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

Optimal. Leaf size=78 \[ \frac{2}{3} \sqrt{3 x^2-x+2}-\frac{\tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{2 \sqrt{13}}-\frac{5 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{6 \sqrt{3}} \]

[Out]

(2*Sqrt[2 - x + 3*x^2])/3 - (5*ArcSinh[(1 - 6*x)/Sqrt[23]])/(6*Sqrt[3]) - ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2
- x + 3*x^2])]/(2*Sqrt[13])

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Rubi [A]  time = 0.0969312, antiderivative size = 78, normalized size of antiderivative = 1., 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 = {1653, 843, 619, 215, 724, 206} \[ \frac{2}{3} \sqrt{3 x^2-x+2}-\frac{\tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{2 \sqrt{13}}-\frac{5 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{6 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2 - x + 3*x^2])/3 - (5*ArcSinh[(1 - 6*x)/Sqrt[23]])/(6*Sqrt[3]) - ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2
- x + 3*x^2])]/(2*Sqrt[13])

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 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 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 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 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 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])

Rubi steps

\begin{align*} \int \frac{1+3 x+4 x^2}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx &=\frac{2}{3} \sqrt{2-x+3 x^2}+\frac{1}{12} \int \frac{16+20 x}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx\\ &=\frac{2}{3} \sqrt{2-x+3 x^2}+\frac{1}{2} \int \frac{1}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx+\frac{5}{6} \int \frac{1}{\sqrt{2-x+3 x^2}} \, dx\\ &=\frac{2}{3} \sqrt{2-x+3 x^2}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+6 x\right )}{6 \sqrt{69}}-\operatorname{Subst}\left (\int \frac{1}{52-x^2} \, dx,x,\frac{9-8 x}{\sqrt{2-x+3 x^2}}\right )\\ &=\frac{2}{3} \sqrt{2-x+3 x^2}-\frac{5 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{6 \sqrt{3}}-\frac{\tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{2-x+3 x^2}}\right )}{2 \sqrt{13}}\\ \end{align*}

Mathematica [A]  time = 0.0371155, size = 78, normalized size = 1. \[ \frac{2}{3} \sqrt{3 x^2-x+2}-\frac{\tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{2 \sqrt{13}}+\frac{5 \sinh ^{-1}\left (\frac{6 x-1}{\sqrt{23}}\right )}{6 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2 - x + 3*x^2])/3 + (5*ArcSinh[(-1 + 6*x)/Sqrt[23]])/(6*Sqrt[3]) - ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[
2 - x + 3*x^2])]/(2*Sqrt[13])

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Maple [A]  time = 0.051, size = 60, normalized size = 0.8 \begin{align*}{\frac{2}{3}\sqrt{3\,{x}^{2}-x+2}}+{\frac{5\,\sqrt{3}}{18}{\it Arcsinh} \left ({\frac{6\,\sqrt{23}}{23} \left ( x-{\frac{1}{6}} \right ) } \right ) }-{\frac{\sqrt{13}}{26}{\it Artanh} \left ({\frac{2\,\sqrt{13}}{13} \left ({\frac{9}{2}}-4\,x \right ){\frac{1}{\sqrt{12\, \left ( x+1/2 \right ) ^{2}-16\,x+5}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*x^2-x+2)^(1/2)+5/18*3^(1/2)*arcsinh(6/23*23^(1/2)*(x-1/6))-1/26*13^(1/2)*arctanh(2/13*(9/2-4*x)*13^(1/2
)/(12*(x+1/2)^2-16*x+5)^(1/2))

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Maxima [A]  time = 1.51391, size = 90, normalized size = 1.15 \begin{align*} \frac{5}{18} \, \sqrt{3} \operatorname{arsinh}\left (\frac{6}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) + \frac{1}{26} \, \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{2}{3} \, \sqrt{3 \, x^{2} - x + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/18*sqrt(3)*arcsinh(6/23*sqrt(23)*x - 1/23*sqrt(23)) + 1/26*sqrt(13)*arcsinh(8/23*sqrt(23)*x/abs(2*x + 1) - 9
/23*sqrt(23)/abs(2*x + 1)) + 2/3*sqrt(3*x^2 - x + 2)

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Fricas [A]  time = 1.65931, size = 289, normalized size = 3.71 \begin{align*} \frac{5}{36} \, \sqrt{3} \log \left (-4 \, \sqrt{3} \sqrt{3 \, x^{2} - x + 2}{\left (6 \, x - 1\right )} - 72 \, x^{2} + 24 \, x - 25\right ) + \frac{1}{52} \, \sqrt{13} \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 ) + \frac{2}{3} \, \sqrt{3 \, x^{2} - x + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/36*sqrt(3)*log(-4*sqrt(3)*sqrt(3*x^2 - x + 2)*(6*x - 1) - 72*x^2 + 24*x - 25) + 1/52*sqrt(13)*log(-(4*sqrt(1
3)*sqrt(3*x^2 - x + 2)*(8*x - 9) + 220*x^2 - 196*x + 185)/(4*x^2 + 4*x + 1)) + 2/3*sqrt(3*x^2 - x + 2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{4 x^{2} + 3 x + 1}{\left (2 x + 1\right ) \sqrt{3 x^{2} - x + 2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.24688, size = 157, normalized size = 2.01 \begin{align*} -\frac{5}{18} \, \sqrt{3} \log \left (-6 \, \sqrt{3} x + \sqrt{3} + 6 \, \sqrt{3 \, x^{2} - x + 2}\right ) + \frac{1}{26} \, \sqrt{13} \log \left (-\frac{{\left | -4 \, \sqrt{3} x - 2 \, \sqrt{13} - 2 \, \sqrt{3} + 4 \, \sqrt{3 \, x^{2} - x + 2} \right |}}{2 \,{\left (2 \, \sqrt{3} x - \sqrt{13} + \sqrt{3} - 2 \, \sqrt{3 \, x^{2} - x + 2}\right )}}\right ) + \frac{2}{3} \, \sqrt{3 \, x^{2} - x + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/18*sqrt(3)*log(-6*sqrt(3)*x + sqrt(3) + 6*sqrt(3*x^2 - x + 2)) + 1/26*sqrt(13)*log(-1/2*abs(-4*sqrt(3)*x -
2*sqrt(13) - 2*sqrt(3) + 4*sqrt(3*x^2 - x + 2))/(2*sqrt(3)*x - sqrt(13) + sqrt(3) - 2*sqrt(3*x^2 - x + 2))) +
2/3*sqrt(3*x^2 - x + 2)