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]

213/128*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)+1/92*(89+219*x)/(2*x^2-x+3)^(1/2)+27/32*(2*x^2-x+3)^(1/2)+5/8*x
*(2*x^2-x+3)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 82, normalized size of antiderivative = 1.00, 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 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 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 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]

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 = 0.13, 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]
)

________________________________________________________________________________________

fricas [A]  time = 0.86, size = 92, normalized size = 1.12 \[ \frac {4899 \, \sqrt {2} {\left (2 \, x^{2} - x + 3\right )} \log \left (4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 8 \, {\left (920 \, x^{3} + 782 \, x^{2} + 2511 \, x + 2575\right )} \sqrt {2 \, x^{2} - x + 3}}{5888 \, {\left (2 \, x^{2} - x + 3\right )}} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.22, size = 62, normalized size = 0.76 \[ \frac {213}{128} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) + \frac {{\left (46 \, {\left (20 \, x + 17\right )} x + 2511\right )} x + 2575}{736 \, \sqrt {2 \, x^{2} - x + 3}} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 98, normalized size = 1.20 \[ \frac {5 x^{3}}{4 \sqrt {2 x^{2}-x +3}}+\frac {17 x^{2}}{16 \sqrt {2 x^{2}-x +3}}+\frac {213 x}{64 \sqrt {2 x^{2}-x +3}}-\frac {213 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{128}+\frac {901}{256 \sqrt {2 x^{2}-x +3}}+\frac {\frac {123 x}{1472}-\frac {123}{5888}}{\sqrt {2 x^{2}-x +3}} \]

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

________________________________________________________________________________________

maxima [A]  time = 0.95, size = 80, normalized size = 0.98 \[ \frac {5 \, x^{3}}{4 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {17 \, x^{2}}{16 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {213}{128} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {2511 \, x}{736 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {2575}{736 \, \sqrt {2 \, x^{2} - x + 3}} \]

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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________