3.1.88 \(\int \frac {(2+3 x+5 x^2)^2}{(3-x+2 x^2)^{3/2}} \, dx\) [88]

Optimal. Leaf size=82 \[ \frac {121 (19-7 x)}{92 \sqrt {3-x+2 x^2}}+\frac {415}{32} \sqrt {3-x+2 x^2}+\frac {25}{8} x \sqrt {3-x+2 x^2}-\frac {223 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{64 \sqrt {2}} \]

[Out]

-223/128*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)+121/92*(19-7*x)/(2*x^2-x+3)^(1/2)+415/32*(2*x^2-x+3)^(1/2)+25/
8*x*(2*x^2-x+3)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1674, 1675, 654, 633, 221} \begin {gather*} \frac {121 (19-7 x)}{92 \sqrt {2 x^2-x+3}}+\frac {25}{8} x \sqrt {2 x^2-x+3}+\frac {415}{32} \sqrt {2 x^2-x+3}-\frac {223 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{64 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(121*(19 - 7*x))/(92*Sqrt[3 - x + 2*x^2]) + (415*Sqrt[3 - x + 2*x^2])/32 + (25*x*Sqrt[3 - x + 2*x^2])/8 - (223
*ArcSinh[(1 - 4*x)/Sqrt[23]])/(64*Sqrt[2])

Rule 221

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 633

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 654

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 1674

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 1675

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 {\left (2+3 x+5 x^2\right )^2}{\left (3-x+2 x^2\right )^{3/2}} \, dx &=\frac {121 (19-7 x)}{92 \sqrt {3-x+2 x^2}}+\frac {2}{23} \int \frac {\frac {1173}{16}+\frac {1955 x}{8}+\frac {575 x^2}{4}}{\sqrt {3-x+2 x^2}} \, dx\\ &=\frac {121 (19-7 x)}{92 \sqrt {3-x+2 x^2}}+\frac {25}{8} x \sqrt {3-x+2 x^2}+\frac {1}{46} \int \frac {-138+\frac {9545 x}{8}}{\sqrt {3-x+2 x^2}} \, dx\\ &=\frac {121 (19-7 x)}{92 \sqrt {3-x+2 x^2}}+\frac {415}{32} \sqrt {3-x+2 x^2}+\frac {25}{8} x \sqrt {3-x+2 x^2}+\frac {223}{64} \int \frac {1}{\sqrt {3-x+2 x^2}} \, dx\\ &=\frac {121 (19-7 x)}{92 \sqrt {3-x+2 x^2}}+\frac {415}{32} \sqrt {3-x+2 x^2}+\frac {25}{8} x \sqrt {3-x+2 x^2}+\frac {223 \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+4 x\right )}{64 \sqrt {46}}\\ &=\frac {121 (19-7 x)}{92 \sqrt {3-x+2 x^2}}+\frac {415}{32} \sqrt {3-x+2 x^2}+\frac {25}{8} x \sqrt {3-x+2 x^2}-\frac {223 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{64 \sqrt {2}}\\ \end {align*}

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Mathematica [A]
time = 0.49, size = 65, normalized size = 0.79 \begin {gather*} \frac {47027-9421 x+16790 x^2+4600 x^3}{736 \sqrt {3-x+2 x^2}}-\frac {223 \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )}{64 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(47027 - 9421*x + 16790*x^2 + 4600*x^3)/(736*Sqrt[3 - x + 2*x^2]) - (223*Log[1 - 4*x + 2*Sqrt[6 - 2*x + 4*x^2]
])/(64*Sqrt[2])

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Maple [A]
time = 0.13, size = 98, normalized size = 1.20

method	result	size
risch	\(\frac {4600 x^{3}+16790 x^{2}-9421 x +47027}{736 \sqrt {2 x^{2}-x +3}}+\frac {223 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{128}\)	\(45\)
trager	\(\frac {4600 x^{3}+16790 x^{2}-9421 x +47027}{736 \sqrt {2 x^{2}-x +3}}+\frac {223 \RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (4 \RootOf \left (\textit {\_Z}^{2}-2\right ) x -\RootOf \left (\textit {\_Z}^{2}-2\right )+4 \sqrt {2 x^{2}-x +3}\right )}{128}\)	\(72\)
default	\(\frac {25 x^{3}}{4 \sqrt {2 x^{2}-x +3}}+\frac {365 x^{2}}{16 \sqrt {2 x^{2}-x +3}}-\frac {223 x}{64 \sqrt {2 x^{2}-x +3}}+\frac {15761}{256 \sqrt {2 x^{2}-x +3}}-\frac {13713 \left (4 x -1\right )}{5888 \sqrt {2 x^{2}-x +3}}+\frac {223 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{128}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2+3*x+2)^2/(2*x^2-x+3)^(3/2),x,method=_RETURNVERBOSE)

[Out]

25/4*x^3/(2*x^2-x+3)^(1/2)+365/16*x^2/(2*x^2-x+3)^(1/2)-223/64*x/(2*x^2-x+3)^(1/2)+15761/256/(2*x^2-x+3)^(1/2)
-13713/5888*(4*x-1)/(2*x^2-x+3)^(1/2)+223/128*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))

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Maxima [A]
time = 0.50, size = 80, normalized size = 0.98 \begin {gather*} \frac {25 \, x^{3}}{4 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {365 \, x^{2}}{16 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {223}{128} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) - \frac {9421 \, x}{736 \, \sqrt {2 \, x^{2} - x + 3}} + \frac {47027}{736 \, \sqrt {2 \, x^{2} - x + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/4*x^3/sqrt(2*x^2 - x + 3) + 365/16*x^2/sqrt(2*x^2 - x + 3) + 223/128*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1
)) - 9421/736*x/sqrt(2*x^2 - x + 3) + 47027/736/sqrt(2*x^2 - x + 3)

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Fricas [A]
time = 1.75, size = 92, normalized size = 1.12 \begin {gather*} \frac {5129 \, \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 (4600 \, x^{3} + 16790 \, x^{2} - 9421 \, x + 47027\right )} \sqrt {2 \, x^{2} - x + 3}}{5888 \, {\left (2 \, x^{2} - x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5888*(5129*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*(4
600*x^3 + 16790*x^2 - 9421*x + 47027)*sqrt(2*x^2 - x + 3))/(2*x^2 - x + 3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (5 x^{2} + 3 x + 2\right )^{2}}{\left (2 x^{2} - x + 3\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.51, size = 62, normalized size = 0.76 \begin {gather*} -\frac {223}{128} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) + \frac {{\left (230 \, {\left (20 \, x + 73\right )} x - 9421\right )} x + 47027}{736 \, \sqrt {2 \, x^{2} - x + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-223/128*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) + 1/736*((230*(20*x + 73)*x - 9421)*x +
47027)/sqrt(2*x^2 - x + 3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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