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]

1/6*arcsinh(1/2*x*6^(1/2))*3^(1/2)-1/22*arctanh(1/11*(4-3*x)*11^(1/2)/(3*x^2+2)^(1/2))*11^(1/2)+2/3*(3*x^2+2)^
(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 67, normalized size of antiderivative = 1.00, 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 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])

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 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 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]))

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*}

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Mathematica [A]  time = 0.03, size = 60, normalized size = 0.90 \[ \frac {1}{66} \left (44 \sqrt {3 x^2+2}-3 \sqrt {11} \tanh ^{-1}\left (\frac {4-3 x}{\sqrt {33 x^2+22}}\right )+11 \sqrt {3} \sinh ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right ) \]

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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fricas [A]  time = 0.91, size = 88, normalized size = 1.31 \[ \frac {1}{12} \, \sqrt {3} \log \left (-\sqrt {3} \sqrt {3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) + \frac {1}{44} \, \sqrt {11} \log \left (-\frac {\sqrt {11} \sqrt {3 \, x^{2} + 2} {\left (3 \, x - 4\right )} + 21 \, x^{2} - 12 \, x + 19}{4 \, x^{2} + 4 \, x + 1}\right ) + \frac {2}{3} \, \sqrt {3 \, x^{2} + 2} \]

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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giac [B]  time = 0.23, size = 99, normalized size = 1.48 \[ -\frac {1}{6} \, \sqrt {3} \log \left (-\sqrt {3} x + \sqrt {3 \, x^{2} + 2}\right ) + \frac {1}{22} \, \sqrt {11} \log \left (-\frac {{\left | -2 \, \sqrt {3} x - \sqrt {11} - \sqrt {3} + 2 \, \sqrt {3 \, x^{2} + 2} \right |}}{2 \, \sqrt {3} x - \sqrt {11} + \sqrt {3} - 2 \, \sqrt {3 \, x^{2} + 2}}\right ) + \frac {2}{3} \, \sqrt {3 \, x^{2} + 2} \]

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)

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maple [A]  time = 0.01, size = 55, normalized size = 0.82 \[ \frac {\sqrt {3}\, \arcsinh \left (\frac {\sqrt {6}\, x}{2}\right )}{6}-\frac {\sqrt {11}\, \arctanh \left (\frac {2 \left (-3 x +4\right ) \sqrt {11}}{11 \sqrt {-12 x +12 \left (x +\frac {1}{2}\right )^{2}+5}}\right )}{22}+\frac {2 \sqrt {3 x^{2}+2}}{3} \]

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*6^(1/2)*x)*3^(1/2)-1/22*11^(1/2)*arctanh(2/11*(4-3*x)*11^(1/2)/(12*(1/2+x)
^2-12*x+5)^(1/2))

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maxima [A]  time = 0.96, size = 58, normalized size = 0.87 \[ \frac {1}{6} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {6} x\right ) + \frac {1}{22} \, \sqrt {11} \operatorname {arsinh}\left (\frac {\sqrt {6} x}{2 \, {\left | 2 \, x + 1 \right |}} - \frac {2 \, \sqrt {6}}{3 \, {\left | 2 \, x + 1 \right |}}\right ) + \frac {2}{3} \, \sqrt {3 \, x^{2} + 2} \]

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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mupad [B]  time = 0.19, size = 61, normalized size = 0.91 \[ \frac {\sqrt {11}\,\left (2\,\ln \left (x+\frac {1}{2}\right )-2\,\ln \left (x-\frac {\sqrt {3}\,\sqrt {11}\,\sqrt {x^2+\frac {2}{3}}}{3}-\frac {4}{3}\right )\right )}{44}+\frac {2\,\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}}{3}+\frac {\sqrt {3}\,\mathrm {asinh}\left (\frac {\sqrt {2}\,\sqrt {3}\,x}{2}\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(11^(1/2)*(2*log(x + 1/2) - 2*log(x - (3^(1/2)*11^(1/2)*(x^2 + 2/3)^(1/2))/3 - 4/3)))/44 + (2*3^(1/2)*(x^2 + 2
/3)^(1/2))/3 + (3^(1/2)*asinh((2^(1/2)*3^(1/2)*x)/2))/6

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {4 x^{2} + 3 x + 1}{\left (2 x + 1\right ) \sqrt {3 x^{2} + 2}}\, dx \]

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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