### 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]

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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Rubi [A]  time = 0.0327453, antiderivative size = 56, normalized size of antiderivative = 1., 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 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]

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{\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*}

Mathematica [A]  time = 0.0283808, size = 56, normalized size = 1. $\frac{\sqrt{x^2+2}}{4}-\frac{1}{16} \sqrt{33} \tanh ^{-1}\left (\frac{8-x}{\sqrt{33} \sqrt{x^2+2}}\right )-\frac{1}{16} \sinh ^{-1}\left (\frac{x}{\sqrt{2}}\right )$

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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Maple [A]  time = 0.043, size = 57, normalized size = 1. \begin{align*}{\frac{1}{16}\sqrt{16\, \left ( x+1/4 \right ) ^{2}-8\,x+31}}-{\frac{1}{16}{\it Arcsinh} \left ({\frac{x\sqrt{2}}{2}} \right ) }-{\frac{\sqrt{33}}{16}{\it Artanh} \left ({\frac{8\,\sqrt{33}}{33} \left ( 4-{\frac{x}{2}} \right ){\frac{1}{\sqrt{16\, \left ( x+1/4 \right ) ^{2}-8\,x+31}}}} \right ) } \end{align*}

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*x*2^(1/2))-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 = 1.53476, size = 72, normalized size = 1.29 \begin{align*} \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 ) \end{align*}

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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Fricas [A]  time = 2.5118, size = 190, normalized size = 3.39 \begin{align*} \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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} + 2}}{4 x + 1}\, dx \end{align*}

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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Giac [A]  time = 1.48047, size = 96, normalized size = 1.71 \begin{align*} \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 ) \end{align*}

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))