### 3.134 $$\int \frac{1+3 x+4 x^2}{(1+2 x)^2 (2+3 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=95 $-\frac{10-97 x}{726 \left (3 x^2+2\right )^{3/2}}-\frac{16 \sqrt{3 x^2+2}}{1331 (2 x+1)}+\frac{887 x+24}{7986 \sqrt{3 x^2+2}}-\frac{32 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{3 x^2+2}}\right )}{1331 \sqrt{11}}$

[Out]

-(10 - 97*x)/(726*(2 + 3*x^2)^(3/2)) + (24 + 887*x)/(7986*Sqrt[2 + 3*x^2]) - (16*Sqrt[2 + 3*x^2])/(1331*(1 + 2
*x)) - (32*ArcTanh[(4 - 3*x)/(Sqrt[11]*Sqrt[2 + 3*x^2])])/(1331*Sqrt[11])

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Rubi [A]  time = 0.169063, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.138, Rules used = {1647, 807, 725, 206} $-\frac{10-97 x}{726 \left (3 x^2+2\right )^{3/2}}-\frac{16 \sqrt{3 x^2+2}}{1331 (2 x+1)}+\frac{887 x+24}{7986 \sqrt{3 x^2+2}}-\frac{32 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{3 x^2+2}}\right )}{1331 \sqrt{11}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(1 + 3*x + 4*x^2)/((1 + 2*x)^2*(2 + 3*x^2)^(5/2)),x]

[Out]

-(10 - 97*x)/(726*(2 + 3*x^2)^(3/2)) + (24 + 887*x)/(7986*Sqrt[2 + 3*x^2]) - (16*Sqrt[2 + 3*x^2])/(1331*(1 + 2
*x)) - (32*ArcTanh[(4 - 3*x)/(Sqrt[11]*Sqrt[2 + 3*x^2])])/(1331*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 )^{5/2}} \, dx &=-\frac{10-97 x}{726 \left (2+3 x^2\right )^{3/2}}-\frac{1}{18} \int \frac{-\frac{798}{121}-\frac{1968 x}{121}-\frac{2328 x^2}{121}}{(1+2 x)^2 \left (2+3 x^2\right )^{3/2}} \, dx\\ &=-\frac{10-97 x}{726 \left (2+3 x^2\right )^{3/2}}+\frac{24+887 x}{7986 \sqrt{2+3 x^2}}+\frac{1}{108} \int \frac{\frac{10368}{1331}+\frac{1728 x}{1331}}{(1+2 x)^2 \sqrt{2+3 x^2}} \, dx\\ &=-\frac{10-97 x}{726 \left (2+3 x^2\right )^{3/2}}+\frac{24+887 x}{7986 \sqrt{2+3 x^2}}-\frac{16 \sqrt{2+3 x^2}}{1331 (1+2 x)}+\frac{32 \int \frac{1}{(1+2 x) \sqrt{2+3 x^2}} \, dx}{1331}\\ &=-\frac{10-97 x}{726 \left (2+3 x^2\right )^{3/2}}+\frac{24+887 x}{7986 \sqrt{2+3 x^2}}-\frac{16 \sqrt{2+3 x^2}}{1331 (1+2 x)}-\frac{32 \operatorname{Subst}\left (\int \frac{1}{11-x^2} \, dx,x,\frac{4-3 x}{\sqrt{2+3 x^2}}\right )}{1331}\\ &=-\frac{10-97 x}{726 \left (2+3 x^2\right )^{3/2}}+\frac{24+887 x}{7986 \sqrt{2+3 x^2}}-\frac{16 \sqrt{2+3 x^2}}{1331 (1+2 x)}-\frac{32 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{2+3 x^2}}\right )}{1331 \sqrt{11}}\\ \end{align*}

Mathematica [A]  time = 0.0632395, size = 91, normalized size = 0.96 $\frac{11 \left (4458 x^4+2805 x^3+4602 x^2+2717 x-446\right )-192 \sqrt{33 x^2+22} \left (6 x^3+3 x^2+4 x+2\right ) \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{33 x^2+22}}\right )}{87846 (2 x+1) \left (3 x^2+2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(1 + 3*x + 4*x^2)/((1 + 2*x)^2*(2 + 3*x^2)^(5/2)),x]

[Out]

(11*(-446 + 2717*x + 4602*x^2 + 2805*x^3 + 4458*x^4) - 192*Sqrt[22 + 33*x^2]*(2 + 4*x + 3*x^2 + 6*x^3)*ArcTanh
[(4 - 3*x)/Sqrt[22 + 33*x^2]])/(87846*(1 + 2*x)*(2 + 3*x^2)^(3/2))

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Maple [A]  time = 0.059, size = 143, normalized size = 1.5 \begin{align*}{\frac{x}{6} \left ( 3\,{x}^{2}+2 \right ) ^{-{\frac{3}{2}}}}+{\frac{x}{6}{\frac{1}{\sqrt{3\,{x}^{2}+2}}}}-{\frac{1}{22} \left ( x+{\frac{1}{2}} \right ) ^{-1} \left ( 3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}} \right ) ^{-{\frac{3}{2}}}}+{\frac{4}{363} \left ( 3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}} \right ) ^{-{\frac{3}{2}}}}-{\frac{10\,x}{121} \left ( 3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}} \right ) ^{-{\frac{3}{2}}}}-{\frac{98\,x}{1331}{\frac{1}{\sqrt{3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}}}}}}+{\frac{16}{1331}{\frac{1}{\sqrt{3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}}}}}}-{\frac{32\,\sqrt{11}}{14641}{\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)^(5/2),x)

[Out]

1/6*x/(3*x^2+2)^(3/2)+1/6*x/(3*x^2+2)^(1/2)-1/22/(x+1/2)/(3*(x+1/2)^2-3*x+5/4)^(3/2)+4/363/(3*(x+1/2)^2-3*x+5/
4)^(3/2)-10/121*x/(3*(x+1/2)^2-3*x+5/4)^(3/2)-98/1331*x/(3*(x+1/2)^2-3*x+5/4)^(1/2)+16/1331/(3*(x+1/2)^2-3*x+5
/4)^(1/2)-32/14641*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 = 1.50504, size = 144, normalized size = 1.52 \begin{align*} \frac{32}{14641} \, \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{743 \, x}{7986 \, \sqrt{3 \, x^{2} + 2}} + \frac{16}{1331 \, \sqrt{3 \, x^{2} + 2}} + \frac{61 \, x}{726 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}} - \frac{1}{11 \,{\left (2 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}} x +{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}\right )}} + \frac{4}{363 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}} \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)^(5/2),x, algorithm="maxima")

[Out]

32/14641*sqrt(11)*arcsinh(1/2*sqrt(6)*x/abs(2*x + 1) - 2/3*sqrt(6)/abs(2*x + 1)) + 743/7986*x/sqrt(3*x^2 + 2)
+ 16/1331/sqrt(3*x^2 + 2) + 61/726*x/(3*x^2 + 2)^(3/2) - 1/11/(2*(3*x^2 + 2)^(3/2)*x + (3*x^2 + 2)^(3/2)) + 4/
363/(3*x^2 + 2)^(3/2)

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Fricas [A]  time = 1.65366, size = 356, normalized size = 3.75 \begin{align*} \frac{96 \, \sqrt{11}{\left (18 \, x^{5} + 9 \, x^{4} + 24 \, x^{3} + 12 \, x^{2} + 8 \, x + 4\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 (4458 \, x^{4} + 2805 \, x^{3} + 4602 \, x^{2} + 2717 \, x - 446\right )} \sqrt{3 \, x^{2} + 2}}{87846 \,{\left (18 \, x^{5} + 9 \, x^{4} + 24 \, x^{3} + 12 \, x^{2} + 8 \, x + 4\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)^(5/2),x, algorithm="fricas")

[Out]

1/87846*(96*sqrt(11)*(18*x^5 + 9*x^4 + 24*x^3 + 12*x^2 + 8*x + 4)*log(-(sqrt(11)*sqrt(3*x^2 + 2)*(3*x - 4) + 2
1*x^2 - 12*x + 19)/(4*x^2 + 4*x + 1)) + 11*(4458*x^4 + 2805*x^3 + 4602*x^2 + 2717*x - 446)*sqrt(3*x^2 + 2))/(1
8*x^5 + 9*x^4 + 24*x^3 + 12*x^2 + 8*x + 4)

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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)**(5/2),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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate((4*x^2 + 3*x + 1)/((3*x^2 + 2)^(5/2)*(2*x + 1)^2), x)