3.6.59 \(\int \frac {\sqrt {2+4 x^2}}{5+4 x} \, dx\) [559]

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

[Out]

-5/8*arcsinh(x*2^(1/2))-1/8*arctanh(1/33*(2-5*x)*66^(1/2)/(2*x^2+1)^(1/2))*33^(1/2)+1/4*(2*x^2+1)^(1/2)*2^(1/2
)

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Rubi [A]
time = 0.03, 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 = {749, 858, 221, 739, 212} \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 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*} \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} \text {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 = 0.17, size = 66, normalized size = 0.99 \begin {gather*} \frac {1}{8} \left (-5 \tanh ^{-1}\left (\frac {x}{\sqrt {\frac {1}{2}+x^2}}\right )+2 \left (\sqrt {2+4 x^2}+\sqrt {33} \tanh ^{-1}\left (\frac {5+4 x-2 \sqrt {2+4 x^2}}{\sqrt {33}}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.44, size = 56, normalized size = 0.84

method	result	size
default	\(\frac {\sqrt {16 \left (x +\frac {5}{4}\right )^{2}-40 x -17}}{8}-\frac {5 \arcsinh \left (x \sqrt {2}\right )}{8}-\frac {\sqrt {33}\, \arctanh \left (\frac {2 \left (4-10 x \right ) \sqrt {33}}{33 \sqrt {16 \left (x +\frac {5}{4}\right )^{2}-40 x -17}}\right )}{8}\)	\(56\)
risch	\(\frac {2 x^{2}+1}{2 \sqrt {4 x^{2}+2}}-\frac {5 \arcsinh \left (x \sqrt {2}\right )}{8}-\frac {\sqrt {33}\, \arctanh \left (\frac {2 \left (4-10 x \right ) \sqrt {33}}{33 \sqrt {16 \left (x +\frac {5}{4}\right )^{2}-40 x -17}}\right )}{8}\)	\(58\)
trager	\(\frac {\sqrt {4 x^{2}+2}}{4}+\frac {5 \ln \left (2 x -\sqrt {4 x^{2}+2}\right )}{8}-\frac {\RootOf \left (\textit {\_Z}^{2}-33\right ) \ln \left (\frac {-10 \RootOf \left (\textit {\_Z}^{2}-33\right ) x +33 \sqrt {4 x^{2}+2}+4 \RootOf \left (\textit {\_Z}^{2}-33\right )}{5+4 x}\right )}{8}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.54, 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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Fricas [A]
time = 3.18, 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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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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (48) = 96\).
time = 1.29, 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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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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