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

Optimal. Leaf size=67 \[ \frac{2}{3} \sqrt{3 x^2+2}-\frac{\tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{3 x^2+2}}\right )}{2 \sqrt{11}}+\frac{\sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{2 \sqrt{3}} \]

[Out]

(2*Sqrt[2 + 3*x^2])/3 + ArcSinh[Sqrt[3/2]*x]/(2*Sqrt[3]) - ArcTanh[(4 - 3*x)/(Sqrt[11]*Sqrt[2 + 3*x^2])]/(2*Sq
rt[11])

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Rubi [A]  time = 0.0806947, antiderivative size = 67, 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 = {1654, 844, 215, 725, 206} \[ \frac{2}{3} \sqrt{3 x^2+2}-\frac{\tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{3 x^2+2}}\right )}{2 \sqrt{11}}+\frac{\sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{2 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2 + 3*x^2])/3 + ArcSinh[Sqrt[3/2]*x]/(2*Sqrt[3]) - ArcTanh[(4 - 3*x)/(Sqrt[11]*Sqrt[2 + 3*x^2])]/(2*Sq
rt[11])

Rule 1654

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (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 + 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 + 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)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 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 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) \sqrt{2+3 x^2}} \, dx &=\frac{2}{3} \sqrt{2+3 x^2}+\frac{1}{12} \int \frac{12+12 x}{(1+2 x) \sqrt{2+3 x^2}} \, dx\\ &=\frac{2}{3} \sqrt{2+3 x^2}+\frac{1}{2} \int \frac{1}{\sqrt{2+3 x^2}} \, dx+\frac{1}{2} \int \frac{1}{(1+2 x) \sqrt{2+3 x^2}} \, dx\\ &=\frac{2}{3} \sqrt{2+3 x^2}+\frac{\sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{2 \sqrt{3}}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{11-x^2} \, dx,x,\frac{4-3 x}{\sqrt{2+3 x^2}}\right )\\ &=\frac{2}{3} \sqrt{2+3 x^2}+\frac{\sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{2 \sqrt{3}}-\frac{\tanh ^{-1}\left (\frac{4-3 x}{\sqrt{11} \sqrt{2+3 x^2}}\right )}{2 \sqrt{11}}\\ \end{align*}

Mathematica [A]  time = 0.0303362, size = 60, normalized size = 0.9 \[ \frac{1}{66} \left (44 \sqrt{3 x^2+2}-3 \sqrt{11} \tanh ^{-1}\left (\frac{4-3 x}{\sqrt{33 x^2+22}}\right )+11 \sqrt{3} \sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(44*Sqrt[2 + 3*x^2] + 11*Sqrt[3]*ArcSinh[Sqrt[3/2]*x] - 3*Sqrt[11]*ArcTanh[(4 - 3*x)/Sqrt[22 + 33*x^2]])/66

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Maple [A]  time = 0.052, size = 55, normalized size = 0.8 \begin{align*}{\frac{2}{3}\sqrt{3\,{x}^{2}+2}}+{\frac{\sqrt{3}}{6}{\it Arcsinh} \left ({\frac{x\sqrt{6}}{2}} \right ) }-{\frac{\sqrt{11}}{22}{\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)^(1/2),x)

[Out]

2/3*(3*x^2+2)^(1/2)+1/6*arcsinh(1/2*x*6^(1/2))*3^(1/2)-1/22*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 = 1.49652, size = 78, normalized size = 1.16 \begin{align*} \frac{1}{6} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{2} \, \sqrt{6} x\right ) + \frac{1}{22} \, \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{2}{3} \, \sqrt{3 \, x^{2} + 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)^(1/2),x, algorithm="maxima")

[Out]

1/6*sqrt(3)*arcsinh(1/2*sqrt(6)*x) + 1/22*sqrt(11)*arcsinh(1/2*sqrt(6)*x/abs(2*x + 1) - 2/3*sqrt(6)/abs(2*x +
1)) + 2/3*sqrt(3*x^2 + 2)

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Fricas [A]  time = 1.57985, size = 240, normalized size = 3.58 \begin{align*} \frac{1}{12} \, \sqrt{3} \log \left (-\sqrt{3} \sqrt{3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) + \frac{1}{44} \, \sqrt{11} \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 ) + \frac{2}{3} \, \sqrt{3 \, x^{2} + 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)^(1/2),x, algorithm="fricas")

[Out]

1/12*sqrt(3)*log(-sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1) + 1/44*sqrt(11)*log(-(sqrt(11)*sqrt(3*x^2 + 2)*(3*x -
4) + 21*x^2 - 12*x + 19)/(4*x^2 + 4*x + 1)) + 2/3*sqrt(3*x^2 + 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 ) \sqrt{3 x^{2} + 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)/(3*x**2+2)**(1/2),x)

[Out]

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

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Giac [B]  time = 1.32843, size = 134, normalized size = 2. \begin{align*} -\frac{1}{6} \, \sqrt{3} \log \left (-\sqrt{3} x + \sqrt{3 \, x^{2} + 2}\right ) + \frac{1}{22} \, \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{2}{3} \, \sqrt{3 \, x^{2} + 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)^(1/2),x, algorithm="giac")

[Out]

-1/6*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2)) + 1/22*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))) + 2/3*sqrt(3*x^2 + 2)