3.211 $$\int \frac{\sqrt{2-x+3 x^2} (1+3 x+4 x^2)}{1+2 x} \, dx$$

Optimal. Leaf size=101 $\frac{2}{9} \left (3 x^2-x+2\right )^{3/2}+\frac{1}{72} (30 x+13) \sqrt{3 x^2-x+2}-\frac{1}{8} \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )-\frac{43 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{144 \sqrt{3}}$

[Out]

((13 + 30*x)*Sqrt[2 - x + 3*x^2])/72 + (2*(2 - x + 3*x^2)^(3/2))/9 - (43*ArcSinh[(1 - 6*x)/Sqrt[23]])/(144*Sqr
t[3]) - (Sqrt[13]*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/8

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Rubi [A]  time = 0.116899, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 32, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.219, Rules used = {1653, 814, 843, 619, 215, 724, 206} $\frac{2}{9} \left (3 x^2-x+2\right )^{3/2}+\frac{1}{72} (30 x+13) \sqrt{3 x^2-x+2}-\frac{1}{8} \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )-\frac{43 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{144 \sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((13 + 30*x)*Sqrt[2 - x + 3*x^2])/72 + (2*(2 - x + 3*x^2)^(3/2))/9 - (43*ArcSinh[(1 - 6*x)/Sqrt[23]])/(144*Sqr
t[3]) - (Sqrt[13]*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/8

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 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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{\sqrt{2-x+3 x^2} \left (1+3 x+4 x^2\right )}{1+2 x} \, dx &=\frac{2}{9} \left (2-x+3 x^2\right )^{3/2}+\frac{1}{36} \int \frac{(48+60 x) \sqrt{2-x+3 x^2}}{1+2 x} \, dx\\ &=\frac{1}{72} (13+30 x) \sqrt{2-x+3 x^2}+\frac{2}{9} \left (2-x+3 x^2\right )^{3/2}-\frac{\int \frac{-3324-1032 x}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx}{1728}\\ &=\frac{1}{72} (13+30 x) \sqrt{2-x+3 x^2}+\frac{2}{9} \left (2-x+3 x^2\right )^{3/2}+\frac{43}{144} \int \frac{1}{\sqrt{2-x+3 x^2}} \, dx+\frac{13}{8} \int \frac{1}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx\\ &=\frac{1}{72} (13+30 x) \sqrt{2-x+3 x^2}+\frac{2}{9} \left (2-x+3 x^2\right )^{3/2}-\frac{13}{4} \operatorname{Subst}\left (\int \frac{1}{52-x^2} \, dx,x,\frac{9-8 x}{\sqrt{2-x+3 x^2}}\right )+\frac{43 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+6 x\right )}{144 \sqrt{69}}\\ &=\frac{1}{72} (13+30 x) \sqrt{2-x+3 x^2}+\frac{2}{9} \left (2-x+3 x^2\right )^{3/2}-\frac{43 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{144 \sqrt{3}}-\frac{1}{8} \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{2-x+3 x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.055795, size = 86, normalized size = 0.85 $\frac{1}{432} \left (6 \sqrt{3 x^2-x+2} \left (48 x^2+14 x+45\right )-54 \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )+43 \sqrt{3} \sinh ^{-1}\left (\frac{6 x-1}{\sqrt{23}}\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(6*Sqrt[2 - x + 3*x^2]*(45 + 14*x + 48*x^2) + 43*Sqrt[3]*ArcSinh[(-1 + 6*x)/Sqrt[23]] - 54*Sqrt[13]*ArcTanh[(9
- 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/432

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Maple [A]  time = 0.054, size = 95, normalized size = 0.9 \begin{align*}{\frac{2}{9} \left ( 3\,{x}^{2}-x+2 \right ) ^{{\frac{3}{2}}}}+{\frac{-5+30\,x}{72}\sqrt{3\,{x}^{2}-x+2}}+{\frac{43\,\sqrt{3}}{432}{\it Arcsinh} \left ({\frac{6\,\sqrt{23}}{23} \left ( x-{\frac{1}{6}} \right ) } \right ) }+{\frac{1}{8}\sqrt{12\, \left ( x+1/2 \right ) ^{2}-16\,x+5}}-{\frac{\sqrt{13}}{8}{\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)*(3*x^2-x+2)^(1/2)/(1+2*x),x)

[Out]

2/9*(3*x^2-x+2)^(3/2)+5/72*(-1+6*x)*(3*x^2-x+2)^(1/2)+43/432*3^(1/2)*arcsinh(6/23*23^(1/2)*(x-1/6))+1/8*(12*(x
+1/2)^2-16*x+5)^(1/2)-1/8*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 = 1.52865, size = 130, normalized size = 1.29 \begin{align*} \frac{2}{9} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}} + \frac{5}{12} \, \sqrt{3 \, x^{2} - x + 2} x + \frac{43}{432} \, \sqrt{3} \operatorname{arsinh}\left (\frac{6}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) + \frac{1}{8} \, \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{13}{72} \, \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)*(3*x^2-x+2)^(1/2)/(1+2*x),x, algorithm="maxima")

[Out]

2/9*(3*x^2 - x + 2)^(3/2) + 5/12*sqrt(3*x^2 - x + 2)*x + 43/432*sqrt(3)*arcsinh(6/23*sqrt(23)*x - 1/23*sqrt(23
)) + 1/8*sqrt(13)*arcsinh(8/23*sqrt(23)*x/abs(2*x + 1) - 9/23*sqrt(23)/abs(2*x + 1)) + 13/72*sqrt(3*x^2 - x +
2)

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Fricas [A]  time = 1.6215, size = 321, normalized size = 3.18 \begin{align*} \frac{1}{72} \,{\left (48 \, x^{2} + 14 \, x + 45\right )} \sqrt{3 \, x^{2} - x + 2} + \frac{43}{864} \, \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}{16} \, \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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(48*x^2 + 14*x + 45)*sqrt(3*x^2 - x + 2) + 43/864*sqrt(3)*log(-4*sqrt(3)*sqrt(3*x^2 - x + 2)*(6*x - 1) -
72*x^2 + 24*x - 25) + 1/16*sqrt(13)*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))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.261, size = 170, normalized size = 1.68 \begin{align*} \frac{1}{72} \,{\left (2 \,{\left (24 \, x + 7\right )} x + 45\right )} \sqrt{3 \, x^{2} - x + 2} - \frac{43}{432} \, \sqrt{3} \log \left (-6 \, \sqrt{3} x + \sqrt{3} + 6 \, \sqrt{3 \, x^{2} - x + 2}\right ) + \frac{1}{8} \, \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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(2*(24*x + 7)*x + 45)*sqrt(3*x^2 - x + 2) - 43/432*sqrt(3)*log(-6*sqrt(3)*x + sqrt(3) + 6*sqrt(3*x^2 - x
+ 2)) + 1/8*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)))