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

Optimal. Leaf size=56 \[ \frac {\sqrt {x^2+2}}{4}-\frac {1}{16} \sqrt {33} \tanh ^{-1}\left (\frac {8-x}{\sqrt {33} \sqrt {x^2+2}}\right )-\frac {1}{16} \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \]

[Out]

-1/16*arcsinh(1/2*x*2^(1/2))-1/16*arctanh(1/33*(8-x)*33^(1/2)/(x^2+2)^(1/2))*33^(1/2)+1/4*(x^2+2)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {735, 844, 215, 725, 206} \[ \frac {\sqrt {x^2+2}}{4}-\frac {1}{16} \sqrt {33} \tanh ^{-1}\left (\frac {8-x}{\sqrt {33} \sqrt {x^2+2}}\right )-\frac {1}{16} \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + x^2]/(1 + 4*x),x]

[Out]

Sqrt[2 + x^2]/4 - ArcSinh[x/Sqrt[2]]/16 - (Sqrt[33]*ArcTanh[(8 - x)/(Sqrt[33]*Sqrt[2 + x^2])])/16

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 735

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] + Dist[(2*p)/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[a*e - c*d*x, x]*(a + c*x^2)^(p - 1),
x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, 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]

Rubi steps

\begin {align*} \int \frac {\sqrt {2+x^2}}{1+4 x} \, dx &=\frac {\sqrt {2+x^2}}{4}+\frac {1}{4} \int \frac {8-x}{(1+4 x) \sqrt {2+x^2}} \, dx\\ &=\frac {\sqrt {2+x^2}}{4}-\frac {1}{16} \int \frac {1}{\sqrt {2+x^2}} \, dx+\frac {33}{16} \int \frac {1}{(1+4 x) \sqrt {2+x^2}} \, dx\\ &=\frac {\sqrt {2+x^2}}{4}-\frac {1}{16} \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )-\frac {33}{16} \operatorname {Subst}\left (\int \frac {1}{33-x^2} \, dx,x,\frac {8-x}{\sqrt {2+x^2}}\right )\\ &=\frac {\sqrt {2+x^2}}{4}-\frac {1}{16} \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right )-\frac {1}{16} \sqrt {33} \tanh ^{-1}\left (\frac {8-x}{\sqrt {33} \sqrt {2+x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 56, normalized size = 1.00 \[ \frac {\sqrt {x^2+2}}{4}-\frac {1}{16} \sqrt {33} \tanh ^{-1}\left (\frac {8-x}{\sqrt {33} \sqrt {x^2+2}}\right )-\frac {1}{16} \sinh ^{-1}\left (\frac {x}{\sqrt {2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + x^2]/(1 + 4*x),x]

[Out]

Sqrt[2 + x^2]/4 - ArcSinh[x/Sqrt[2]]/16 - (Sqrt[33]*ArcTanh[(8 - x)/(Sqrt[33]*Sqrt[2 + x^2])])/16

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fricas [A]  time = 1.30, size = 62, normalized size = 1.11 \[ \frac {1}{16} \, \sqrt {33} \log \left (-\frac {\sqrt {33} {\left (x - 8\right )} + \sqrt {x^{2} + 2} {\left (\sqrt {33} + 33\right )} + x - 8}{4 \, x + 1}\right ) + \frac {1}{4} \, \sqrt {x^{2} + 2} + \frac {1}{16} \, \log \left (-x + \sqrt {x^{2} + 2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)^(1/2)/(1+4*x),x, algorithm="fricas")

[Out]

1/16*sqrt(33)*log(-(sqrt(33)*(x - 8) + sqrt(x^2 + 2)*(sqrt(33) + 33) + x - 8)/(4*x + 1)) + 1/4*sqrt(x^2 + 2) +
1/16*log(-x + sqrt(x^2 + 2))

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giac [A]  time = 0.25, size = 71, normalized size = 1.27 \[ \frac {1}{16} \, \sqrt {33} \log \left (\frac {{\left | -4 \, x - \sqrt {33} + 4 \, \sqrt {x^{2} + 2} - 1 \right |}}{{\left | -4 \, x + \sqrt {33} + 4 \, \sqrt {x^{2} + 2} - 1 \right |}}\right ) + \frac {1}{4} \, \sqrt {x^{2} + 2} + \frac {1}{16} \, \log \left (-x + \sqrt {x^{2} + 2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)^(1/2)/(1+4*x),x, algorithm="giac")

[Out]

1/16*sqrt(33)*log(abs(-4*x - sqrt(33) + 4*sqrt(x^2 + 2) - 1)/abs(-4*x + sqrt(33) + 4*sqrt(x^2 + 2) - 1)) + 1/4
*sqrt(x^2 + 2) + 1/16*log(-x + sqrt(x^2 + 2))

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maple [A]  time = 0.05, size = 57, normalized size = 1.02 \[ -\frac {\arcsinh \left (\frac {\sqrt {2}\, x}{2}\right )}{16}-\frac {\sqrt {33}\, \arctanh \left (\frac {8 \left (-\frac {x}{2}+4\right ) \sqrt {33}}{33 \sqrt {-8 x +16 \left (x +\frac {1}{4}\right )^{2}+31}}\right )}{16}+\frac {\sqrt {-8 x +16 \left (x +\frac {1}{4}\right )^{2}+31}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*(x+1/4)^2-8*x+31)^(1/2)-1/16*arcsinh(1/2*2^(1/2)*x)-1/16*33^(1/2)*arctanh(8/33*(4-1/2*x)*33^(1/2)/(16
*(x+1/4)^2-8*x+31)^(1/2))

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maxima [A]  time = 2.99, size = 53, normalized size = 0.95 \[ \frac {1}{16} \, \sqrt {33} \operatorname {arsinh}\left (\frac {\sqrt {2} x}{2 \, {\left | 4 \, x + 1 \right |}} - \frac {4 \, \sqrt {2}}{{\left | 4 \, x + 1 \right |}}\right ) + \frac {1}{4} \, \sqrt {x^{2} + 2} - \frac {1}{16} \, \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {2} x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2)^(1/2)/(1+4*x),x, algorithm="maxima")

[Out]

1/16*sqrt(33)*arcsinh(1/2*sqrt(2)*x/abs(4*x + 1) - 4*sqrt(2)/abs(4*x + 1)) + 1/4*sqrt(x^2 + 2) - 1/16*arcsinh(
1/2*sqrt(2)*x)

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mupad [B]  time = 0.41, size = 49, normalized size = 0.88 \[ \frac {\sqrt {x^2+2}}{4}-\frac {\mathrm {asinh}\left (\frac {\sqrt {2}\,x}{2}\right )}{16}+\frac {\sqrt {33}\,\left (132\,\ln \left (x+\frac {1}{4}\right )-132\,\ln \left (x-\sqrt {33}\,\sqrt {x^2+2}-8\right )\right )}{2112} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + 2)^(1/2)/(4*x + 1),x)

[Out]

(x^2 + 2)^(1/2)/4 - asinh((2^(1/2)*x)/2)/16 + (33^(1/2)*(132*log(x + 1/4) - 132*log(x - 33^(1/2)*(x^2 + 2)^(1/
2) - 8)))/2112

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x**2 + 2)/(4*x + 1), x)

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