {2}\,\left (2\,x-\frac {1}{2}\right )}{2}\right )}{256}+\frac {\left (\frac {x}{2}-\frac {1}{8}\right )\,\sqrt {2\,x^2-x+3}}{8}+\frac {73\,\sqrt {2\,x^2-x+3}\,\left (32\,x^2-4\,x+45\right )}{1536}+\frac {5\,x\,{\left (2\,x^2-x+3\right )}^{3/2}}{8}+\frac {1679\,\sqrt {2}\,\ln \left (2\,\sqrt {2\,x^2-x+3}+\frac {\sqrt {2}\,\left (4\,x-1\right )}{2}\right )}{2048} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(23*2^(1/2)*log((2*x^2 - x + 3)^(1/2) + (2^(1/2)*(2*x - 1/2))/2))/256 + ((x/2 - 1/8)*(2*x^2 - x + 3)^(1/2))/8
+ (73*(2*x^2 - x + 3)^(1/2)*(32*x^2 - 4*x + 45))/1536 + (5*x*(2*x^2 - x + 3)^(3/2))/8 + (1679*2^(1/2)*log(2*(2
*x^2 - x + 3)^(1/2) + (2^(1/2)*(4*x - 1))/2))/2048

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**2+3*x+2)*(2*x**2-x+3)**(1/2),x)

[Out]

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

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