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

Optimal. Leaf size=82 \[ \frac{5}{8} \sqrt{2 x^2-x+3} x+\frac{27}{32} \sqrt{2 x^2-x+3}+\frac{219 x+89}{92 \sqrt{2 x^2-x+3}}+\frac{213 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{64 \sqrt{2}} \]

[Out]

(89 + 219*x)/(92*Sqrt[3 - x + 2*x^2]) + (27*Sqrt[3 - x + 2*x^2])/32 + (5*x*Sqrt[3 - x + 2*x^2])/8 + (213*ArcSi
nh[(1 - 4*x)/Sqrt[23]])/(64*Sqrt[2])

________________________________________________________________________________________

Rubi [A]  time = 0.0551878, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {1660, 1661, 640, 619, 215} \[ \frac{5}{8} \sqrt{2 x^2-x+3} x+\frac{27}{32} \sqrt{2 x^2-x+3}+\frac{219 x+89}{92 \sqrt{2 x^2-x+3}}+\frac{213 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{64 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(89 + 219*x)/(92*Sqrt[3 - x + 2*x^2]) + (27*Sqrt[3 - x + 2*x^2])/32 + (5*x*Sqrt[3 - x + 2*x^2])/8 + (213*ArcSi
nh[(1 - 4*x)/Sqrt[23]])/(64*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 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 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{2+x+3 x^2-x^3+5 x^4}{\left (3-x+2 x^2\right )^{3/2}} \, dx &=\frac{89+219 x}{92 \sqrt{3-x+2 x^2}}+\frac{2}{23} \int \frac{-\frac{345}{16}+\frac{69 x}{8}+\frac{115 x^2}{4}}{\sqrt{3-x+2 x^2}} \, dx\\ &=\frac{89+219 x}{92 \sqrt{3-x+2 x^2}}+\frac{5}{8} x \sqrt{3-x+2 x^2}+\frac{1}{46} \int \frac{-\frac{345}{2}+\frac{621 x}{8}}{\sqrt{3-x+2 x^2}} \, dx\\ &=\frac{89+219 x}{92 \sqrt{3-x+2 x^2}}+\frac{27}{32} \sqrt{3-x+2 x^2}+\frac{5}{8} x \sqrt{3-x+2 x^2}-\frac{213}{64} \int \frac{1}{\sqrt{3-x+2 x^2}} \, dx\\ &=\frac{89+219 x}{92 \sqrt{3-x+2 x^2}}+\frac{27}{32} \sqrt{3-x+2 x^2}+\frac{5}{8} x \sqrt{3-x+2 x^2}-\frac{213 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+4 x\right )}{64 \sqrt{46}}\\ &=\frac{89+219 x}{92 \sqrt{3-x+2 x^2}}+\frac{27}{32} \sqrt{3-x+2 x^2}+\frac{5}{8} x \sqrt{3-x+2 x^2}+\frac{213 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{64 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.128901, size = 55, normalized size = 0.67 \[ \frac{920 x^3+782 x^2+2511 x+2575}{736 \sqrt{2 x^2-x+3}}-\frac{213 \sinh ^{-1}\left (\frac{4 x-1}{\sqrt{23}}\right )}{64 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2575 + 2511*x + 782*x^2 + 920*x^3)/(736*Sqrt[3 - x + 2*x^2]) - (213*ArcSinh[(-1 + 4*x)/Sqrt[23]])/(64*Sqrt[2]
)

________________________________________________________________________________________

Maple [A]  time = 0.051, size = 98, normalized size = 1.2 \begin{align*}{\frac{5\,{x}^{3}}{4}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}+{\frac{17\,{x}^{2}}{16}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}+{\frac{213\,x}{64}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}+{\frac{901}{256}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}+{\frac{-123+492\,x}{5888}{\frac{1}{\sqrt{2\,{x}^{2}-x+3}}}}-{\frac{213\,\sqrt{2}}{128}{\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^4-x^3+3*x^2+x+2)/(2*x^2-x+3)^(3/2),x)

[Out]

5/4*x^3/(2*x^2-x+3)^(1/2)+17/16*x^2/(2*x^2-x+3)^(1/2)+213/64*x/(2*x^2-x+3)^(1/2)+901/256/(2*x^2-x+3)^(1/2)+123
/5888*(-1+4*x)/(2*x^2-x+3)^(1/2)-213/128*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))

________________________________________________________________________________________

Maxima [A]  time = 1.54162, size = 108, normalized size = 1.32 \begin{align*} \frac{5 \, x^{3}}{4 \, \sqrt{2 \, x^{2} - x + 3}} + \frac{17 \, x^{2}}{16 \, \sqrt{2 \, x^{2} - x + 3}} - \frac{213}{128} \, \sqrt{2} \operatorname{arsinh}\left (\frac{1}{23} \, \sqrt{23}{\left (4 \, x - 1\right )}\right ) + \frac{2511 \, x}{736 \, \sqrt{2 \, x^{2} - x + 3}} + \frac{2575}{736 \, \sqrt{2 \, x^{2} - x + 3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*x^3/sqrt(2*x^2 - x + 3) + 17/16*x^2/sqrt(2*x^2 - x + 3) - 213/128*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1))
+ 2511/736*x/sqrt(2*x^2 - x + 3) + 2575/736/sqrt(2*x^2 - x + 3)

________________________________________________________________________________________

Fricas [A]  time = 1.35129, size = 244, normalized size = 2.98 \begin{align*} \frac{4899 \, \sqrt{2}{\left (2 \, x^{2} - x + 3\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 (920 \, x^{3} + 782 \, x^{2} + 2511 \, x + 2575\right )} \sqrt{2 \, x^{2} - x + 3}}{5888 \,{\left (2 \, x^{2} - x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5888*(4899*sqrt(2)*(2*x^2 - x + 3)*log(4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25) + 8*(92
0*x^3 + 782*x^2 + 2511*x + 2575)*sqrt(2*x^2 - x + 3))/(2*x^2 - x + 3)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.15944, size = 84, normalized size = 1.02 \begin{align*} \frac{213}{128} \, \sqrt{2} \log \left (-2 \, \sqrt{2}{\left (\sqrt{2} x - \sqrt{2 \, x^{2} - x + 3}\right )} + 1\right ) + \frac{{\left (46 \,{\left (20 \, x + 17\right )} x + 2511\right )} x + 2575}{736 \, \sqrt{2 \, x^{2} - x + 3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

213/128*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) + 1/736*((46*(20*x + 17)*x + 2511)*x + 2
575)/sqrt(2*x^2 - x + 3)