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]

(-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])

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Rubi [A]  time = 0.039791, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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]

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 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 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 \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*}

Mathematica [A]  time = 0.0534298, 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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Maple [A]  time = 0.048, size = 64, normalized size = 0.8 \begin{align*}{\frac{5\,x}{8} \left ( 2\,{x}^{2}-x+3 \right ) ^{{\frac{3}{2}}}}+{\frac{73}{96} \left ( 2\,{x}^{2}-x+3 \right ) ^{{\frac{3}{2}}}}+{\frac{-81+324\,x}{512}\sqrt{2\,{x}^{2}-x+3}}+{\frac{1863\,\sqrt{2}}{2048}{\it Arcsinh} \left ({\frac{4\,\sqrt{23}}{23} \left ( x-{\frac{1}{4}} \right ) } \right ) } \end{align*}

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

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Maxima [A]  time = 1.50358, size = 101, normalized size = 1.23 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 1.64869, size = 207, normalized size = 2.52 \begin{align*} \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 ) \end{align*}

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

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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Giac [A]  time = 1.16374, size = 85, normalized size = 1.04 \begin{align*} \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 ) \end{align*}

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)