3.1000 \(\int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=59 \[ \frac {c \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3} \]

[Out]

1/2*c*arctanh(x*b^(1/2)/(c*x^4+b*x^2)^(1/2))/b^(3/2)-1/2*(c*x^4+b*x^2)^(1/2)/b/x^3

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Rubi [A]  time = 0.06, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3, 2025, 2008, 206} \[ \frac {c \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[b*x^2 + c*x^4]/(2*b*x^3) + (c*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(2*b^(3/2))

Rule 3

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

Rule 2008

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx &=\int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx\\ &=-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}-\frac {c \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{2 b}\\ &=-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}+\frac {c \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+c x^4}}\right )}{2 b}\\ &=-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}+\frac {c \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 68, normalized size = 1.15 \[ \frac {c \sqrt {x^2 \left (b+c x^2\right )} \left (\frac {\tanh ^{-1}\left (\sqrt {\frac {c x^2}{b}+1}\right )}{2 \sqrt {\frac {c x^2}{b}+1}}-\frac {b}{2 c x^2}\right )}{b^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[x^2*(b + c*x^2)]*(-1/2*b/(c*x^2) + ArcTanh[Sqrt[1 + (c*x^2)/b]]/(2*Sqrt[1 + (c*x^2)/b])))/(b^2*x)

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fricas [A]  time = 0.90, size = 133, normalized size = 2.25 \[ \left [\frac {\sqrt {b} c x^{3} \log \left (-\frac {c x^{3} + 2 \, b x + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} b}{4 \, b^{2} x^{3}}, -\frac {\sqrt {-b} c x^{3} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + \sqrt {c x^{4} + b x^{2}} b}{2 \, b^{2} x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(sqrt(b)*c*x^3*log(-(c*x^3 + 2*b*x + 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) - 2*sqrt(c*x^4 + b*x^2)*b)/(b^2*
x^3), -1/2*(sqrt(-b)*c*x^3*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + sqrt(c*x^4 + b*x^2)*b)/(b^2*x^
3)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-4,[1
,0,0]%%%}+%%%{-2,[0,1,2]%%%},0,%%%{1,[0,2,4]%%%}] at parameters values [64.3995612673,65,-85]Warning, choosing
root of [1,0,%%%{-4,[1,0,0]%%%}+%%%{-2,[0,1,2]%%%},0,%%%{1,[0,2,4]%%%}] at parameters values [66.1769613782,9
3,91]-1/2/b/x*sqrt(b*(1/x)^2+c)-2*c/4/b/sqrt(b)*ln(abs(sqrt(b*(1/x)^2+c)-sqrt(b)/x))

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maple [A]  time = 0.01, size = 73, normalized size = 1.24 \[ -\frac {\sqrt {c \,x^{2}+b}\, \left (-b c \,x^{2} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )+\sqrt {c \,x^{2}+b}\, b^{\frac {3}{2}}\right )}{2 \sqrt {c \,x^{4}+b \,x^{2}}\, b^{\frac {5}{2}} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(c*x^4+b*x^2)^(1/2),x)

[Out]

-1/2/x*(c*x^2+b)^(1/2)*(-c*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^2*b+(c*x^2+b)^(1/2)*b^(3/2))/(c*x^4+b*x^2)^(1
/2)/b^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{4} + b x^{2}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(c*x^4 + b*x^2)*x^2), x)

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mupad [B]  time = 4.64, size = 76, normalized size = 1.29 \[ -\frac {\left (\frac {\sqrt {c}\,x^2\,\sqrt {c+\frac {b}{x^2}}}{2\,b}+\frac {c^{3/2}\,x^3\,\mathrm {asin}\left (\frac {\sqrt {b}\,1{}\mathrm {i}}{\sqrt {c}\,x}\right )\,1{}\mathrm {i}}{2\,b^{3/2}}\right )\,\sqrt {\frac {b}{c\,x^2}+1}}{x\,\sqrt {c\,x^4+b\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(b*x^2 + c*x^4)^(1/2)),x)

[Out]

-(((c^(1/2)*x^2*(c + b/x^2)^(1/2))/(2*b) + (c^(3/2)*x^3*asin((b^(1/2)*1i)/(c^(1/2)*x))*1i)/(2*b^(3/2)))*(b/(c*
x^2) + 1)^(1/2))/(x*(b*x^2 + c*x^4)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral(1/(x**2*sqrt(x**2*(b + c*x**2))), x)

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