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

Optimal. Leaf size=40 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{1-c^4 x^4}}{c x \sqrt{\frac{1}{c^2 x^2}+1}}\right )}{c} \]

[Out]

-(ArcTanh[Sqrt[1 - c^4*x^4]/(c*Sqrt[1 + 1/(c^2*x^2)]*x)]/c)

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Rubi [A]  time = 0.0722516, antiderivative size = 44, normalized size of antiderivative = 1.1, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {1448, 1252, 848, 63, 208} \[ -\frac{x \sqrt{\frac{1}{c^2 x^2}+1} \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{\sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1448

Int[((d_) + (e_.)*(x_)^(mn_.))^(q_)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Dist[(e^IntPart[q]*(d + e*x
^mn)^FracPart[q])/(x^(mn*FracPart[q])*(1 + d/(x^mn*e))^FracPart[q]), Int[x^(mn*q)*(1 + d/(x^mn*e))^q*(a + c*x^
n2)^p, x], x] /; FreeQ[{a, c, d, e, mn, p, q}, x] && EqQ[n2, -2*mn] &&  !IntegerQ[p] &&  !IntegerQ[q] && PosQ[
n2]

Rule 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0555167, size = 44, normalized size = 1.1 \[ -\frac{x \sqrt{\frac{1}{c^2 x^2}+1} \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{\sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.061, size = 101, normalized size = 2.5 \begin{align*} -{\frac{x{\it csgn} \left ({c}^{-1} \right ) }{ \left ({c}^{2}{x}^{2}+1 \right ) c}\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}{x}^{2}}}}\sqrt{-{c}^{4}{x}^{4}+1}\ln \left ( 2\,{\frac{1}{x{c}^{2}} \left ({\it csgn} \left ({c}^{-1} \right ) c\sqrt{-{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}}}}+1 \right ) } \right ){\frac{1}{\sqrt{-{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{1}{c^{2} x^{2}} + 1}}{\sqrt{-c^{4} x^{4} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.8348, size = 255, normalized size = 6.38 \begin{align*} -\frac{\log \left (\frac{c^{2} x^{2} + \sqrt{-c^{4} x^{4} + 1} c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c^{2} x^{2} + 1}\right ) - \log \left (-\frac{c^{2} x^{2} - \sqrt{-c^{4} x^{4} + 1} c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c^{2} x^{2} + 1}\right )}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{1 + \frac{1}{c^{2} x^{2}}}}{\sqrt{- \left (c x - 1\right ) \left (c x + 1\right ) \left (c^{2} x^{2} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1095, size = 78, normalized size = 1.95 \begin{align*} \frac{{\left (\log \left (\sqrt{2} + 1\right ) - \log \left (\sqrt{2} - 1\right ) - \log \left (\sqrt{-c^{2} x^{2} + 1} + 1\right ) + \log \left (-\sqrt{-c^{2} x^{2} + 1} + 1\right )\right )}{\left | c \right |} \mathrm{sgn}\left (x\right )}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(log(sqrt(2) + 1) - log(sqrt(2) - 1) - log(sqrt(-c^2*x^2 + 1) + 1) + log(-sqrt(-c^2*x^2 + 1) + 1))*abs(c)*
sgn(x)/c^2