\(\int \frac {\sqrt {2+x^2}}{1+4 x} \, dx\) [558]

Optimal result

Integrand size = 17, antiderivative size = 56 \[ \int \frac {\sqrt {2+x^2}}{1+4 x} \, dx=\frac {\sqrt {2+x^2}}{4}-\frac {1}{16} \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right )-\frac {1}{16} \sqrt {33} \text {arctanh}\left (\frac {8-x}{\sqrt {33} \sqrt {2+x^2}}\right ) \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {749, 858, 221, 739, 212} \[ \int \frac {\sqrt {2+x^2}}{1+4 x} \, dx=-\frac {1}{16} \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right )-\frac {1}{16} \sqrt {33} \text {arctanh}\left (\frac {8-x}{\sqrt {33} \sqrt {x^2+2}}\right )+\frac {\sqrt {x^2+2}}{4} \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 221

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 739

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 749

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 858

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*} \text {integral}& = \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} \text {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] (verified)

Time = 0.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt {2+x^2}}{1+4 x} \, dx=\frac {1}{16} \left (4 \sqrt {2+x^2}+2 \sqrt {33} \text {arctanh}\left (\frac {1+4 x-4 \sqrt {2+x^2}}{\sqrt {33}}\right )+\log \left (-x+\sqrt {2+x^2}\right )\right ) \]

[In]

Integrate[Sqrt[2 + x^2]/(1 + 4*x),x]

[Out]

(4*Sqrt[2 + x^2] + 2*Sqrt[33]*ArcTanh[(1 + 4*x - 4*Sqrt[2 + x^2])/Sqrt[33]] + Log[-x + Sqrt[2 + x^2]])/16

Maple [A] (verified)

Time = 1.98 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.89

method	result	size
risch	\(\frac {\sqrt {x^{2}+2}}{4}-\frac {\operatorname {arcsinh}\left (\frac {\sqrt {2}\, x}{2}\right )}{16}-\frac {\sqrt {33}\, \operatorname {arctanh}\left (\frac {8 \left (4-\frac {x}{2}\right ) \sqrt {33}}{33 \sqrt {16 \left (x +\frac {1}{4}\right )^{2}-8 x +31}}\right )}{16}\)	\(50\)
default	\(\frac {\sqrt {16 \left (x +\frac {1}{4}\right )^{2}-8 x +31}}{16}-\frac {\operatorname {arcsinh}\left (\frac {\sqrt {2}\, x}{2}\right )}{16}-\frac {\sqrt {33}\, \operatorname {arctanh}\left (\frac {8 \left (4-\frac {x}{2}\right ) \sqrt {33}}{33 \sqrt {16 \left (x +\frac {1}{4}\right )^{2}-8 x +31}}\right )}{16}\)	\(57\)
trager	\(\frac {\sqrt {x^{2}+2}}{4}-\frac {\ln \left (x +\sqrt {x^{2}+2}\right )}{16}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-33\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-33\right ) x -8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-33\right )+33 \sqrt {x^{2}+2}}{1+4 x}\right )}{16}\)	\(66\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {2+x^2}}{1+4 x} \, dx=\frac {1}{16} \, \sqrt {33} \log \left (-\frac {\sqrt {33} {\left (x - 8\right )} + \sqrt {x^{2} + 2} {\left (\sqrt {33} + 33\right )} + x - 8}{4 \, x + 1}\right ) + \frac {1}{4} \, \sqrt {x^{2} + 2} + \frac {1}{16} \, \log \left (-x + \sqrt {x^{2} + 2}\right ) \]

[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))

Sympy [F]

\[ \int \frac {\sqrt {2+x^2}}{1+4 x} \, dx=\int \frac {\sqrt {x^{2} + 2}}{4 x + 1}\, dx \]

[In]

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

[Out]

Integral(sqrt(x**2 + 2)/(4*x + 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {2+x^2}}{1+4 x} \, dx=\frac {1}{16} \, \sqrt {33} \operatorname {arsinh}\left (\frac {\sqrt {2} x}{2 \, {\left | 4 \, x + 1 \right |}} - \frac {4 \, \sqrt {2}}{{\left | 4 \, x + 1 \right |}}\right ) + \frac {1}{4} \, \sqrt {x^{2} + 2} - \frac {1}{16} \, \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {2} x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt {2+x^2}}{1+4 x} \, dx=\frac {1}{16} \, \sqrt {33} \log \left (\frac {{\left | -4 \, x - \sqrt {33} + 4 \, \sqrt {x^{2} + 2} - 1 \right |}}{{\left | -4 \, x + \sqrt {33} + 4 \, \sqrt {x^{2} + 2} - 1 \right |}}\right ) + \frac {1}{4} \, \sqrt {x^{2} + 2} + \frac {1}{16} \, \log \left (-x + \sqrt {x^{2} + 2}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 9.48 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {2+x^2}}{1+4 x} \, dx=\frac {\sqrt {x^2+2}}{4}-\frac {\mathrm {asinh}\left (\frac {\sqrt {2}\,x}{2}\right )}{16}+\frac {\sqrt {33}\,\left (132\,\ln \left (x+\frac {1}{4}\right )-132\,\ln \left (x-\sqrt {33}\,\sqrt {x^2+2}-8\right )\right )}{2112} \]

[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