3.1016 \(\int \frac{1}{\sqrt{a+b x^2+(2+2 c-2 (1+c)) x^4}} \, dx\)

Optimal. Leaf size=25 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{\sqrt{b}} \]

[Out]

ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]]/Sqrt[b]

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Rubi [A]  time = 0.0058784, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {5, 217, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*x^2 + (2 + 2*c - 2*(1 + c))*x^4],x]

[Out]

ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]]/Sqrt[b]

Rule 5

Int[(u_.)*((a_.) + (c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(a + b*x^n)^p, x] /; FreeQ[{
a, b, c, n, p}, x] && EqQ[j, 2*n] && EqQ[c, 0]

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

Mathematica [A]  time = 0.0045132, size = 25, normalized size = 1. \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + b*x^2 + (2 + 2*c - 2*(1 + c))*x^4],x]

[Out]

ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]]/Sqrt[b]

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Maple [A]  time = 0.042, size = 21, normalized size = 0.8 \begin{align*}{\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a)^(1/2),x)

[Out]

ln(x*b^(1/2)+(b*x^2+a)^(1/2))/b^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58478, size = 153, normalized size = 6.12 \begin{align*} \left [\frac{\log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right )}{2 \, \sqrt{b}}, -\frac{\sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right )}{b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/2*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a)/sqrt(b), -sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a))/b]

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Sympy [A]  time = 1.01721, size = 17, normalized size = 0.68 \begin{align*} \frac{\operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{\sqrt{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a)**(1/2),x)

[Out]

asinh(sqrt(b)*x/sqrt(a))/sqrt(b)

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Giac [A]  time = 1.21282, size = 31, normalized size = 1.24 \begin{align*} -\frac{\log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{\sqrt{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/sqrt(b)