3.559 \(\int \frac{\sqrt{2+4 x^2}}{5+4 x} \, dx\)

Optimal. Leaf size=67 \[ \frac{\sqrt{2 x^2+1}}{2 \sqrt{2}}-\frac{1}{8} \sqrt{33} \tanh ^{-1}\left (\frac{\sqrt{\frac{2}{33}} (2-5 x)}{\sqrt{2 x^2+1}}\right )-\frac{5}{8} \sinh ^{-1}\left (\sqrt{2} x\right ) \]

[Out]

Sqrt[1 + 2*x^2]/(2*Sqrt[2]) - (5*ArcSinh[Sqrt[2]*x])/8 - (Sqrt[33]*ArcTanh[(Sqrt[2/33]*(2 - 5*x))/Sqrt[1 + 2*x
^2]])/8

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Rubi [A]  time = 0.0439505, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {735, 844, 215, 725, 206} \[ \frac{\sqrt{2 x^2+1}}{2 \sqrt{2}}-\frac{1}{8} \sqrt{33} \tanh ^{-1}\left (\frac{\sqrt{\frac{2}{33}} (2-5 x)}{\sqrt{2 x^2+1}}\right )-\frac{5}{8} \sinh ^{-1}\left (\sqrt{2} x\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + 4*x^2]/(5 + 4*x),x]

[Out]

Sqrt[1 + 2*x^2]/(2*Sqrt[2]) - (5*ArcSinh[Sqrt[2]*x])/8 - (Sqrt[33]*ArcTanh[(Sqrt[2/33]*(2 - 5*x))/Sqrt[1 + 2*x
^2]])/8

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+4 x^2}}{5+4 x} \, dx &=\frac{\sqrt{1+2 x^2}}{2 \sqrt{2}}+\frac{1}{4} \int \frac{8-20 x}{(5+4 x) \sqrt{2+4 x^2}} \, dx\\ &=\frac{\sqrt{1+2 x^2}}{2 \sqrt{2}}-\frac{5}{4} \int \frac{1}{\sqrt{2+4 x^2}} \, dx+\frac{33}{4} \int \frac{1}{(5+4 x) \sqrt{2+4 x^2}} \, dx\\ &=\frac{\sqrt{1+2 x^2}}{2 \sqrt{2}}-\frac{5}{8} \sinh ^{-1}\left (\sqrt{2} x\right )-\frac{33}{4} \operatorname{Subst}\left (\int \frac{1}{132-x^2} \, dx,x,\frac{8-20 x}{\sqrt{2+4 x^2}}\right )\\ &=\frac{\sqrt{1+2 x^2}}{2 \sqrt{2}}-\frac{5}{8} \sinh ^{-1}\left (\sqrt{2} x\right )-\frac{1}{8} \sqrt{33} \tanh ^{-1}\left (\frac{\sqrt{\frac{2}{33}} (2-5 x)}{\sqrt{1+2 x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0450026, size = 57, normalized size = 0.85 \[ \frac{1}{4} \sqrt{4 x^2+2}-\frac{1}{8} \sqrt{33} \tanh ^{-1}\left (\frac{2-5 x}{\sqrt{33 x^2+\frac{33}{2}}}\right )-\frac{5}{8} \sinh ^{-1}\left (\sqrt{2} x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + 4*x^2]/(5 + 4*x),x]

[Out]

Sqrt[2 + 4*x^2]/4 - (5*ArcSinh[Sqrt[2]*x])/8 - (Sqrt[33]*ArcTanh[(2 - 5*x)/Sqrt[33/2 + 33*x^2]])/8

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Maple [A]  time = 0.042, size = 56, normalized size = 0.8 \begin{align*}{\frac{1}{8}\sqrt{16\, \left ( x+5/4 \right ) ^{2}-40\,x-17}}-{\frac{5\,{\it Arcsinh} \left ( x\sqrt{2} \right ) }{8}}-{\frac{\sqrt{33}}{8}{\it Artanh} \left ({\frac{ \left ( 8-20\,x \right ) \sqrt{33}}{33}{\frac{1}{\sqrt{16\, \left ( x+5/4 \right ) ^{2}-40\,x-17}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2+2)^(1/2)/(5+4*x),x)

[Out]

1/8*(16*(x+5/4)^2-40*x-17)^(1/2)-5/8*arcsinh(x*2^(1/2))-1/8*33^(1/2)*arctanh(2/33*(4-10*x)*33^(1/2)/(16*(x+5/4
)^2-40*x-17)^(1/2))

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Maxima [A]  time = 1.94091, size = 73, normalized size = 1.09 \begin{align*} \frac{1}{8} \, \sqrt{33} \operatorname{arsinh}\left (\frac{5 \, \sqrt{2} x}{{\left | 4 \, x + 5 \right |}} - \frac{2 \, \sqrt{2}}{{\left | 4 \, x + 5 \right |}}\right ) + \frac{1}{4} \, \sqrt{4 \, x^{2} + 2} - \frac{5}{8} \, \operatorname{arsinh}\left (\sqrt{2} x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+2)^(1/2)/(5+4*x),x, algorithm="maxima")

[Out]

1/8*sqrt(33)*arcsinh(5*sqrt(2)*x/abs(4*x + 5) - 2*sqrt(2)/abs(4*x + 5)) + 1/4*sqrt(4*x^2 + 2) - 5/8*arcsinh(sq
rt(2)*x)

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Fricas [A]  time = 2.47994, size = 212, normalized size = 3.16 \begin{align*} \frac{1}{8} \, \sqrt{33} \log \left (-\frac{2 \, \sqrt{33}{\left (5 \, x - 2\right )} + \sqrt{4 \, x^{2} + 2}{\left (5 \, \sqrt{33} + 33\right )} + 50 \, x - 20}{4 \, x + 5}\right ) + \frac{1}{4} \, \sqrt{4 \, x^{2} + 2} + \frac{5}{8} \, \log \left (-2 \, x + \sqrt{4 \, x^{2} + 2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+2)^(1/2)/(5+4*x),x, algorithm="fricas")

[Out]

1/8*sqrt(33)*log(-(2*sqrt(33)*(5*x - 2) + sqrt(4*x^2 + 2)*(5*sqrt(33) + 33) + 50*x - 20)/(4*x + 5)) + 1/4*sqrt
(4*x^2 + 2) + 5/8*log(-2*x + sqrt(4*x^2 + 2))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \sqrt{2} \int \frac{\sqrt{2 x^{2} + 1}}{4 x + 5}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**2+2)**(1/2)/(5+4*x),x)

[Out]

sqrt(2)*Integral(sqrt(2*x**2 + 1)/(4*x + 5), x)

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Giac [B]  time = 1.40396, size = 142, normalized size = 2.12 \begin{align*} \frac{1}{16} \, \sqrt{2}{\left (5 \, \sqrt{2} \log \left (-\sqrt{2} x + \sqrt{2 \, x^{2} + 1}\right ) + \sqrt{66} \log \left (-\frac{{\left | -4 \, \sqrt{2} x - \sqrt{66} - 5 \, \sqrt{2} + 4 \, \sqrt{2 \, x^{2} + 1} \right |}}{4 \, \sqrt{2} x - \sqrt{66} + 5 \, \sqrt{2} - 4 \, \sqrt{2 \, x^{2} + 1}}\right ) + 4 \, \sqrt{2 \, x^{2} + 1}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+2)^(1/2)/(5+4*x),x, algorithm="giac")

[Out]

1/16*sqrt(2)*(5*sqrt(2)*log(-sqrt(2)*x + sqrt(2*x^2 + 1)) + sqrt(66)*log(-abs(-4*sqrt(2)*x - sqrt(66) - 5*sqrt
(2) + 4*sqrt(2*x^2 + 1))/(4*sqrt(2)*x - sqrt(66) + 5*sqrt(2) - 4*sqrt(2*x^2 + 1))) + 4*sqrt(2*x^2 + 1))