### 3.128 $$\int \frac{1+3 x+4 x^2}{(1+2 x)^2 (2+3 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=75 $-\frac{10-97 x}{242 \sqrt{3 x^2+2}}-\frac{4 \sqrt{3 x^2+2}}{121 (2 x+1)}+\frac{4 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{3 x^2+2}}\right )}{121 \sqrt{11}}$

[Out]

-(10 - 97*x)/(242*Sqrt[2 + 3*x^2]) - (4*Sqrt[2 + 3*x^2])/(121*(1 + 2*x)) + (4*ArcTanh[(4 - 3*x)/(Sqrt[11]*Sqrt
[2 + 3*x^2])])/(121*Sqrt[11])

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Rubi [A]  time = 0.0772505, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.138, Rules used = {1647, 807, 725, 206} $-\frac{10-97 x}{242 \sqrt{3 x^2+2}}-\frac{4 \sqrt{3 x^2+2}}{121 (2 x+1)}+\frac{4 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{3 x^2+2}}\right )}{121 \sqrt{11}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(10 - 97*x)/(242*Sqrt[2 + 3*x^2]) - (4*Sqrt[2 + 3*x^2])/(121*(1 + 2*x)) + (4*ArcTanh[(4 - 3*x)/(Sqrt[11]*Sqrt
[2 + 3*x^2])])/(121*Sqrt[11])

Rule 1647

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

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 725

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

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+3 x^2\right )^{3/2}} \, dx &=-\frac{10-97 x}{242 \sqrt{2+3 x^2}}-\frac{1}{6} \int \frac{-\frac{72}{121}+\frac{120 x}{121}}{(1+2 x)^2 \sqrt{2+3 x^2}} \, dx\\ &=-\frac{10-97 x}{242 \sqrt{2+3 x^2}}-\frac{4 \sqrt{2+3 x^2}}{121 (1+2 x)}-\frac{4}{121} \int \frac{1}{(1+2 x) \sqrt{2+3 x^2}} \, dx\\ &=-\frac{10-97 x}{242 \sqrt{2+3 x^2}}-\frac{4 \sqrt{2+3 x^2}}{121 (1+2 x)}+\frac{4}{121} \operatorname{Subst}\left (\int \frac{1}{11-x^2} \, dx,x,\frac{4-3 x}{\sqrt{2+3 x^2}}\right )\\ &=-\frac{10-97 x}{242 \sqrt{2+3 x^2}}-\frac{4 \sqrt{2+3 x^2}}{121 (1+2 x)}+\frac{4 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{2+3 x^2}}\right )}{121 \sqrt{11}}\\ \end{align*}

Mathematica [A]  time = 0.0419811, size = 71, normalized size = 0.95 $\frac{11 \left (170 x^2+77 x-26\right )+8 (2 x+1) \sqrt{33 x^2+22} \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{33 x^2+22}}\right )}{2662 (2 x+1) \sqrt{3 x^2+2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(11*(-26 + 77*x + 170*x^2) + 8*(1 + 2*x)*Sqrt[22 + 33*x^2]*ArcTanh[(4 - 3*x)/Sqrt[22 + 33*x^2]])/(2662*(1 + 2*
x)*Sqrt[2 + 3*x^2])

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Maple [A]  time = 0.057, size = 98, normalized size = 1.3 \begin{align*}{\frac{x}{2}{\frac{1}{\sqrt{3\,{x}^{2}+2}}}}-{\frac{1}{22} \left ( x+{\frac{1}{2}} \right ) ^{-1}{\frac{1}{\sqrt{3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}}}}}}-{\frac{2}{121}{\frac{1}{\sqrt{3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}}}}}}-{\frac{18\,x}{121}{\frac{1}{\sqrt{3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}}}}}}+{\frac{4\,\sqrt{11}}{1331}{\it Artanh} \left ({\frac{ \left ( 8-6\,x \right ) \sqrt{11}}{11}{\frac{1}{\sqrt{12\, \left ( x+1/2 \right ) ^{2}-12\,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)^2/(3*x^2+2)^(3/2),x)

[Out]

1/2*x/(3*x^2+2)^(1/2)-1/22/(x+1/2)/(3*(x+1/2)^2-3*x+5/4)^(1/2)-2/121/(3*(x+1/2)^2-3*x+5/4)^(1/2)-18/121*x/(3*(
x+1/2)^2-3*x+5/4)^(1/2)+4/1331*11^(1/2)*arctanh(2/11*(4-3*x)*11^(1/2)/(12*(x+1/2)^2-12*x+5)^(1/2))

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Maxima [A]  time = 1.48201, size = 113, normalized size = 1.51 \begin{align*} -\frac{4}{1331} \, \sqrt{11} \operatorname{arsinh}\left (\frac{\sqrt{6} x}{2 \,{\left | 2 \, x + 1 \right |}} - \frac{2 \, \sqrt{6}}{3 \,{\left | 2 \, x + 1 \right |}}\right ) + \frac{85 \, x}{242 \, \sqrt{3 \, x^{2} + 2}} - \frac{2}{121 \, \sqrt{3 \, x^{2} + 2}} - \frac{1}{11 \,{\left (2 \, \sqrt{3 \, x^{2} + 2} x + \sqrt{3 \, x^{2} + 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+2)^(3/2),x, algorithm="maxima")

[Out]

-4/1331*sqrt(11)*arcsinh(1/2*sqrt(6)*x/abs(2*x + 1) - 2/3*sqrt(6)/abs(2*x + 1)) + 85/242*x/sqrt(3*x^2 + 2) - 2
/121/sqrt(3*x^2 + 2) - 1/11/(2*sqrt(3*x^2 + 2)*x + sqrt(3*x^2 + 2))

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Fricas [A]  time = 1.70027, size = 266, normalized size = 3.55 \begin{align*} \frac{4 \, \sqrt{11}{\left (6 \, x^{3} + 3 \, x^{2} + 4 \, x + 2\right )} \log \left (\frac{\sqrt{11} \sqrt{3 \, x^{2} + 2}{\left (3 \, x - 4\right )} - 21 \, x^{2} + 12 \, x - 19}{4 \, x^{2} + 4 \, x + 1}\right ) + 11 \,{\left (170 \, x^{2} + 77 \, x - 26\right )} \sqrt{3 \, x^{2} + 2}}{2662 \,{\left (6 \, x^{3} + 3 \, x^{2} + 4 \, 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+2)^(3/2),x, algorithm="fricas")

[Out]

1/2662*(4*sqrt(11)*(6*x^3 + 3*x^2 + 4*x + 2)*log((sqrt(11)*sqrt(3*x^2 + 2)*(3*x - 4) - 21*x^2 + 12*x - 19)/(4*
x^2 + 4*x + 1)) + 11*(170*x^2 + 77*x - 26)*sqrt(3*x^2 + 2))/(6*x^3 + 3*x^2 + 4*x + 2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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+2)**(3/2),x)

[Out]

Timed out

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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} + 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+2)^(3/2),x, algorithm="giac")

[Out]

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