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

Optimal. Leaf size=105 \[ \frac {4 (18982-20383 x)}{1587 \sqrt {2 x^2-x+3}}+\frac {5}{4} x \sqrt {2 x^2-x+3}+\frac {247}{16} \sqrt {2 x^2-x+3}-\frac {4 (346-533 x)}{69 \left (2 x^2-x+3\right )^{3/2}}-\frac {1471 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{32 \sqrt {2}} \]

[Out]

-4/69*(346-533*x)/(2*x^2-x+3)^(3/2)-1471/64*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)+4/1587*(18982-20383*x)/(2*x
^2-x+3)^(1/2)+247/16*(2*x^2-x+3)^(1/2)+5/4*x*(2*x^2-x+3)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1660, 1661, 640, 619, 215} \[ \frac {4 (18982-20383 x)}{1587 \sqrt {2 x^2-x+3}}+\frac {5}{4} x \sqrt {2 x^2-x+3}+\frac {247}{16} \sqrt {2 x^2-x+3}-\frac {4 (346-533 x)}{69 \left (2 x^2-x+3\right )^{3/2}}-\frac {1471 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{32 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(346 - 533*x))/(69*(3 - x + 2*x^2)^(3/2)) + (4*(18982 - 20383*x))/(1587*Sqrt[3 - x + 2*x^2]) + (247*Sqrt[3
- x + 2*x^2])/16 + (5*x*Sqrt[3 - x + 2*x^2])/4 - (1471*ArcSinh[(1 - 4*x)/Sqrt[23]])/(32*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)^2 \left (2+x+3 x^2-x^3+5 x^4\right )}{\left (3-x+2 x^2\right )^{5/2}} \, dx &=-\frac {4 (346-533 x)}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac {2}{69} \int \frac {-145-\frac {1725 x}{2}+2415 x^2+\frac {3657 x^3}{2}+345 x^4}{\left (3-x+2 x^2\right )^{3/2}} \, dx\\ &=-\frac {4 (346-533 x)}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac {4 (18982-20383 x)}{1587 \sqrt {3-x+2 x^2}}+\frac {4 \int \frac {\frac {33327}{2}+\frac {46023 x}{4}+\frac {7935 x^2}{4}}{\sqrt {3-x+2 x^2}} \, dx}{1587}\\ &=-\frac {4 (346-533 x)}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac {4 (18982-20383 x)}{1587 \sqrt {3-x+2 x^2}}+\frac {5}{4} x \sqrt {3-x+2 x^2}+\frac {\int \frac {\frac {242811}{4}+\frac {391989 x}{8}}{\sqrt {3-x+2 x^2}} \, dx}{1587}\\ &=-\frac {4 (346-533 x)}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac {4 (18982-20383 x)}{1587 \sqrt {3-x+2 x^2}}+\frac {247}{16} \sqrt {3-x+2 x^2}+\frac {5}{4} x \sqrt {3-x+2 x^2}+\frac {1471}{32} \int \frac {1}{\sqrt {3-x+2 x^2}} \, dx\\ &=-\frac {4 (346-533 x)}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac {4 (18982-20383 x)}{1587 \sqrt {3-x+2 x^2}}+\frac {247}{16} \sqrt {3-x+2 x^2}+\frac {5}{4} x \sqrt {3-x+2 x^2}+\frac {1471 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+4 x\right )}{32 \sqrt {46}}\\ &=-\frac {4 (346-533 x)}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac {4 (18982-20383 x)}{1587 \sqrt {3-x+2 x^2}}+\frac {247}{16} \sqrt {3-x+2 x^2}+\frac {5}{4} x \sqrt {3-x+2 x^2}-\frac {1471 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{32 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.69, size = 65, normalized size = 0.62 \[ \frac {126960 x^5+1440996 x^4-3764360 x^3+8639625 x^2-6410082 x+6663133}{25392 \left (2 x^2-x+3\right )^{3/2}}-\frac {1471 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{32 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6663133 - 6410082*x + 8639625*x^2 - 3764360*x^3 + 1440996*x^4 + 126960*x^5)/(25392*(3 - x + 2*x^2)^(3/2)) - (
1471*ArcSinh[(1 - 4*x)/Sqrt[23]])/(32*Sqrt[2])

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fricas [A]  time = 0.86, size = 122, normalized size = 1.16 \[ \frac {2334477 \, \sqrt {2} {\left (4 \, x^{4} - 4 \, x^{3} + 13 \, x^{2} - 6 \, x + 9\right )} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 8 \, {\left (126960 \, x^{5} + 1440996 \, x^{4} - 3764360 \, x^{3} + 8639625 \, x^{2} - 6410082 \, x + 6663133\right )} \sqrt {2 \, x^{2} - x + 3}}{203136 \, {\left (4 \, x^{4} - 4 \, x^{3} + 13 \, x^{2} - 6 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/203136*(2334477*sqrt(2)*(4*x^4 - 4*x^3 + 13*x^2 - 6*x + 9)*log(-4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32
*x^2 + 16*x - 25) + 8*(126960*x^5 + 1440996*x^4 - 3764360*x^3 + 8639625*x^2 - 6410082*x + 6663133)*sqrt(2*x^2
- x + 3))/(4*x^4 - 4*x^3 + 13*x^2 - 6*x + 9)

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giac [A]  time = 0.22, size = 71, normalized size = 0.68 \[ -\frac {1471}{64} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) + \frac {{\left ({\left (4 \, {\left (1587 \, {\left (20 \, x + 227\right )} x - 941090\right )} x + 8639625\right )} x - 6410082\right )} x + 6663133}{25392 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1471/64*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) + 1/25392*(((4*(1587*(20*x + 227)*x - 9
41090)*x + 8639625)*x - 6410082)*x + 6663133)/(2*x^2 - x + 3)^(3/2)

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maple [B]  time = 0.02, size = 180, normalized size = 1.71 \[ \frac {5 x^{5}}{\left (2 x^{2}-x +3\right )^{\frac {3}{2}}}+\frac {227 x^{4}}{4 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}-\frac {1471 x^{3}}{48 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}+\frac {19073 x^{2}}{64 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}-\frac {32257 x}{512 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}-\frac {1471 x}{32 \sqrt {2 x^{2}-x +3}}+\frac {1471 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{64}-\frac {162931 \left (4 x -1\right )}{50784 \sqrt {2 x^{2}-x +3}}-\frac {753223 \left (4 x -1\right )}{141312 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}+\frac {577397}{2048 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}-\frac {1471}{128 \sqrt {2 x^{2}-x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-162931/50784*(4*x-1)/(2*x^2-x+3)^(1/2)-753223/141312*(4*x-1)/(2*x^2-x+3)^(3/2)+5*x^5/(2*x^2-x+3)^(3/2)+227/4*
x^4/(2*x^2-x+3)^(3/2)-1471/48*x^3/(2*x^2-x+3)^(3/2)+19073/64*x^2/(2*x^2-x+3)^(3/2)-32257/512*x/(2*x^2-x+3)^(3/
2)+1471/64*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))-1471/32/(2*x^2-x+3)^(1/2)*x+577397/2048/(2*x^2-x+3)^(3/2)-14
71/128/(2*x^2-x+3)^(1/2)

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maxima [B]  time = 0.98, size = 219, normalized size = 2.09 \[ \frac {5 \, x^{5}}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {227 \, x^{4}}{4 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {1471}{50784} \, x {\left (\frac {284 \, x}{\sqrt {2 \, x^{2} - x + 3}} - \frac {3174 \, x^{2}}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {71}{\sqrt {2 \, x^{2} - x + 3}} + \frac {805 \, x}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {3243}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}\right )} + \frac {1471}{64} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {104441}{25392} \, \sqrt {2 \, x^{2} - x + 3} - \frac {383581 \, x}{12696 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {321 \, x^{2}}{{\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {15965}{4232 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {4147 \, x}{46 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {42883}{138 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*x^5/(2*x^2 - x + 3)^(3/2) + 227/4*x^4/(2*x^2 - x + 3)^(3/2) + 1471/50784*x*(284*x/sqrt(2*x^2 - x + 3) - 3174
*x^2/(2*x^2 - x + 3)^(3/2) - 71/sqrt(2*x^2 - x + 3) + 805*x/(2*x^2 - x + 3)^(3/2) - 3243/(2*x^2 - x + 3)^(3/2)
) + 1471/64*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1)) - 104441/25392*sqrt(2*x^2 - x + 3) - 383581/12696*x/sqrt(
2*x^2 - x + 3) + 321*x^2/(2*x^2 - x + 3)^(3/2) - 15965/4232/sqrt(2*x^2 - x + 3) - 4147/46*x/(2*x^2 - x + 3)^(3
/2) + 42883/138/(2*x^2 - x + 3)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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