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

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

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Rubi [A]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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 ) \end {gather*}

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 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+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*}

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Mathematica [A]  time = 0.04, size = 57, normalized size = 0.85 \begin {gather*} \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 ) \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.26, size = 93, normalized size = 1.39 \begin {gather*} \frac {\sqrt {2 x^2+1}}{2 \sqrt {2}}+\frac {5}{8} \log \left (\sqrt {2} \sqrt {2 x^2+1}-2 x\right )+\frac {1}{4} \sqrt {33} \tanh ^{-1}\left (-2 \sqrt {\frac {2}{33}} \sqrt {2 x^2+1}+\frac {4 x}{\sqrt {33}}+\frac {5}{\sqrt {33}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 + 2*x^2]/(2*Sqrt[2]) + (Sqrt[33]*ArcTanh[5/Sqrt[33] + (4*x)/Sqrt[33] - 2*Sqrt[2/33]*Sqrt[1 + 2*x^2]])/4
+ (5*Log[-2*x + Sqrt[2]*Sqrt[1 + 2*x^2]])/8

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fricas [A]  time = 0.41, size = 75, normalized size = 1.12 \begin {gather*} \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 {gather*}

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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giac [B]  time = 0.30, size = 105, normalized size = 1.57 \begin {gather*} \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 {gather*}

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))

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maple [A]  time = 0.06, size = 56, normalized size = 0.84 \begin {gather*} -\frac {5 \arcsinh \left (\sqrt {2}\, x \right )}{8}-\frac {\sqrt {33}\, \arctanh \left (\frac {2 \left (-10 x +4\right ) \sqrt {33}}{33 \sqrt {-40 x +16 \left (x +\frac {5}{4}\right )^{2}-17}}\right )}{8}+\frac {\sqrt {-40 x +16 \left (x +\frac {5}{4}\right )^{2}-17}}{8} \end {gather*}

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

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maxima [A]  time = 3.04, size = 54, normalized size = 0.81 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.16, size = 48, normalized size = 0.72 \begin {gather*} \frac {\sqrt {x^2+\frac {1}{2}}}{2}-\frac {5\,\mathrm {asinh}\left (\sqrt {2}\,x\right )}{8}+\frac {\sqrt {33}\,\left (132\,\ln \left (x+\frac {5}{4}\right )-132\,\ln \left (x-\frac {\sqrt {33}\,\sqrt {x^2+\frac {1}{2}}}{5}-\frac {2}{5}\right )\right )}{1056} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2 + 2)^(1/2)/(4*x + 5),x)

[Out]

(x^2 + 1/2)^(1/2)/2 - (5*asinh(2^(1/2)*x))/8 + (33^(1/2)*(132*log(x + 5/4) - 132*log(x - (33^(1/2)*(x^2 + 1/2)
^(1/2))/5 - 2/5)))/1056

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \sqrt {2} \int \frac {\sqrt {2 x^{2} + 1}}{4 x + 5}\, dx \end {gather*}

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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