### 3.250 $$\int \frac{1+3 x+4 x^2}{(1+2 x)^2 (2-x+3 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=87 $-\frac{2 (197-837 x)}{3887 \sqrt{3 x^2-x+2}}-\frac{4 \sqrt{3 x^2-x+2}}{169 (2 x+1)}+\frac{2 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{169 \sqrt{13}}$

[Out]

(-2*(197 - 837*x))/(3887*Sqrt[2 - x + 3*x^2]) - (4*Sqrt[2 - x + 3*x^2])/(169*(1 + 2*x)) + (2*ArcTanh[(9 - 8*x)
/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/(169*Sqrt[13])

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Rubi [A]  time = 0.092342, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 32, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {1646, 806, 724, 206} $-\frac{2 (197-837 x)}{3887 \sqrt{3 x^2-x+2}}-\frac{4 \sqrt{3 x^2-x+2}}{169 (2 x+1)}+\frac{2 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{169 \sqrt{13}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*(197 - 837*x))/(3887*Sqrt[2 - x + 3*x^2]) - (4*Sqrt[2 - x + 3*x^2])/(169*(1 + 2*x)) + (2*ArcTanh[(9 - 8*x)
/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/(169*Sqrt[13])

Rule 1646

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

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(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] && EqQ[Sim
plify[m + 2*p + 3], 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 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)^2 \left (2-x+3 x^2\right )^{3/2}} \, dx &=-\frac{2 (197-837 x)}{3887 \sqrt{2-x+3 x^2}}+\frac{2}{23} \int \frac{\frac{184}{169}-\frac{230 x}{169}}{(1+2 x)^2 \sqrt{2-x+3 x^2}} \, dx\\ &=-\frac{2 (197-837 x)}{3887 \sqrt{2-x+3 x^2}}-\frac{4 \sqrt{2-x+3 x^2}}{169 (1+2 x)}-\frac{2}{169} \int \frac{1}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx\\ &=-\frac{2 (197-837 x)}{3887 \sqrt{2-x+3 x^2}}-\frac{4 \sqrt{2-x+3 x^2}}{169 (1+2 x)}+\frac{4}{169} \operatorname{Subst}\left (\int \frac{1}{52-x^2} \, dx,x,\frac{9-8 x}{\sqrt{2-x+3 x^2}}\right )\\ &=-\frac{2 (197-837 x)}{3887 \sqrt{2-x+3 x^2}}-\frac{4 \sqrt{2-x+3 x^2}}{169 (1+2 x)}+\frac{2 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{2-x+3 x^2}}\right )}{169 \sqrt{13}}\\ \end{align*}

Mathematica [A]  time = 0.0459473, size = 74, normalized size = 0.85 $\frac{2 \left (1536 x^2+489 x-289\right )}{3887 (2 x+1) \sqrt{3 x^2-x+2}}+\frac{2 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{169 \sqrt{13}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(-289 + 489*x + 1536*x^2))/(3887*(1 + 2*x)*Sqrt[2 - x + 3*x^2]) + (2*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 -
x + 3*x^2])])/(169*Sqrt[13])

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Maple [A]  time = 0.055, size = 109, normalized size = 1.3 \begin{align*}{\frac{-2+12\,x}{23}{\frac{1}{\sqrt{3\,{x}^{2}-x+2}}}}-{\frac{1}{169}{\frac{1}{\sqrt{3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}}}}}}-{\frac{-82+492\,x}{3887}{\frac{1}{\sqrt{3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}}}}}}+{\frac{2\,\sqrt{13}}{2197}{\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 ) }-{\frac{1}{26} \left ( x+{\frac{1}{2}} \right ) ^{-1}{\frac{1}{\sqrt{3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/23*(-1+6*x)/(3*x^2-x+2)^(1/2)-1/169/(3*(x+1/2)^2-4*x+5/4)^(1/2)-82/3887*(-1+6*x)/(3*(x+1/2)^2-4*x+5/4)^(1/2)
+2/2197*13^(1/2)*arctanh(2/13*(9/2-4*x)*13^(1/2)/(12*(x+1/2)^2-16*x+5)^(1/2))-1/26/(x+1/2)/(3*(x+1/2)^2-4*x+5/
4)^(1/2)

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Maxima [A]  time = 1.50075, size = 130, normalized size = 1.49 \begin{align*} -\frac{2}{2197} \, \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{1536 \, x}{3887 \, \sqrt{3 \, x^{2} - x + 2}} - \frac{279}{3887 \, \sqrt{3 \, x^{2} - x + 2}} - \frac{1}{13 \,{\left (2 \, \sqrt{3 \, x^{2} - x + 2} x + \sqrt{3 \, x^{2} - x + 2}\right )}} \end{align*}

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)^(3/2),x, algorithm="maxima")

[Out]

-2/2197*sqrt(13)*arcsinh(8/23*sqrt(23)*x/abs(2*x + 1) - 9/23*sqrt(23)/abs(2*x + 1)) + 1536/3887*x/sqrt(3*x^2 -
x + 2) - 279/3887/sqrt(3*x^2 - x + 2) - 1/13/(2*sqrt(3*x^2 - x + 2)*x + sqrt(3*x^2 - x + 2))

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Fricas [A]  time = 1.15101, size = 285, normalized size = 3.28 \begin{align*} \frac{23 \, \sqrt{13}{\left (6 \, x^{3} + x^{2} + 3 \, x + 2\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 ) + 26 \,{\left (1536 \, x^{2} + 489 \, x - 289\right )} \sqrt{3 \, x^{2} - x + 2}}{50531 \,{\left (6 \, x^{3} + x^{2} + 3 \, x + 2\right )}} \end{align*}

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)^(3/2),x, algorithm="fricas")

[Out]

1/50531*(23*sqrt(13)*(6*x^3 + x^2 + 3*x + 2)*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)) + 26*(1536*x^2 + 489*x - 289)*sqrt(3*x^2 - x + 2))/(6*x^3 + x^2 + 3*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 )^{2} \left (3 x^{2} - x + 2\right )^{\frac{3}{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)**2/(3*x**2-x+2)**(3/2),x)

[Out]

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

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

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)^(3/2),x, algorithm="giac")

[Out]

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