### 3.133 $$\int \frac{1+3 x+4 x^2}{(1+2 x) (2+3 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=73 $-\frac{38-21 x}{198 \left (3 x^2+2\right )^{3/2}}+\frac{95 x+24}{726 \sqrt{3 x^2+2}}-\frac{8 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{3 x^2+2}}\right )}{121 \sqrt{11}}$

[Out]

-(38 - 21*x)/(198*(2 + 3*x^2)^(3/2)) + (24 + 95*x)/(726*Sqrt[2 + 3*x^2]) - (8*ArcTanh[(4 - 3*x)/(Sqrt[11]*Sqrt
[2 + 3*x^2])])/(121*Sqrt[11])

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Rubi [A]  time = 0.0850621, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.172, Rules used = {1647, 823, 12, 725, 206} $-\frac{38-21 x}{198 \left (3 x^2+2\right )^{3/2}}+\frac{95 x+24}{726 \sqrt{3 x^2+2}}-\frac{8 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{3 x^2+2}}\right )}{121 \sqrt{11}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(38 - 21*x)/(198*(2 + 3*x^2)^(3/2)) + (24 + 95*x)/(726*Sqrt[2 + 3*x^2]) - (8*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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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) \left (2+3 x^2\right )^{5/2}} \, dx &=-\frac{38-21 x}{198 \left (2+3 x^2\right )^{3/2}}-\frac{1}{18} \int \frac{-\frac{78}{11}-\frac{84 x}{11}}{(1+2 x) \left (2+3 x^2\right )^{3/2}} \, dx\\ &=-\frac{38-21 x}{198 \left (2+3 x^2\right )^{3/2}}+\frac{24+95 x}{726 \sqrt{2+3 x^2}}+\frac{\int \frac{864}{11 (1+2 x) \sqrt{2+3 x^2}} \, dx}{1188}\\ &=-\frac{38-21 x}{198 \left (2+3 x^2\right )^{3/2}}+\frac{24+95 x}{726 \sqrt{2+3 x^2}}+\frac{8}{121} \int \frac{1}{(1+2 x) \sqrt{2+3 x^2}} \, dx\\ &=-\frac{38-21 x}{198 \left (2+3 x^2\right )^{3/2}}+\frac{24+95 x}{726 \sqrt{2+3 x^2}}-\frac{8}{121} \operatorname{Subst}\left (\int \frac{1}{11-x^2} \, dx,x,\frac{4-3 x}{\sqrt{2+3 x^2}}\right )\\ &=-\frac{38-21 x}{198 \left (2+3 x^2\right )^{3/2}}+\frac{24+95 x}{726 \sqrt{2+3 x^2}}-\frac{8 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{2+3 x^2}}\right )}{121 \sqrt{11}}\\ \end{align*}

Mathematica [A]  time = 0.0514794, size = 58, normalized size = 0.79 $\frac{855 x^3+216 x^2+801 x-274}{2178 \left (3 x^2+2\right )^{3/2}}-\frac{8 \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{33 x^2+22}}\right )}{121 \sqrt{11}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-274 + 801*x + 216*x^2 + 855*x^3)/(2178*(2 + 3*x^2)^(3/2)) - (8*ArcTanh[(4 - 3*x)/Sqrt[22 + 33*x^2]])/(121*Sq
rt[11])

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Maple [B]  time = 0.056, size = 133, normalized size = 1.8 \begin{align*} -{\frac{2}{9} \left ( 3\,{x}^{2}+2 \right ) ^{-{\frac{3}{2}}}}+{\frac{x}{12} \left ( 3\,{x}^{2}+2 \right ) ^{-{\frac{3}{2}}}}+{\frac{x}{12}{\frac{1}{\sqrt{3\,{x}^{2}+2}}}}+{\frac{1}{33} \left ( 3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}} \right ) ^{-{\frac{3}{2}}}}+{\frac{x}{44} \left ( 3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}} \right ) ^{-{\frac{3}{2}}}}+{\frac{23\,x}{484}{\frac{1}{\sqrt{3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}}}}}}+{\frac{4}{121}{\frac{1}{\sqrt{3\, \left ( x+1/2 \right ) ^{2}-3\,x+{\frac{5}{4}}}}}}-{\frac{8\,\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)/(3*x^2+2)^(5/2),x)

[Out]

-2/9/(3*x^2+2)^(3/2)+1/12*x/(3*x^2+2)^(3/2)+1/12*x/(3*x^2+2)^(1/2)+1/33/(3*(x+1/2)^2-3*x+5/4)^(3/2)+1/44*x/(3*
(x+1/2)^2-3*x+5/4)^(3/2)+23/484*x/(3*(x+1/2)^2-3*x+5/4)^(1/2)+4/121/(3*(x+1/2)^2-3*x+5/4)^(1/2)-8/1331*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 = 1.49518, size = 109, normalized size = 1.49 \begin{align*} \frac{8}{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{95 \, x}{726 \, \sqrt{3 \, x^{2} + 2}} + \frac{4}{121 \, \sqrt{3 \, x^{2} + 2}} + \frac{7 \, x}{66 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}} - \frac{19}{99 \,{\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)/(3*x^2+2)^(5/2),x, algorithm="maxima")

[Out]

8/1331*sqrt(11)*arcsinh(1/2*sqrt(6)*x/abs(2*x + 1) - 2/3*sqrt(6)/abs(2*x + 1)) + 95/726*x/sqrt(3*x^2 + 2) + 4/
121/sqrt(3*x^2 + 2) + 7/66*x/(3*x^2 + 2)^(3/2) - 19/99/(3*x^2 + 2)^(3/2)

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Fricas [A]  time = 1.56274, size = 273, normalized size = 3.74 \begin{align*} \frac{72 \, \sqrt{11}{\left (9 \, x^{4} + 12 \, x^{2} + 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 (855 \, x^{3} + 216 \, x^{2} + 801 \, x - 274\right )} \sqrt{3 \, x^{2} + 2}}{23958 \,{\left (9 \, x^{4} + 12 \, x^{2} + 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)/(3*x^2+2)^(5/2),x, algorithm="fricas")

[Out]

1/23958*(72*sqrt(11)*(9*x^4 + 12*x^2 + 4)*log(-(sqrt(11)*sqrt(3*x^2 + 2)*(3*x - 4) + 21*x^2 - 12*x + 19)/(4*x^
2 + 4*x + 1)) + 11*(855*x^3 + 216*x^2 + 801*x - 274)*sqrt(3*x^2 + 2))/(9*x^4 + 12*x^2 + 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)/(3*x**2+2)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.27288, size = 123, normalized size = 1.68 \begin{align*} \frac{8}{1331} \, \sqrt{11} \log \left (-\frac{{\left | -2 \, \sqrt{3} x - \sqrt{11} - \sqrt{3} + 2 \, \sqrt{3 \, x^{2} + 2} \right |}}{2 \, \sqrt{3} x - \sqrt{11} + \sqrt{3} - 2 \, \sqrt{3 \, x^{2} + 2}}\right ) + \frac{9 \,{\left ({\left (95 \, x + 24\right )} x + 89\right )} x - 274}{2178 \,{\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)/(3*x^2+2)^(5/2),x, algorithm="giac")

[Out]

8/1331*sqrt(11)*log(-abs(-2*sqrt(3)*x - sqrt(11) - sqrt(3) + 2*sqrt(3*x^2 + 2))/(2*sqrt(3)*x - sqrt(11) + sqrt
(3) - 2*sqrt(3*x^2 + 2))) + 1/2178*(9*((95*x + 24)*x + 89)*x - 274)/(3*x^2 + 2)^(3/2)