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

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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 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]

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*}

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Mathematica [A]  time = 0.00, size = 25, normalized size = 1.00 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}} \]

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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fricas [A]  time = 0.80, size = 59, normalized size = 2.36 \[ \left [\frac {\log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right )}{2 \, \sqrt {b}}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right )}{b}\right ] \]

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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giac [A]  time = 0.18, size = 23, normalized size = 0.92 \[ -\frac {\log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{\sqrt {b}} \]

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)

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maple [A]  time = 0.00, size = 21, normalized size = 0.84 \[ \frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.02, size = 13, normalized size = 0.52 \[ \frac {\operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.12, size = 20, normalized size = 0.80 \[ \frac {\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x^2)^(1/2),x)

[Out]

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

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sympy [A]  time = 1.20, size = 17, normalized size = 0.68 \[ \frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{\sqrt {b}} \]

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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