3.244 \(\int \frac {1+3 x+4 x^2}{(1+2 x)^2 \sqrt {2-x+3 x^2}} \, dx\)

Optimal. Leaf size=83 \[ -\frac {\sqrt {3 x^2-x+2}}{13 (2 x+1)}+\frac {9 \tanh ^{-1}\left (\frac {9-8 x}{2 \sqrt {13} \sqrt {3 x^2-x+2}}\right )}{26 \sqrt {13}}-\frac {\sinh ^{-1}\left (\frac {1-6 x}{\sqrt {23}}\right )}{\sqrt {3}} \]

[Out]

-1/3*arcsinh(1/23*(1-6*x)*23^(1/2))*3^(1/2)+9/338*arctanh(1/26*(9-8*x)*13^(1/2)/(3*x^2-x+2)^(1/2))*13^(1/2)-1/
13*(3*x^2-x+2)^(1/2)/(1+2*x)

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Rubi [A]  time = 0.10, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1650, 843, 619, 215, 724, 206} \[ -\frac {\sqrt {3 x^2-x+2}}{13 (2 x+1)}+\frac {9 \tanh ^{-1}\left (\frac {9-8 x}{2 \sqrt {13} \sqrt {3 x^2-x+2}}\right )}{26 \sqrt {13}}-\frac {\sinh ^{-1}\left (\frac {1-6 x}{\sqrt {23}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 3*x + 4*x^2)/((1 + 2*x)^2*Sqrt[2 - x + 3*x^2]),x]

[Out]

-Sqrt[2 - x + 3*x^2]/(13*(1 + 2*x)) - ArcSinh[(1 - 6*x)/Sqrt[23]]/Sqrt[3] + (9*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*S
qrt[2 - x + 3*x^2])])/(26*Sqrt[13])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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]

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 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1650

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomia
lQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + b*x + c*
x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^
(m + 1)*(a + b*x + c*x^2)^p*ExpandToSum[(m + 1)*(c*d^2 - b*d*e + a*e^2)*Q + c*d*R*(m + 1) - b*e*R*(m + p + 2)
- c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {1+3 x+4 x^2}{(1+2 x)^2 \sqrt {2-x+3 x^2}} \, dx &=-\frac {\sqrt {2-x+3 x^2}}{13 (1+2 x)}-\frac {1}{13} \int \frac {-\frac {17}{2}-26 x}{(1+2 x) \sqrt {2-x+3 x^2}} \, dx\\ &=-\frac {\sqrt {2-x+3 x^2}}{13 (1+2 x)}-\frac {9}{26} \int \frac {1}{(1+2 x) \sqrt {2-x+3 x^2}} \, dx+\int \frac {1}{\sqrt {2-x+3 x^2}} \, dx\\ &=-\frac {\sqrt {2-x+3 x^2}}{13 (1+2 x)}+\frac {9}{13} \operatorname {Subst}\left (\int \frac {1}{52-x^2} \, dx,x,\frac {9-8 x}{\sqrt {2-x+3 x^2}}\right )+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+6 x\right )}{\sqrt {69}}\\ &=-\frac {\sqrt {2-x+3 x^2}}{13 (1+2 x)}-\frac {\sinh ^{-1}\left (\frac {1-6 x}{\sqrt {23}}\right )}{\sqrt {3}}+\frac {9 \tanh ^{-1}\left (\frac {9-8 x}{2 \sqrt {13} \sqrt {2-x+3 x^2}}\right )}{26 \sqrt {13}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 82, normalized size = 0.99 \[ -\frac {\sqrt {3 x^2-x+2}}{13 (2 x+1)}+\frac {9 \tanh ^{-1}\left (\frac {9-8 x}{2 \sqrt {13} \sqrt {3 x^2-x+2}}\right )}{26 \sqrt {13}}+\frac {\sinh ^{-1}\left (\frac {6 x-1}{\sqrt {23}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 3*x + 4*x^2)/((1 + 2*x)^2*Sqrt[2 - x + 3*x^2]),x]

[Out]

-1/13*Sqrt[2 - x + 3*x^2]/(1 + 2*x) + ArcSinh[(-1 + 6*x)/Sqrt[23]]/Sqrt[3] + (9*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*
Sqrt[2 - x + 3*x^2])])/(26*Sqrt[13])

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fricas [A]  time = 0.87, size = 123, normalized size = 1.48 \[ \frac {338 \, \sqrt {3} {\left (2 \, x + 1\right )} \log \left (-4 \, \sqrt {3} \sqrt {3 \, x^{2} - x + 2} {\left (6 \, x - 1\right )} - 72 \, x^{2} + 24 \, x - 25\right ) + 27 \, \sqrt {13} {\left (2 \, x + 1\right )} \log \left (\frac {4 \, \sqrt {13} \sqrt {3 \, x^{2} - x + 2} {\left (8 \, x - 9\right )} - 220 \, x^{2} + 196 \, x - 185}{4 \, x^{2} + 4 \, x + 1}\right ) - 156 \, \sqrt {3 \, x^{2} - x + 2}}{2028 \, {\left (2 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+3*x+1)/(1+2*x)^2/(3*x^2-x+2)^(1/2),x, algorithm="fricas")

[Out]

1/2028*(338*sqrt(3)*(2*x + 1)*log(-4*sqrt(3)*sqrt(3*x^2 - x + 2)*(6*x - 1) - 72*x^2 + 24*x - 25) + 27*sqrt(13)
*(2*x + 1)*log((4*sqrt(13)*sqrt(3*x^2 - x + 2)*(8*x - 9) - 220*x^2 + 196*x - 185)/(4*x^2 + 4*x + 1)) - 156*sqr
t(3*x^2 - x + 2))/(2*x + 1)

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giac [A]  time = 0.28, size = 48, normalized size = 0.58 \[ \frac {1}{26} \, \sqrt {3} \mathrm {sgn}\left (\frac {1}{2 \, x + 1}\right ) - \frac {\sqrt {-\frac {8}{2 \, x + 1} + \frac {13}{{\left (2 \, x + 1\right )}^{2}} + 3}}{26 \, \mathrm {sgn}\left (\frac {1}{2 \, x + 1}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+3*x+1)/(1+2*x)^2/(3*x^2-x+2)^(1/2),x, algorithm="giac")

[Out]

1/26*sqrt(3)*sgn(1/(2*x + 1)) - 1/26*sqrt(-8/(2*x + 1) + 13/(2*x + 1)^2 + 3)/sgn(1/(2*x + 1))

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maple [A]  time = 0.01, size = 67, normalized size = 0.81 \[ \frac {\sqrt {3}\, \arcsinh \left (\frac {6 \sqrt {23}\, \left (x -\frac {1}{6}\right )}{23}\right )}{3}+\frac {9 \sqrt {13}\, \arctanh \left (\frac {2 \left (-4 x +\frac {9}{2}\right ) \sqrt {13}}{13 \sqrt {-16 x +12 \left (x +\frac {1}{2}\right )^{2}+5}}\right )}{338}-\frac {\sqrt {-4 x +3 \left (x +\frac {1}{2}\right )^{2}+\frac {5}{4}}}{26 \left (x +\frac {1}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2+3*x+1)/(2*x+1)^2/(3*x^2-x+2)^(1/2),x)

[Out]

1/3*3^(1/2)*arcsinh(6/23*23^(1/2)*(x-1/6))+9/338*13^(1/2)*arctanh(2/13*(-4*x+9/2)*13^(1/2)/(-16*x+12*(x+1/2)^2
+5)^(1/2))-1/26/(x+1/2)*(-4*x+3*(x+1/2)^2+5/4)^(1/2)

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maxima [A]  time = 0.98, size = 74, normalized size = 0.89 \[ \frac {1}{3} \, \sqrt {3} \operatorname {arsinh}\left (\frac {6}{23} \, \sqrt {23} x - \frac {1}{23} \, \sqrt {23}\right ) - \frac {9}{338} \, \sqrt {13} \operatorname {arsinh}\left (\frac {8 \, \sqrt {23} x}{23 \, {\left | 2 \, x + 1 \right |}} - \frac {9 \, \sqrt {23}}{23 \, {\left | 2 \, x + 1 \right |}}\right ) - \frac {\sqrt {3 \, x^{2} - x + 2}}{13 \, {\left (2 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+3*x+1)/(1+2*x)^2/(3*x^2-x+2)^(1/2),x, algorithm="maxima")

[Out]

1/3*sqrt(3)*arcsinh(6/23*sqrt(23)*x - 1/23*sqrt(23)) - 9/338*sqrt(13)*arcsinh(8/23*sqrt(23)*x/abs(2*x + 1) - 9
/23*sqrt(23)/abs(2*x + 1)) - 1/13*sqrt(3*x^2 - x + 2)/(2*x + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + 4*x^2 + 1)/((2*x + 1)^2*(3*x^2 - x + 2)^(1/2)),x)

[Out]

int((3*x + 4*x^2 + 1)/((2*x + 1)^2*(3*x^2 - x + 2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {4 x^{2} + 3 x + 1}{\left (2 x + 1\right )^{2} \sqrt {3 x^{2} - x + 2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**2+3*x+1)/(1+2*x)**2/(3*x**2-x+2)**(1/2),x)

[Out]

Integral((4*x**2 + 3*x + 1)/((2*x + 1)**2*sqrt(3*x**2 - x + 2)), x)

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