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

Optimal. Leaf size=103 \[ \frac {5}{6} \sqrt {2 x^2-x+3} x^2+\frac {193}{48} \sqrt {2 x^2-x+3} x+\frac {33}{64} \sqrt {2 x^2-x+3}+\frac {373 x-53}{23 \sqrt {2 x^2-x+3}}+\frac {3111 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{128 \sqrt {2}} \]

[Out]

3111/256*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)+1/23*(-53+373*x)/(2*x^2-x+3)^(1/2)+33/64*(2*x^2-x+3)^(1/2)+193
/48*x*(2*x^2-x+3)^(1/2)+5/6*x^2*(2*x^2-x+3)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {1660, 1661, 640, 619, 215} \[ \frac {5}{6} \sqrt {2 x^2-x+3} x^2+\frac {193}{48} \sqrt {2 x^2-x+3} x+\frac {33}{64} \sqrt {2 x^2-x+3}-\frac {53-373 x}{23 \sqrt {2 x^2-x+3}}+\frac {3111 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{128 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(53 - 373*x)/(23*Sqrt[3 - x + 2*x^2]) + (33*Sqrt[3 - x + 2*x^2])/64 + (193*x*Sqrt[3 - x + 2*x^2])/48 + (5*x^2
*Sqrt[3 - x + 2*x^2])/6 + (3111*ArcSinh[(1 - 4*x)/Sqrt[23]])/(128*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 {(5+2 x) \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{3/2}} \, dx &=-\frac {53-373 x}{23 \sqrt {3-x+2 x^2}}+\frac {2}{23} \int \frac {-\frac {575}{4}+161 x^2+\frac {115 x^3}{2}}{\sqrt {3-x+2 x^2}} \, dx\\ &=-\frac {53-373 x}{23 \sqrt {3-x+2 x^2}}+\frac {5}{6} x^2 \sqrt {3-x+2 x^2}+\frac {1}{69} \int \frac {-\frac {1725}{2}-345 x+\frac {4439 x^2}{4}}{\sqrt {3-x+2 x^2}} \, dx\\ &=-\frac {53-373 x}{23 \sqrt {3-x+2 x^2}}+\frac {193}{48} x \sqrt {3-x+2 x^2}+\frac {5}{6} x^2 \sqrt {3-x+2 x^2}+\frac {1}{276} \int \frac {-\frac {27117}{4}+\frac {2277 x}{8}}{\sqrt {3-x+2 x^2}} \, dx\\ &=-\frac {53-373 x}{23 \sqrt {3-x+2 x^2}}+\frac {33}{64} \sqrt {3-x+2 x^2}+\frac {193}{48} x \sqrt {3-x+2 x^2}+\frac {5}{6} x^2 \sqrt {3-x+2 x^2}-\frac {3111}{128} \int \frac {1}{\sqrt {3-x+2 x^2}} \, dx\\ &=-\frac {53-373 x}{23 \sqrt {3-x+2 x^2}}+\frac {33}{64} \sqrt {3-x+2 x^2}+\frac {193}{48} x \sqrt {3-x+2 x^2}+\frac {5}{6} x^2 \sqrt {3-x+2 x^2}-\frac {3111 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+4 x\right )}{128 \sqrt {46}}\\ &=-\frac {53-373 x}{23 \sqrt {3-x+2 x^2}}+\frac {33}{64} \sqrt {3-x+2 x^2}+\frac {193}{48} x \sqrt {3-x+2 x^2}+\frac {5}{6} x^2 \sqrt {3-x+2 x^2}+\frac {3111 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{128 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 60, normalized size = 0.58 \[ \frac {7360 x^4+31832 x^3-2162 x^2+122607 x-3345}{4416 \sqrt {2 x^2-x+3}}-\frac {3111 \sinh ^{-1}\left (\frac {4 x-1}{\sqrt {23}}\right )}{128 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3345 + 122607*x - 2162*x^2 + 31832*x^3 + 7360*x^4)/(4416*Sqrt[3 - x + 2*x^2]) - (3111*ArcSinh[(-1 + 4*x)/Sqr
t[23]])/(128*Sqrt[2])

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fricas [A]  time = 0.86, size = 97, normalized size = 0.94 \[ \frac {214659 \, \sqrt {2} {\left (2 \, x^{2} - x + 3\right )} \log \left (4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 8 \, {\left (7360 \, x^{4} + 31832 \, x^{3} - 2162 \, x^{2} + 122607 \, x - 3345\right )} \sqrt {2 \, x^{2} - x + 3}}{35328 \, {\left (2 \, x^{2} - x + 3\right )}} \]

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)^(3/2),x, algorithm="fricas")

[Out]

1/35328*(214659*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*
(7360*x^4 + 31832*x^3 - 2162*x^2 + 122607*x - 3345)*sqrt(2*x^2 - x + 3))/(2*x^2 - x + 3)

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giac [A]  time = 0.22, size = 67, normalized size = 0.65 \[ \frac {3111}{256} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) + \frac {{\left (46 \, {\left (4 \, {\left (40 \, x + 173\right )} x - 47\right )} x + 122607\right )} x - 3345}{4416 \, \sqrt {2 \, x^{2} - x + 3}} \]

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)^(3/2),x, algorithm="giac")

[Out]

3111/256*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) + 1/4416*((46*(4*(40*x + 173)*x - 47)*x
+ 122607)*x - 3345)/sqrt(2*x^2 - x + 3)

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maple [A]  time = 0.01, size = 115, normalized size = 1.12 \[ \frac {5 x^{4}}{3 \sqrt {2 x^{2}-x +3}}+\frac {173 x^{3}}{24 \sqrt {2 x^{2}-x +3}}-\frac {47 x^{2}}{96 \sqrt {2 x^{2}-x +3}}+\frac {3111 x}{128 \sqrt {2 x^{2}-x +3}}-\frac {3111 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{256}+\frac {\frac {10185 x}{2944}-\frac {10185}{11776}}{\sqrt {2 x^{2}-x +3}}+\frac {55}{512 \sqrt {2 x^{2}-x +3}} \]

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)^(3/2),x)

[Out]

10185/11776*(4*x-1)/(2*x^2-x+3)^(1/2)-3111/256*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))+5/3/(2*x^2-x+3)^(1/2)*x^
4+173/24/(2*x^2-x+3)^(1/2)*x^3-47/96/(2*x^2-x+3)^(1/2)*x^2+3111/128/(2*x^2-x+3)^(1/2)*x+55/512/(2*x^2-x+3)^(1/
2)

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maxima [A]  time = 0.97, size = 97, normalized size = 0.94 \[ \frac {5 \, x^{4}}{3 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {173 \, x^{3}}{24 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {47 \, x^{2}}{96 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {3111}{256} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {40869 \, x}{1472 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {1115}{1472 \, \sqrt {2 \, x^{2} - x + 3}} \]

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)^(3/2),x, algorithm="maxima")

[Out]

5/3*x^4/sqrt(2*x^2 - x + 3) + 173/24*x^3/sqrt(2*x^2 - x + 3) - 47/96*x^2/sqrt(2*x^2 - x + 3) - 3111/256*sqrt(2
)*arcsinh(1/23*sqrt(23)*(4*x - 1)) + 40869/1472*x/sqrt(2*x^2 - x + 3) - 1115/1472/sqrt(2*x^2 - x + 3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)**(3/2),x)

[Out]

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

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