3.202 \(\int (\frac{10}{\sqrt{-4+x^2}}+\frac{1}{\sqrt{-1+x^2}}) \, dx\)

Optimal. Leaf size=27 \[ 10 \tanh ^{-1}\left (\frac{x}{\sqrt{x^2-4}}\right )+\tanh ^{-1}\left (\frac{x}{\sqrt{x^2-1}}\right ) \]

[Out]

10*ArcTanh[x/Sqrt[-4 + x^2]] + ArcTanh[x/Sqrt[-1 + x^2]]

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Rubi [A]  time = 0.0080556, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {217, 206} \[ 10 \tanh ^{-1}\left (\frac{x}{\sqrt{x^2-4}}\right )+\tanh ^{-1}\left (\frac{x}{\sqrt{x^2-1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[10/Sqrt[-4 + x^2] + 1/Sqrt[-1 + x^2],x]

[Out]

10*ArcTanh[x/Sqrt[-4 + x^2]] + ArcTanh[x/Sqrt[-1 + x^2]]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 \left (\frac{10}{\sqrt{-4+x^2}}+\frac{1}{\sqrt{-1+x^2}}\right ) \, dx &=10 \int \frac{1}{\sqrt{-4+x^2}} \, dx+\int \frac{1}{\sqrt{-1+x^2}} \, dx\\ &=10 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{x}{\sqrt{-4+x^2}}\right )+\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{x}{\sqrt{-1+x^2}}\right )\\ &=10 \tanh ^{-1}\left (\frac{x}{\sqrt{-4+x^2}}\right )+\tanh ^{-1}\left (\frac{x}{\sqrt{-1+x^2}}\right )\\ \end{align*}

Mathematica [B]  time = 0.0088356, size = 71, normalized size = 2.63 \[ -5 \log \left (1-\frac{x}{\sqrt{x^2-4}}\right )+5 \log \left (\frac{x}{\sqrt{x^2-4}}+1\right )-\frac{1}{2} \log \left (1-\frac{x}{\sqrt{x^2-1}}\right )+\frac{1}{2} \log \left (\frac{x}{\sqrt{x^2-1}}+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[10/Sqrt[-4 + x^2] + 1/Sqrt[-1 + x^2],x]

[Out]

-5*Log[1 - x/Sqrt[-4 + x^2]] + 5*Log[1 + x/Sqrt[-4 + x^2]] - Log[1 - x/Sqrt[-1 + x^2]]/2 + Log[1 + x/Sqrt[-1 +
x^2]]/2

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Maple [A]  time = 0.003, size = 24, normalized size = 0.9 \begin{align*} \ln \left ( x+\sqrt{{x}^{2}-1} \right ) +10\,\ln \left ( x+\sqrt{{x}^{2}-4} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10/(x^2-4)^(1/2)+1/(x^2-1)^(1/2),x)

[Out]

ln(x+(x^2-1)^(1/2))+10*ln(x+(x^2-4)^(1/2))

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Maxima [A]  time = 0.951793, size = 42, normalized size = 1.56 \begin{align*} \log \left (2 \, x + 2 \, \sqrt{x^{2} - 1}\right ) + 10 \, \log \left (2 \, x + 2 \, \sqrt{x^{2} - 4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10/(x^2-4)^(1/2)+1/(x^2-1)^(1/2),x, algorithm="maxima")

[Out]

log(2*x + 2*sqrt(x^2 - 1)) + 10*log(2*x + 2*sqrt(x^2 - 4))

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Fricas [A]  time = 1.55508, size = 74, normalized size = 2.74 \begin{align*} -\log \left (-x + \sqrt{x^{2} - 1}\right ) - 10 \, \log \left (-x + \sqrt{x^{2} - 4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10/(x^2-4)^(1/2)+1/(x^2-1)^(1/2),x, algorithm="fricas")

[Out]

-log(-x + sqrt(x^2 - 1)) - 10*log(-x + sqrt(x^2 - 4))

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Sympy [A]  time = 0.198941, size = 8, normalized size = 0.3 \begin{align*} 10 \operatorname{acosh}{\left (\frac{x}{2} \right )} + \operatorname{acosh}{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10/(x**2-4)**(1/2)+1/(x**2-1)**(1/2),x)

[Out]

10*acosh(x/2) + acosh(x)

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Giac [A]  time = 1.09578, size = 42, normalized size = 1.56 \begin{align*} -\log \left ({\left | -x + \sqrt{x^{2} - 1} \right |}\right ) - 10 \, \log \left ({\left | -x + \sqrt{x^{2} - 4} \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10/(x^2-4)^(1/2)+1/(x^2-1)^(1/2),x, algorithm="giac")

[Out]

-log(abs(-x + sqrt(x^2 - 1))) - 10*log(abs(-x + sqrt(x^2 - 4)))